algebra-lemma-generic-points-geometrically-reduced

Let $k$ be a field and let $S$ be a $k$-algebra. Assume that $S$ is reduced and that $S_{\mathfrak p}$ is geometrically reduced for every minimal prime $\mathfrak p$ of $S$. Then $S$ is geometrically reduced.


algebra-lemma-separable-algebraic-diagonal

Let $k'/k$ be a separable algebraic extension. Then there exists a multiplicative subset $S \subset k' \otimes_k k'$ such that the multiplication map $k' \otimes_k k' \to k'$ is identified with $k' \otimes_k k' \to S^{-1}(k' \otimes_k k')$.


algebra-lemma-geometrically-reduced-over-separable-algebraic

Let $k \subset k'$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically reduced over $k$ if and only if it is geometrically reduced over $k'$.


algebra-lemma-characterize-separable-field-extensions

Let $k$ be a field of characteristic $p > 0$. Let $k \subset K$ be a field extension. The following are equivalent: \begin{enumerate} \item $K$ is separable over $k$, \item the ring $K \otimes_k k^{1/p}$ is reduced, and \item $K$ is geometrically reduced over $k$. \end{enumerate}


algebra-lemma-separably-generated-separable

A separably generated field extension is separable.


algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension

Let $k$ be a field. Let $S$ be a $k$-algebra. The following are equivalent: \begin{enumerate} \item $k' \otimes_k S$ is reduced for every finite purely inseparable extension $k'$ of $k$, \item $k^{1/p} \otimes_k S$ is reduced, \item $k^{perf} \otimes_k S$ is reduced, where $k^{perf}$ is the perfect closure of $k$, \item $\overline{k} \otimes_k S$ is reduced, where $\overline{k}$ is the algebraic closure of $k$, and \item $S$ is geometrically reduced over $k$. \end{enumerate}


algebra-lemma-make-separable

Let $k \subset K$ be a finitely generated field extension. There exists a diagram $$ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } $$ where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is a separable field extension. In this situation we can assume that $K' = k'K$ is the compositum, and also that $K' = (k' \otimes_k K)_{red}$.


algebra-lemma-perfection

\begin{slogan} Every field has a unique perfect closure. \end{slogan} For every field $k$ there exists a purely inseparable extension $k \subset k'$ such that $k'$ is perfect. The field extension $k \subset k'$ is unique up to unique isomorphism.


algebra-lemma-perfect-reduced

Let $k$ be a perfect field. Any reduced $k$ algebra is geometrically reduced over $k$. Let $R$, $S$ be $k$-algebras. Assume both $R$ and $S$ are reduced. Then the $k$-algebra $R \otimes_k S$ is reduced.


algebra-lemma-surjective-locally-nilpotent-kernel

Let $\varphi : R \to S$ be a surjective map with locally nilpotent kernel. Then $\varphi$ induces a homeomorphism of spectra and isomorphisms on residue fields. For any ring map $R \to R'$ the ring map $R' \to R' \otimes_R S$ is surjective with locally nilpotent kernel.


algebra-lemma-powers-field

\begin{reference} \cite[Lemma 3.1.6]{Alper-adequate} \end{reference} Let $k \subset k'$ be a field extension. The following are equivalent \begin{enumerate} \item for each $x \in k'$ there exists an $n > 0$ such that $x^n \in k$, and \item $k' = k$ or $k$ and $k'$ have characteristic $p > 0$ and either $k'/k$ is a purely inseparable extension or $k$ and $k'$ are algebraic extensions of $\mathbf{F}_p$. \end{enumerate}


algebra-lemma-powers

Let $\varphi : R \to S$ be a ring map. If \begin{enumerate} \item for any $x \in S$ there exists $n > 0$ such that $x^n$ is in the image of $\varphi$, and \item $\Ker(\varphi)$ is locally nilpotent, \end{enumerate} then $\varphi$ induces a homeomorphism on spectra and induces residue field extensions satisfying the equivalent conditions of Lemma \ref{lemma-powers-field}.


algebra-lemma-2-3-ring-map

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item[(a)] $S$ is generated as an $R$-algebra by elements $x$ such that $x^2, x^3 \in \varphi(R)$, and \item[(b)] $\Ker(\varphi)$ is locally nilpotent, \end{enumerate} Then $\varphi$ induces isomorphisms on residue fields and a homeomorphism of spectra. For any ring map $R \to R'$ the ring map $R' \to R' \otimes_R S$ also satisfies (a) and (b).


algebra-lemma-help-with-powers

Let $p$ be a prime number. Let $n, m > 0$ be two integers. There exists an integer $a$ such that $(x + y)^{p^a}, p^a(x + y) \in \mathbf{Z}[x^{p^n}, p^nx, y^{p^m}, p^my]$.


algebra-lemma-p-ring-map-field

Let $k \subset k'$ be a field extension. Let $p$ be a prime number. The following are equivalent \begin{enumerate} \item $k'$ is generated as a field extension of $k$ by elements $x$ such that there exists an $n > 0$ with $x^{p^n} \in k$ and $p^nx \in k$, and \item $k = k'$ or the characteristic of $k$ and $k'$ is $p$ and $k'/k$ is purely inseparable. \end{enumerate}


algebra-lemma-p-ring-map

Let $\varphi : R \to S$ be a ring map. Let $p$ be a prime number. Assume \begin{enumerate} \item[(a)] $S$ is generated as an $R$-algebra by elements $x$ such that there exists an $n > 0$ with $x^{p^n} \in \varphi(R)$ and $p^nx \in \varphi(R)$, and \item[(b)] $\Ker(\varphi)$ is locally nilpotent, \end{enumerate} Then $\varphi$ induces a homeomorphism of spectra and induces residue field extensions satisfying the equivalent conditions of Lemma \ref{lemma-p-ring-map-field}. For any ring map $R \to R'$ the ring map $R' \to R' \otimes_R S$ also satisfies (a) and (b).


algebra-lemma-radicial

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ induces an injective map of spectra, \item $\varphi$ induces purely inseparable residue field extensions. \end{enumerate} Then for any ring map $R \to R'$ properties (1) and (2) are true for $R' \to R' \otimes_R S$.


algebra-lemma-radicial-integral

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ is integral, \item $\varphi$ induces an injective map of spectra, \item $\varphi$ induces purely inseparable residue field extensions. \end{enumerate} Then $\varphi$ induces a homeomorphism from $\Spec(S)$ onto a closed subset of $\Spec(R)$ and for any ring map $R \to R'$ properties (1), (2), (3) are true for $R' \to R' \otimes_R S$.


algebra-lemma-radicial-integral-bijective

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ is integral, \item $\varphi$ induces an bijective map of spectra, \item $\varphi$ induces purely inseparable residue field extensions. \end{enumerate} Then $\varphi$ induces a homeomorphism on spectra and for any ring map $R \to R'$ properties (1), (2), (3) are true for $R' \to R' \otimes_R S$.


algebra-lemma-universally-bijective

Let $\varphi : R \to S$ be a ring map such that \begin{enumerate} \item the kernel of $\varphi$ is locally nilpotent, and \item $S$ is generated as an $R$-algebra by elements $x$ such that there exist $n > 0$ and a polynomial $P(T) \in R[T]$ whose image in $S[T]$ is $(T - x)^n$. \end{enumerate} Then $\Spec(S) \to \Spec(R)$ is a homeomorphism and $R \to S$ induces purely inseparable extensions of residue fields. Moreover, conditions (1) and (2) remain true on arbitrary base change.


algebra-lemma-flat-fibres-irreducible

Let $R \to S$ be a ring map. Assume \begin{enumerate} \item[(a)] $\Spec(R)$ is irreducible, \item[(b)] $R \to S$ is flat, \item[(c)] $R \to S$ is of finite presentation, \item[(d)] the fibre rings $S \otimes_R \kappa(\mathfrak p)$ have irreducible spectra for a dense collection of primes $\mathfrak p$ of $R$. \end{enumerate} Then $\Spec(S)$ is irreducible. This is true more generally with (b) $+$ (c) replaced by ``the map $\Spec(S) \to \Spec(R)$ is open''.


algebra-lemma-separably-closed-irreducible

Let $k$ be a separably closed field. Let $R$, $S$ be $k$-algebras. If $R$, $S$ have a unique minimal prime, so does $R \otimes_k S$.


algebra-lemma-geometrically-irreducible

Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent \begin{enumerate} \item for every field extension $k \subset k'$ the spectrum of $R \otimes_k k'$ is irreducible, \item for every finite separable field extension $k \subset k'$ the spectrum of $R \otimes_k k'$ is irreducible, \item the spectrum of $R \otimes_k \overline{k}$ is irreducible where $\overline{k}$ is the separable algebraic closure of $k$, and \item the spectrum of $R \otimes_k \overline{k}$ is irreducible where $\overline{k}$ is the algebraic closure of $k$. \end{enumerate}


algebra-lemma-separably-closed-irreducible-implies-geometric

Let $k$ be a field. Let $R$ be a $k$-algebra. If $k$ is separably algebraically closed then $R$ is geometrically irreducible over $k$ if and only if the spectrum of $R$ is irreducible.


algebra-lemma-subalgebra-geometrically-irreducible

Let $k$ be a field. Let $S$ be a $k$-algebra. \begin{enumerate} \item If $S$ is geometrically irreducible over $k$ so is every $k$-subalgebra. \item If all finitely generated $k$-subalgebras of $S$ are geometrically irreducible, then $S$ is geometrically irreducible. \item A directed colimit of geometrically irreducible $k$-algebras is geometrically irreducible. \end{enumerate}


algebra-lemma-geometrically-irreducible-any-base-change

Let $k$ be a field. Let $S$ be a geometrically irreducible $k$-algebra. Let $R$ be any $k$-algebra. The map $$ \Spec(R \otimes_k S) \longrightarrow \Spec(R) $$ induces a bijection on irreducible components.


algebra-lemma-field-extension-geometrically-irreducible

Let $K/k$ be a field extension. If $k$ is algebraically closed in $K$, then $K$ is geometrically irreducible over $k$.


algebra-lemma-geometrically-irreducible-transitive

Let $K/k$ be a geometrically irreducible field extension. Let $S$ be a geometrically irreducible $K$-algebra. Then $S$ is geometrically irreducible over $k$.


algebra-lemma-geometrically-irreducible-base-change-transcendental

Let $K/k$ be a field extension. The following are equivalent \begin{enumerate} \item $K$ is geometrically irreducible over $k$, and \item the induced extension $K(t)/k(t)$ of purely transcendental extensions is geometrically irreducible. \end{enumerate}


algebra-lemma-geometrically-irreducible-add-transcendental

Let $K/L/M$ be a tower of fields with $L/M$ geometrically irreducible. Let $x \in K$ be transcendental over $L$. Then $L(x)/M(x)$ is geometrically irreducible.


algebra-lemma-geometrically-irreducible-separable-elements

Let $K/k$ be a field extension. The following are equivalent \begin{enumerate} \item $K/k$ is geometrically irreducible, and \item every element $\alpha \in K$ separably algebraic over $k$ is in $k$. \end{enumerate}


algebra-lemma-make-geometrically-irreducible

Let $K/k$ be a field extension. Consider the subextension $K/k'/k$ consisting of elements separably algebraic over $k$. Then $K$ is geometrically irreducible over $k'$. If $K/k$ is a finitely generated field extension, then $[k' : k] < \infty$.


algebra-lemma-Galois-orbit

Let $k \subset K$ be an extension of fields. Let $k \subset \overline{k}$ be a separable algebraic closure. Then $\text{Gal}(\overline{k}/k)$ acts transitively on the primes of $\overline{k} \otimes_k K$.


algebra-lemma-separably-closed-connected

Let $k$ be a separably algebraically closed field. Let $R$, $S$ be $k$-algebras. If $\Spec(R)$, and $\Spec(S)$ are connected, then so is $\Spec(R \otimes_k S)$.


algebra-lemma-separably-closed-connected-implies-geometric

Let $k$ be a field. Let $R$ be a $k$-algebra. If $k$ is separably algebraically closed then $R$ is geometrically connected over $k$ if and only if the spectrum of $R$ is connected.


algebra-lemma-subalgebra-geometrically-connected

Let $k$ be a field. Let $S$ be a $k$-algebra. \begin{enumerate} \item If $S$ is geometrically connected over $k$ so is every $k$-subalgebra. \item If all finitely generated $k$-subalgebras of $S$ are geometrically connected, then $S$ is geometrically connected. \item A directed colimit of geometrically connected $k$-algebras is geometrically connected. \end{enumerate}


algebra-lemma-geometrically-integral

Let $k$ be a field. Let $S$ be a $k$-algebra. In this case $S$ is geometrically integral over $k$ if and only if $S$ is geometrically irreducible as well as geometrically reduced over $k$.


algebra-lemma-geometrically-integral-any-integral-base-change

Let $k$ be a field. Let $S$ be a geometrically integral $k$-algebra. Let $R$ be a $k$-algebra and an integral domain. Then $R \otimes_k S$ is an integral domain.


algebra-lemma-dominate

Let $K$ be a field. Let $A \subset K$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.


algebra-lemma-valuation-ring-x-or-x-inverse

Let $A$ be a valuation ring with maximal ideal $\mathfrak m$ and fraction field $K$. Let $x \in K$. Then either $x \in A$ or $x^{-1} \in A$ or both.


algebra-lemma-x-or-x-inverse-valuation-ring

Let $A \subset K$ be a subring of a field $K$ such that for all $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both. Then $A$ is a valuation ring with fraction field $K$.


