algebra-lemma-add-trivial-complex

Suppose $R$ is a ring. Let $$ \ldots \xrightarrow{\varphi_{i + 1}} R^{n_i} \xrightarrow{\varphi_i} R^{n_{i-1}} \xrightarrow{\varphi_{i-1}} \ldots $$ be a complex of finite free $R$-modules. Suppose that for some $i$ some matrix coefficient of the map $\varphi_i$ is invertible. Then the displayed complex is isomorphic to the direct sum of a complex $$ \ldots \to R^{n_{i + 2}} \xrightarrow{\varphi_{i + 2}} R^{n_{i + 1}} \to R^{n_i - 1} \to R^{n_{i - 1} - 1} \to R^{n_{i - 2}} \xrightarrow{\varphi_{i - 2}} R^{n_{i - 3}} \to \ldots $$ and the complex $\ldots \to 0 \to R \to R \to 0 \to \ldots$ where the map $R \to R$ is the identity map.


algebra-lemma-exact-depth-zero-local

In Situation \ref{situation-complex}. Suppose $R$ is a local Noetherian ring with maximal ideal $\mathfrak m$. Assume $\mathfrak m \in \text{Ass}(R)$, in other words $R$ has depth $0$. Suppose that $0 \to R^{n_e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_e}, \ldots, R^{n_1}$. Then the complex is isomorphic to a direct sum of trivial complexes.


algebra-lemma-trivial-case-exact

In Situation \ref{situation-complex}, suppose the complex is isomorphic to a direct sum of trivial complexes. Then we have \begin{enumerate} \item the maps $\varphi_i$ have rank $r_i = n_i - n_{i + 1} + \ldots + (-1)^{e-i-1} n_{e-1} + (-1)^{e-i} n_e$, \item for all $i$, $1 \leq i \leq e - 1$ we have $\text{rank}(\varphi_{i + 1}) + \text{rank}(\varphi_i) = n_i$, \item each $I(\varphi_i) = R$. \end{enumerate}


algebra-lemma-div-x-exact-one-less

In Situation \ref{situation-complex}. Suppose $R$ is a local ring with maximal ideal $\mathfrak m$. Suppose that $0 \to R^{n_e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_e}, \ldots, R^{n_1}$. Let $x \in \mathfrak m$ be a nonzerodivisor. The complex $0 \to (R/xR)^{n_e} \to \ldots \to (R/xR)^{n_1}$ is exact at $(R/xR)^{n_e}, \ldots, (R/xR)^{n_2}$.


algebra-lemma-acyclic

\begin{reference} \cite[Lemma 1.8]{Peskine-Szpiro} \end{reference} Let $R$ be a local Noetherian ring. Let $0 \to M_e \to M_{e-1} \to \ldots \to M_0$ be a complex of finite $R$-modules. Assume $\text{depth}(M_i) \geq i$. Let $i$ be the largest index such that the complex is not exact at $M_i$. If $i > 0$ then $\Ker(M_i \to M_{i-1})/\Im(M_{i + 1} \to M_i)$ has depth $\geq 1$.


algebra-lemma-good-element

Notation and assumptions as above. If $g$ is good with respect to $(M, f_1, \ldots, f_d)$, then (a) $g$ is a nonzerodivisor on $M$, and (b) $M/gM$ is Cohen-Macaulay with maximal regular sequence $f_1, \ldots, f_{d - 1}$.


algebra-lemma-CM-one-g

Let $R$ be a Noetherian local ring. Let $M$ be a Cohen-Macaulay module over $R$. Suppose $g \in \mathfrak m$ is such that $\dim(\text{Supp}(M) \cap V(g)) = \dim(\text{Supp}(M)) - 1$. Then (a) $g$ is a nonzerodivisor on $M$, and (b) $M/gM$ is Cohen-Macaulay of depth one less.


algebra-lemma-CM-over-quotient

Let $R \to S$ be a surjective homomorphism of Noetherian local rings. Let $N$ be a finite $S$-module. Then $N$ is Cohen-Macaulay as an $S$-module if and only if $N$ is Cohen-Macaulay as an $R$-module.


algebra-lemma-CM-ass-minimal-support

\begin{reference} \cite[Chapter 0, Proposition 16.5.4]{EGA} \end{reference} Let $R$ be a Noetherian local ring. Let $M$ be a finite Cohen-Macaulay $R$-module. If $\mathfrak p \in \text{Ass}(M)$, then $\dim(R/\mathfrak p) = \dim(\text{Supp}(M))$ and $\mathfrak p$ is a minimal prime in the support of $M$. In particular, $M$ has no embedded associated primes.


algebra-lemma-maximal-chain-maximal-CM

\begin{slogan} In a local Cohen-Macaulay ring, any maximal chain of prime ideals has length equal to the dimension. \end{slogan} Let $R$ be a Noetherian local ring. Assume there exists a Cohen-Macaulay module $M$ with $\Spec(R) = \text{Supp}(M)$. Then any maximal chain of ideals $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n$ has length $n = \dim(R)$.


algebra-lemma-dim-formula-maximal-CM

Suppose $R$ is a Noetherian local ring. Assume there exists a Cohen-Macaulay module $M$ with $\Spec(R) = \text{Supp}(M)$. Then for a prime $\mathfrak p \subset R$ we have $$ \dim(R) = \dim(R_{\mathfrak p}) + \dim(R/\mathfrak p). $$


algebra-lemma-localize-CM-module

Suppose $R$ is a Noetherian local ring. Let $M$ be a Cohen-Macaulay module over $R$. For any prime $\mathfrak p \subset R$ the module $M_{\mathfrak p}$ is Cohen-Macaulay over $R_\mathfrak p$.


algebra-lemma-maximal-CM-polynomial-algebra

Let $R$ be a Noetherian ring. Let $M$ be a Cohen-Macaulay module over $R$. Then $M \otimes_R R[x_1, \ldots, x_n]$ is a Cohen-Macaulay module over $R[x_1, \ldots, x_n]$.


algebra-lemma-reformulate-CM

\begin{slogan} Regular sequences in Cohen-Macaulay local rings are characterized by cutting out something of the correct dimension. \end{slogan} Let $R$ be a Noetherian local Cohen-Macaulay ring with maximal ideal $\mathfrak m $. Let $x_1, \ldots, x_c \in \mathfrak m$ be elements. Then $$ x_1, \ldots, x_c \text{ is a regular sequence } \Leftrightarrow \dim(R/(x_1, \ldots, x_c)) = \dim(R) - c $$ If so $x_1, \ldots, x_c$ can be extended to a regular sequence of length $\dim(R)$ and each quotient $R/(x_1, \ldots, x_i)$ is a Cohen-Macaulay ring of dimension $\dim(R) - i$.


algebra-lemma-maximal-chain-CM

Let $R$ be Noetherian local. Suppose $R$ is Cohen-Macaulay of dimension $d$. Any maximal chain of ideals $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n$ has length $n = d$.


algebra-lemma-CM-dim-formula

Suppose $R$ is a Noetherian local Cohen-Macaulay ring of dimension $d$. For any prime $\mathfrak p \subset R$ we have $$ \dim(R) = \dim(R_{\mathfrak p}) + \dim(R/\mathfrak p). $$


algebra-lemma-localize-CM

Suppose $R$ is a Cohen-Macaulay local ring. For any prime $\mathfrak p \subset R$ the ring $R_{\mathfrak p}$ is Cohen-Macaulay as well.


algebra-lemma-CM-polynomial-algebra

Suppose $R$ is a Noetherian Cohen-Macaulay ring. Any polynomial algebra over $R$ is Cohen-Macaulay.


algebra-lemma-dimension-shift

Let $R$ be a Noetherian local Cohen-Macaulay ring of dimension $d$. Let $0 \to K \to R^{\oplus n} \to M \to 0$ be an exact sequence of $R$-modules. Then either $M = 0$, or $\text{depth}(K) > \text{depth}(M)$, or $\text{depth}(K) = \text{depth}(M) = d$.


algebra-lemma-mcm-resolution

Let $R$ be a local Noetherian Cohen-Macaulay ring of dimension $d$. Let $M$ be a finite $R$ module of depth $e$. There exists an exact complex $$ 0 \to K \to F_{d-e-1} \to \ldots \to F_0 \to M \to 0 $$ with each $F_i$ finite free and $K$ maximal Cohen-Macaulay.


algebra-lemma-find-sequence-image-regular

Let $\varphi : A \to B$ be a map of local rings. Assume that $B$ is Noetherian and Cohen-Macaulay and that $\mathfrak m_B = \sqrt{\varphi(\mathfrak m_A) B}$. Then there exists a sequence of elements $f_1, \ldots, f_{\dim(B)}$ in $A$ such that $\varphi(f_1), \ldots, \varphi(f_{\dim(B)})$ is a regular sequence in $B$.


algebra-lemma-catenary

A ring $R$ is catenary if and only if the topological space $\Spec(R)$ is catenary (see Topology, Definition \ref{topology-definition-catenary}).


algebra-lemma-localization-catenary

Any localization of a catenary ring is catenary. Any localization of a Noetherian universally catenary ring is universally catenary.


algebra-lemma-universally-catenary

Let $A$ be a Noetherian universally catenary ring. Any $A$-algebra essentially of finite type over $A$ is universally catenary.


algebra-lemma-catenary-check-local

Let $R$ be a ring. The following are equivalent \begin{enumerate} \item $R$ is catenary, \item $R_\mathfrak p$ is catenary for all prime ideals $\mathfrak p$, \item $R_\mathfrak m$ is catenary for all maximal ideals $\mathfrak m$. \end{enumerate} Assume $R$ is Noetherian. The following are equivalent \begin{enumerate} \item $R$ is universally catenary, \item $R_\mathfrak p$ is universally catenary for all prime ideals $\mathfrak p$, \item $R_\mathfrak m$ is universally catenary for all maximal ideals $\mathfrak m$. \end{enumerate}


algebra-lemma-quotient-catenary

Any quotient of a catenary ring is catenary. Any quotient of a Noetherian universally catenary ring is universally catenary.


algebra-lemma-catenary-check-irreducible

Let $R$ be a Noetherian ring. \begin{enumerate} \item $R$ is catenary if and only if $R/\mathfrak p$ is catenary for every minimal prime $\mathfrak p$. \item $R$ is universally catenary if and only if $R/\mathfrak p$ is universally catenary for every minimal prime $\mathfrak p$. \end{enumerate}


algebra-lemma-CM-ring-catenary

A Noetherian Cohen-Macaulay ring is universally catenary. More generally, if $R$ is a Noetherian ring and $M$ is a Cohen-Macaulay $R$-module with $\text{Supp}(M) = \Spec(R)$, then $R$ is universally catenary.


algebra-lemma-regular-graded

Let $(R, \mathfrak m, \kappa)$ be a regular local ring of dimension $d$. The graded ring $\bigoplus \mathfrak m^n / \mathfrak m^{n + 1}$ is isomorphic to the graded polynomial algebra $\kappa[X_1, \ldots, X_d]$.


algebra-lemma-regular-domain

Any regular local ring is a domain.


algebra-lemma-regular-ring-CM

Let $R$ be a regular local ring and let $x_1, \ldots, x_d$ be a minimal set of generators for the maximal ideal $\mathfrak m$. Then $x_1, \ldots, x_d$ is a regular sequence, and each $R/(x_1, \ldots, x_c)$ is a regular local ring of dimension $d - c$. In particular $R$ is Cohen-Macaulay.


algebra-lemma-regular-quotient-regular

Let $R$ be a regular local ring. Let $I \subset R$ be an ideal such that $R/I$ is a regular local ring as well. Then there exists a minimal set of generators $x_1, \ldots, x_d$ for the maximal ideal $\mathfrak m$ of $R$ such that $I = (x_1, \ldots, x_c)$ for some $0 \leq c \leq d$.


algebra-lemma-free-mod-x

Let $R$ be a Noetherian local ring. Let $x \in \mathfrak m$. Let $M$ be a finite $R$-module such that $x$ is a nonzerodivisor on $M$ and $M/xM$ is free over $R/xR$. Then $M$ is free over $R$.


algebra-lemma-regular-mcm-free

Let $R$ be a regular local ring. Any maximal Cohen-Macaulay module over $R$ is free.


algebra-lemma-regular-mod-x

Suppose $R$ is a Noetherian local ring. Let $x \in \mathfrak m$ be a nonzerodivisor such that $R/xR$ is a regular local ring. Then $R$ is a regular local ring. More generally, if $x_1, \ldots, x_r$ is a regular sequence in $R$ such that $R/(x_1, \ldots, x_r)$ is a regular local ring, then $R$ is a regular local ring.


algebra-lemma-colimit-regular

Let $(R_i, \varphi_{ii'})$ be a directed system of local rings whose transition maps are local ring maps. If each $R_i$ is a regular local ring and $R = \colim R_i$ is Noetherian, then $R$ is a regular local ring.


algebra-lemma-epimorphism

Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is an epimorphism, \item the two ring maps $S \to S \otimes_R S$ are equal, \item either of the ring maps $S \to S \otimes_R S$ is an isomorphism, and \item the ring map $S \otimes_R S \to S$ is an isomorphism. \end{enumerate}


algebra-lemma-composition-epimorphism

The composition of two epimorphisms of rings is an epimorphism.


algebra-lemma-base-change-epimorphism

If $R \to S$ is an epimorphism of rings and $R \to R'$ is any ring map, then $R' \to R' \otimes_R S$ is an epimorphism.


algebra-lemma-permanence-epimorphism

If $A \to B \to C$ are ring maps and $A \to C$ is an epimorphism, so is $B \to C$.