algebra-lemma-colimit-valuation-rings

\begin{slogan} Valuation rings are stable under filtered direct limits \end{slogan} Let $I$ be a directed set. Let $(A_i, \varphi_{ij})$ be a system of valuation rings over $I$. Then $A = \colim A_i$ is a valuation ring.


algebra-lemma-valuation-ring-cap-field

Let $K \subset L$ be an extension of fields. If $B \subset L$ is a valuation ring, then $A = K \cap B$ is a valuation ring.


algebra-lemma-valuation-ring-cap-field-finite

Let $K \subset L$ be an algebraic extension of fields. If $B \subset L$ is a valuation ring with fraction field $L$ and not a field, then $A = K \cap B$ is a valuation ring and not a field.


algebra-lemma-make-valuation-rings

Let $A$ be a valuation ring. For any prime ideal $\mathfrak p \subset A$ the quotient $A/\mathfrak p$ is a valuation ring. The same is true for the localization $A_\mathfrak p$ and in fact any localization of $A$.


algebra-lemma-stack-valuation-rings

Let $A'$ be a valuation ring with residue field $K$. Let $A$ be a valuation ring with fraction field $K$. Then $C = \{\lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A\}$ is a valuation ring.


algebra-lemma-valuation-ring-normal

Let $A$ be a valuation ring. Then $A$ is a normal domain.


algebra-lemma-find-valuation-rings

Let $A$ be a normal domain with fraction field $K$. \begin{enumerate} \item For every $x \in K$, $x \not \in A$ there exists a valuation ring $A \subset V \subset K$ with fraction field $K$ such that $x \not \in V$. \item If $A$ is local, we can moreover choose $V$ which dominates $A$. \end{enumerate} In other words, $A$ is the intersection of all valuation rings in $K$ containing $A$ and if $A$ is local, then $A$ is the intersection of all valuation rings in $K$ dominating $A$.


algebra-lemma-valuation-group

Let $A$ be a valuation ring with field of fractions $K$. Set $\Gamma = K^*/A^*$ (with group law written additively). For $\gamma, \gamma' \in \Gamma$ define $\gamma \geq \gamma'$ if and only if $\gamma - \gamma'$ is in the image of $A - \{0\} \to \Gamma$. Then $(\Gamma, \geq)$ is a totally ordered abelian group.


algebra-lemma-properties-valuation

Let $A$ be a valuation ring. The valuation $v : A -\{0\} \to \Gamma_{\geq 0}$ has the following properties: \begin{enumerate} \item $v(a) = 0 \Leftrightarrow a \in A^*$, \item $v(ab) = v(a) + v(b)$, \item $v(a + b) \geq \min(v(a), v(b))$. \end{enumerate}


algebra-lemma-characterize-valuation-ring

Let $A$ be a ring. The following are equivalent \begin{enumerate} \item $A$ is a valuation ring, \item $A$ is a local domain and every finitely generated ideal of $A$ is principal. \end{enumerate}


algebra-lemma-valuation-valuation-ring

Let $(\Gamma, \geq)$ be a totally ordered abelian group. Let $K$ be a field. Let $v : K^* \to \Gamma$ be a homomorphism of abelian groups such that $v(a + b) \geq \min(v(a), v(b))$ for $a, b \in K$ with $a, b, a + b$ not zero. Then $$ A = \{ x \in K \mid x = 0 \text{ or } v(x) \geq 0 \} $$ is a valuation ring with value group $\Im(v) \subset \Gamma$, with maximal ideal $$ \mathfrak m = \{ x \in K \mid x = 0 \text{ or } v(x) > 0 \} $$ and with group of units $$ A^* = \{ x \in K^* \mid v(x) = 0 \}. $$


algebra-lemma-ideals-valuation-ring

Let $A$ be a valuation ring. Ideals in $A$ correspond $1 - 1$ with ideals of $\Gamma$. This bijection is inclusion preserving, and maps prime ideals to prime ideals.


algebra-lemma-valuation-ring-Noetherian-discrete

A valuation ring is Noetherian if and only if it is a discrete valuation ring or a field.


algebra-lemma-Noetherian-basic

Let $R$ be a Noetherian ring. Any finite $R$-module is of finite presentation. Any submodule of a finite $R$-module is finite. The ascending chain condition holds for $R$-submodules of a finite $R$-module.


algebra-lemma-Artin-Rees

Suppose that $R$ is Noetherian, $I \subset R$ an ideal. Let $N \subset M$ be finite $R$-modules. There exists a constant $c > 0$ such that $I^n M \cap N  =  I^{n-c}(I^cM \cap N)$ for all $n \geq c$.


algebra-lemma-map-AR

Suppose that $0 \to K \to M \xrightarrow{f} N$ is an exact sequence of finitely generated modules over a Noetherian ring $R$. Let $I \subset R$ be an ideal. Then there exists a $c$ such that $$ f^{-1}(I^nN) = K + I^{n-c}f^{-1}(I^cN) \quad\text{and}\quad f(M) \cap I^nN \subset f(I^{n - c}M) $$ for all $n \geq c$.


algebra-lemma-intersect-powers-ideal-module-zero

Let $R$ be a Noetherian local ring. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $R$-module. Then $\bigcap_{n \geq 0} I^nM = 0$.


algebra-lemma-intersection-powers-ideal-module

Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $N = \bigcap_n I^n M$. \begin{enumerate} \item For every prime $\mathfrak p$, $I \subset \mathfrak p$ there exists a $f \in R$, $f \not \in \mathfrak p$ such that $N_f = 0$. \item If $I$ is contained in the Jacobson radical of $R$, then $N = 0$. \end{enumerate}


algebra-lemma-Artin-Tate

Let $R$ be a Noetherian ring. Let $S$ be a finitely generated $R$-algebra. If $T \subset S$ is an $R$-subalgebra such that $S$ is finitely generated as a $T$-module, then $T$ is of finite type over $R$.


algebra-lemma-finite-length-finite

\begin{slogan} Modules of finite length are finite. \end{slogan} Let $R$ be a ring. Let $M$ be an $R$-module. If $\text{length}_R(M) < \infty$ then $M$ is a finite $R$-module.


algebra-lemma-length-additive

\begin{slogan} Length is additive in short exact sequences. \end{slogan} If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules over $R$ then the length of $M$ is the sum of the lengths of $M'$ and $M''$.


algebra-lemma-length-infinite

Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $M$ be an $R$-module. \begin{enumerate} \item If $M$ is a finite module and $\mathfrak m^n M \not = 0$ for all $n\geq 0$, then $\text{length}_R(M) = \infty$. \item If $M$ has finite length then $\mathfrak m^nM = 0$ for some $n$. \end{enumerate}


algebra-lemma-length-independent

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. We always have $\text{length}_R(M) \geq \text{length}_S(M)$. If $R \to S$ is surjective then equality holds.


algebra-lemma-dimension-is-length

Let $R$ be a ring with maximal ideal $\mathfrak m$. Suppose that $M$ is an $R$-module with $\mathfrak m M  =  0$. Then the length of $M$ as an $R$-module agrees with the dimension of $M$ as a $R/\mathfrak m$ vector space. The length is finite if and only if $M$ is a finite $R$-module.


algebra-lemma-length-finite

Let $R$ be a ring with finitely generated maximal ideal $\mathfrak m$. (For example $R$ Noetherian.) Suppose that $M$ is a finite $R$-module with $\mathfrak m^n M  =  0$ for some $n$. Then $\text{length}_R(M) < \infty$.


algebra-lemma-characterize-length-1

Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent: \begin{enumerate} \item $M$ is simple, \item $\text{length}_R(M) = 1$, and \item $M \cong R/\mathfrak m$ for some maximal ideal $\mathfrak m \subset R$. \end{enumerate}


algebra-lemma-simple-pieces

Let $R$ be a ring. Let $M$ be a finite length $R$-module. Choose any maximal chain of submodules $$ 0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_n = M $$ with $M_i \not = M_{i-1}$, $i = 1, \ldots, n$. Then \begin{enumerate} \item $n = \text{length}_R(M)$, \item each $M_i/M_{i-1}$ is simple, \item each $M_i/M_{i-1}$ is of the form $R/\mathfrak m_i$ for some maximal ideal $\mathfrak m_i$, \item given a maximal ideal $\mathfrak m \subset R$ we have $$ \# \{i \mid \mathfrak m_i = \mathfrak m\} = \text{length}_{R_{\mathfrak m}} (M_{\mathfrak m}). $$ \end{enumerate}


algebra-lemma-pushdown-module

Let $A$ be a local ring with maximal ideal $\mathfrak m$. Let $B$ be a semi-local ring with maximal ideals $\mathfrak m_i$, $i = 1, \ldots, n$. Suppose that $A \to B$ is a homomorphism such that each $\mathfrak m_i$ lies over $\mathfrak m$ and such that $$ [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] < \infty. $$ Let $M$ be a $B$-module of finite length. Then $$ \text{length}_A(M) = \sum\nolimits_{i = 1, \ldots, n} [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] \text{length}_{B_{\mathfrak m_i}}(M_{\mathfrak m_i}), $$ in particular $\text{length}_A(M) < \infty$.


algebra-lemma-pullback-module

Let $A \to B$ be a flat local homomorphism of local rings. Then for any $A$-module $M$ we have $$ \text{length}_A(M) \text{length}_B(B/\mathfrak m_AB) = \text{length}_B(M \otimes_A B). $$ In particular, if $\text{length}_B(B/\mathfrak m_AB) < \infty$ then $M$ has finite length if and only if $M \otimes_A B$ has finite length.


algebra-lemma-pullback-transitive

Let $A \to B \to C$ be flat local homomorphisms of local rings. Then $$ \text{length}_B(B/\mathfrak m_A B) \text{length}_C(C/\mathfrak m_B C) = \text{length}_C(C/\mathfrak m_A C) $$


algebra-lemma-finite-dimensional-algebra

Suppose $R$ is a finite dimensional algebra over a field. Then $R$ is Artinian.


algebra-lemma-artinian-finite-nr-max

If $R$ is Artinian then $R$ has only finitely many maximal ideals.


algebra-lemma-artinian-radical-nilpotent

Let $R$ be Artinian. The Jacobson radical of $R$ is a nilpotent ideal.


algebra-lemma-product-local

Any ring with finitely many maximal ideals and locally nilpotent Jacobson radical is the product of its localizations at its maximal ideals. Also, all primes are maximal.


algebra-lemma-artinian-finite-length

A ring $R$ is Artinian if and only if it has finite length as a module over itself. Any such ring $R$ is both Artinian and Noetherian, any prime ideal of $R$ is a maximal ideal, and $R$ is equal to the (finite) product of its localizations at its maximal ideals.


algebra-lemma-composition-essentially-of-finite-type

The class of ring maps which are essentially of finite type is preserved under composition. Similarly for essentially of finite presentation.


algebra-lemma-base-change-essentially-of-finite-type

The class of ring maps which are essentially of finite type is preserved by base change. Similarly for essentially of finite presentation.


algebra-lemma-essentially-of-finite-type-into-artinian-local

Let $R \to S$ be a ring map. Assume $S$ is an Artinian local ring with maximal ideal $\mathfrak m$. Then \begin{enumerate} \item $R \to S$ is finite if and only if $R \to S/\mathfrak m$ is finite, \item $R \to S$ is of finite type if and only if $R \to S/\mathfrak m$ is of finite type. \item $R \to S$ is essentially of finite type if and only if the composition $R \to S/\mathfrak m$ is essentially of finite type. \end{enumerate}


algebra-lemma-localization-at-closed-point-special-fibre

Let $\varphi : R \to S$ be essentially of finite type with $R$ and $S$ local (but not necessarily $\varphi$ local). Then there exists an $n$ and a maximal ideal $\mathfrak m \subset R[x_1, \ldots, x_n]$ lying over $\mathfrak m_R$ such that $S$ is a localization of a quotient of $R[x_1, \ldots, x_n]_\mathfrak m$.


algebra-lemma-length-K0

If $R$ is an Artinian local ring then the length function defines a natural abelian group homomorphism $\text{length}_R : K'_0(R) \to \mathbf{Z}$.


algebra-lemma-K0-product

Let $R = R_1 \times R_2$. Then $K_0(R) = K_0(R_1) \times K_0(R_2)$ and $K'_0(R) = K'_0(R_1) \times K'_0(R_2)$


Let $R$ be an Artinian local ring. The map $\text{length}_R : K'_0(R) \to \mathbf{Z}$ of Lemma \ref{lemma-length-K0} is an isomorphism.


algebra-lemma-K0-local

Let $(R, \mathfrak m)$ be a local ring. Every finite projective $R$-module is finite free. The map $\text{rank}_R : K_0(R) \to \mathbf{Z}$ defined by $[M] \to \text{rank}_R(M)$ is well defined and an isomorphism.


algebra-lemma-graded-NAK

Let $S$ be a graded ring. Let $M$ be a graded $S$-module. \begin{enumerate} \item If $S_+M = M$ and $M$ is finite, then $M = 0$. \item If $N, N' \subset M$ are graded submodules, $M = N + S_+N'$, and $N'$ is finite, then $M = N$. \item If $N \to M$ is a map of graded modules, $N/S_+N \to M/S_+M$ is surjective, and $M$ is finite, then $N \to M$ is surjective. \item If $x_1, \ldots, x_n \in M$ are homogeneous and generate $M/S_+M$ and $M$ is finite, then $x_1, \ldots, x_n$ generate $M$. \end{enumerate}


algebra-lemma-uple-generated-degree-1

Let $S$ be a graded ring, which is finitely generated over $S_0$. Then for all sufficiently divisible $d$ the algebra $S^{(d)}$ is generated in degree $1$ over $S_0$.


algebra-lemma-Z-graded

Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous invertible element of positive degree. Then the set $G \subset \Spec(S)$ of $\mathbf{Z}$-graded primes of $S$ (with induced topology) maps homeomorphically to $\Spec(S_0)$.


algebra-lemma-topology-proj

Let $S = \oplus_{d \geq 0} S_d$ be a graded ring. \begin{enumerate} \item The sets $D_{+}(f)$ are open in $\text{Proj}(S)$. \item We have $D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$. \item Let $g = g_0 + \ldots + g_m$ be an element of $S$ with $g_i \in S_i$. Then $$ D(g) \cap \text{Proj}(S) = (D(g_0) \cap \text{Proj}(S)) \cup \bigcup\nolimits_{i \geq 1} D_{+}(g_i). $$ \item Let $g_0\in S_0$ be a homogeneous element of degree $0$. Then $$ D(g_0) \cap \text{Proj}(S) = \bigcup\nolimits_{f \in S_d, \ d\geq 1} D_{+}(g_0 f). $$ \item The open sets $D_{+}(f)$ form a basis for the topology of $\text{Proj}(S)$. \item Let $f \in S$ be homogeneous of positive degree. The ring $S_f$ has a natural $\mathbf{Z}$-grading. The ring maps $S \to S_f \leftarrow S_{(f)}$ induce homeomorphisms $$ D_{+}(f) \leftarrow \{\mathbf{Z}\text{-graded primes of }S_f\} \to \Spec(S_{(f)}). $$ \item There exists an $S$ such that $\text{Proj}(S)$ is not quasi-compact. \item The sets $V_{+}(I)$ are closed. \item Any closed subset $T \subset \text{Proj}(S)$ is of the form $V_{+}(I)$ for some homogeneous ideal $I \subset S$. \item For any graded ideal $I \subset S$ we have $V_{+}(I) = \emptyset$ if and only if $S_{+} \subset \sqrt{I}$. \end{enumerate}


algebra-lemma-proj-prime

Let $S$ be a graded ring. Let $M$ be a graded $S$-module. Let $\mathfrak p$ be an element of $\text{Proj}(S)$. Let $f \in S$ be a homogeneous element of positive degree such that $f \not \in \mathfrak p$, i.e., $\mathfrak p \in D_{+}(f)$. Let $\mathfrak p' \subset S_{(f)}$ be the element of $\Spec(S_{(f)})$ corresponding to $\mathfrak p$ as in Lemma \ref{lemma-topology-proj}. Then $S_{(\mathfrak p)} = (S_{(f)})_{\mathfrak p'}$ and compatibly $M_{(\mathfrak p)} = (M_{(f)})_{\mathfrak p'}$.


algebra-lemma-graded-silly

Suppose $S$ is a graded ring, $\mathfrak p_i$, $i = 1, \ldots, r$ homogeneous prime ideals and $I \subset S_{+}$ a graded ideal. Assume $I \not\subset \mathfrak p_i$ for all $i$. Then there exists a homogeneous element $x\in I$ of positive degree such that $x\not\in \mathfrak p_i$ for all $i$.


algebra-lemma-smear-out

Let $S$ be a graded ring. Let $\mathfrak p \subset S$ be a prime. Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a prime ideal of $S$.


algebra-lemma-graded-ring-minimal-prime

Let $S$ be a graded ring. \begin{enumerate} \item Any minimal prime of $S$ is a homogeneous ideal of $S$. \item Given a homogeneous ideal $I \subset S$ any minimal prime over $I$ is homogeneous. \end{enumerate}


algebra-lemma-dehomogenize-finite-type

Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$ be homogeneous. Assume that $S$ is of finite type over $R$. Then \begin{enumerate} \item the ring $S_{(f)}$ is of finite type over $R$, and \item for any finite graded $S$-module $M$ the module $M_{(f)}$ is a finite $S_{(f)}$-module. \end{enumerate}


algebra-lemma-homogenize

Let $R$ be a ring. Let $R'$ be a finite type $R$-algebra, and let $M$ be a finite $R'$-module. There exists a graded $R$-algebra $S$, a graded $S$-module $N$ and an element $f \in S$ homogeneous of degree $1$ such that \begin{enumerate} \item $R' \cong S_{(f)}$ and $M \cong N_{(f)}$ (as modules), \item $S_0 = R$ and $S$ is generated by finitely many elements of degree $1$ over $R$, and \item $N$ is a finite $S$-module. \end{enumerate}


algebra-lemma-S-plus-generated

Let $S$ be a graded ring. A set of homogeneous elements $f_i \in S_{+}$ generates $S$ as an algebra over $S_0$ if and only if they generate $S_{+}$ as an ideal of $S$.


algebra-lemma-graded-Noetherian

A graded ring $S$ is Noetherian if and only if $S_0$ is Noetherian and $S_{+}$ is finitely generated as an ideal of $S$.


algebra-lemma-numerical-polynomial-functorial

If $A \to A'$ is a homomorphism of abelian groups and if $f : n \mapsto f(n) \in A$ is a numerical polynomial, then so is the composition.


algebra-lemma-numerical-polynomial

Suppose that $f: n \mapsto f(n) \in A$ is defined for all $n$ sufficiently large and suppose that $n \mapsto f(n) - f(n-1)$ is a numerical polynomial. Then $f$ is a numerical polynomial.


algebra-lemma-graded-module-fg

If $M$ is a finitely generated graded $S$-module, and if $S$ is finitely generated over $S_0$, then each $M_n$ is a finite $S_0$-module.


algebra-lemma-quotient-smaller-d

Let $k$ be a field. Suppose that $I \subset k[X_1, \ldots, X_d]$ is a nonzero graded ideal. Let $M = k[X_1, \ldots, X_d]/I$. Then the numerical polynomial $n \mapsto \dim_k(M_n)$ (see Example \ref{example-hilbert-function}) has degree $ < d - 1$ (or is zero if $d = 1$).


algebra-lemma-differ-finite

Suppose that $M' \subset M$ are finite $R$-modules with finite length quotient. Then there exists a constants $c_1, c_2$ such that for all $n \geq c_2$ we have $$ c_1 + \chi_{I, M'}(n - c_2) \leq \chi_{I, M}(n) \leq c_1 + \chi_{I, M'}(n) $$


algebra-lemma-hilbert-ses

Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of finite $R$-modules. Then there exists a submodule $N \subset M'$ with finite colength $l$ and $c \geq 0$ such that $$ \chi_{I, M}(n) = \chi_{I, M''}(n) + \chi_{I, N}(n - c) + l $$ and $$ \varphi_{I, M}(n) = \varphi_{I, M''}(n) + \varphi_{I, N}(n - c) $$ for all $n \geq c$.


algebra-lemma-hilbert-change-I

Suppose that $I$, $I'$ are two ideals of definition for the Noetherian local ring $R$. Let $M$ be a finite $R$-module. There exists a constant $a$ such that $\chi_{I, M}(n) \leq \chi_{I', M}(an)$ for $n \geq 1$.


algebra-lemma-differ-finite-chi

Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal of definition. Let $M$ be a finite $R$-module which does not have finite length. If $M' \subset M$ is a submodule with finite colength, then $\chi_{I, M} - \chi_{I, M'}$ is a polynomial of degree $<$ degree of either polynomial.


algebra-lemma-hilbert-ses-chi

Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal of definition. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of finite $R$-modules. Then \begin{enumerate} \item if $M'$ does not have finite length, then $\chi_{I, M} - \chi_{I, M''} - \chi_{I, M'}$ is a numerical polynomial of degree $<$ the degree of $\chi_{I, M'}$, \item $\max\{ \deg(\chi_{I, M'}), \deg(\chi_{I, M''}) \} = \deg(\chi_{I, M})$, and \item $\max\{d(M'), d(M'')\} = d(M)$, \end{enumerate}


algebra-lemma-dimension-height

The Krull dimension of $R$ is the supremum of the heights of its (maximal) primes.


algebra-lemma-Noetherian-dimension-0

A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.


algebra-lemma-dimension-0-d-0

Let $R$ be a Noetherian local ring. Then $\dim(R) = 0 \Leftrightarrow d(R) = 0$.


algebra-lemma-height-1

Let $R$ be a local Noetherian ring. The following are equivalent: \begin{enumerate} \item \label{item-dim-1} $\dim(R) = 1$, \item \label{item-d-1} $d(R) = 1$, \item \label{item-Vx} there exists an $x \in \mathfrak m$, $x$ not nilpotent such that $V(x) = \{\mathfrak m\}$, \item \label{item-x} there exists an $x \in \mathfrak m$, $x$ not nilpotent such that $\mathfrak m = \sqrt{(x)}$, and \item \label{item-ideal-1} there exists an ideal of definition generated by $1$ element, and no ideal of definition is generated by $0$ elements. \end{enumerate}


algebra-lemma-minimal-over-1

Let $R$ be a Noetherian ring. Let $x \in R$. \begin{enumerate} \item If $\mathfrak p$ is minimal over $(x)$ then the height of $\mathfrak p$ is $0$ or $1$. \item If $\mathfrak p, \mathfrak q \in \Spec(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, x)$, then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$. \end{enumerate}


algebra-lemma-elements-generate-ideal-definition

Let $(R, \mathfrak m)$ be a Noetherian local ring. Suppose $x_1, \ldots, x_d \in \mathfrak m$ generate an ideal of definition and $d = \dim(R)$. Then $\dim(R/(x_1, \ldots, x_i)) = d - i$ for all $i = 1, \ldots, d$.


algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens

Let $R$ be a Noetherian local domain of dimension $\geq 2$. A nonempty open subset $U \subset \Spec(R)$ is infinite.


algebra-lemma-finite-type-algebra-finite-nr-primes

Let $S$ be a nonzero finite type algebra over a field $k$. Then $\dim(S) = 0$ if and only if $S$ has finitely many primes.


algebra-lemma-noetherian-dim-1-Jacobson

Noetherian Jacobson rings. \begin{enumerate} \item Any Noetherian domain $R$ of dimension $1$ with infinitely many primes is Jacobson. \item Any Noetherian ring such that every prime $\mathfrak p$ is either maximal or contained in infinitely many prime ideals is Jacobson. \end{enumerate}


algebra-lemma-filter-Noetherian-module

Let $R$ be a Noetherian ring, and let $M$ be a finite $R$-module. There exists a filtration by $R$-submodules $$ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M $$ such that each quotient $M_i/M_{i-1}$ is isomorphic to $R/\mathfrak p_i$ for some prime ideal $\mathfrak p_i$ of $R$.


algebra-lemma-filter-primes-in-support

Let $R$, $M$, $M_i$, $\mathfrak p_i$ as in Lemma \ref{lemma-filter-Noetherian-module}. Then $\text{Supp}(M) = \bigcup V(\mathfrak p_i)$ and in particular $\mathfrak p_i \in \text{Supp}(M)$.


algebra-lemma-support-point

Suppose that $R$ is a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{Supp}(M) = \{ \mathfrak m\}$ if and only if $M$ has finite length over $R$.


algebra-lemma-Noetherian-power-ideal-kills-module

Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Then $I^nM = 0$ for some $n \geq 0$ if and only if $\text{Supp}(M) \subset V(I)$.


algebra-lemma-filter-minimal-primes-in-support

Let $R$, $M$, $M_i$, $\mathfrak p_i$ as in Lemma \ref{lemma-filter-Noetherian-module}. The minimal elements of the set $\{\mathfrak p_i\}$ are the minimal elements of $\text{Supp}(M)$. The number of times a minimal prime $\mathfrak p$ occurs is $$ \#\{i \mid \mathfrak p_i = \mathfrak p\} = \text{length}_{R_\mathfrak p} M_{\mathfrak p}. $$


algebra-lemma-support-dimension-d

Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Then $d(M) = \dim(\text{Supp}(M))$.


algebra-lemma-ass-support

Let $R$ be a ring. Let $M$ be an $R$-module. Then $\text{Ass}(M) \subset \text{Supp}(M)$.


algebra-lemma-ass

Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{Ass}(M') \subset \text{Ass}(M)$ and $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. Also $\text{Ass}(M' \oplus M'') = \text{Ass}(M') \cup \text{Ass}(M'')$.


algebra-lemma-ass-filter

Let $R$ be a ring, and $M$ an $R$-module. Suppose there exists a filtration by $R$-submodules $$ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M $$ such that each quotient $M_i/M_{i-1}$ is isomorphic to $R/\mathfrak p_i$ for some prime ideal $\mathfrak p_i$ of $R$. Then $\text{Ass}(M) \subset \{\mathfrak p_1, \ldots, \mathfrak p_n\}$.


algebra-lemma-finite-ass

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Then $\text{Ass}(M)$ is finite.


algebra-lemma-ass-zero

\begin{slogan} Over a Noetherian ring each nonzero module has an associated prime. \end{slogan} Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. Then $$ M = (0) \Leftrightarrow \text{Ass}(M) = \emptyset. $$


algebra-lemma-ass-minimal-prime-support

Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. Any $\mathfrak p \in \text{Supp}(M)$ which is minimal among the elements of $\text{Supp}(M)$ is an element of $\text{Ass}(M)$.


algebra-lemma-ass-zero-divisors

Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. The union $\bigcup_{\mathfrak q \in \text{Ass}(M)} \mathfrak q$ is the set of elements of $R$ which are zerodivisors on $M$.