algebra-lemma-epimorphism-local

Let $R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item $R \to S$ is an epimorphism, and \item $R_{\mathfrak p} \to S_{\mathfrak p}$ is an epimorphism for each prime $\mathfrak p$ of $R$. \end{enumerate}


algebra-lemma-finite-epimorphism-surjective

\begin{slogan} A ring map is surjective if and only if it is a finite epimorphism. \end{slogan} Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is an epimorphism and finite, and \item $R \to S$ is surjective. \end{enumerate}


algebra-lemma-faithfully-flat-epimorphism

A faithfully flat epimorphism is an isomorphism.


algebra-lemma-epimorphism-over-field

If $k \to S$ is an epimorphism and $k$ is a field, then $S = k$ or $S = 0$.


algebra-lemma-relations

Let $R$ be a ring. Let $M$, $N$ be $R$-modules. Let $\{x_i\}_{i \in I}$ be a set of generators of $M$. Let $\{y_j\}_{j \in J}$ be a set of generators of $N$. Let $\{m_j\}_{j \in J}$ be a family of elements of $M$ with $m_j = 0$ for all but finitely many $j$. Then $$ \sum\nolimits_{j \in J} m_j \otimes y_j = 0 \text{ in } M \otimes_R N $$ is equivalent to the following: There exist $a_{i, j} \in R$ with $a_{i, j} = 0$ for all but finitely many pairs $(i, j)$ such that \begin{align*} m_j & = \sum\nolimits_{i \in I} a_{i, j} x_i \quad\text{for all } j \in J, \\ 0 & = \sum\nolimits_{j \in J} a_{i, j} y_j \quad\text{for all } i \in I. \end{align*}


algebra-lemma-kernel-difference-projections

Let $\varphi : R \to S$ be a ring map. Let $g \in S$. The following are equivalent: \begin{enumerate} \item $g \otimes 1 = 1 \otimes g$ in $S \otimes_R S$, and \item there exist $n \geq 0$ and elements $y_i, z_j \in S$ and $x_{i, j} \in R$ for $1 \leq i, j \leq n$ such that \begin{enumerate} \item $g = \sum_{i, j \leq n} x_{i, j} y_i z_j$, \item for each $j$ we have $\sum x_{i, j}y_i \in \varphi(R)$, and \item for each $i$ we have $\sum x_{i, j}z_j \in \varphi(R)$. \end{enumerate} \end{enumerate}


algebra-lemma-epimorphism-modules

Let $R \to S$ be an epimorphism of rings. Let $N_1, N_2$ be $S$-modules. Then $\Hom_S(N_1, N_2) = \Hom_R(N_1, N_2)$. In other words, the restriction functor $\text{Mod}_S \to \text{Mod}_R$ is fully faithful.


algebra-lemma-pure

Let $R$ be a ring. Let $I \subset R$ be an ideal. The following are equivalent: \begin{enumerate} \item $I$ is pure, \item for every ideal $J \subset R$ we have $J \cap I = IJ$, \item for every finitely generated ideal $J \subset R$ we have $J \cap I = JI$, \item for every $x \in R$ we have $(x) \cap I = xI$, \item for every $x \in I$ we have $x = yx$ for some $y \in I$, \item for every $x_1, \ldots, x_n \in I$ there exists a $y \in I$ such that $x_i = yx_i$ for all $i = 1, \ldots, n$, \item for every prime $\mathfrak p$ of $R$ we have $IR_{\mathfrak p} = 0$ or $IR_{\mathfrak p} = R_{\mathfrak p}$, \item $\text{Supp}(I) = \Spec(R) \setminus V(I)$, \item $I$ is the kernel of the map $R \to (1 + I)^{-1}R$, \item $R/I \cong S^{-1}R$ as $R$-algebras for some multiplicative subset $S$ of $R$, and \item $R/I \cong (1 + I)^{-1}R$ as $R$-algebras. \end{enumerate}


algebra-lemma-pure-ideal-determined-by-zero-set

\begin{slogan} Pure ideals are determined by their vanishing locus. \end{slogan} Let $R$ be a ring. If $I, J \subset R$ are pure ideals, then $V(I) = V(J)$ implies $I = J$.


algebra-lemma-pure-open-closed-specializations

Let $R$ be a ring. The rule $I \mapsto V(I)$ determines a bijection $$ \{I \subset R \text{ pure}\} \leftrightarrow \{Z \subset \Spec(R)\text{ closed and closed under generalizations}\} $$


algebra-lemma-finitely-generated-pure-ideal

Let $R$ be a ring. Let $I \subset R$ be an ideal. The following are equivalent \begin{enumerate} \item $I$ is pure and finitely generated, \item $I$ is generated by an idempotent, \item $I$ is pure and $V(I)$ is open, and \item $R/I$ is a projective $R$-module. \end{enumerate}


algebra-lemma-finite-flat-module-finitely-presented

Let $R$ be a ring. The following are equivalent: \begin{enumerate} \item every $Z \subset \Spec(R)$ which is closed and closed under generalizations is also open, and \item any finite flat $R$-module is finite locally free. \end{enumerate}


algebra-lemma-Schanuel

Let $R$ be a ring. Let $M$ be an $R$-module. Suppose that $$ 0 \to K \xrightarrow{c_1} P_1 \xrightarrow{p_1} M \to 0 \quad\text{and}\quad 0 \to L \xrightarrow{c_2} P_2 \xrightarrow{p_2} M \to 0 $$ are two short exact sequences, with $P_i$ projective. Then $K \oplus P_2 \cong L \oplus P_1$. More precisely, there exist a commutative diagram $$ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} \ar[d] & P_1 \oplus P_2 \ar[r]_{(p_1, 0)} \ar[d] & M \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(0, p_2)} & M \ar[r] & 0 } $$ whose vertical arrows are isomorphisms.


algebra-lemma-independent-resolution

Let $R$ be a ring. Suppose that $M$ is an $R$-module of projective dimension $d$. Suppose that $F_e \to F_{e-1} \to \ldots \to F_0 \to M \to 0$ is exact with $F_i$ projective and $e \geq d - 1$. Then the kernel of $F_e \to F_{e-1}$ is projective (or the kernel of $F_0 \to M$ is projective in case $e = 0$).


algebra-lemma-what-kind-of-resolutions

Let $R$ be a ring. Let $M$ be an $R$-module. Let $d \geq 0$. The following are equivalent \begin{enumerate} \item $M$ has projective dimension $\leq d$, \item there exists a resolution $0 \to P_d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_i$ projective, \item for some resolution $\ldots \to P_2 \to P_1 \to P_0 \to M \to 0$ with $P_i$ projective we have $\Ker(P_{d - 1} \to P_{d - 2})$ is projective if $d \geq 2$, or $\Ker(P_0 \to M)$ is projective if $d = 1$, or $M$ is projective if $d = 0$, \item for any resolution $\ldots \to P_2 \to P_1 \to P_0 \to M \to 0$ with $P_i$ projective we have $\Ker(P_{d - 1} \to P_{d - 2})$ is projective if $d \geq 2$, or $\Ker(P_0 \to M)$ is projective if $d = 1$, or $M$ is projective if $d = 0$. \end{enumerate}


algebra-lemma-what-kind-of-resolutions-local

Let $R$ be a local ring. Let $M$ be an $R$-module. Let $d \geq 0$. The equivalent conditions (1) -- (4) of Lemma \ref{lemma-what-kind-of-resolutions} are also equivalent to \begin{enumerate} \item[(5)] there exists a resolution $0 \to P_d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_i$ free. \end{enumerate}


algebra-lemma-what-kind-of-resolutions-Noetherian

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $d \geq 0$. The equivalent conditions (1) -- (4) of Lemma \ref{lemma-what-kind-of-resolutions} are also equivalent to \begin{enumerate} \item[(6)] there exists a resolution $0 \to P_d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_i$ finite projective. \end{enumerate}


algebra-lemma-what-kind-of-resolutions-Noetherian-local

Let $R$ be a local Noetherian ring. Let $M$ be a finite $R$-module. Let $d \geq 0$. The equivalent conditions (1) -- (4) of Lemma \ref{lemma-what-kind-of-resolutions}, condition (5) of Lemma \ref{lemma-what-kind-of-resolutions-local}, and condition (6) of Lemma \ref{lemma-what-kind-of-resolutions-Noetherian} are also equivalent to \begin{enumerate} \item[(7)] there exists a resolution $0 \to F_d \to F_{d - 1} \to \ldots \to F_0 \to M \to 0$ with $F_i$ finite free. \end{enumerate}


algebra-lemma-projective-dimension-ext

Let $R$ be a ring. Let $M$ be an $R$-module. Let $n \geq 0$. The following are equivalent \begin{enumerate} \item $M$ has projective dimension $\leq n$, \item $\Ext^i_R(M, N) = 0$ for all $R$-modules $N$ and all $i \geq n + 1$, and \item $\Ext^{n + 1}_R(M, N) = 0$ for all $R$-modules $N$. \end{enumerate}


algebra-lemma-exact-sequence-projective-dimension

Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. \begin{enumerate} \item If $M$ has projective dimension $\leq n$ and $M''$ has projective dimension $\leq n + 1$, then $M'$ has projective dimension $\leq n$. \item If $M'$ and $M''$ have projective dimension $\leq n$ then $M$ has projective dimension $\leq n$. \item If $M'$ has projective dimension $\leq n$ and $M$ has projective dimension $\leq n + 1$ then $M''$ has projective dimension $\leq n + 1$. \end{enumerate}


algebra-lemma-colimit-projective-dimension

Let $R$ be a ring. Suppose we have a module $M = \bigcup_{e \in E} M_e$ where the $M_e$ are submodules well-ordered by inclusion. Assume the quotients $M_e/\bigcup\nolimits_{e' < e} M_{e'}$ have projective dimension $\leq n$. Then $M$ has projective dimension $\leq n$.


algebra-lemma-finite-gl-dim

Let $R$ be a ring. The following are equivalent \begin{enumerate} \item $R$ has finite global dimension $\leq n$, \item every finite $R$-module has projective dimension $\leq n$, and \item every cyclic $R$-module $R/I$ has projective dimension $\leq n$. \end{enumerate}


algebra-lemma-localize-finite-gl-dim

Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. \begin{enumerate} \item If $M$ has projective dimension $\leq n$, then $S^{-1}M$ has projective dimension $\leq n$ over $S^{-1}R$. \item If $R$ has finite global dimension $\leq n$, then $S^{-1}R$ has finite global dimension $\leq n$. \end{enumerate}


algebra-lemma-finite-gl-dim-primes

Let $R$ be a Noetherian ring. Then $R$ has finite global dimension if and only if there exists an integer $n$ such that for all maximal ideals $\mathfrak m$ of $R$ the ring $R_{\mathfrak m}$ has global dimension $\leq n$.


algebra-lemma-length-resolution-residue-field

Suppose that $R$ is a Noetherian local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. In this case the projective dimension of $\kappa$ is $\geq \dim_\kappa \mathfrak m / \mathfrak m^2$.


algebra-lemma-dim-gl-dim

Let $R$ be a Noetherian local ring. Suppose that the residue field $\kappa$ has finite projective dimension $n$ over $R$. In this case $\dim(R) \geq n$.


algebra-lemma-localization-of-regular-local-is-regular

A Noetherian local ring $R$ is a regular local ring if and only if it has finite global dimension. In this case $R_{\mathfrak p}$ is a regular local ring for all primes $\mathfrak p$.


algebra-lemma-finite-gl-dim-finite-dim-regular

Let $R$ be a Noetherian ring. The following are equivalent: \begin{enumerate} \item $R$ has finite global dimension $n$, \item there exists an integer $n$ such that all the localizations $R_{\mathfrak m}$ at maximal ideals are regular of dimension $\leq n$ with equality for at least one $\mathfrak m$, and \item there exists an integer $n$ such that all the localizations $R_{\mathfrak p}$ at prime ideals are regular of dimension $\leq n$ with equality for at least one $\mathfrak p$. \end{enumerate}


algebra-lemma-flat-under-regular

Let $R \to S$ be a local homomorphism of local Noetherian rings. Assume that $R \to S$ is flat and that $S$ is regular. Then $R$ is regular.


algebra-lemma-dimension-going-up

Suppose $R \to S$ is a ring map satisfying either going up, see Definition \ref{definition-going-up-down}, or going down see Definition \ref{definition-going-up-down}. Assume in addition that $\Spec(S) \to \Spec(R)$ is surjective. Then $\dim(R) \leq \dim(S)$.


algebra-lemma-going-up-maximal-on-top

Suppose that $R \to S$ is a ring map with the going up property, see Definition \ref{definition-going-up-down}. If $\mathfrak q \subset S$ is a maximal ideal. Then the inverse image of $\mathfrak q$ in $R$ is a maximal ideal too.


algebra-lemma-integral-dim-up

Suppose that $R \to S$ is a ring map such that $S$ is integral over $R$. Then $\dim (R) \geq \dim(S)$, and every closed point of $\Spec(S)$ maps to a closed point of $\Spec(R)$.


algebra-lemma-integral-sub-dim-equal

Suppose $R \subset S$ and $S$ integral over $R$. Then $\dim(R) = \dim(S)$.


algebra-lemma-dimension-base-fibre-total

Let $R \to S$ be a homomorphism of Noetherian rings. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$. Then $$ \dim(S_{\mathfrak q}) \leq \dim(R_{\mathfrak p}) + \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}). $$


algebra-lemma-dimension-base-fibre-equals-total

Let $R \to S$ be a homomorphism of Noetherian rings. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$. Assume the going down property holds for $R \to S$ (for example if $R \to S$ is flat, see Lemma \ref{lemma-flat-going-down}). Then $$ \dim(S_{\mathfrak q}) = \dim(R_{\mathfrak p}) + \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}). $$


algebra-lemma-flat-over-regular-with-regular-fibre

Let $R \to S$ be a local homomorphism of local Noetherian rings. Assume \begin{enumerate} \item $R$ is regular, \item $S/\mathfrak m_RS$ is regular, and \item $R \to S$ is flat. \end{enumerate} Then $S$ is regular.


algebra-lemma-finite-flat-over-regular-CM

Let $R \to S$ be a local homomorphism of Noetherian local rings. Assume $R$ Cohen-Macaulay. If $S$ is finite flat over $R$, or if $S$ is flat over $R$ and $\dim(S) \leq \dim(R)$, then $S$ is Cohen-Macaulay and $\dim(R) = \dim(S)$.


algebra-lemma-dimension-formula

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak p$ of $R$. Assume that \begin{enumerate} \item $R$ is Noetherian, \item $R \to S$ is of finite type, \item $R$, $S$ are domains, and \item $R \subset S$. \end{enumerate} Then we have $$ \text{height}(\mathfrak q) \leq \text{height}(\mathfrak p) + \text{trdeg}_R(S) - \text{trdeg}_{\kappa(\mathfrak p)} \kappa(\mathfrak q) $$ with equality if $R$ is universally catenary.


algebra-lemma-finite-in-codim-1

Let $A \to B$ be a ring map. Assume \begin{enumerate} \item $A \subset B$ is an extension of domains, \item the induced extension of fraction fields is finite, \item $A$ is Noetherian, and \item $A \to B$ is of finite type. \end{enumerate} Let $\mathfrak p \subset A$ be a prime of height $1$. Then there are at most finitely many primes of $B$ lying over $\mathfrak p$ and they all have height $1$.


algebra-lemma-dim-affine-space

Let $\mathfrak m$ be a maximal ideal in $k[x_1, \ldots, x_n]$. The ideal $\mathfrak m$ is generated by $n$ elements. The dimension of $k[x_1, \ldots, x_n]_{\mathfrak m}$ is $n$. Hence $k[x_1, \ldots, x_n]_{\mathfrak m}$ is a regular local ring of dimension $n$.


algebra-lemma-dimension-height-polynomial-ring

Let $k$ be a field. Let $\mathfrak p \subset \mathfrak q \subset k[x_1, \ldots, x_n]$ be a pair of primes. Any maximal chain of primes between $\mathfrak p$ and $\mathfrak q$ has length $\text{height}(\mathfrak q) - \text{height}(\mathfrak p)$.


algebra-lemma-dimension-spell-it-out

Let $k$ be a field. Let $S$ be a finite type $k$-algebra which is an integral domain. Then $\dim(S) = \dim(S_{\mathfrak m})$ for any maximal ideal $\mathfrak m$ of $S$. In words: every maximal chain of primes has length equal to the dimension of $S$.


algebra-lemma-dimension-at-a-point-finite-type-over-field

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $X = \Spec(S)$. Let $\mathfrak p \subset S$ be a prime ideal and let $x \in X$ be the corresponding point. The following numbers are equal \begin{enumerate} \item $\dim_x(X)$, \item $\max \dim(Z)$ where the maximum is over those irreducible components $Z$ of $X$ passing through $x$, and \item $\min \dim(S_{\mathfrak m})$ where the minimum is over maximal ideals $\mathfrak m$ with $\mathfrak p \subset \mathfrak m$. \end{enumerate}


algebra-lemma-dimension-closed-point-finite-type-field

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $X = \Spec(S)$. Let $\mathfrak m \subset S$ be a maximal ideal and let $x \in X$ be the associated closed point. Then $\dim_x(X) = \dim(S_{\mathfrak m})$.


algebra-lemma-disjoint-decomposition-CM-algebra

Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Assume that $S$ is Cohen-Macaulay. Then $\Spec(S) = \coprod T_d$ is a finite disjoint union of open and closed subsets $T_d$ with $T_d$ equidimensional (see Topology, Definition \ref{topology-definition-equidimensional}) of dimension $d$. Equivalently, $S$ is a product of rings $S_d$, $d = 0, \ldots, \dim(S)$ such that every maximal ideal $\mathfrak m$ of $S_d$ has height $d$.


algebra-lemma-helper

Let $n \in \mathbf{N}$. Let $N$ be a finite nonempty set of multi-indices $\nu = (\nu_1, \ldots, \nu_n)$. Given $e = (e_1, \ldots, e_n)$ we set $e \cdot \nu = \sum e_i\nu_i$. Then for $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_n$ we have: If $\nu, \nu' \in N$ then $$ (e \cdot \nu = e \cdot \nu') \Leftrightarrow (\nu = \nu') $$


algebra-lemma-helper-polynomial

Let $R$ be a ring. Let $g \in R[x_1, \ldots, x_n]$ be an element which is nonconstant, i.e., $g \not \in R$. For $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_n = 1$ the polynomial $$ g(x_1 + x_n^{e_1}, x_2 + x_n^{e_2}, \ldots, x_{n - 1} + x_n^{e_{n - 1}}, x_n) = ax_n^d + \text{lower order terms in }x_n $$ where $d > 0$ and $a \in R$ is one of the nonzero coefficients of $g$.


algebra-lemma-one-relation

Let $k$ be a field. Let $S = k[x_1, \ldots, x_n]/I$ for some proper ideal $I$. If $I \not = 0$, then there exist $y_1, \ldots, y_{n-1} \in k[x_1, \ldots, x_n]$ such that $S$ is finite over $k[y_1, \ldots, y_{n-1}]$. Moreover we may choose $y_i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots, x_n]$ generated by $x_1, \ldots, x_n$.


algebra-lemma-Noether-normalization

Let $k$ be a field. Let $S = k[x_1, \ldots, x_n]/I$ for some ideal $I$. If $I \neq (1)$, there exist $r\geq 0$, and $y_1, \ldots, y_r \in k[x_1, \ldots, x_n]$ such that (a) the map $k[y_1, \ldots, y_r] \to S$ is injective, and (b) the map $k[y_1, \ldots, y_r] \to S$ is finite. In this case the integer $r$ is the dimension of $S$. Moreover we may choose $y_i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots, x_n]$ generated by $x_1, \ldots, x_n$.


algebra-lemma-Noether-normalization-at-point

Let $k$ be a field. Let $S$ be a finite type $k$ algebra and denote $X = \Spec(S)$. Let $\mathfrak q$ be a prime of $S$, and let $x \in X$ be the corresponding point. There exists a $g \in S$, $g \not \in \mathfrak q$ such that $\dim(S_g) = \dim_x(X) =: d$ and such that there exists a finite injective map $k[y_1, \ldots, y_d] \to S_g$.


algebra-lemma-refined-Noether-normalization

Let $k$ be a field. Let $\mathfrak q \subset k[x_1, \ldots, x_n]$ be a prime ideal. Set $r = \text{trdeg}_k\ \kappa(\mathfrak q)$. Then there exists a finite ring map $\varphi : k[y_1, \ldots, y_n] \to k[x_1, \ldots, x_n]$ such that $\varphi^{-1}(\mathfrak q) = (y_{r + 1}, \ldots, y_n)$.


algebra-lemma-Noether-normalization-over-a-domain

Let $R \to S$ be an injective finite type ring map. Assume $R$ is a domain. Then there exists an integer $d$ and a factorization $$ R \to R[y_1, \ldots, y_d] \to S' \to S $$ by injective maps such that $S'$ is finite over $R[y_1, \ldots, y_d]$ and such that $S'_f \cong S_f$ for some nonzero $f \in R$.


algebra-lemma-dimension-prime-polynomial-ring

Let $k$ be a field. Let $S$ be a finite type $k$ algebra which is an integral domain. Let $K$ be the field of fractions of $S$. Let $r = \text{trdeg}(K/k)$ be the transcendence degree of $K$ over $k$. Then $\dim(S) = r$. Moreover, the local ring of $S$ at every maximal ideal has dimension $r$.


algebra-lemma-tr-deg-specialization

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset \mathfrak q' \subset S$ be distinct prime ideals. Then $\text{trdeg}_k\ \kappa(\mathfrak q') < \text{trdeg}_k\ \kappa(\mathfrak q)$.


algebra-lemma-dimension-at-a-point-finite-type-field

Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $X = \Spec(S)$. Let $\mathfrak p \subset S$ be a prime ideal, and let $x \in X$ be the corresponding point. Then we have $$ \dim_x(X) = \dim(S_{\mathfrak p}) + \text{trdeg}_k\ \kappa(\mathfrak p). $$


algebra-lemma-codimension

Let $k$ be a field. Let $S' \to S$ be a surjection of finite type $k$ algebras. Let $\mathfrak p \subset S$ be a prime ideal, and let $\mathfrak p'$ be the corresponding prime ideal of $S'$. Let $X = \Spec(S)$, resp.\ $X' = \Spec(S')$, and let $x \in X$, resp. $x'\in X'$ be the point corresponding to $\mathfrak p$, resp.\ $\mathfrak p'$. Then $$ \dim_{x'} X' - \dim_x X = \text{height}(\mathfrak p') - \text{height}(\mathfrak p). $$


algebra-lemma-dimension-preserved-field-extension

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension. Then $\dim(S) = \dim(K \otimes_k S)$.


algebra-lemma-dimension-at-a-point-preserved-field-extension

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Set $X = \Spec(S)$. Let $k \subset K$ be a field extension. Set $S_K = K \otimes_k S$, and $X_K = \Spec(S_K)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$ and let $\mathfrak q_K \subset S_K$ be a prime corresponding to $x_K \in X_K$ lying over $\mathfrak q$. Then $\dim_x X = \dim_{x_K} X_K$.


algebra-lemma-inequalities-under-field-extension

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension. Set $S_K = K \otimes_k S$. Let $\mathfrak q \subset S$ be a prime and let $\mathfrak q_K \subset S_K$ be a prime lying over $\mathfrak q$. Then $$ \dim (S_K \otimes_S \kappa(\mathfrak q))_{\mathfrak q_K} = \dim (S_K)_{\mathfrak q_K} - \dim S_\mathfrak q = \text{trdeg}_k \kappa(\mathfrak q) - \text{trdeg}_K \kappa(\mathfrak q_K) $$ Moreover, given $\mathfrak q$ we can always choose $\mathfrak q_K$ such that the number above is zero.


algebra-lemma-generic-flatness-Noetherian

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume \begin{enumerate} \item $R$ is Noetherian, \item $R$ is a domain, \item $R \to S$ is of finite type, and \item $M$ is a finite type $S$-module. \end{enumerate} Then there exists a nonzero $f \in R$ such that $M_f$ is a free $R_f$-module.


algebra-lemma-generic-flatness-finitely-presented

\begin{slogan} Generic freeness. \end{slogan} Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume \begin{enumerate} \item $R$ is a domain, \item $R \to S$ is of finite presentation, and \item $M$ is an $S$-module of finite presentation. \end{enumerate} Then there exists a nonzero $f \in R$ such that $M_f$ is a free $R_f$-module.


algebra-lemma-generic-flatness

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume \begin{enumerate} \item $R$ is a domain, \item $R \to S$ is of finite type, and \item $M$ is a finite type $S$-module. \end{enumerate} Then there exists a nonzero $f \in R$ such that \begin{enumerate} \item[(a)] $M_f$ and $S_f$ are free as $R_f$-modules, and \item[(b)] $S_f$ is a finitely presented $R_f$-algebra and $M_f$ is a finitely presented $S_f$-module. \end{enumerate}


algebra-lemma-generic-flatness-locus-extension

Let $R \to S$ be a ring map. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $S$-modules. Then $$ U(R \to S, M_1) \cap U(R \to S, M_3) \subset U(R \to S, M_2). $$


algebra-lemma-generic-flatness-locus-localize

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $f \in R$. Using the identification $\Spec(R_f) = D(f)$ we have $U(R_f \to S_f, M_f) = D(f) \cap U(R \to S, M)$.


algebra-lemma-generic-flatness-locus-reduce

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $U \subset \Spec(R)$ be a dense open. Assume there is a covering $U = \bigcup_{i \in I} D(f_i)$ of opens such that $U(R_{f_i} \to S_{f_i}, M_{f_i})$ is dense in $D(f_i)$ for each $i \in I$. Then $U(R \to S, M)$ is dense in $\Spec(R)$.