algebra-lemma-ass-functorial

Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then $\Spec(\varphi)(\text{Ass}_S(M)) \subset \text{Ass}_R(M)$.


algebra-lemma-ass-functorial-Noetherian

Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. If $S$ is Noetherian, then $\Spec(\varphi)(\text{Ass}_S(M)) = \text{Ass}_R(M)$.


algebra-lemma-ass-quotient-ring

Let $R$ be a ring. Let $I$ be an ideal. Let $M$ be an $R/I$-module. Via the canonical injection $\Spec(R/I) \to \Spec(R)$ we have $\text{Ass}_{R/I}(M) = \text{Ass}_R(M)$.


algebra-lemma-associated-primes-localize

Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p \subset R$ be a prime. \begin{enumerate} \item If $\mathfrak p \in \text{Ass}(M)$ then $\mathfrak pR_{\mathfrak p} \in \text{Ass}(M_{\mathfrak p})$. \item If $\mathfrak p$ is finitely generated then the converse holds as well. \end{enumerate}


algebra-lemma-localize-ass

Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Via the canonical injection $\Spec(S^{-1}R) \to \Spec(R)$ we have \begin{enumerate} \item $\text{Ass}_R(S^{-1}M) = \text{Ass}_{S^{-1}R}(S^{-1}M)$, \item $\text{Ass}_R(M) \cap \Spec(S^{-1}R) \subset \text{Ass}_R(S^{-1}M)$, and \item if $R$ is Noetherian this inclusion is an equality. \end{enumerate}


algebra-lemma-localize-ass-nonzero-divisors

Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Assume that every $s \in S$ is a nonzerodivisor on $M$. Then $$ \text{Ass}_R(M) = \text{Ass}_R(S^{-1}M). $$


algebra-lemma-ideal-nonzerodivisor

Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $I \subset \mathfrak m$ be an ideal. Let $M$ be a finite $R$-module. The following are equivalent: \begin{enumerate} \item There exists an $x \in I$ which is not a zerodivisor on $M$. \item We have $I \not \subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$. \end{enumerate}


algebra-lemma-zero-at-ass-zero

Let $R$ be a ring. Let $M$ be an $R$-module. If $R$ is Noetherian the map $$ M \longrightarrow \prod\nolimits_{\mathfrak p \in \text{Ass}(M)} M_{\mathfrak p} $$ is injective.


algebra-lemma-symbolic-power-associated

Let $R$ be a Noetherian ring. Let $\mathfrak p$ be a prime ideal. Let $n > 0$. Then $\text{Ass}(R/\mathfrak p^{(n)}) = \{\mathfrak p\}$.


algebra-lemma-symbolic-power-flat-extension

Let $R \to S$ be flat ring map. Let $\mathfrak p \subset R$ be a prime such that $\mathfrak q = \mathfrak p S$ is a prime of $S$. Then $\mathfrak p^{(n)} S = \mathfrak q^{(n)}$.


algebra-lemma-compare-relative-assassins

Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Let $A$, $A'$, $A_{fin}$, $B$, and $B_{fin}$ be the subsets of $\Spec(S)$ introduced above. \begin{enumerate} \item We always have $A = A'$. \item We always have $A_{fin} \subset A$, $B_{fin} \subset B$, $A_{fin} \subset A'_{fin} \subset B_{fin}$ and $A \subset B$. \item If $S$ is Noetherian, then $A = A_{fin}$ and $B = B_{fin}$. \item If $N$ is flat over $R$, then $A = A_{fin} = A'_{fin}$ and $B = B_{fin}$. \item If $R$ is Noetherian and $N$ is flat over $R$, then all of the sets are equal, i.e., $A = A' = A_{fin} = A'_{fin} = B = B_{fin}$. \end{enumerate}


algebra-lemma-bourbaki

Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. If $N$ is flat as $R$-module, then $$ \text{Ass}_S(M \otimes_R N) \supset \bigcup\nolimits_{\mathfrak p \in \text{Ass}_R(M)} \text{Ass}_S(N/\mathfrak pN) $$ and if $R$ is Noetherian then we have equality.


algebra-lemma-post-bourbaki

Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume $N$ is flat as an $R$-module and $R$ is a domain with fraction field $K$. Then $$ \text{Ass}_S(N) = \text{Ass}_S(N \otimes_R K) = \text{Ass}_{S \otimes_R K}(N \otimes_R K) $$ via the canonical inclusion $\Spec(S \otimes_R K) \subset \Spec(S)$.


algebra-lemma-bourbaki-fibres

Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. Assume $N$ is flat as $R$-module. Then $$ \text{Ass}_S(M \otimes_R N) \supset \bigcup\nolimits_{\mathfrak p \in \text{Ass}_R(M)} \text{Ass}_{S \otimes_R \kappa(\mathfrak p)}(N \otimes_R \kappa(\mathfrak p)) $$ where we use Remark \ref{remark-fundamental-diagram} to think of the spectra of fibre rings as subsets of $\Spec(S)$. If $R$ is Noetherian then this inclusion is an equality.


algebra-lemma-weakly-ass-local

Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime of $R$. The following are equivalent: \begin{enumerate} \item $\mathfrak p$ is weakly associated to $M$, \item $\mathfrak pR_{\mathfrak p}$ is weakly associated to $M_{\mathfrak p}$, and \item $M_{\mathfrak p}$ contains an element whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$. \end{enumerate}


algebra-lemma-reduced-weakly-ass-minimal

For a reduced ring the weakly associated primes of the ring are the minimal primes.


algebra-lemma-weakly-ass

Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ and $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$.


algebra-lemma-weakly-ass-zero

\begin{slogan} Every nonzero module has a weakly associated prime. \end{slogan} Let $R$ be a ring. Let $M$ be an $R$-module. Then $$ M = (0) \Leftrightarrow \text{WeakAss}(M) = \emptyset $$


algebra-lemma-weakly-ass-support

Let $R$ be a ring. Let $M$ be an $R$-module. Then $$ \text{Ass}(M) \subset \text{WeakAss}(M) \subset \text{Supp}(M). $$


algebra-lemma-weakly-ass-zero-divisors

Let $R$ be a ring. Let $M$ be an $R$-module. The union $\bigcup_{\mathfrak q \in \text{WeakAss}(M)} \mathfrak q$ is the set elements of $R$ which are zerodivisors on $M$.


algebra-lemma-weakly-ass-minimal-prime-support

Let $R$ be a ring. Let $M$ be an $R$-module. Any $\mathfrak p \in \text{Supp}(M)$ which is minimal among the elements of $\text{Supp}(M)$ is an element of $\text{WeakAss}(M)$.


algebra-lemma-ass-weakly-ass

Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime ideal of $R$ which is finitely generated. Then $$ \mathfrak p \in \text{Ass}(M) \Leftrightarrow \mathfrak p \in \text{WeakAss}(M). $$ In particular, if $R$ is Noetherian, then $\text{Ass}(M) = \text{WeakAss}(M)$.


algebra-lemma-weakly-ass-reverse-functorial

Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then we have $\Spec(\varphi)(\text{WeakAss}_S(M)) \supset \text{WeakAss}_R(M)$.


algebra-lemma-weakly-ass-finite-ring-map

Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Denote $f : \Spec(S) \to \Spec(R)$ the associated map on spectra. If $\varphi$ is a finite ring map, then $$ \text{WeakAss}_R(M) = f(\text{WeakAss}_S(M)). $$


algebra-lemma-localize-weakly-ass

Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Via the canonical injection $\Spec(S^{-1}R) \to \Spec(R)$ we have $\text{WeakAss}_R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$ and $$ \text{WeakAss}(M) \cap \Spec(S^{-1}R) = \text{WeakAss}(S^{-1}M). $$


algebra-lemma-localize-weakly-ass-nonzero-divisors

Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Assume that every $s \in S$ is a nonzerodivisor on $M$. Then $$ \text{WeakAss}(M) = \text{WeakAss}(S^{-1}M). $$


algebra-lemma-zero-at-weakly-ass-zero

Let $R$ be a ring. Let $M$ be an $R$-module. The map $$ M \longrightarrow \prod\nolimits_{\mathfrak p \in \text{WeakAss}(M)} M_{\mathfrak p} $$ is injective.


algebra-lemma-weakly-ass-change-fields

Let $K/k$ be a field extension. Let $R$ be a $k$-algebra. Let $M$ be an $R$-module. Let $\mathfrak q \subset R \otimes_k K$ be a prime lying over $\mathfrak p \subset R$. If $\mathfrak q$ is weakly associated to $M \otimes_k K$, then $\mathfrak p$ is weakly associated to $M$.


algebra-lemma-remove-embedded-primes

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Consider the set of $R$-submodules $$ \{ K \subset M \mid \text{Supp}(K) \text{ nowhere dense in } \text{Supp}(M) \}. $$ This set has a maximal element $K$ and the quotient $M' = M/K$ has the following properties \begin{enumerate} \item $\text{Supp}(M) = \text{Supp}(M')$, \item $M'$ has no embedded associated primes, \item for any $f \in R$ which is contained in all embedded associated primes of $M$ we have $M_f \cong M'_f$. \end{enumerate}


algebra-lemma-remove-embedded-primes-localize

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. For any $f \in R$ we have $(M')_f = (M_f)'$ where $M \to M'$ and $M_f \to (M_f)'$ are the quotients constructed in Lemma \ref{lemma-remove-embedded-primes}.


algebra-lemma-no-embedded-primes-endos

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module without embedded associated primes. Let $I = \{x \in R \mid xM = 0\}$. Then the ring $R/I$ has no embedded primes.


algebra-lemma-permute-xi

Let $R$ be a local Noetherian ring. Let $M$ be a finite $R$-module. Let $x_1, \ldots, x_c$ be an $M$-regular sequence. Then any permutation of the $x_i$ is a regular sequence as well.


algebra-lemma-flat-increases-depth

Let $R, S$ be local rings. Let $R \to S$ be a flat local ring homomorphism. Let $x_1, \ldots, x_r$ be a sequence in $R$. Let $M$ be an $R$-module. The following are equivalent \begin{enumerate} \item $x_1, \ldots, x_r$ is an $M$-regular sequence in $R$, and \item the images of $x_1, \ldots, x_r$ in $S$ form a $M \otimes_R S$-regular sequence. \end{enumerate}


algebra-lemma-regular-sequence-in-neighbourhood

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots, x_r$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-regular sequence. Then there exists a $g \in R$, $g \not \in \mathfrak p$ such that the image of $x_1, \ldots, x_r$ in $R_g$ forms an $M_g$-regular sequence.


algebra-lemma-join-regular-sequences

Let $A$ be a ring. Let $I$ be an ideal generated by a regular sequence $f_1, \ldots, f_n$ in $A$. Let $g_1, \ldots, g_m \in A$ be elements whose images $\overline{g}_1, \ldots, \overline{g}_m$ form a regular sequence in $A/I$. Then $f_1, \ldots, f_n, g_1, \ldots, g_m$ is a regular sequence in $A$.


algebra-lemma-regular-sequence-short-exact-sequence

Let $R$ be a ring. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $R$-modules. Let $f_1, \ldots, f_r \in R$. If $f_1, \ldots, f_r$ is $M_1$-regular and $M_3$-regular, then $f_1, \ldots, f_r$ is $M_2$-regular.


algebra-lemma-regular-sequence-powers

Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots, f_r \in R$ and $e_1, \ldots, e_r > 0$ integers. Then $f_1, \ldots, f_r$ is an $M$-regular sequence if and only if $f_1^{e_1}, \ldots, f_r^{e_r}$ is an $M$-regular sequence.


algebra-lemma-regular-sequence-in-polynomial-ring

Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$ which do not generate the unit ideal. The following are equivalent: \begin{enumerate} \item any permutation of $f_1, \ldots, f_r$ is a regular sequence, \item any subsequence of $f_1, \ldots, f_r$ (in the given order) is a regular sequence, and \item $f_1x_1, \ldots, f_rx_r$ is a regular sequence in the polynomial ring $R[x_1, \ldots, x_r]$. \end{enumerate}


algebra-lemma-regular-quasi-regular

Let $R$ be a ring. \begin{enumerate} \item A regular sequence $f_1, \ldots, f_c$ of $R$ is a quasi-regular sequence. \item Suppose that $M$ is an $R$-module and that $f_1, \ldots, f_c$ is an $M$-regular sequence. Then $f_1, \ldots, f_c$ is an $M$-quasi-regular sequence. \end{enumerate}


algebra-lemma-flat-base-change-quasi-regular

Let $R \to R'$ be a flat ring map. Let $M$ be an $R$-module. Suppose that $f_1, \ldots, f_r \in R$ form an $M$-quasi-regular sequence. Then the images of $f_1, \ldots, f_r$ in $R'$ form a $M \otimes_R R'$-quasi-regular sequence.


algebra-lemma-quasi-regular-sequence-in-neighbourhood

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots, x_c$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-quasi-regular sequence. Then there exists a $g \in R$, $g \not \in \mathfrak p$ such that the image of $x_1, \ldots, x_c$ in $R_g$ forms an $M_g$-quasi-regular sequence.


algebra-lemma-truncate-quasi-regular

Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots, f_c \in R$ be an $M$-quasi-regular sequence. For any $i$ the sequence $\overline{f}_{i + 1}, \ldots, \overline{f}_c$ of $\overline{R} = R/(f_1, \ldots, f_i)$ is an $\overline{M} = M/(f_1, \ldots, f_i)M$-quasi-regular sequence.