algebra-lemma-generic-flatness-reduced

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume \begin{enumerate} \item $R \to S$ is of finite type, \item $M$ is a finite $S$-module, and \item $R$ is reduced. \end{enumerate} Then there exists a subset $U \subset \Spec(R)$ such that \begin{enumerate} \item $U$ is open and dense in $\Spec(R)$, \item for every $u \in U$ there exists an $f \in R$ such that $u \in D(f) \subset U$ and such that we have \begin{enumerate} \item $M_f$ and $S_f$ are free over $R_f$, \item $S_f$ is a finitely presented $R_f$-algebra, and \item $M_f$ is a finitely presented $S_f$-module. \end{enumerate} \end{enumerate}


algebra-lemma-dominate-by-dimension-1

Let $R$ be a local Noetherian domain with fraction field $K$. Assume $R$ is not a field. Then there exist $R \subset R' \subset K$ with \begin{enumerate} \item $R'$ local Noetherian of dimension $1$, \item $R \to R'$ a local ring map, i.e., $R'$ dominates $R$, and \item $R \to R'$ essentially of finite type. \end{enumerate}


algebra-lemma-hart-serre-loc-thm

\begin{reference} This is taken from a forthcoming paper by J\'anos Koll\'ar entitled ``Variants of normality for Noetherian schemes''. \end{reference} Let $(R, \mathfrak m)$ be a local Noetherian ring. Then exactly one of the following holds: \begin{enumerate} \item $(R, \mathfrak m)$ is Artinian, \item $(R, \mathfrak m)$ is regular of dimension $1$, \item $\text{depth}(R) \geq 2$, or \item there exists a finite ring map $R \to R'$ which is not an isomorphism whose kernel and cokernel are annihilated by a power of $\mathfrak m$ such that $\mathfrak m$ is not an associated prime of $R'$. \end{enumerate}


algebra-lemma-nonregular-dimension-one

Let $R$ be a local ring with maximal ideal $\mathfrak m$. Assume $R$ is Noetherian, has dimension $1$, and that $\dim(\mathfrak m/\mathfrak m^2) > 1$. Then there exists a ring map $R \to R'$ such that \begin{enumerate} \item $R \to R'$ is finite, \item $R \to R'$ is not an isomorphism, \item the kernel and cokernel of $R \to R'$ are annihilated by a power of $\mathfrak m$, and \item $\mathfrak m$ is not an associated prime of $R'$. \end{enumerate}


algebra-lemma-characterize-dvr

Let $A$ be a ring. The following are equivalent. \begin{enumerate} \item The ring $A$ is a discrete valuation ring. \item The ring $A$ is a valuation ring and Noetherian but not a field. \item The ring $A$ is a regular local ring of dimension $1$. \item The ring $A$ is a Noetherian local domain with maximal ideal $\mathfrak m$ generated by a single nonzero element. \item The ring $A$ is a Noetherian local normal domain of dimension $1$. \end{enumerate} In this case if $\pi$ is a generator of the maximal ideal of $A$, then every element of $A$ can be uniquely written as $u\pi^n$, where $u \in A$ is a unit.


algebra-lemma-finite-length

Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is local Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_R(R/xR) < \infty$ and $$ \text{length}_R(M/xM) \leq r \cdot \text{length}_R(R/xR). $$


algebra-lemma-finite-extension-residue-fields-dimension-1

Let $R \to S$ be a homomorphism of domains inducing an injection of fraction fields $K \subset L$. If $R$ is Noetherian local of dimension $1$ and $[L : K] < \infty$ then \begin{enumerate} \item each prime ideal $\mathfrak n_i$ of $S$ lying over the maximal ideal $\mathfrak m$ of $R$ is maximal, \item there are finitely many of these, and \item $[\kappa(\mathfrak n_i) : \kappa(\mathfrak m)] < \infty$ for each $i$. \end{enumerate}


algebra-lemma-finite-length-global

Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_R(M/xM) < \infty$.


algebra-lemma-krull-akizuki

Let $R$ be a domain with fraction field $K$. Let $K \subset L$ be a finite extension of fields. Assume $R$ is Noetherian and $\dim(R) = 1$. In this case any ring $A$ with $R \subset A \subset L$ is Noetherian.


algebra-lemma-exists-dvr

Let $R$ be a Noetherian local domain with fraction field $K$. Assume that $R$ is not a field. Let $K \subset L$ be a finitely generated field extension. Then there exists discrete valuation ring $A$ with fraction field $L$ which dominates $R$.


algebra-lemma-easy-divisibility

Let $R$ be a domain. Let $x, y \in R$. Then $x$, $y$ are associates if and only if $(x) = (y)$.


algebra-lemma-factorization-exists

Let $R$ be a domain. Consider the following conditions: \begin{enumerate} \item The ring $R$ satisfies the ascending chain condition for principal ideals. \item Every nonzero, nonunit element $a \in R$ has a factorization $a = b_1 \ldots b_k$ with each $b_i$ an irreducible element of $R$. \end{enumerate} Then (1) implies (2).


algebra-lemma-characterize-UFD

Let $R$ be a domain. Assume every nonzero, nonunit factors into irreducibles. Then $R$ is a UFD if and only if every irreducible element is prime.


algebra-lemma-characterize-UFD-height-1

Let $R$ be a Noetherian domain. Then $R$ is a UFD if and only if every height $1$ prime ideal is principal.


algebra-lemma-invert-prime-elements

\begin{reference} \cite[Lemma 2]{Nagata-UFD} \end{reference} Let $A$ be a domain. Let $S \subset A$ be a multiplicative subset generated by prime elements. Let $x \in A$ be irreducible. Then \begin{enumerate} \item the image of $x$ in $S^{-1}A$ is irreducible or a unit, and \item $x$ is prime if and only if the image of $x$ in $S^{-1}A$ is a unit or a prime element in $S^{-1}A$. \end{enumerate} Moreover, then $A$ is a UFD if and only if every element of $A$ has a factorization into irreducibles and $S^{-1}A$ is a UFD.


algebra-lemma-UFD-ascending-chain-condition-principal-ideals

A UFD satisfies the ascending chain condition for principal ideals.


algebra-lemma-factoring-in-polynomial

Let $R$ be a domain. Assume $R$ has the ascending chain condition for principal ideals. Then the same property holds for a polynomial ring over $R$.


algebra-lemma-polynomial-ring-UFD

A polynomial ring over a UFD is a UFD. In particular, if $k$ is a field, then $k[x_1, \ldots, x_n]$ is a UFD.


algebra-lemma-UFD-normal

A unique factorization domain is normal.


algebra-lemma-PID-UFD

A principal ideal domain is a unique factorization domain.


algebra-lemma-PID-dedekind

A PID is a Dedekind domain.


algebra-lemma-product-ideals-principal

\begin{slogan} A product of ideals is an invertible module iff both factors are. \end{slogan} Let $A$ be a ring. Let $I$ and $J$ be nonzero ideals of $A$ such that $IJ = (f)$ for some nonzerodivisor $f \in A$. Then $I$ and $J$ are finitely generated ideals and finitely locally free of rank $1$ as $A$-modules.


algebra-lemma-characterize-Dedekind

Let $R$ be a ring. The following are equivalent \begin{enumerate} \item $R$ is a Dedekind domain, \item $R$ is a Noetherian domain, and for every maximal ideal $\mathfrak m$ the local ring $R_{\mathfrak m}$ is a discrete valuation ring, and \item $R$ is a Noetherian, normal domain, and $\dim(R) \leq 1$. \end{enumerate}


algebra-lemma-integral-closure-Dedekind

Let $A$ be a Noetherian domain of dimension $1$ with fraction field $K$. Let $K \subset L$ be a finite extension. Let $B$ be the integral closure of $A$ in $L$. Then $B$ is a Dedekind domain and $\Spec(B) \to \Spec(A)$ is surjective, has finite fibres, and induces finite residue field extensions.


algebra-lemma-ord-additive

Let $R$ be a semi-local Noetherian ring of dimension $1$. If $a, b \in R$ are nonzerodivisors then $$ \text{length}_R(R/(ab)) = \text{length}_R(R/(a)) + \text{length}_R(R/(b)) $$ and these lengths are finite.


algebra-lemma-compare-lattices

Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. \begin{enumerate} \item If $M$ is a lattice in $V$ and $M \subset M' \subset V$ is an $R$-submodule of $V$ containing $M$ then the following are equivalent \begin{enumerate} \item $M'$ is a lattice, \item $\text{length}_R(M'/M)$ is finite, and \item $M'$ is finitely generated. \end{enumerate} \item If $M$ is a lattice in $V$ and $M' \subset M$ is an $R$-submodule of $M$ then $M'$ is a lattice if and only if $\text{length}_R(M/M')$ is finite. \item If $M$, $M'$ are lattices in $V$, then so are $M \cap M'$ and $M + M'$. \item If $M \subset M' \subset M'' \subset V$ are lattices in $V$ then $$ \text{length}_R(M''/M) = \text{length}_R(M'/M) + \text{length}_R(M''/M'). $$ \item If $M$, $M'$, $N$, $N'$ are lattices in $V$ and $N \subset M \cap M'$, $M + M' \subset N'$, then we have \begin{eqnarray*} & & \text{length}_R(M/M \cap M') - \text{length}_R(M'/M \cap M')\\ & = & \text{length}_R(M/N) - \text{length}_R(M'/N) \\ & = & \text{length}_R(M + M' / M') - \text{length}_R(M + M'/M) \\ & = & \text{length}_R(N' / M') - \text{length}_R(N'/M) \end{eqnarray*} \end{enumerate}


algebra-lemma-properties-distance-function

Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. This distance function has the property that $$ d(M, M'') = d(M, M') + d(M', M'') $$ whenever given three lattices $M$, $M'$, $M''$ of $V$. In particular we have $d(M, M') = - d(M', M)$.


algebra-lemma-order-vanishing-determinant

Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. Let $\varphi : V \to V$ be a $K$-linear isomorphism. For any lattice $M \subset V$ we have $$ d(M, \varphi(M)) = \text{ord}_R(\det(\varphi)) $$


algebra-lemma-finite-extension-dim-1

Let $A \to B$ be a ring map. Assume \begin{enumerate} \item $A$ is a Noetherian local domain of dimension $1$, \item $A \subset B$ is a finite extension of domains. \end{enumerate} Let $L/K$ be the corresponding finite extension of fraction fields. Let $y \in L^*$ and $x = \text{Nm}_{L/K}(y)$. In this situation $B$ is semi-local. Let $\mathfrak m_i$, $i = 1, \ldots, n$ be the maximal ideals of $B$. Then $$ \text{ord}_A(x) = \sum\nolimits_i [\kappa(\mathfrak m_i) : \kappa(\mathfrak m_A)] \text{ord}_{B_{\mathfrak m_i}}(y) $$ where $\text{ord}$ is defined as in Definition \ref{definition-ord}.


algebra-lemma-isolated-point

Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $\mathfrak q$ be a prime of $S$. The following are equivalent: \begin{enumerate} \item $\mathfrak q$ is an isolated point of $\Spec(S)$, \item $S_{\mathfrak q}$ is finite over $k$, \item there exists a $g \in S$, $g \not\in \mathfrak q$ such that $D(g) = \{ \mathfrak q \}$, \item $\dim_{\mathfrak q} \Spec(S) = 0$, \item $\mathfrak q$ is a closed point of $\Spec(S)$ and $\dim(S_{\mathfrak q}) = 0$, and \item the field extension $k \subset \kappa(\mathfrak q)$ is finite and $\dim(S_{\mathfrak q}) = 0$. \end{enumerate} In this case $S = S_{\mathfrak q} \times S'$ for some finite type $k$-algebra $S'$. Also, the element $g$ as in (3) has the property $S_{\mathfrak q} = S_g$.


algebra-lemma-isolated-point-fibre

\begin{slogan} Equivalent conditions for isolated points in fibres \end{slogan} Let $R \to S$ be a ring map of finite type. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $F = \Spec(S \otimes_R \kappa(\mathfrak p))$ be the fibre of $\Spec(S) \to \Spec(R)$, see Remark \ref{remark-fundamental-diagram}. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent \begin{enumerate} \item $\overline{\mathfrak q}$ is an isolated point of $F$, \item $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa(\mathfrak p)$, \item there exists a $g \in S$, $g \not \in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$, \item $\dim_{\overline{\mathfrak q}}(F) = 0$, \item $\overline{\mathfrak q}$ is a closed point of $F$ and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$. \end{enumerate}


algebra-lemma-quasi-finite

Let $R \to S$ be a finite type ring map. Then $R \to S$ is quasi-finite if and only if for all primes $\mathfrak p \subset R$ the fibre $S \otimes_R \kappa(\mathfrak p)$ is finite over $\kappa(\mathfrak p)$.


algebra-lemma-quasi-finite-local

Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $f \in R$, $f \not \in \mathfrak p$ and $g \in S$, $g \not \in \mathfrak q$. Then $R \to S$ is quasi-finite at $\mathfrak q$ if and only if $R_f \to S_{fg}$ is quasi-finite at $\mathfrak qS_{fg}$.


algebra-lemma-four-rings

Let $$ \xymatrix{ S \ar[r] & S' & & \mathfrak q \ar@{-}[r] & \mathfrak q' \\ R \ar[u] \ar[r] &  R' \ar[u] & & \mathfrak p \ar@{-}[r] \ar@{-}[u] & \mathfrak p' \ar@{-}[u] } $$ be a commutative diagram of rings with primes as indicated. Assume $R \to S$ of finite type, and $S \otimes_R R' \to S'$ surjective. If $R \to S$ is quasi-finite at $\mathfrak q$, then $R' \to S'$ is quasi-finite at $\mathfrak q'$.


algebra-lemma-quasi-finite-composition

A composition of quasi-finite ring maps is quasi-finite.


algebra-lemma-quasi-finite-base-change

Let $R \to S$ be a ring map of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes_R S$. \begin{enumerate} \item The set $\{\mathfrak q' \mid R' \to S' \text{ quasi-finite at }\mathfrak q'\}$ is the inverse image of the corresponding set of $\Spec(S)$ under the canonical map $\Spec(S') \to \Spec(S)$. \item If $\Spec(R') \to \Spec(R)$ is surjective, then $R \to S$ is quasi-finite if and only if $R' \to S'$ is quasi-finite. \item Any base change of a quasi-finite ring map is quasi-finite. \end{enumerate}


algebra-lemma-quasi-finite-permanence

Let $A \to B$ and $B \to C$ be ring homomorphisms such that $A \to C$ is of finite type. Let $\mathfrak r$ be a prime of $C$ lying over $\mathfrak q \subset B$ and $\mathfrak p \subset A$. If $A \to C$ is quasi-finite at $\mathfrak r$, then $B \to C$ is quasi-finite at $\mathfrak r$.


algebra-lemma-generically-finite

Let $R \to S$ be a ring map of finite type. Let $\mathfrak p \subset R$ be a minimal prime. Assume that there are at most finitely many primes of $S$ lying over $\mathfrak p$. Then there exists a $g \in R$, $g \not \in \mathfrak p$ such that the ring map $R_g \to S_g$ is finite.


algebra-lemma-make-integral-trivial

Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi(a_0) + \varphi(a_1)t + \ldots + \varphi(a_n) t^n = 0$. Then $\varphi(a_n)t$ is integral over $R$.


algebra-lemma-make-integral-trick

Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume that (a) $t$ is integral over $R[x]$, and (b) there exists a monic $p \in R[x]$ such that $t \varphi(p) \in \Im(\varphi)$. Then there exists a $q \in R[x]$ such that $t - \varphi(q)$ is integral over $R$.


algebra-lemma-combine-lemmas

Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume $t$ is integral over $R[x]$. Let $p \in R[x]$, $p = a_0 + a_1x + \ldots + a_k x^k$ such that $t \varphi(p) \in \Im(\varphi)$. Then there exists a $q \in R[x]$ and $n \geq 0$ such that $\varphi(a_k)^n t - \varphi(q)$ is integral over $R$.


algebra-lemma-leading-coefficient-in-J

In Situation \ref{situation-one-transcendental-element}. Suppose $u \in S$, $a_0, \ldots, a_k \in R$, $u \varphi(a_0 + a_1x + \ldots + a_k x^k) \in J$. Then there exists an $m \geq 0$ such that $u \varphi(a_k)^m \in J$.


algebra-lemma-all-coefficients-in-J

In Situation \ref{situation-one-transcendental-element}. Suppose $u \in S$, $a_0, \ldots, a_k \in R$, $u \varphi(a_0 + a_1x + \ldots + a_k x^k) \in \sqrt{J}$. Then $u \varphi(a_i) \in \sqrt{J}$ for all $i$.


algebra-lemma-reduced-strongly-transcendental-minimal-prime

Suppose $R \subset S$ is an inclusion of reduced rings and suppose that $x \in S$ is strongly transcendental over $R$. Let $\mathfrak q \subset S$ be a minimal prime and let $\mathfrak p = R \cap \mathfrak q$. Then the image of $x$ in $S/\mathfrak q$ is strongly transcendental over the subring $R/\mathfrak p$.


algebra-lemma-domains-transcendental-not-quasi-finite

Suppose $R\subset S$ is an inclusion of domains and let $x \in S$. Assume $x$ is (strongly) transcendental over $R$ and that $S$ is finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.


algebra-lemma-reduced-strongly-transcendental-not-quasi-finite

Suppose $R \subset S$ is an inclusion of reduced rings. Assume $x \in S$ be strongly transcendental over $R$, and $S$ finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.


algebra-lemma-quasi-finite-monogenic

Let $R$ be a ring. Let $S = R[x]/I$. Let $\mathfrak q \subset S$ be a prime. Assume $R \to S$ is quasi-finite at $\mathfrak q$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then there exists an element $g \in S'$, $g \not\in \mathfrak q$ such that $S'_g \cong S_g$.


algebra-lemma-quasi-finite-open

Let $R \to S$ be a finite type ring map. The set of points $\mathfrak q$ of $\Spec(S)$ at which $S/R$ is quasi-finite is open in $\Spec(S)$.


algebra-lemma-quasi-finite-open-integral-closure

Let $R \to S$ be a finite type ring map. Suppose that $S$ is quasi-finite over $R$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then \begin{enumerate} \item $\Spec(S) \to \Spec(S')$ is a homeomorphism onto an open subset, \item if $g \in S'$ and $D(g)$ is contained in the image of the map, then $S'_g \cong S_g$, and \item there exists a finite $R$-algebra $S'' \subset S'$ such that (1) and (2) hold for the ring map $S'' \to S$. \end{enumerate}


algebra-lemma-quasi-finite-extension-dim-1

Let $A \subset B$ be an extension of domains. Assume \begin{enumerate} \item $A$ is a local Noetherian ring of dimension $1$, \item $A \to B$ is of finite type, and \item the induced extension $L/K$ of fraction fields is finite. \end{enumerate} Then $B$ is semi-local. Let $x \in \mathfrak m_A$, $x \not = 0$. Let $\mathfrak m_i$, $i = 1, \ldots, n$ be the maximal ideals of $B$. Then $$ [L : K]\text{ord}_A(x) \geq \sum\nolimits_i [\kappa(\mathfrak m_i) : \kappa(\mathfrak m_A)] \text{ord}_{B_{\mathfrak m_i}}(x) $$ where $\text{ord}$ is defined as in Definition \ref{definition-ord}. We have equality if and only if $A \to B$ is finite.


algebra-lemma-essentially-finite-type-fibre-dim-zero

Let $(R, \mathfrak m_R) \to (S, \mathfrak m_S)$ be a local homomorphism of local rings. Assume \begin{enumerate} \item $R \to S$ is essentially of finite type, \item $\kappa(\mathfrak m_R) \subset \kappa(\mathfrak m_S)$ is finite, and \item $\dim(S/\mathfrak m_RS) = 0$. \end{enumerate} Then $S$ is the localization of a finite $R$-algebra.


algebra-lemma-completion-at-quasi-finite-prime

Let $R \to S$ be a ring map, $\mathfrak q$ a prime of $S$ lying over $\mathfrak p$ in $R$. If \begin{enumerate} \item $R$ is Noetherian, \item $R \to S$ is of finite type, and \item $R \to S$ is quasi-finite at $\mathfrak q$, \end{enumerate} then $R_\mathfrak p^\wedge \otimes_R S = S_\mathfrak q^\wedge \times B$ for some $R_\mathfrak p^\wedge$-algebra $B$.


algebra-lemma-quasi-finite-over-polynomial-algebra

Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime. Suppose that $\dim_{\mathfrak q}(S/R) = n$. There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_g$ is quasi-finite over a polynomial algebra $R[t_1, \ldots, t_n]$.


algebra-lemma-refined-quasi-finite-over-polynomial-algebra

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume \begin{enumerate} \item $R \to S$ is of finite type, \item $\dim_{\mathfrak q}(S/R) = n$, and \item $\text{trdeg}_{\kappa(\mathfrak p)}\kappa(\mathfrak q) = r$. \end{enumerate} Then there exist $f \in R$, $f \not \in \mathfrak p$, $g \in S$, $g \not\in \mathfrak q$ and a quasi-finite ring map $$ \varphi : R_f[x_1, \ldots, x_n] \longrightarrow S_g $$ such that $\varphi^{-1}(\mathfrak qS_g) = (\mathfrak p, x_{r + 1}, \ldots, x_n)R_f[x_{r + 1}, \ldots, x_n]$


algebra-lemma-dimension-inequality-quasi-finite

Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. If $R \to S$ is quasi-finite at $\mathfrak q$, then $\dim(S_{\mathfrak q}) \leq \dim(R_{\mathfrak p})$.


algebra-lemma-dimension-quasi-finite-over-polynomial-algebra

\begin{slogan} A quasi-finite cover of affine n-space has dimension at most n. \end{slogan} Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Suppose there is a quasi-finite $k$-algebra map $k[t_1, \ldots, t_n] \subset S$. Then $\dim(S) \leq n$.


algebra-lemma-dimension-fibres-bounded-open-upstairs

Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime. Suppose that $\dim_{\mathfrak q}(S/R) = n$. There exists an open neighbourhood $V$ of $\mathfrak q$ in $\Spec(S)$ such that $\dim_{\mathfrak q'}(S/R) \leq n$ for all $\mathfrak q' \in V$.


algebra-lemma-dimension-fibres-bounded-open-upstairs-base-change

Let $R \to S$ be a finite type ring map. Let $R \to R'$ be any ring map. Set $S' = R' \otimes_R S$ and denote $f : \Spec(S') \to \Spec(S)$ the associated map on spectra. Let $n \geq 0$. The inverse image $f^{-1}(\{\mathfrak q \in \Spec(S) \mid \dim_{\mathfrak q}(S/R) \leq n\})$ is equal to $\{\mathfrak q' \in \Spec(S') \mid \dim_{\mathfrak q'}(S'/R') \leq n\}$.


algebra-lemma-dimension-fibres-bounded-quasi-compact-open-upstairs

Let $R \to S$ be a ring homomorphism of finite presentation. Let $n \geq 0$. The set $$ V_n = \{\mathfrak q \in \Spec(S) \mid \dim_{\mathfrak q}(S/R) \leq n\} $$ is a quasi-compact open subset of $\Spec(S)$.


algebra-lemma-finite-type-domain-over-valuation-ring-dim-fibres

Let $R$ be a valuation ring with residue field $k$ and field of fractions $K$. Let $S$ be a domain containing $R$ such that $S$ is of finite type over $R$. If $S \otimes_R k$ is not the zero ring then $$ \dim(S \otimes_R k) = \dim(S \otimes_R K) $$ In fact, $\Spec(S \otimes_R k)$ is equidimensional.


algebra-lemma-finite-type-descends

Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes_R S$. Then $R \to S$ is of finite type if and only if $R' \to S'$ is of finite type.


algebra-lemma-finite-presentation-descends

Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes_R S$. Then $R \to S$ is of finite presentation if and only if $R' \to S'$ is of finite presentation.


algebra-lemma-construct-fp-module

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \subset R$ be a multiplicative subset. Set $R' = S^{-1}(R/I) = S^{-1}R/S^{-1}I$. \begin{enumerate} \item For any finite $R'$-module $M'$ there exists a finite $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$. \item For any finitely presented $R'$-module $M'$ there exists a finitely presented $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$. \end{enumerate}


algebra-lemma-construct-fp-module-from-localization

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module. \begin{enumerate} \item If $S^{-1}M$ is a finite $S^{-1}R$-module then there exists a finite $R$-module $M'$ and a map $M' \to M$ which induces an isomorphism $S^{-1}M' \to S^{-1}M$. \item If $S^{-1}M$ is a finitely presented $S^{-1}R$-module then there exists an $R$-module $M'$ of finite presentation and a map $M' \to M$ which induces an isomorphism $S^{-1}M' \to S^{-1}M$. \end{enumerate}


algebra-lemma-construct-fp-module-from-stalk

Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $M$ be an $R$-module. \begin{enumerate} \item If $M_{\mathfrak p}$ is a finite $R_{\mathfrak p}$-module then there exists a finite $R$-module $M'$ and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$. \item If $M_{\mathfrak p}$ is a finitely presented $R_{\mathfrak p}$-module then there exists an $R$-module $M'$ of finite presentation and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$. \end{enumerate}


algebra-lemma-local-isomorphism

Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Assume \begin{enumerate} \item $S$ is of finite presentation over $R$, \item $\varphi$ induces an isomorphism $R_\mathfrak p \cong S_\mathfrak q$. \end{enumerate} Then there exist $f \in R$, $f \not \in \mathfrak p$ and an $R_f$-algebra $C$ such that $S_f \cong R_f \times C$ as $R_f$-algebras.


algebra-lemma-isomorphic-local-rings

Let $R$ be a ring. Let $S$, $S'$ be of finite presentation over $R$. Let $\mathfrak q \subset S$ and $\mathfrak q' \subset S'$ be primes. If $S_{\mathfrak q} \cong S'_{\mathfrak q'}$ as $R$-algebras, then there exist $g \in S$, $g \not \in \mathfrak q$ and $g' \in S'$, $g' \not \in \mathfrak q'$ such that $S_g \cong S'_{g'}$ as $R$-algebras.


algebra-lemma-finite-type-mod-nilpotent

Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $S$ be an $R$-algebra such that $R/I \to S/IS$ is of finite type. Then $R \to S$ is of finite type.


algebra-lemma-surjective-mod-locally-nilpotent

Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $S \to S'$ be an $R$-algebra map such that $S \to S'/IS'$ is surjective and such that $S'$ is of finite type over $R$. Then $S \to S'$ is surjective.


algebra-lemma-isomorphism-modulo-ideal

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Let $IS \subset \mathfrak q \subset S$ be a prime ideal. Assume that \begin{enumerate} \item $S \to S'$ is surjective, \item $S_\mathfrak q/IS_\mathfrak q \to S'_\mathfrak q/IS'_\mathfrak q$ is an isomorphism, \item $S$ is of finite type over $R$, \item $S'$ of finite presentation over $R$, and \item $S'_\mathfrak q$ is flat over $R$. \end{enumerate} Then $S_g \to S'_g$ is an isomorphism for some $g \in S$, $g \not \in \mathfrak q$.


algebra-lemma-isomorphism-modulo-locally-nilpotent

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Assume that \begin{enumerate} \item $I$ is locally nilpotent, \item $S/IS \to S'/IS'$ is an isomorphism, \item $S$ is of finite type over $R$, \item $S'$ of finite presentation over $R$, and \item $S'$ is flat over $R$. \end{enumerate} Then $S \to S'$ is an isomorphism.


algebra-lemma-ring-colimit-fp-category

Let $R \to A$ be a ring map. Consider the category $\mathcal{I}$ of all diagrams of $R$-algebra maps $A' \to A$ with $A'$ finitely presented over $R$. Then $\mathcal{I}$ is filtered, and the colimit of the $A'$ over $\mathcal{I}$ is isomorphic to $A$.