algebra-lemma-quasi-regular-regular

Let $(R, \mathfrak m)$ be a local Noetherian ring. Let $M$ be a nonzero finite $R$-module. Let $f_1, \ldots, f_c \in \mathfrak m$ be an $M$-quasi-regular sequence. Then $f_1, \ldots, f_c$ is an $M$-regular sequence.


algebra-lemma-quasi-regular-on-quotient

Let $R$ be a ring. Let $J = (f_1, \ldots, f_r)$ be an ideal of $R$. Let $M$ be an $R$-module. Set $\overline{R} = R/\bigcap_{n \geq 0} J^n$, $\overline{M} = M/\bigcap_{n \geq 0} J^nM$, and denote $\overline{f}_i$ the image of $f_i$ in $\overline{R}$. Then $f_1, \ldots, f_r$ is $M$-quasi-regular if and only if $\overline{f}_1, \ldots, \overline{f}_r$ is $\overline{M}$-quasi-regular.


algebra-lemma-affine-blowup

Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$. Let $R' = R[\frac{I}{a}]$ be the affine blowup algebra. Then \begin{enumerate} \item the image of $a$ in $R'$ is a nonzerodivisor, \item $IR' = aR'$, and \item $(R')_a = R_a$. \end{enumerate}


algebra-lemma-blowup-base-change

Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal and $a \in I$. Set $J = IS$ and let $b \in J$ be the image of $a$. Then $S[\frac{J}{b}]$ is the quotient of $S \otimes_R R[\frac{I}{a}]$ by the ideal of elements annihilated by some power of $b$.


algebra-lemma-affine-blowup-quotient-description

Let $R$ be a ring. Let $I = (a_1, \ldots, a_n)$ be an ideal of $R$. Let $a = a_1$. Then there is a surjection $$ R[x_2, \ldots, x_n]/(a x_2 - a_2, \ldots, a x_n - a_n) \longrightarrow \textstyle{R[\frac{I}{a}]} $$ whose kernel is the $a$-power torsion in the source.


algebra-lemma-blowup-in-principal

Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$. Set $R' = R[\frac{I}{a}]$. If $f \in R$ is such that $V(f) = V(I)$, then $f$ maps to a nonzerodivisor in $R'$ and $R'_f = R'_a = R_a$.


algebra-lemma-blowup-add-principal

Let $R$ be a ring, $I \subset R$ an ideal, $a \in I$, and $f \in R$. Set $R' = R[\frac{I}{a}]$ and $R'' = R[\frac{fI}{fa}]$. Then there is a surjective $R$-algebra map $R' \to R''$ whose kernel is the set of $f$-power torsion elements of $R'$.


algebra-lemma-blowup-reduced

\begin{slogan} Being reduced is invariant under blowup \end{slogan} If $R$ is reduced then every (affine) blowup algebra of $R$ is reduced.


algebra-lemma-blowup-domain

Let $R$ be a domain, $I \subset R$ an ideal, and $a \in I$ a nonzero element. Then the affine blowup algebra $R[\frac{I}{a}]$ is a domain.


algebra-lemma-valuation-ring-colimit-affine-blowups

Let $(R, \mathfrak m)$ be a local domain with fraction field $K$. Let $R \subset A \subset K$ be a valuation ring which dominates $R$. Then $$ A = \colim R[\textstyle{\frac{I}{a}}] $$ is a directed colimit of affine blowups $R \to R[\frac{I}{a}]$ with the following properties \begin{enumerate} \item $a \in I \subset \mathfrak m$, \item $I$ is finitely generated, and \item the fibre ring of $R \to R[\frac{I}{a}]$ at $\mathfrak m$ is not zero. \end{enumerate}


algebra-lemma-resolution-by-finite-free

Let $R$ be a ring. Let $M$ be an $R$-module. \begin{enumerate} \item There exists an exact complex $$ \ldots \to F_2 \to F_1 \to F_0 \to M \to 0. $$ with $F_i$ free $R$-modules. \item If $R$ is Noetherian and $M$ finite over $R$, then we can choose the complex such that $F_i$ is finite free. In other words, we can find an exact complex $$ \ldots \to R^{\oplus n_2} \to R^{\oplus n_1} \to R^{\oplus n_0} \to M \to 0. $$ \end{enumerate}


algebra-lemma-homotopic-equal-homology

Any two homotopic maps of complexes induce the same maps on (co)homology groups.


algebra-lemma-compare-resolutions

Let $R$ be a ring. Let $M \to N$ be a map of $R$-modules. Let $N_\bullet \to N$ be an arbitrary resolution. Let $$ \ldots \to F_2 \to F_1 \to F_0 \to M $$ be a complex of $R$-modules where each $F_i$ is a free $R$-module. Then \begin{enumerate} \item there exists a map of complexes $F_\bullet \to N_\bullet$ such that $$ \xymatrix{ F_0 \ar[r] \ar[d] & M \ar[d] \\ N_0 \ar[r] & N } $$ is commutative, and \item any two maps $\alpha, \beta : F_\bullet \to N_\bullet$ as in (1) are homotopic. \end{enumerate}


algebra-lemma-ext-welldefined

Let $R$ be a ring. Let $M_1, M_2, N$ be $R$-modules. Suppose that $F_{\bullet}$ is a free resolution of the module $M_1$, and $G_{\bullet}$ is a free resolution of the module $M_2$. Let $\varphi : M_1 \to M_2$ be a module map. Let $\alpha : F_{\bullet} \to G_{\bullet}$ be a map of complexes inducing $\varphi$ on $M_1 = \Coker(d_{F, 1}) \to M_2 = \Coker(d_{G, 1})$, see Lemma \ref{lemma-compare-resolutions}. Then the induced maps $$ H^i(\alpha) : H^i(\Hom_R(F_{\bullet}, N)) \longrightarrow H^i(\Hom_R(G_{\bullet}, N)) $$ are independent of the choice of $\alpha$. If $\varphi$ is an isomorphism, so are all the maps $H^i(\alpha)$. If $M_1 = M_2$, $F_\bullet = G_\bullet$, and $\varphi$ is the identity, so are all the maps $H_i(\alpha)$.


algebra-lemma-long-exact-seq-ext

Let $R$ be a ring. Let $M$ be an $R$-module. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence. Then we get a long exact sequence $$ \begin{matrix} 0 \to \Hom_R(M, N') \to \Hom_R(M, N) \to \Hom_R(M, N'') \\ \phantom{0\ } \to \Ext^1_R(M, N') \to \Ext^1_R(M, N) \to \Ext^1_R(M, N'') \to \ldots \end{matrix} $$


algebra-lemma-reverse-long-exact-seq-ext

Let $R$ be a ring. Let $N$ be an $R$-module. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence. Then we get a long exact sequence $$ \begin{matrix} 0 \to \Hom_R(M'', N) \to \Hom_R(M, N) \to \Hom_R(M', N) \\ \phantom{0\ } \to \Ext^1_R(M'', N) \to \Ext^1_R(M, N) \to \Ext^1_R(M', N) \to \ldots \end{matrix} $$


algebra-lemma-annihilate-ext

Let $R$ be a ring. Let $M$, $N$ be $R$-modules. Any $x\in R$ such that either $xN = 0$, or $xM = 0$ annihilates each of the modules $\Ext^i_R(M, N)$.


algebra-lemma-ext-noetherian

Let $R$ be a Noetherian ring. Let $M$, $N$ be finite $R$-modules. Then $\Ext^i_R(M, N)$ is a finite $R$-module for all $i$.


algebra-lemma-depth-weak-sequence

Let $R$ be a ring, $I \subset R$ an ideal, and $M$ a finite $R$-module. Then $\text{depth}_I(M)$ is equal to the supremum of the lengths of sequences $f_1, \ldots, f_r \in I$ such that $f_i$ is a nonzerodivisor on $M/(f_1, \ldots, f_{i - 1})M$.


algebra-lemma-bound-depth

Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module. Then $\dim(\text{Supp}(M)) \geq \text{depth}(M)$.


algebra-lemma-depth-finite-noetherian

Let $R$ be a Noetherian ring, $I \subset R$ an ideal, and $M$ a finite nonzero $R$-module such that $IM \not = M$. Then $\text{depth}_I(M) < \infty$.


algebra-lemma-depth-ext

Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{depth}(M)$ is equal to the smallest integer $i$ such that $\Ext^i_R(R/\mathfrak m, M)$ is nonzero.


algebra-lemma-depth-in-ses

Let $R$ be a local Noetherian ring. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence of nonzero finite $R$-modules. \begin{enumerate} \item $\text{depth}(N) \geq \min\{\text{depth}(N'), \text{depth}(N'')\}$ \item $\text{depth}(N'') \geq \min\{\text{depth}(N), \text{depth}(N') - 1\}$ \item $\text{depth}(N') \geq \min\{\text{depth}(N), \text{depth}(N'') + 1\}$ \end{enumerate}


algebra-lemma-depth-drops-by-one

Let $R$ be a local Noetherian ring and $M$ a nonzero finite $R$-module. \begin{enumerate} \item If $x \in \mathfrak m$ is a nonzerodivisor on $M$, then $\text{depth}(M/xM) = \text{depth}(M) - 1$. \item Any $M$-regular sequence $x_1, \ldots, x_r$ can be extended to an $M$-regular sequence of length $\text{depth}(M)$. \end{enumerate}


algebra-lemma-inherit-minimal-primes

Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. Let $x \in \mathfrak m$, $\mathfrak p \in \text{Ass}(M)$, and $\mathfrak q$ minimal over $\mathfrak p + (x)$. Then $\mathfrak q \in \text{Ass}(M/x^nM)$ for some $n \geq 1$.


algebra-lemma-depth-dim-associated-primes

Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. For $\mathfrak p \in \text{Ass}(M)$ we have $\dim(R/\mathfrak p) \geq \text{depth}(M)$.


algebra-lemma-depth-localization

Let $R$ be a local Noetherian ring and $M$ a finite $R$-module. For a prime ideal $\mathfrak p \subset R$ we have $\text{depth}(M_\mathfrak p) + \dim(R/\mathfrak p) \geq \text{depth}(M)$.


algebra-lemma-depth-goes-down-finite

Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $R \to S$ be a finite ring map. Let $\mathfrak m_1, \ldots, \mathfrak m_n$ be the maximal ideals of $S$. Let $N$ be a finite $S$-module. Then $$ \min\nolimits_{i = 1, \ldots, n} \text{depth}(N_{\mathfrak m_i}) = \text{depth}_\mathfrak m(N) $$


algebra-lemma-flat-base-change-ext

Given a flat ring map $R \to R'$, an $R$-module $M$, and an $R'$-module $N'$ the natural map $$ \Ext^i_{R'}(M \otimes_R R', N') \to \text{Ext}^i_R(M, N') $$ is an isomorphism for $i \geq 0$.


algebra-lemma-split-injection-after-completion

Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal contained in the Jacobson radical of $R$. Let $N \to M$ be a homomorphism of finite $R$-modules. Suppose that there exists arbitrarily large $n$ such that $N/I^nN \to M/I^nM$ is a split injection. Then $N \to M$ is a split injection.


algebra-lemma-long-exact-sequence-tor

Let $R$ be a ring and let $M$ be an $R$-module. Suppose that $0 \to N' \to N \to N'' \to 0$ is a short exact sequence of $R$-modules. There exists a long exact sequence $$ \text{Tor}_1^R(M, N') \to \text{Tor}_1^R(M, N) \to \text{Tor}_1^R(M, N'') \to M \otimes_R N' \to M \otimes_R N \to M \otimes_R N'' \to 0 $$


algebra-lemma-no-spectral-sequence

Let $(A_{\bullet, \bullet}, d, \delta)$ be a double complex such that \begin{enumerate} \item Each row $A_{\bullet, j}$ is a resolution of $R(A)_j$. \item Each column $A_{i, \bullet}$ is a resolution of $U(A)_i$. \end{enumerate} Then there are canonical isomorphisms $$ H_i(R(A)_\bullet) \cong H_i(U(A)_\bullet). $$ The isomorphisms are functorial with respect to morphisms of double complexes with the properties above.


algebra-lemma-tor-noetherian

Let $R$ be a Noetherian ring. Let $M$, $N$ be finite $R$-modules. Then $\text{Tor}_p^R(M, N)$ is a finite $R$-module for all $p$.


algebra-lemma-characterize-flat

Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent: \begin{enumerate} \item The module $M$ is flat over $R$. \item For all $i > 0$ the functor $\text{Tor}_i^R(M, -)$ is zero. \item The functor $\text{Tor}_1^R(M, -)$ is zero. \item For all ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$. \item For all finitely generated ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$. \end{enumerate}


algebra-lemma-flat-base-change-tor

Given a flat ring map $R \to R'$ and $R$-modules $M$, $N$ the natural $R$-module map $\text{Tor}_i^R(M, N)\otimes_R R' \to \text{Tor}_i^{R'}(M \otimes_R R', N \otimes_R R')$ is an isomorphism for all $i$.


algebra-lemma-tor-commutes-filtered-colimits

Let $R$ be a ring. Let $M = \colim M_i$ be a filtered colimit of $R$-modules. Let $N$ be an $R$-module. Then $\text{Tor}_n^R(M, N) = \colim \text{Tor}_n^R(M_i, N)$ for all $n$.


algebra-lemma-characterize-projective

Let $R$ be a ring. Let $P$ be an $R$-module. The following are equivalent \begin{enumerate} \item $P$ is projective, \item $P$ is a direct summand of a free $R$-module, and \item $\Ext^1_R(P, M) = 0$ for every $R$-module $M$. \end{enumerate}


algebra-lemma-direct-sum-projective

A direct sum of projective modules is projective.