algebra-lemma-ring-colimit-fp

Let $R \to A$ be a ring map. There exists a directed system $A_\lambda$ of $R$-algebras of finite presentation such that $A = \colim_\lambda A_\lambda$. If $A$ is of finite type over $R$ we may arrange it so that all the transition maps in the system of $A_\lambda$ are surjective.


algebra-lemma-characterize-finite-presentation

Let $\varphi : R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $\varphi$ is of finite presentation, \item for every directed system $A_\lambda$ of $R$-algebras the map $$ \colim_\lambda \Hom_R(S, A_\lambda) \longrightarrow \Hom_R(S, \colim_\lambda A_\lambda) $$ is bijective, and \item for every directed system $A_\lambda$ of $R$-algebras the map $$ \colim_\lambda \Hom_R(S, A_\lambda) \longrightarrow \Hom_R(S, \colim_\lambda A_\lambda) $$ is surjective. \end{enumerate}


algebra-lemma-when-colimit

Let $R \to \Lambda$ be a ring map. Let $\mathcal{E}$ be a set of $R$-algebras such that each $A \in \mathcal{E}$ is of finite presentation over $R$. Then the following two statements are equivalent \begin{enumerate} \item $\Lambda$ is a filtered colimit of elements of $\mathcal{E}$, and \item for any $R$ algebra map $A \to \Lambda$ with $A$ of finite presentation over $R$ we can find a factorization $A \to B \to \Lambda$ with $B \in \mathcal{E}$. \end{enumerate}


algebra-lemma-module-map-property-in-colimit

Let $A$ be a ring and let $M, N$ be $A$-modules. Suppose that $R = \colim_{i \in I} R_i$ is a directed colimit of $A$-algebras. \begin{enumerate} \item If $M$ is a finite $A$-module, and $u, u' : M \to N$ are $A$-module maps such that $u \otimes 1 = u' \otimes 1 : M \otimes_A R \to N \otimes_A R$ then for some $i$ we have $u \otimes 1 = u' \otimes 1 : M \otimes_A R_i \to N \otimes_A R_i$. \item If $N$ is a finite $A$-module and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes_A R \to N \otimes_A R$ is surjective, then for some $i$ the map $u \otimes 1 : M \otimes_A R_i \to N \otimes_A R_i$ is surjective. \item If $N$ is a finitely presented $A$-module, and $v : N \otimes_A R \to M \otimes_A R$ is an $R$-module map, then there exists an $i$ and an $R_i$-module map $v_i : N \otimes_A R_i \to M \otimes_A R_i$ such that $v = v_i \otimes 1$. \item If $M$ is a finite $A$-module, $N$ is a finitely presented $A$-module, and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes_A R \to N \otimes_A R$ is an isomorphism, then for some $i$ the map $u \otimes 1 : M \otimes_A R_i \to N \otimes_A R_i$ is an isomorphism. \end{enumerate}


algebra-lemma-colimit-category-fp-modules

Suppose that $R = \colim_{\lambda \in \Lambda} R_\lambda$ is a directed colimit of rings. Then the category of finitely presented $R$-modules is the colimit of the categories of finitely presented $R_\lambda$-modules. More precisely \begin{enumerate} \item Given a finitely presented $R$-module $M$ there exists a $\lambda \in \Lambda$ and a finitely presented $R_\lambda$-module $M_\lambda$ such that $M \cong M_\lambda \otimes_{R_\lambda} R$. \item Given a $\lambda \in \Lambda$, finitely presented $R_\lambda$-modules $M_\lambda, N_\lambda$, and an $R$-module map $\varphi : M_\lambda \otimes_{R_\lambda} R \to N_\lambda \otimes_{R_\lambda} R$, then there exists a $\mu \geq \lambda$ and an $R_\mu$-module map $\varphi_\mu : M_\lambda \otimes_{R_\lambda} R_\mu \to N_\lambda \otimes_{R_\lambda} R_\mu$ such that $\varphi = \varphi_\mu \otimes 1_R$. \item Given a $\lambda \in \Lambda$, finitely presented $R_\lambda$-modules $M_\lambda, N_\lambda$, and $R$-module maps $\varphi_\lambda, \psi_\lambda : M_\lambda \to N_\lambda$ such that $\varphi \otimes 1_R = \psi \otimes 1_R$, then $\varphi \otimes 1_{R_\mu} = \psi \otimes 1_{R_\mu}$ for some $\mu \geq \lambda$. \end{enumerate}


algebra-lemma-algebra-map-property-in-colimit

Let $A$ be a ring and let $B, C$ be $A$-algebras. Suppose that $R = \colim_{i \in I} R_i$ is a directed colimit of $A$-algebras. \begin{enumerate} \item If $B$ is a finite type $A$-algebra, and $u, u' : B \to C$ are $A$-algebra maps such that $u \otimes 1 = u' \otimes 1 : B \otimes_A R \to C \otimes_A R$ then for some $i$ we have $u \otimes 1 = u' \otimes 1 : B \otimes_A R_i \to C \otimes_A R_i$. \item If $C$ is a finite type $A$-algebra and $u : B \to C$ is an $A$-algebra map such that $u \otimes 1 : B \otimes_A R \to C \otimes_A R$ is surjective, then for some $i$ the map $u \otimes 1 : B \otimes_A R_i \to C \otimes_A R_i$ is surjective. \item If $C$ is of finite presentation over $A$ and $v : C \otimes_A R \to B \otimes_A R$ is an $R$-algebra map, then there exists an $i$ and an $R_i$-algebra map $v_i : C \otimes_A R_i \to B \otimes_A R_i$ such that $v = v_i \otimes 1$. \item If $B$ is a finite type $A$-algebra, $C$ is a finitely presented $A$-algebra, and $u \otimes 1 : B \otimes_A R \to C \otimes_A R$ is an isomorphism, then for some $i$ the map $u \otimes 1 : B \otimes_A R_i \to C \otimes_A R_i$ is an isomorphism. \end{enumerate}


algebra-lemma-colimit-category-fp-algebras

Suppose that $R = \colim_{\lambda \in \Lambda} R_\lambda$ is a directed colimit of rings. Then the category of finitely presented $R$-algebras is the colimit of the categories of finitely presented $R_\lambda$-algebras. More precisely \begin{enumerate} \item Given a finitely presented $R$-algebra $A$ there exists a $\lambda \in \Lambda$ and a finitely presented $R_\lambda$-algebra $A_\lambda$ such that $A \cong A_\lambda \otimes_{R_\lambda} R$. \item Given a $\lambda \in \Lambda$, finitely presented $R_\lambda$-algebras $A_\lambda, B_\lambda$, and an $R$-algebra map $\varphi : A_\lambda \otimes_{R_\lambda} R \to B_\lambda \otimes_{R_\lambda} R$, then there exists a $\mu \geq \lambda$ and an $R_\mu$-algebra map $\varphi_\mu : A_\lambda \otimes_{R_\lambda} R_\mu \to B_\lambda \otimes_{R_\lambda} R_\mu$ such that $\varphi = \varphi_\mu \otimes 1_R$. \item Given a $\lambda \in \Lambda$, finitely presented $R_\lambda$-algebras $A_\lambda, B_\lambda$, and $R_\lambda$-algebra maps $\varphi_\lambda, \psi_\lambda : A_\lambda \to B_\lambda$ such that $\varphi \otimes 1_R = \psi \otimes 1_R$, then $\varphi \otimes 1_{R_\mu} = \psi \otimes 1_{R_\mu}$ for some $\mu \geq \lambda$. \end{enumerate}


algebra-lemma-limit-no-condition-local

Suppose $R \to S$ is a local homomorphism of local rings. There exists a directed set $(\Lambda, \leq)$, and a system of local homomorphisms $R_\lambda \to S_\lambda$ of local rings such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. \item Each $R_\lambda$ is essentially of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is essentially of finite type over $R_\lambda$. \end{enumerate}


algebra-lemma-limit-essentially-finite-type

Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite type over $R$. Then there exists a directed set $(\Lambda, \leq)$, and a system of local homomorphisms $R_\lambda \to S_\lambda$ of local rings such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. \item Each $R_\lambda$ is essentially of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is essentially of finite type over $R_\lambda$. \item For each $\lambda \leq \mu$ the map $S_\lambda \otimes_{R_\lambda} R_\mu \to S_\mu$ presents $S_\mu$ as the localization of a quotient of $S_\lambda \otimes_{R_\lambda} R_\mu$. \end{enumerate}


algebra-lemma-limit-essentially-finite-presentation

Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite presentation over $R$. Then there exists a directed set $(\Lambda, \leq)$, and a system of local homomorphism $R_\lambda \to S_\lambda$ of local rings such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. \item Each $R_\lambda$ is essentially of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is essentially of finite type over $R_\lambda$. \item For each $\lambda \leq \mu$ the map $S_\lambda \otimes_{R_\lambda} R_\mu \to S_\mu$ presents $S_\mu$ as the localization of $S_\lambda \otimes_{R_\lambda} R_\mu$ at a prime ideal. \end{enumerate}


algebra-lemma-limit-module-essentially-finite-presentation

Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite presentation over $R$. Let $M$ be a finitely presented $S$-module. Then there exists a directed set $(\Lambda, \leq)$, and a system of local homomorphisms $R_\lambda \to S_\lambda$ of local rings together with $S_\lambda$-modules $M_\lambda$, such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. The colimit of the system $M_\lambda$ is $M$. \item Each $R_\lambda$ is essentially of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is essentially of finite type over $R_\lambda$. \item Each $M_\lambda$ is finite over $S_\lambda$. \item For each $\lambda \leq \mu$ the map $S_\lambda \otimes_{R_\lambda} R_\mu \to S_\mu$ presents $S_\mu$ as the localization of $S_\lambda \otimes_{R_\lambda} R_\mu$ at a prime ideal. \item For each $\lambda \leq \mu$ the map $M_\lambda \otimes_{S_\lambda} S_\mu \to M_\mu$ is an isomorphism. \end{enumerate}


algebra-lemma-limit-no-condition

Suppose $R \to S$ is a ring map. Then there exists a directed set $(\Lambda, \leq)$, and a system of ring maps $R_\lambda \to S_\lambda$ such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. \item Each $R_\lambda$ is of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is of finite type over $R_\lambda$. \end{enumerate}


algebra-lemma-limit-integral

Suppose $R \to S$ is a ring map. Assume that $S$ is integral over $R$. Then there exists a directed set $(\Lambda, \leq)$, and a system of ring maps $R_\lambda \to S_\lambda$ such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. \item Each $R_\lambda$ is of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is of finite over $R_\lambda$. \end{enumerate}


algebra-lemma-limit-module-finite-presentation

Suppose $R \to S$ is a ring map. Assume that $S$ is of finite presentation over $R$. Let $M$ be a finitely presented $S$-module. Then there exists a directed set $(\Lambda, \leq)$, and a system of ring maps $R_\lambda \to S_\lambda$ together with $S_\lambda$-modules $M_\lambda$, such that \begin{enumerate} \item The colimit of the system $R_\lambda \to S_\lambda$ is equal to $R \to S$. The colimit of the system $M_\lambda$ is $M$. \item Each $R_\lambda$ is of finite type over $\mathbf{Z}$. \item Each $S_\lambda$ is of finite type over $R_\lambda$. \item Each $M_\lambda$ is finite over $S_\lambda$. \item For each $\lambda \leq \mu$ the map $S_\lambda \otimes_{R_\lambda} R_\mu \to S_\mu$ is an isomorphism. \item For each $\lambda \leq \mu$ the map $M_\lambda \otimes_{S_\lambda} S_\mu \to M_\mu$ is an isomorphism. \end{enumerate} In particular, for every $\lambda \in \Lambda$ we have $$ M = M_\lambda \otimes_{S_\lambda} S = M_\lambda \otimes_{R_\lambda} R. $$


algebra-lemma-CM-over-regular-flat

\begin{slogan} Miracle flatness \end{slogan} Let $R \to S$ be a local homomorphism of Noetherian local rings. Assume \begin{enumerate} \item $R$ is regular, \item $S$ Cohen-Macaulay, \item $\dim(S) = \dim(R) + \dim(S/\mathfrak m_R S)$. \end{enumerate} Then $R \to S$ is flat.


algebra-lemma-flat-over-regular

Let $R \to S$ be a homomorphism of Noetherian local rings. Assume that $R$ is a regular local ring and that a regular system of parameters maps to a regular sequence in $S$. Then $R \to S$ is flat.


algebra-lemma-colimit-eventually-flat

Let $R \to S$, $M$, $\Lambda$, $R_\lambda \to S_\lambda$, $M_\lambda$ be as in Lemma \ref{lemma-limit-module-essentially-finite-presentation}. Assume that $M$ is flat over $R$. Then for some $\lambda \in \Lambda$ the module $M_\lambda$ is flat over $R_\lambda$.


algebra-lemma-mod-injective-general

Suppose that $R \to S$ is a local homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $u : M \to N$ be a map of $S$-modules. Assume \begin{enumerate} \item $S$ is essentially of finite presentation over $R$, \item $M$, $N$ are finitely presented over $S$, \item $N$ is flat over $R$, and \item $\overline{u} : M/\mathfrak mM \to N/\mathfrak mN$ is injective. \end{enumerate} Then $u$ is injective, and $N/u(M)$ is flat over $R$.


algebra-lemma-grothendieck-general

Suppose that $R \to S$ is a local ring homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose \begin{enumerate} \item $S$ is essentially of finite presentation over $R$, \item $S$ is flat over $R$, and \item $f \in S$ is a nonzerodivisor in $S/{\mathfrak m}S$. \end{enumerate} Then $S/fS$ is flat over $R$, and $f$ is a nonzerodivisor in $S$.


algebra-lemma-grothendieck-regular-sequence-general

Suppose that $R \to S$ is a local ring homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose \begin{enumerate} \item $R \to S$ is essentially of finite presentation, \item $R \to S$ is flat, and \item $f_1, \ldots, f_c$ is a sequence of elements of $S$ such that the images $\overline{f}_1, \ldots, \overline{f}_c$ form a regular sequence in $S/{\mathfrak m}S$. \end{enumerate} Then $f_1, \ldots, f_c$ is a regular sequence in $S$ and each of the quotients $S/(f_1, \ldots, f_i)$ is flat over $R$.


algebra-lemma-criterion-flatness-fibre

Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume \begin{enumerate} \item The ring maps $R \to S$ and $R \to S'$ are essentially of finite presentation. \item The module $M$ is of finite presentation over $S'$. \item The module $M$ is not zero. \item The module $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module. \item The module $M$ is a flat $R$-module. \end{enumerate} Then $S$ is flat over $R$ and $M$ is a flat $S$-module.


algebra-lemma-criterion-flatness-fibre-fp-over-ft

Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume \begin{enumerate} \item $R \to S'$ is essentially of finite presentation, \item $R \to S$ is essentially of finite type, \item $M$ is of finite presentation over $S'$, \item $M$ is not zero, \item $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module, and \item $M$ is a flat $R$-module. \end{enumerate} Then $S$ is essentially of finite presentation and flat over $R$ and $M$ is a flat $S$-module.


algebra-lemma-criterion-flatness-fibre-locally-nilpotent

Let $$ \xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] } $$ be a commutative diagram in the category of rings. Let $I \subset R$ be a locally nilpotent ideal and $M$ an $S'$-module. Assume \begin{enumerate} \item $R \to S$ is of finite type, \item $R \to S'$ is of finite presentation, \item $M$ is a finitely presented $S'$-module, \item $M/IM$ is flat as a $S/IS$-module, and \item $M$ is flat as an $R$-module. \end{enumerate} Then $M$ is a flat $S$-module and $S_\mathfrak q$ is flat and essentially of finite presentation over $R$ for every $\mathfrak q \subset S$ such that $M \otimes_S \kappa(\mathfrak q)$ is nonzero.


algebra-lemma-CM-dim-finite-type

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $f_1, \ldots, f_i$ be elements of $S$. Assume that $S$ is Cohen-Macaulay and equidimensional of dimension $d$, and that $\dim V(f_1, \ldots, f_i) \leq d - i$. Then equality holds and $f_1, \ldots, f_i$ forms a regular sequence in $S_{\mathfrak q}$ for every prime $\mathfrak q$ of $V(f_1, \ldots, f_i)$.


algebra-lemma-open-regular-sequence

Suppose that $R \to S$ is a ring map which is finite type, flat. Let $d$ be an integer such that all fibres $S \otimes_R \kappa(\mathfrak p)$ are Cohen-Macaulay and equidimensional of dimension $d$. Let $f_1, \ldots, f_i$ be elements of $S$. The set $$ \{ \mathfrak q \in V(f_1, \ldots, f_i) \mid f_1, \ldots, f_i \text{ are a regular sequence in } S_{\mathfrak q}/\mathfrak p S_{\mathfrak q} \text{ where }\mathfrak p = R \cap \mathfrak q \} $$ is open in $V(f_1, \ldots, f_i)$.


algebra-lemma-exact-on-fibres-open

Let $R \to S$ is a ring map. Consider a finite homological complex of finite free $S$-modules: $$ F_{\bullet} : 0 \to S^{n_e} \xrightarrow{\varphi_e} S^{n_{e-1}} \xrightarrow{\varphi_{e-1}} \ldots \xrightarrow{\varphi_{i + 1}} S^{n_i} \xrightarrow{\varphi_i} S^{n_{i-1}} \xrightarrow{\varphi_{i-1}} \ldots \xrightarrow{\varphi_1} S^{n_0} $$ For every prime $\mathfrak q$ of $S$ consider the complex $\overline{F}_{\bullet, \mathfrak q} = F_{\bullet, \mathfrak q} \otimes_R \kappa(\mathfrak p)$ where $\mathfrak p$ is inverse image of $\mathfrak q$ in $R$. Assume there exists an integer $d$ such that $R \to S$ is finite type, flat with fibres $S \otimes_R \kappa(\mathfrak p)$ Cohen-Macaulay of dimension $d$. The set $$ \{\mathfrak q \in \Spec(S) \mid \overline{F}_{\bullet, \mathfrak q}\text{ is exact}\} $$ is open in $\Spec(S)$.


algebra-lemma-where-CM

Let $S$ be a finite type algebra over a field $k$. Let $\varphi : k[y_1, \ldots, y_d] \to S$ be a quasi-finite ring map. As subsets of $\Spec(S)$ we have $$ \{ \mathfrak q \mid S_{\mathfrak q} \text{ flat over }k[y_1, \ldots, y_d]\} = \{ \mathfrak q \mid S_{\mathfrak q} \text{ CM and }\dim_{\mathfrak q}(S/k) = d\} $$ For notation see Definition \ref{definition-relative-dimension}.


algebra-lemma-finite-type-over-field-CM-open

Let $S$ be a finite type algebra over a field $k$. The set of primes $\mathfrak q$ such that $S_{\mathfrak q}$ is Cohen-Macaulay is open in $S$.


algebra-lemma-generic-CM

Let $k$ be a field. Let $S$ be a finite type $k$ algebra. The set of Cohen-Macaulay primes forms a dense open $U \subset \Spec(S)$.


algebra-lemma-dim-not-zero-exists-nonzerodivisor-nonunit

Let $k$ be a field. Let $S$ be a finite type $k$ algebra. If $\dim(S) > 0$, then there exists an element $f \in S$ which is a nonzerodivisor and a nonunit.


algebra-lemma-finite-presentation-flat-CM-locus-open

Let $R$ be a ring. Let $R \to S$ be of finite presentation and flat. For any $d \geq 0$ the set $$ \left\{ \begin{matrix} \mathfrak q \in \Spec(S) \text{ such that setting }\mathfrak p = R \cap \mathfrak q \text{ the fibre ring}\\ S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \text{ is Cohen-Macaulay} \text{ and } \dim_{\mathfrak q}(S/R) = d \end{matrix} \right\} $$ is open in $\Spec(S)$.


algebra-lemma-generic-CM-flat-finite-presentation

Let $R$ be a ring. Let $R \to S$ be flat of finite presentation. The set of primes $\mathfrak q$ such that the fibre ring $S_{\mathfrak q} \otimes_R \kappa(\mathfrak p)$, with $\mathfrak p = R \cap \mathfrak q$ is Cohen-Macaulay is open and dense in every fibre of $\Spec(S) \to \Spec(R)$.


algebra-lemma-extend-field-CM-locus

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension, and set $S_K = K \otimes_k S$. Let $\mathfrak q \subset S$ be a prime of $S$. Let $\mathfrak q_K \subset S_K$ be a prime of $S_K$ lying over $\mathfrak q$. Then $S_{\mathfrak q}$ is Cohen-Macaulay if and only if $(S_K)_{\mathfrak q_K}$ is Cohen-Macaulay.


algebra-lemma-CM-locus-commutes-base-change

Let $R$ be a ring. Let $R \to S$ be of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes_R S$. Denote $f : \Spec(S') \to \Spec(S)$ the map associated to the ring map $S \to S'$. Set $W$ equal to the set of primes $\mathfrak q$ such that the fibre ring $S_{\mathfrak q} \otimes_R \kappa(\mathfrak p)$, $\mathfrak p = R \cap \mathfrak q$ is Cohen-Macaulay, and let $W'$ denote the analogue for $S'/R'$. Then $W' = f^{-1}(W)$.


algebra-lemma-relative-dimension-CM

Let $R$ be a ring. Let $R \to S$ be a ring map which is (a) flat, (b) of finite presentation, (c) has Cohen-Macaulay fibres. Then we can write $S = S_0 \times \ldots \times S_n$ as a product of $R$-algebras $S_d$ such that each $S_d$ satisfies (a), (b), (c) and has all fibres equidimensional of dimension $d$.


algebra-lemma-universal-omega

\begin{slogan} Maps out of the module of differentials are the same as derivations. \end{slogan} The module of differentials of $S$ over $R$ has the following universal property. The map $$ \Hom_S(\Omega_{S/R}, M) \longrightarrow \text{Der}_R(S, M), \quad \alpha \longmapsto \alpha \circ \text{d} $$ is an isomorphism of functors.


algebra-lemma-colimit-differentials

Let $I$ be a directed set. Let $(R_i \to S_i, \varphi_{ii'})$ be a system of ring maps over $I$, see Categories, Section \ref{categories-section-posets-limits}. Then we have $$ \Omega_{S/R} = \colim_i \Omega_{S_i/R_i}. $$ where $R \to S = \colim (R_i \to S_i)$.


algebra-lemma-trivial-differential-surjective

Suppose that $R \to S$ is surjective. Then $\Omega_{S/R} = 0$.


algebra-lemma-differential-surjective

In diagram (\ref{equation-functorial-omega}), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega_{S/R} \to \Omega_{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi(a) \in \beta(R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)


algebra-lemma-exact-sequence-differentials

Let $A \to B \to C$ be ring maps. Then there is a canonical exact sequence $$ C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0 $$ of $C$-modules.


algebra-lemma-differentials-localize

Let $\varphi : A \to B$ be a ring map. \begin{enumerate} \item If $S \subset A$ is a multiplicative subset mapping to invertible elements of $B$, then $\Omega_{B/A} = \Omega_{B/S^{-1}A}$. \item If $S \subset B$ is a multiplicative subset then $S^{-1}\Omega_{B/A} = \Omega_{S^{-1}B/A}$. \end{enumerate}


algebra-lemma-differential-seq

In diagram (\ref{equation-functorial-omega}), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Then there is a canonical exact sequence of $S'$-modules $$ I/I^2 \longrightarrow \Omega_{S/R} \otimes_S S' \longrightarrow \Omega_{S'/R} \longrightarrow 0 $$ The leftmost map is characterized by the rule that $f \in I$ maps to $\text{d}f \otimes 1$.


algebra-lemma-differential-seq-split

In diagram (\ref{equation-functorial-omega}), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Moreover, assume that there exists an $R$-algebra map $S' \to S$ which is a right inverse to $S \to S'$. Then the exact sequence of $S'$-modules of Lemma \ref{lemma-differential-seq} turns into a short exact sequence $$ 0 \longrightarrow I/I^2 \longrightarrow \Omega_{S/R} \otimes_S S' \longrightarrow \Omega_{S'/R} \longrightarrow 0 $$ which is even a split short exact sequence.


algebra-lemma-differential-mod-power-ideal

Let $R \to S$ be a ring map. Let $I \subset S$ be an ideal. Let $n \geq 1$ be an integer. Set $S' = S/I^{n + 1}$. The map $\Omega_{S/R} \to \Omega_{S'/R}$ induces an isomorphism $$ \Omega_{S/R} \otimes_S S/I^n \longrightarrow \Omega_{S'/R} \otimes_{S'} S/I^n. $$


algebra-lemma-differentials-base-change

Suppose that we have ring maps $R \to R'$ and $R \to S$. Set $S' = S \otimes_R R'$, so that we obtain a diagram (\ref{equation-functorial-omega}). Then the canonical map defined above induces an isomorphism $\Omega_{S/R} \otimes_R R' = \Omega_{S'/R'}$.


algebra-lemma-differentials-diagonal

Let $R \to S$ be a ring map. Let $J = \Ker(S \otimes_R S \to S)$ be the kernel of the multiplication map. There is a canonical isomorphism of $S$-modules $\Omega_{S/R} \to J/J^2$, $a \text{d} b \mapsto a \otimes b - ab \otimes 1$.


algebra-lemma-differentials-polynomial-ring

If $S = R[x_1, \ldots, x_n]$, then $\Omega_{S/R}$ is a finite free $S$-module with basis $\text{d}x_1, \ldots, \text{d}x_n$.


algebra-lemma-differentials-finitely-presented

Suppose $R \to S$ is of finite presentation. Then $\Omega_{S/R}$ is a finitely presented $S$-module.


algebra-lemma-differentials-finitely-generated

Suppose $R \to S$ is of finite type. Then $\Omega_{S/R}$ is finitely generated $S$-module.


algebra-lemma-de-rham-complex

Let $A \to B$ be a ring map. Let $\pi : \Omega_{B/A} \to \Omega$ be a surjective $B$-module map. Denote $\text{d} : B \to \Omega$ the composition of $\pi$ with the universal derivation $\text{d}_{B/A} : B \to \Omega_{B/A}$. Set $\Omega^i = \wedge_B^i(\Omega)$. Assume that the kernel of $\pi$ is generated, as a $B$-module, by elements $\omega \in \Omega_{B/A}$ such that $\text{d}_{B/A}(\omega) \in \Omega_{B/A}^2$ maps to zero in $\Omega^2$. Then there is a de Rham complex $$ \Omega^0 \to \Omega^1 \to \Omega^2 \to \ldots $$ whose differential is defined by the rule $$ \text{d} : \Omega^p \to \Omega^{p + 1},\quad \text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_p\right) = \text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_p $$


algebra-lemma-composition-differential-operators

Let $R \to S$ be a ring map. Let $L, M, N$ be $S$-modules. If $D : L \to M$ and $D' : M \to N$ are differential operators of order $k$ and $k'$, then $D' \circ D$ is a differential operator of order $k + k'$.