algebra-lemma-lift-projective-module

Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $\overline{P}$ be a projective $R/I$-module. Then there exists a projective $R$-module $P$ such that $P/IP \cong \overline{P}$.


algebra-lemma-lift-projective

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume \begin{enumerate} \item $I$ is nilpotent, \item $M/IM$ is a projective $R/I$-module, \item $M$ is a flat $R$-module. \end{enumerate} Then $M$ is a projective $R$-module.


algebra-lemma-finite-projective

Let $R$ be a ring and let $M$ be an $R$-module. The following are equivalent \begin{enumerate} \item $M$ is finitely presented and $R$-flat, \item $M$ is finite projective, \item $M$ is a direct summand of a finite free $R$-module, \item $M$ is finitely presented and for all $\mathfrak p \in \Spec(R)$ the localization $M_{\mathfrak p}$ is free, \item $M$ is finitely presented and for all maximal ideals $\mathfrak m \subset R$ the localization $M_{\mathfrak m}$ is free, \item $M$ is finite and locally free, \item $M$ is finite locally free, and \item $M$ is finite, for every prime $\mathfrak p$ the module $M_{\mathfrak p}$ is free, and the function $$ \rho_M : \Spec(R) \to \mathbf{Z}, \quad \mathfrak p \longmapsto \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p) $$ is locally constant in the Zariski topology. \end{enumerate}


algebra-lemma-finite-projective-reduced

Let $R$ be a reduced ring and let $M$ be an $R$-module. Then the equivalent conditions of Lemma \ref{lemma-finite-projective} are also equivalent to \begin{enumerate} \item[(9)] $M$ is finite and the function $\rho_M : \Spec(R) \to \mathbf{Z}$, $\mathfrak p \mapsto \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p)$ is locally constant in the Zariski topology. \end{enumerate}


algebra-lemma-finite-flat-local

(Warning: see Remark \ref{remark-warning}.) Suppose $R$ is a local ring, and $M$ is a finite flat $R$-module. Then $M$ is finite free.


algebra-lemma-finite-projective-descends

Let $R \to S$ be a flat local homomorphism of local rings. Let $M$ be a finite $R$-module. Then $M$ is finite projective over $R$ if and only if $M \otimes_R S$ is finite projective over $S$.


algebra-lemma-locally-free-semi-local-free

Let $R$ be a semi-local ring. Let $M$ be a finite locally free module. If $M$ has constant rank, then $M$ is free. In particular, if $R$ has connected spectrum, then $M$ is free.


algebra-lemma-semi-local-module-basis-in-submodule

Let $R$ be a local ring with maximal ideal $\mathfrak m$ and infinite residue field. Let $R \to S$ be a ring map. Let $M$ be an $S$-module and let $N \subset M$ be an $R$-submodule. Assume \begin{enumerate} \item $S$ is semi-local and $\mathfrak mS$ is contained in the Jacobson radical of $S$, \item $M$ is a finite free $S$-module, and \item $N$ generates $M$ as an $S$-module. \end{enumerate} Then $N$ contains an $S$-basis of $M$.


algebra-lemma-evaluation-map-iso-finite-projective

Let $R$ be ring. Let $L$, $M$, $N$ be $R$-modules. The canonical map $$ \Hom_R(M, N) \otimes_R L \to \Hom_R(M, N \otimes_R L) $$ is an isomorphism if $M$ is finite projective.


algebra-lemma-map-between-finite

Let $R$ be a ring. Let $\varphi : M \to N$ be a map of $R$-modules with $N$ a finite $R$-module. Then we have the equality \begin{align*} U & = \{\mathfrak p \subset R \mid \varphi_{\mathfrak p} : M_{\mathfrak p} \to N_{\mathfrak p} \text{ is surjective}\} \\ & = \{\mathfrak p \subset R \mid \varphi \otimes \kappa(\mathfrak p) : M \otimes \kappa(\mathfrak p) \to N \otimes \kappa(\mathfrak p) \text{ is surjective}\} \end{align*} and $U$ is an open subset of $\Spec(R)$. Moreover, for any $f \in R$ such that $D(f) \subset U$ the map $M_f \to N_f$ is surjective.


algebra-lemma-map-between-finitely-presented

Let $R$ be a ring. Let $\varphi : M \to N$ be a map of $R$-modules with $M$ finite and $N$ finitely presented. Then $$ U = \{\mathfrak p \subset R \mid \varphi_{\mathfrak p} : M_{\mathfrak p} \to N_{\mathfrak p} \text{ is an isomorphism}\} $$ is an open subset of $\Spec(R)$.


algebra-lemma-cokernel-flat

Let $R$ be a ring. Let $\varphi : P_1 \to P_2$ be a map of finite projective modules. Then \begin{enumerate} \item The set $U$ of primes $\mathfrak p \in \Spec(R)$ such that $\varphi \otimes \kappa(\mathfrak p)$ is injective is open and for any $f\in R$ such that $D(f) \subset U$ we have \begin{enumerate} \item $P_{1, f} \to P_{2, f}$ is injective, and \item the module $\Coker(\varphi)_f$ is finite projective over $R_f$. \end{enumerate} \item The set $W$ of primes $\mathfrak p \in \Spec(R)$ such that $\varphi \otimes \kappa(\mathfrak p)$ is surjective is open and for any $f\in R$ such that $D(f) \subset W$ we have \begin{enumerate} \item $P_{1, f} \to P_{2, f}$ is surjective, and \item the module $\Ker(\varphi)_f$ is finite projective over $R_f$. \end{enumerate} \item The set $V$ of primes $\mathfrak p \in \Spec(R)$ such that $\varphi \otimes \kappa(\mathfrak p)$ is an isomorphism is open and for any $f\in R$ such that $D(f) \subset V$ the map $\varphi : P_{1, f} \to P_{2, f}$ is an isomorphism of modules over $R_f$. \end{enumerate}


algebra-lemma-flat-factors-free

Let $M$ be an $R$-module.  The following are equivalent: \begin{enumerate} \item $M$ is flat. \item If $f: R^n \to M$ is a module map and $x \in \Ker(f)$, then there are module maps $h: R^n \to R^m$ and $g: R^m \to M$ such that $f = g \circ h$ and $x \in \Ker(h)$. \item Suppose $f: R^n \to M$ is a module map, $N \subset \Ker(f)$ any submodule, and $h: R^n \to R^{m}$ a map such that $N \subset \Ker(h)$ and $f$ factors through $h$.  Then given any $x \in \Ker(f)$ we can find a map $h': R^n \to R^{m'}$ such that $N + Rx \subset \Ker(h')$ and $f$ factors through $h'$. \item If $f: R^n \to M$ is a module map and $N \subset \Ker(f)$ is a finitely generated submodule, then there are module maps $h: R^n \to R^m$ and $g: R^m \to M$ such that $f = g \circ h$ and $N \subset \Ker(h)$. \end{enumerate}


algebra-lemma-flat-factors-fp

Let $M$ be an $R$-module.  Then $M$ is flat if and only if the following condition holds: if $P$ is a finitely presented $R$-module and $f: P \to M$ a module map, then there is a free finite $R$-module $F$ and module maps $h: P \to F$ and $g: F \to M$ such that $f = g \circ h$.


algebra-lemma-flat-surjective-hom

Let $M$ be an $R$-module.  Then $M$ is flat if and only if the following condition holds: for every finitely presented $R$-module $P$, if $N \to M$ is a surjective $R$-module map, then the induced map $\Hom_R(P, N) \to \Hom_R(P, M)$ is surjective.


algebra-lemma-universally-exact-split

Let $$ 0 \to M_1 \to M_2 \to M_3 \to 0 $$ be an exact sequence of $R$-modules.  Suppose $M_3$ is of finite presentation. Then $$ 0 \to M_1 \to M_2 \to M_3 \to 0 $$ is universally exact if and only if it is split.


algebra-lemma-flat-universally-injective

Let $M$ be an $R$-module.  Then $M$ is flat if and only if any exact sequence of $R$-modules $$ 0 \to M_1 \to M_2 \to M \to 0 $$ is universally exact.


algebra-lemma-ui-flat-domain

Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules, and suppose $M_2$ is flat. Then $M_1$ and $M_3$ are flat.


algebra-lemma-universally-injective-tensor

Let $R$ be a ring. Let $M \to M'$ be a universally injective $R$-module map. Then for any $R$-module $N$ the map $M \otimes_R N \to M' \otimes_R N$ is universally injective.


algebra-lemma-composition-universally-injective

Let $R$ be a ring. A composition of universally injective $R$-module maps is universally injective.


algebra-lemma-universally-injective-permanence

Let $R$ be a ring. Let $M \to M'$ and $M' \to M''$ be $R$-module maps. If their composition $M \to M''$ is universally injective, then $M \to M'$ is universally injective.


algebra-lemma-faithfully-flat-universally-injective

Let $R \to S$ be a faithfully flat ring map. Then $R \to S$ is universally injective as a map of $R$-modules. In particular $R \cap IS = I$ for any ideal $I \subset R$.


algebra-lemma-universally-injective-check-stalks

Let $R \to S$ be a ring map. Let $M \to M'$ be a map of $S$-modules. The following are equivalent \begin{enumerate} \item $M \to M'$ is universally injective as a map of $R$-modules, \item for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R$-modules, \item for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R$-modules, \item for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak q$ in $R$, and \item for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak m$ in $R$. \end{enumerate}


algebra-lemma-universally-injective-localize

Let $\varphi : A \to B$ be a ring map. Let $S \subset A$ and $S' \subset B$ be multiplicative subsets such that $\varphi(S) \subset S'$. Let $M \to M'$ be a map of $B$-modules. \begin{enumerate} \item If $M \to M'$ is universally injective as a map of $A$-modules, then $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules and as a map of $S^{-1}A$-modules. \item If $M$ and $M'$ are $(S')^{-1}B$-modules, then $M \to M'$ is universally injective as a map of $A$-modules if and only if it is universally injective as a map of $S^{-1}A$-modules. \end{enumerate}


algebra-lemma-check-universally-injective-into-flat

Let $R$ be a ring and let $M \to M'$ be a map of $R$-modules. If $M'$ is flat, then $M \to M'$ is universally injective if and only if $M/IM \to M'/IM'$ is injective for every finitely generated ideal $I$ of $R$.


algebra-lemma-finite-projective-again

Let $M$ be an $R$-module.  Then $M$ is finite projective if and only if $M$ is finitely presented and flat.


algebra-lemma-descend-properties-modules

Let $R \to S$ be a faithfully flat ring map. Let $M$ be an $R$-module. Then \begin{enumerate} \item if the $S$-module $M \otimes_R S$ is of finite type, then $M$ is of finite type, \item if the $S$-module $M \otimes_R S$ is of finite presentation, then $M$ is of finite presentation, \item if the $S$-module $M \otimes_R S$ is flat, then $M$ is flat, and \item add more here as needed. \end{enumerate}


algebra-lemma-direct-sum-devissage

Let $M$ be an $R$-module.  If $(M_{\alpha})_{\alpha \in S}$ is a direct sum d\'evissage of $M$, then $M \cong \bigoplus_{\alpha + 1 \in S} M_{\alpha + 1}/M_{\alpha}$.


algebra-lemma-Kaplansky-devissage

Let $M$ be an $R$-module.  Then $M$ is a direct sum of countably generated $R$-modules if and only if it admits a Kaplansky d\'evissage.


algebra-lemma-projective-free

Let $R$ be a ring.  Then every projective $R$-module is free if and only if every countably generated projective $R$-module is free.


algebra-lemma-freeness-criteria

Let $M$ be a countably generated $R$-module.  Suppose any direct summand $N$ of $M$ satisfies: any element of $N$ is contained in a free direct summand of $N$.  Then $M$ is free.


algebra-lemma-projective-freeness-criteria

Let $P$ be a projective module over a local ring $R$.  Then any element of $P$ is contained in a free direct summand of $P$.


algebra-lemma-ML-limit-nonempty

Let $(A_i, \varphi_{ji})$ be a directed inverse system over $I$.  Suppose $I$ is countable.  If $(A_i, \varphi_{ji})$ is Mittag-Leffler and the $A_i$ are nonempty, then $\lim A_i$ is nonempty.


algebra-lemma-ML-exact-sequence

Let $$ 0 \to A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i \to 0 $$ be an exact sequence of directed inverse systems of abelian groups over $I$. Suppose $I$ is countable.  If $(A_i)$ is Mittag-Leffler, then $$ 0 \to \lim A_i \to \lim B_i \to \lim C_i\to 0 $$ is exact.


algebra-lemma-Mittag-Leffler

Let $R$ be a ring. Let $0 \to K_i \to L_i \to M_i \to 0$ be short exact sequences of $R$-modules, $i \geq 1$ which fit into maps of short exact sequences $$ \xymatrix{ 0 \ar[r] & K_i \ar[r] & L_i \ar[r] & M_i \ar[r] & 0 \\ 0 \ar[r] & K_{i + 1} \ar[r] \ar[u] & L_{i + 1} \ar[r] \ar[u] & M_{i + 1} \ar[r] \ar[u] & 0} $$ If for every $i$ there exists a $c = c(i) \geq i$ such that $\Im(K_c \to K_i) = \Im(K_j \to K_i)$ for all $j \geq c$, then the sequence $$ 0 \to \lim K_i \to \lim L_i \to \lim M_i \to 0 $$ is exact.