algebra-lemma-module-principal-parts

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $k \geq 0$. There exists an $S$-module $P^k_{S/R}(M)$ and a canonical isomorphism $$ \text{Diff}^k_{S/R}(M, N) = \Hom_S(P^k_{S/R}(M), N) $$ functorial in the $S$-module $N$.


algebra-lemma-sequence-of-principal-parts

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. There is a canonical short exact sequence $$ 0 \to \Omega_{S/R} \otimes_S M \to P^1_{S/R}(M) \to M \to 0 $$ functorial in $M$ called the {\it sequence of principal parts}.


algebra-lemma-differentials-de-rham-complex-order-1

Let $A \to B$ be a ring map. The differentials $\text{d} : \Omega^i_{B/A}  \to \Omega^{i + 1}_{B/A}$ are differential operators of order $1$.


algebra-lemma-check-differential-operators

Let $A \to B$ be a ring map. Let $g_i \in B$, $i \in I$ be a set of generators for $B$ as an $A$-algebra. Let $M, N$ be $B$-modules. Let $D : M \to N$ be an $A$-linear map. In order to show that $D$ is a differential operator of order $k$ it suffices to show that $D \circ g_i - g_i \circ D$ is a differential operator of order $k - 1$ for $i \in I$.


algebra-lemma-invert-system-differential-operators

Let $A \to B$ be a ring map. Let $M, N$ be $B$-modules. Let $S \subset B$ be a multiplicative subset. Any differential operator $D : M \to N$ of order $k$ extends uniquely to a differential operator $E : S^{-1}M \to S^{-1}N$ of order $k$.


algebra-lemma-base-change-differential-operators

Let $R \to A$ and $R \to B$ be ring maps. Let $M$ and $M'$ be $A$-modules. Let $D : M \to M'$ be a differential operator of order $k$ with respect to $R \to A$. Let $N$ be any $B$-module. Then the map $$ D \otimes \text{id}_N : M \otimes_R N \to M' \otimes_R N $$ is a differential operator of order $k$ with respect to $B \to A \otimes_R B$.


algebra-lemma-NL-homotopy

Suppose given a diagram (\ref{equation-functoriality-NL}). Let $\alpha : P \to S$ and $\alpha' : P' \to S'$ be presentations. \begin{enumerate} \item There exists a morphism of presentations from $\alpha$ to $\alpha'$. \item Any two morphisms of presentations induce homotopic morphisms of complexes $\NL(\alpha) \to \NL(\alpha')$. \item The construction is compatible with compositions of morphisms of presentations (see proof for exact statement). \item If $R \to R'$ and $S \to S'$ are isomorphisms, then for any map $\varphi$ of presentations from $\alpha$ to $\alpha'$ the induced map $\NL(\alpha) \to \NL(\alpha')$ is a homotopy equivalence and a quasi-isomorphism. \end{enumerate} In particular, comparing $\alpha$ to the canonical presentation (\ref{equation-canonical-presentation}) we conclude there is a quasi-isomorphism $\NL(\alpha) \to \NL_{S/R}$ well defined up to homotopy and compatible with all functorialities (up to homotopy).


algebra-lemma-NL-polynomial-algebra

Let $A \to B$ be a polynomial algebra. Then $\NL_{B/A}$ is homotopy equivalent to the chain complex $(0 \to \Omega_{B/A})$ with $\Omega_{B/A}$ in degree $0$.


algebra-lemma-exact-sequence-NL

Let $A \to B \to C$ be ring maps. Choose a presentation $\alpha : A[x_s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_t, t \in T] \to C$ with kernel $J$. Let $\gamma : A[x_s, y_t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram $$ \xymatrix{ 0 \ar[r] & \Omega_{A[x_s]/A} \otimes C \ar[r] & \Omega_{A[x_s, y_t]/A} \otimes C \ar[r] & \Omega_{B[y_t]/B} \otimes C \ar[r] & 0 \\ & I/I^2 \otimes C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & 0 } $$ with exact rows. We get the following exact sequence of homology groups $$ H_1(\NL_{B/A} \otimes_B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0 $$ of $C$-modules extending the sequence of Lemma \ref{lemma-exact-sequence-differentials}. If $\text{Tor}_1^B(\Omega_{B/A}, C) = 0$, then $H_1(\NL_{B/A} \otimes_B C) = H_1(L_{B/A}) \otimes_B C$.


algebra-lemma-NL-surjection

Let $A \to B$ be a surjective ring map with kernel $I$. Then $\NL_{B/A}$ is homotopy equivalent to the chain complex $(I/I^2 \to 0)$ with $I/I^2$ in degree $1$. In particular $H_1(L_{B/A}) = I/I^2$.


algebra-lemma-application-NL

Let $A \to B \to C$ be ring maps. Assume $A \to C$ is surjective (so also $B \to C$ is). Denote $I = \Ker(A \to C)$ and $J = \Ker(B \to C)$. Then the sequence $$ I/I^2 \to J/J^2 \to \Omega_{B/A} \otimes_B B/J \to 0 $$ is exact.


algebra-lemma-change-base-NL

Let $R \to S$ be a ring map. Let $\alpha : P \to S$ be a presentation. Let $R \to R'$ be a flat ring map. Let $\alpha' : P \otimes_R R' \to S' = S \otimes_R R'$ be the induced presentation. Then $\NL(\alpha) \otimes_R R' = \NL(\alpha) \otimes_S S' = \NL(\alpha')$. In particular, the canonical map $$ \NL_{S/R} \otimes_S S' \longrightarrow \NL_{S \otimes_R R'/R'} $$ is a homotopy equivalence if $R \to R'$ is flat.


algebra-lemma-colimits-NL

Let $R_i \to S_i$ be a system of ring maps over the directed set $I$. Set $R = \colim R_i$ and $S = \colim S_i$. Then $\NL_{S/R} = \colim \NL_{S_i/R_i}$.


algebra-lemma-NL-localize-bottom

Let $S \subset A$ is a multiplicative subset of $A$. Let $S^{-1}A \to B$ be a ring map. Then $\NL_{B/A} \to \NL_{B/S^{-1}A}$ is a homotopy equivalence.


algebra-lemma-principal-localization-NL

\begin{slogan} The formation of the naive cotangent complex commutes with localization at an element. \end{slogan} Let $A \to B$ be a ring map. Let $g \in B$. Suppose $\alpha : P \to B$ is a presentation with kernel $I$. Then a presentation of $B_g$ over $A$ is the map $$ \beta : P[x] \longrightarrow B_g $$ extending $\alpha$ and sending $x$ to $1/g$. The kernel $J$ of $\beta$ is generated by $I$ and the element $f x - 1$ where $f \in P$ is an element mapped to $g \in B$ by $\alpha$. In this situation we have \begin{enumerate} \item $J/J^2 = (I/I^2)_g \oplus B_g (f x - 1)$, \item $\Omega_{P[x]/A} \otimes_{P[x]} B_g = \Omega_{P/A} \otimes_P B_g \oplus B_g \text{d}x$, \item $\NL(\beta) \cong \NL(\alpha) \otimes_B B_g \oplus (B_g \xrightarrow{g} B_g)$ \end{enumerate} Hence the canonical map $\NL_{B/A} \otimes_B B_g \to \NL_{B_g/A}$ is a homotopy equivalence.


algebra-lemma-localize-NL

Let $A \to B$ be a ring map. Let $S \subset B$ be a multiplicative subset. The canonical map $\NL_{B/A} \otimes_B S^{-1}B \to \NL_{S^{-1}B/A}$ is a quasi-isomorphism.


algebra-lemma-sum-two-terms

Let $R$ be a ring. Let $A_1 \to A_0$, and $B_1 \to B_0$ be two term complexes. Suppose that there exist morphisms of complexes $\varphi : A_\bullet \to B_\bullet$ and $\psi : B_\bullet \to A_\bullet$ such that $\varphi \circ \psi$ and $\psi \circ \varphi$ are homotopic to the identity maps. Then $A_1 \oplus B_0 \cong B_1 \oplus A_0$ as $R$-modules.


algebra-lemma-conormal-module

Let $R \to S$ be a ring map of finite type. For any presentations $\alpha : R[x_1, \ldots, x_n] \to S$, and $\beta : R[y_1, \ldots, y_m] \to S$ we have $$ I/I^2 \oplus S^{\oplus m} \cong J/J^2 \oplus S^{\oplus n} $$ as $S$-modules where $I = \Ker(\alpha)$ and $J = \Ker(\beta)$.


algebra-lemma-conormal-module-localize

Let $R \to S$ be a ring map of finite type. Let $g \in S$. For any presentations $\alpha : R[x_1, \ldots, x_n] \to S$, and $\beta : R[y_1, \ldots, y_m] \to S_g$ we have $$ (I/I^2)_g \oplus S^{\oplus m}_g \cong J/J^2 \oplus S_g^{\oplus n} $$ as $S_g$-modules where $I = \Ker(\alpha)$ and $J = \Ker(\beta)$.


algebra-lemma-localize-lci

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $g \in S$. \begin{enumerate} \item If $S$ is a global complete intersection so is $S_g$. \item If $S$ is a local complete intersection so is $S_g$. \end{enumerate}


algebra-lemma-lci-CM

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. If $S$ is a local complete intersection, then $S$ is a Cohen-Macaulay ring.


algebra-lemma-lci

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. Choose any presentation $S = k[x_1, \ldots, x_n]/I$. Let $\mathfrak q'$ be the prime of $k[x_1, \ldots, x_n]$ corresponding to $\mathfrak q$. Set $c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$, in other words $\dim_{\mathfrak q}(S) = n - c$ (see Lemma \ref{lemma-codimension}). The following are equivalent \begin{enumerate} \item There exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a global complete intersection over $k$. \item The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots, x_n]_{\mathfrak q'}$ can be generated by $c$ elements. \item The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by $c$ elements over $S_{\mathfrak q}$. \item The conormal module $(I/I^2)_{\mathfrak q}$ is a free $S_{\mathfrak q}$-module of rank $c$. \item The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence in the regular local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$. \end{enumerate} In this case any $c$ elements of $I_{\mathfrak q'}$ which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence in the local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.


algebra-lemma-ci-well-defined

Let $A \to B \to C$ be surjective local ring homomorphisms. Assume $A$ and $B$ are regular local rings. The following are equivalent \begin{enumerate} \item $\Ker(A \to C)$ is generated by a regular sequence, \item $\Ker(A \to C)$ is generated by $\dim(A) - \dim(C)$ elements, \item $\Ker(B \to C)$ is generated by a regular sequence, and \item $\Ker(B \to C)$ is generated by $\dim(B) - \dim(C)$ elements. \end{enumerate}


algebra-lemma-lci-local

Let $k$ be a field. Let $S$ be a local $k$-algebra essentially of finite type over $k$. The following are equivalent: \begin{enumerate} \item $S$ is a complete intersection over $k$, \item for any surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\Ker(R \to S)$ can be generated by a regular sequence, \item for some surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\Ker(R \to S)$ can be generated by $\dim(R) - \dim(S)$ elements, \item there exists a global complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$, and \item there exists a local complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$. \end{enumerate}


algebra-lemma-lci-at-prime

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. The following are equivalent: \begin{enumerate} \item The local ring $S_{\mathfrak q}$ is a complete intersection ring (Definition \ref{definition-lci-local-ring}). \item There exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a local complete intersection over $k$. \item There exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a global complete intersection over $k$. \item For any presentation $S = k[x_1, \ldots, x_n]/I$ with $\mathfrak q' \subset k[x_1, \ldots, x_n]$ corresponding to $\mathfrak q$ any of the equivalent conditions (1) -- (5) of Lemma \ref{lemma-lci} hold. \end{enumerate}


algebra-lemma-lci-global

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. The following are equivalent: \begin{enumerate} \item The ring $S$ is a local complete intersection over $k$. \item All local rings of $S$ are complete intersection rings over $k$. \item All localizations of $S$ at maximal ideals are complete intersection rings over $k$. \end{enumerate}


algebra-lemma-lci-field-change-local

Let $k \subset K$ be a field extension. Let $S$ be a finite type algebra over $k$. Let $\mathfrak q_K$ be a prime of $S_K = K \otimes_k S$ and let $\mathfrak q$ be the corresponding prime of $S$. Then $S_{\mathfrak q}$ is a complete intersection over $k$ (Definition \ref{definition-lci-local-ring}) if and only if $(S_K)_{\mathfrak q_K}$ is a complete intersection over $K$.


algebra-lemma-lci-field-change

Let $k \to K$ be a field extension. Let $S$ be a finite type $k$-algebra. Then $S$ is a local complete intersection over $k$ if and only if $S \otimes_k K$ is a local complete intersection over $K$.


algebra-lemma-lci-permanence-initial

Let $$ \xymatrix{ B & S \ar[l] \\ A \ar[u] & R \ar[l] \ar[u] } $$ be a commutative square of local rings. Assume \begin{enumerate} \item $R$ and $\overline{S} = S/\mathfrak m_R S$ are regular local rings, \item $A = R/I$ and $B = S/J$ for some ideals $I$, $J$, \item $J \subset S$ and $\overline{J} = J/\mathfrak m_R \cap J \subset \overline{S}$ are generated by regular sequences, and \item $A \to B$ and $R \to S$ are flat. \end{enumerate} Then $I$ is generated by a regular sequence.


algebra-lemma-base-change-syntomic

Any base change of a syntomic map is syntomic.