algebra-lemma-domination-fp

Let $f: M \to N$ and $g: M \to M'$ be maps of $R$-modules. Then $g$ dominates $f$ if and only if for any finitely presented $R$-module $Q$, we have $\Ker(f \otimes_R \text{id}_Q) \subset \Ker(g \otimes_R \text{id}_Q)$.


algebra-lemma-domination-universally-injective

Let $f : M \to N$ and $g : M \to M'$ be maps of $R$-modules. Consider the pushout of $f$ and $g$, $$ \xymatrix{ M  \ar[r]_f \ar[d]_g & N \ar[d]^{g'} \\ M' \ar[r]^{f'} & N' } $$ Then $g$ dominates $f$ if and only if $f'$ is universally injective.


algebra-lemma-domination

Let $f: M \to N$ and $g: M \to M'$ be maps of $R$-modules. Suppose $\Coker(f)$ is of finite presentation.  Then $g$ dominates $f$ if and only if $g$ factors through $f$, i.e.\ there exists a module map $h: N \to M'$ such that $g = h \circ f$.


algebra-lemma-tensor-ML-modules

If $R$ is a ring and $M$, $N$ are Mittag-Leffler modules over $R$, then $M \otimes_R N$ is a Mittag-Leffler module.


algebra-lemma-restrict-ML-modules

Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. If $M$ is a Mittag-Leffler module over $S$ then $M$ is a Mittag-Leffler module over $R$.


algebra-lemma-mod-ideal-ML-modules

Let $R$ be a ring. Let $S = R/I$ for some finitely generated ideal $I$. Let $M$ be an $S$-module. Then $M$ is a Mittag-Leffler module over $R$ if and only if $M$ is a Mittag-Leffler module over $S$.


algebra-lemma-kernel-tensored-fp

Let $M$ be an $R$-module, $P$ a finitely presented $R$-module, and $f: P \to M$ a map.  Let $Q$ be an $R$-module and suppose $x \in \Ker(P \otimes Q \to M \otimes Q)$.  Then there exists a finitely presented $R$-module $P'$ and a map $f': P \to P'$ such that $f$ factors through $f'$ and $x \in \Ker(P \otimes Q \to P' \otimes Q)$.


algebra-lemma-minimal-contains

Let $M$ be a flat Mittag-Leffler module over $R$. Let $F$ be an $R$-module and let $x \in F \otimes_R M$. Then there exists a smallest submodule $F' \subset F$ such that $x \in F' \otimes_R M$.


algebra-lemma-pure-submodule-ML

Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules.  Then: \begin{enumerate} \item If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler. \item If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler. \end{enumerate}


algebra-lemma-colimit-universally-injective-ML

If $M = \colim M_i$ is the colimit of a directed system of Mittag-Leffler $R$-modules $M_i$ with universally injective transition maps, then $M$ is Mittag-Leffler.


algebra-lemma-direct-sum-ML

If $M = \bigoplus_{i \in I} M_i$ is a direct sum of $R$-modules, then $M$ is Mittag-Leffler if and only if each $M_i$ is Mittag-Leffler.


algebra-lemma-coherent

Let $R$ be a ring. \begin{enumerate} \item A finite submodule of a coherent module is coherent. \item Let $\varphi : N \to M$ be a homomorphism from a finite module to a coherent module. Then $\Ker(\varphi)$ is finite. \item Let $\varphi : N \to M$ be a homomorphism of coherent modules. Then $\Ker(\varphi)$ and $\Coker(\varphi)$ are coherent modules. \item Given a short exact sequence of $R$-modules $0 \to M_1 \to M_2 \to M_3 \to 0$ if two out of three are coherent so is the third. \end{enumerate}


algebra-lemma-coherent-ring

Let $R$ be a ring. If $R$ is coherent, then a module is coherent if and only if it is finitely presented.


algebra-lemma-Noetherian-coherent

A Noetherian ring is a coherent ring.


algebra-lemma-flat-ML-criterion

Let $M$ be a flat $R$-module. The following are equivalent \begin{enumerate} \item $M$ is Mittag-Leffler, and \item if $F$ is a finite free $R$-module and $x \in F \otimes_R M$, then there exists a smallest submodule $F'$ of $F$ such that $x \in F' \otimes_R M$. \end{enumerate}


algebra-lemma-product-over-Noetherian-ring

Let $R$ be a Noetherian ring and $A$ a set. Then $M = R^A$ is a flat and Mittag-Leffler $R$-module.


algebra-lemma-power-series-ML

Let $R$ be a Noetherian ring and $n$ a positive integer.  Then the $R$-module $M = R[[t_1, \ldots, t_n]]$ is flat and Mittag-Leffler.


algebra-lemma-ML-countable-colimit

Let $M$ be an $R$-module.  Write $M = \colim_{i \in I} M_i$ where $(M_i, f_{ij})$ is a directed system of finitely presented $R$-modules.  If $M$ is Mittag-Leffler and countably generated, then there is a directed countable subset $I' \subset I$ such that $M \cong \colim_{i \in I'} M_i$.


algebra-lemma-ML-countable

Let $R$ be a ring. Let $M$ be an $R$-module. Assume $M$ is Mittag-Leffler and countably generated. For any $R$-module map $f : P \to M$ with $P$ finitely generated there exists an endomorphism $\alpha : M \to M$ such that \begin{enumerate} \item $\alpha : M \to M$ factors through a finitely presented $R$-module, and \item $\alpha \circ f = f$. \end{enumerate}


algebra-lemma-countgen-projective

Let $M$ be an $R$-module.  If $M$ is flat, Mittag-Leffler, and countably generated, then $M$ is projective.


algebra-lemma-ML-ui-descent

Let $f: M \to N$ be universally injective map of $R$-modules.  Suppose $M$ is a direct sum of countably generated $R$-modules, and suppose $N$ is flat and Mittag-Leffler.  Then $M$ is projective.


algebra-lemma-ascend-properties-modules

Let $R \to S$ be a ring map.  Let $M$ be an $R$-module.  Then: \begin{enumerate} \item If $M$ is flat, then the $S$-module $M \otimes_R S$ is flat. \item If $M$ is Mittag-Leffler, then the $S$-module $M \otimes_R S$ is Mittag-Leffler. \item If $M$ is a direct sum of countably generated $R$-modules, then the $S$-module $M \otimes_R S$ is a direct sum of countably generated $S$-modules. \item If $M$ is projective, then the $S$-module $M \otimes_R S$ is projective. \end{enumerate}


algebra-lemma-ffdescent-ML

\begin{reference} Email from Juan Pablo Acosta Lopez dated 12/20/14. \end{reference} Let $R \to S$ be a faithfully flat ring map. Let $M$ be an $R$-module. If the $S$-module $M \otimes_R S$ is Mittag-Leffler, then $M$ is Mittag-Leffler.


algebra-lemma-ffdescent-countable-projectivity

Let $R \to S$ be a faithfully flat ring map.  Let $M$ be an $R$-module.  If the $S$-module $M \otimes_R S$ is countably generated and projective, then $M$ is countably generated and projective.


algebra-lemma-lift-countably-generated-submodule

Let $R \to S$ be a ring map, let $M$ be an $R$-module, and let $Q$ be a countably generated $S$-submodule of $M \otimes_R S$.  Then there exists a countably generated $R$-submodule $P$ of $M$ such that $\Im(P \otimes_R S \to M \otimes_R S)$ contains $Q$.


algebra-lemma-adapted-submodule

Let $R \to S$ be a ring map, and let $M$ be an $R$-module.  Suppose $M \otimes_R S = \bigoplus_{i \in I} Q_i$ is a direct sum of countably generated $S$-modules $Q_i$.  If $N$ is a countably generated submodule of $M$, then there is a countably generated submodule $N'$ of $M$ such that $N' \supset N$ and $\Im(N' \otimes_R S \to M \otimes_R S) = \bigoplus_{i \in I'} Q_i$ for some subset $I' \subset I$.


algebra-lemma-completion-generalities

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\varphi : M \to N$ be a map of $R$-modules. \begin{enumerate} \item If $M/IM \to N/IN$ is surjective, then $M^\wedge \to N^\wedge$ is surjective. \item If $M \to N$ is surjective, then $M^\wedge \to N^\wedge$ is surjective. \item If $0 \to K \to M \to N \to 0$ is a short exact sequence of $R$-modules and $N$ is flat, then $0 \to K^\wedge \to M^\wedge \to N^\wedge \to 0$ is a short exact sequence. \item The map $M \otimes_R R^\wedge \to M^\wedge$ is surjective for any finite $R$-module $M$. \end{enumerate}


algebra-lemma-hathat-finitely-generated

\begin{reference} \cite[Theorem 15]{Matlis}. The slick proof given here is from an email of Bjorn Poonen dated Nov 5, 2016. \end{reference} Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $M$ be an $R$-module. Then \begin{enumerate} \item the completion $M^\wedge$ is $I$-adically complete, and \item $I^nM^\wedge = \Ker(M^\wedge \to M/I^nM) = (I^nM)^\wedge$ for all $n \geq 1$. \end{enumerate} In particular $R^\wedge$ is $I$-adically complete, $I^nR^\wedge = (I^n)^\wedge$, and $R^\wedge/I^nR^\wedge = R/I^n$.


algebra-lemma-completion-differ-by-torsion

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $0 \to M \to N \to Q \to 0$ be an exact sequence of $R$-modules such that $Q$ is annihilated by a power of $I$. Then completion produces an exact sequence $0 \to M^\wedge \to N^\wedge \to Q \to 0$.


algebra-lemma-radical-completion

Let $R$ be a ring, let $I \subset R$ be an ideal, and let $R^\wedge = \lim R/I^n$. \begin{enumerate} \item any element of $R^\wedge$ which maps to a unit of $R/I$ is a unit, \item any element of $1 + I$ maps to an invertible element of $R^\wedge$, \item any element of $1 + IR^\wedge$ is invertible in $R^\wedge$, and \item the ideals $IR^\wedge$ and $\Ker(R^\wedge \to R/I)$ are contained in the Jacobson radical of $R^\wedge$. \end{enumerate}


algebra-lemma-when-surjective-to-completion

Let $A$ be a ring. Let $I = (f_1, \ldots, f_r)$ be a finitely generated ideal. If $M \to \lim M/f_i^nM$ is surjective for each $i$, then $M \to \lim M/I^nM$ is surjective.


algebra-lemma-complete-by-sub

Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. If $M$ is $J$-adically complete and $I$ is finitely generated, then $M$ is $I$-adically complete.


algebra-lemma-change-ideal-completion

Let $R$ be a ring. Let $I$, $J$ be ideals of $R$. Assume there exist integers $c, d > 0$ such that $I^c \subset J$ and $J^d \subset I$. Then completion with respect to $I$ agrees with completion with respect to $J$ for any $R$-module. In particular an $R$-module $M$ is $I$-adically complete if and only if it is $J$-adically complete.


algebra-lemma-quotient-complete

Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $I$-adically complete $R$-module, and let $K \subset M$ be an $R$-submodule. The following are equivalent \begin{enumerate} \item $K = \bigcap (K + I^nM)$ and \item $M/K$ is $I$-adically complete. \end{enumerate}


algebra-lemma-when-finite-module-complete-over-complete-ring

Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $R$-module. If (a) $R$ is $I$-adically complete, (b) $M$ is a finite $R$-module, and (c) $\bigcap I^nM = (0)$, then $M$ is $I$-adically complete.


algebra-lemma-finite-over-complete-ring

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume \begin{enumerate} \item $R$ is $I$-adically complete, \item $\bigcap_{n \geq 1} I^nM = (0)$, and \item $M/IM$ is a finite $R/I$-module. \end{enumerate} Then $M$ is a finite $R$-module.


algebra-lemma-completion-tensor

Let $I$ be an ideal of a Noetherian ring $R$. Denote ${}^\wedge$ completion with respect to $I$. \begin{enumerate} \item If $K \to N$ is an injective map of finite $R$-modules, then the map on completions $K^\wedge \to N^\wedge$ is injective. \item If $0 \to K \to N \to M \to 0$ is a short exact sequence of finite $R$-modules, then $0 \to K^\wedge \to N^\wedge \to M^\wedge \to 0$ is a short exact sequence. \item If $M$ is a finite $R$-module, then $M^\wedge = M \otimes_R R^\wedge$. \end{enumerate}


algebra-lemma-completion-flat

Let $I$ be a ideal of a Noetherian ring $R$. Denote ${}^\wedge$ completion with respect to $I$. \begin{enumerate} \item The ring map $R \to R^\wedge$ is flat. \item The functor $M \mapsto M^\wedge$ is exact on the category of finitely generated $R$-modules. \end{enumerate}


algebra-lemma-completion-faithfully-flat

Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $I \subset \mathfrak m$ be an ideal. Denote $R^\wedge$ the completion of $R$ with respect to $I$. The ring map $R \to R^\wedge$ is faithfully flat. In particular the completion with respect to $\mathfrak m$, namely $\lim_n R/\mathfrak m^n$ is faithfully flat.


algebra-lemma-completion-complete

Let $R$ be a Noetherian ring. Let $I$ be an ideal of $R$. Let $M$ be an $R$-module. Then the completion $M^\wedge$ of $M$ with respect to $I$ is $I$-adically complete, $I^n M^\wedge = (I^nM)^\wedge$, and $M^\wedge/I^nM^\wedge = M/I^nM$.


algebra-lemma-completion-Noetherian

Let $I$ be an ideal of a ring $R$. Assume \begin{enumerate} \item $R/I$ is a Noetherian ring, \item $I$ is finitely generated. \end{enumerate} Then the completion $R^\wedge$ of $R$ with respect to $I$ is a Noetherian ring complete with respect to $IR^\wedge$.


algebra-lemma-completion-Noetherian-Noetherian

Let $R$ be a Noetherian ring. Let $I$ be an ideal of $R$. The completion $R^\wedge$ of $R$ with respect to $I$ is Noetherian.


algebra-lemma-finite-after-completion

Let $R \to S$ be a local homomorphism of local rings $(R, \mathfrak m)$ and $(S, \mathfrak n)$. Let $R^\wedge$, resp.\ $S^\wedge$ be the completion of $R$, resp.\ $S$ with respect to $\mathfrak m$, resp.\ $\mathfrak n$. If $\mathfrak m$ and $\mathfrak n$ are finitely generated and $\dim_{\kappa(\mathfrak m)} S/\mathfrak mS < \infty$, then \begin{enumerate} \item $S^\wedge$ is equal to the $\mathfrak m$-adic completion of $S$, and \item $S^\wedge$ is a finite $R^\wedge$-module. \end{enumerate}


algebra-lemma-completion-finite-extension

Let $R$ be a Noetherian ring. Let $R \to S$ be a finite ring map. Let $\mathfrak p \subset R$ be a prime and let $\mathfrak q_1, \ldots, \mathfrak q_m$ be the primes of $S$ lying over $\mathfrak p$ (Lemma \ref{lemma-finite-finite-fibres}). Then $$ R_\mathfrak p^\wedge \otimes_R S = (S_\mathfrak p)^\wedge = S_{\mathfrak q_1}^\wedge \times \ldots \times S_{\mathfrak q_m}^\wedge $$ where the $(S_\mathfrak p)^\wedge$ is the completion with respect to $\mathfrak p$ and the local rings $R_\mathfrak p$ and $S_{\mathfrak q_i}$ are completed with respect to their maximal ideals.


algebra-lemma-split-completed-sequence

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $0 \to K \to P \to M \to 0$ be a short exact sequence of $R$-modules. If $M$ is flat over $R$ and $M/IM$ is a projective $R/I$-module, then the sequence of $I$-adic completions $$ 0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0 $$ is a split exact sequence.


algebra-lemma-complete-modulo-nilpotent

Let $A$ be a Noetherian ring. Let $I, J \subset A$ be ideals. If $A$ is $I$-adically complete and $A/I$ is $J$-adically complete, then $A$ is $J$-adically complete.


algebra-lemma-limit-complete-pre

Let $I \subset A$ be a finitely generated ideal of a ring. Let $(M_n)$ be an inverse system of $A$-modules with $I^n M_n = 0$. Then $M = \lim M_n$ is $I$-adically complete.


algebra-lemma-limit-complete

Let $I \subset A$ be a finitely generated ideal of a ring. Let $(M_n)$ be an inverse system of $A$-modules with $M_n = M_{n + 1}/I^nM_{n + 1}$. Then $M/I^nM = M_n$ and $M$ is $I$-adically complete.


algebra-lemma-finiteness-graded

Let $A$ be a Noetherian graded ring. Let $I \subset A_+$ be a homogeneous ideal. Let $(N_n)$ be an inverse system of finite graded $A$-modules with $N_n = N_{n + 1}/I^n N_{n + 1}$. Then there is a finite graded $A$-module $N$ such that $N_n = N/I^nN$ as graded modules for all $n$.


algebra-lemma-daniel-litt

Let $A$ be a graded ring. Let $I \subset A_+$ be a homogeneous ideal. Denote $A' = \lim A/I^n$. Let $(G_n)$ be an inverse system of graded $A$-modules with $G_n$ annihilated by $I^n$. Let $M$ be a graded $A$-module and let $\varphi_n : M \to G_n$ be a compatible system of graded $A$-module maps. If the induced map $$ \varphi : M \otimes_A A' \longrightarrow \lim G_n $$ is an isomorphism, then $M_d \to \lim G_{n, d}$ is an isomorphism for all $d \in \mathbf{Z}$.


algebra-lemma-mod-injective

Suppose that $R \to S$ is a local homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $M$ be a flat $R$-module and $N$ a finite $S$-module. Let $u : N \to M$ be a map of $R$-modules. If $\overline{u} : N/\mathfrak m N \to M/\mathfrak m M$ is injective then $u$ is injective. In this case $M/u(N)$ is flat over $R$.


algebra-lemma-grothendieck

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


algebra-lemma-grothendieck-regular-sequence

Suppose that $R \to S$ is a flat and local ring homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose $f_1, \ldots, f_c$ is a sequence of elements of $S$ such that the images $\overline{f}_1, \ldots, \overline{f}_c$ form a regular sequence in $S/{\mathfrak m}S$. 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-free-fibre-flat-free

Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $\mathfrak m$ be the maximal ideal of $R$. Let $M$ be a finite $S$-module. Suppose that (a) $M/\mathfrak mM$ is a free $S/\mathfrak mS$-module, and (b) $M$ is flat over $R$. Then $M$ is free and $S$ is flat over $R$.


algebra-lemma-complex-exact-mod

Let $R \to S$ be a local homomorphism of local Noetherian rings. Let $\mathfrak m$ be the maximal ideal of $R$. Let $0 \to F_e \to F_{e-1} \to \ldots \to F_0$ be a finite complex of finite $S$-modules. Assume that each $F_i$ is $R$-flat, and that the complex $0 \to F_e/\mathfrak m F_e \to F_{e-1}/\mathfrak m F_{e-1} \to \ldots \to F_0 / \mathfrak m F_0$ is exact. Then $0 \to F_e \to F_{e-1} \to \ldots \to F_0$ is exact, and moreover the module $\Coker(F_1 \to F_0)$ is $R$-flat.


algebra-lemma-prepare-local-criterion-flatness

Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa = R/\mathfrak m$. Let $M$ be an $R$-module. If $\text{Tor}_1^R(\kappa, M) = 0$, then for every finite length $R$-module $N$ we have $\text{Tor}_1^R(N, M) = 0$.


algebra-lemma-local-criterion-flatness

Let $R \to S$ be a local homomorphism of local Noetherian rings. Let $\mathfrak m$ be the maximal ideal of $R$, and let $\kappa = R/\mathfrak m$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^R(\kappa, M) = 0$, then $M$ is flat over $R$.


algebra-lemma-what-does-it-mean

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. If $M/IM$ is flat over $R/I$ and $\text{Tor}_1^R(R/I, M) = 0$ then \begin{enumerate} \item $M/I^nM$ is flat over $R/I^n$ for all $n \geq 1$, and \item for any module $N$ which is annihilated by $I^m$ for some $m \geq 0$ we have $\text{Tor}_1^R(N, M) = 0$. \end{enumerate} In particular, if $I$ is nilpotent, then $M$ is flat over $R$.


algebra-lemma-what-does-it-mean-again

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. \begin{enumerate} \item If $M/IM$ is flat over $R/I$ and $M \otimes_R I/I^2 \to IM/I^2M$ is injective, then $M/I^2M$ is flat over $R/I^2$. \item If $M/IM$ is flat over $R/I$ and $M \otimes_R I^n/I^{n + 1} \to I^nM/I^{n + 1}M$ is injective for $n = 1, \ldots, k$, then $M/I^{k + 1}M$ is flat over $R/I^{k + 1}$. \end{enumerate}


algebra-lemma-variant-local-criterion-flatness

Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $I \not = R$ be an ideal in $R$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^R(M, R/I) = 0$ and $M/IM$ is flat over $R/I$, then $M$ is flat over $R$.


algebra-lemma-flat-module-powers

Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Assume \begin{enumerate} \item $R$ is a Noetherian ring, \item $S$ is a Noetherian ring, \item $M$ is a finite $S$-module, and \item for each $n \geq 1$ the module $M/I^n M$ is flat over $R/I^n$. \end{enumerate} Then for every $\mathfrak q \in V(IS)$ the localization $M_{\mathfrak q}$ is flat over $R$. In particular, if $S$ is local and $IS$ is contained in its maximal ideal, then $M$ is flat over $R$.


algebra-lemma-surjective-on-tor-one

Let $R \to R' \to R''$ be ring maps. Let $M$ be an $R$-module. Suppose that $M \otimes_R R'$ is flat over $R'$. Then the natural map $\text{Tor}_1^R(M, R') \otimes_{R'} R'' \to \text{Tor}_1^R(M, R'')$ is onto.


algebra-lemma-surjective-on-tor-one-trivial

Let $R \to R'$ be a ring map. Let $I \subset R$ be an ideal and $I' = IR'$. Let $M$ be an $R$-module and set $M' = M \otimes_R R'$. The natural map $\text{Tor}_1^R(R'/I', M) \to \text{Tor}_1^{R'}(R'/I', M')$ is surjective.


algebra-lemma-another-variant-local-criterion-flatness

Let $$ \xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] } $$ be a commutative diagram of local homomorphisms of local Noetherian rings. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $S$-module. Denote $I' = IR'$ and $M' = M \otimes_S S'$. Assume that \begin{enumerate} \item $S'$ is a localization of the tensor product $S \otimes_R R'$, \item $M/IM$ is flat over $R/I$, \item $\text{Tor}_1^R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$ is zero. \end{enumerate} Then $M'$ is flat over $R'$.


algebra-lemma-criterion-flatness-fibre-Noetherian

Let $R$, $S$, $S'$ be Noetherian local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $\mathfrak m \subset R$ be the maximal ideal. Let $M$ be an $S'$-module. Assume \begin{enumerate} \item The module $M$ is finite over $S'$. \item The module $M$ is not zero. \item The module $M/\mathfrak m M$ is a flat $S/\mathfrak m S$-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-base-change-flat-up-down

Let $$ \xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] } $$ be a commutative diagram of local homomorphisms of local rings. Assume that $S'$ is a localization of the tensor product $S \otimes_R R'$. Let $M$ be an $S$-module and set $M' = S' \otimes_S M$. \begin{enumerate} \item If $M$ is flat over $R$ then $M'$ is flat over $R'$. \item If $M'$ is flat over $R'$ and $R \to R'$ is flat then $M$ is flat over $R$. \end{enumerate} In particular we have \begin{enumerate} \item[(3)] If $S$ is flat over $R$ then $S'$ is flat over $R'$. \item[(4)] If $R' \to S'$ and $R \to R'$ are flat then $S$ is flat over $R$. \end{enumerate}


algebra-lemma-yet-another-variant-local-criterion-flatness

Consider a commutative diagram of local rings and local homomorphisms $$ \xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] } $$ Let $M$ be a finite $S$-module. Assume that \begin{enumerate} \item the horizontal arrows are flat ring maps \item $M$ is flat over $R$, \item $\mathfrak m_R R' = \mathfrak m_{R'}$, \item $R'$ and $S'$ are Noetherian. \end{enumerate} Then $M' = M \otimes_R S'$ is flat over $R'$.


algebra-lemma-local-artinian-basis-when-flat

Let $(R, \mathfrak m)$ be a local ring with nilpotent maximal ideal $\mathfrak m$. Let $M$ be a flat $R$-module. If $A$ is a set and $x_\alpha \in M$, $\alpha \in A$ is a collection of elements of $M$, then the following are equivalent: \begin{enumerate} \item $\{\overline{x}_\alpha\}_{\alpha \in A}$ forms a basis for the vector space $M/\mathfrak mM$ over $R/\mathfrak m$, and \item $\{x_\alpha\}_{\alpha \in A}$ forms a basis for $M$ over $R$. \end{enumerate}


algebra-lemma-local-artinian-characterize-flat

Let $R$ be a local ring with nilpotent maximal ideal. Let $M$ be an $R$-module. The following are equivalent \begin{enumerate} \item $M$ is flat over $R$, \item $M$ is a free $R$-module, and \item $M$ is a projective $R$-module. \end{enumerate}


algebra-lemma-lift-basis

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be a collection of elements of $M$. Assume \begin{enumerate} \item $I$ is nilpotent, \item $\{\overline{x}_\alpha\}_{\alpha \in A}$ forms a basis for $M/IM$ over $R/I$, and \item $\text{Tor}_1^R(R/I, M) = 0$. \end{enumerate} Then $M$ is free on $\{x_\alpha\}_{\alpha \in A}$ over $R$.


algebra-lemma-prepare-lift-flatness

Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume \begin{enumerate} \item $M/IM$ is flat over $R/I$, and \item $R' \otimes_R M$ is flat over $R'$. \end{enumerate} Set $I_2 = \varphi^{-1}(\varphi(I^2)R')$. Then $M/I_2M$ is flat over $R/I_2$.


algebra-lemma-lift-flatness

Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume \begin{enumerate} \item $I$ is nilpotent, \item $R \to R'$ is injective, \item $M/IM$ is flat over $R/I$, and \item $R' \otimes_R M$ is flat over $R'$. \end{enumerate} Then $M$ is flat over $R$.


algebra-lemma-descent-flatness-injective-map-artinian-rings

Let $R \to S$ be a ring map. Let $M$ be an $R$-module. Assume \begin{enumerate} \item $R$ is Artinian \item $R \to S$ is injective, and \item $M \otimes_R S$ is a flat $S$-module. \end{enumerate} Then $M$ is a flat $R$-module.


algebra-lemma-criterion-flatness-fibre-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 nilpotent ideal and $M$ an $S'$-module. Assume \begin{enumerate} \item The module $M/IM$ is a flat $S/IS$-module. \item The module $M$ is a flat $R$-module. \end{enumerate} Then $M$ is a flat $S$-module and $S_{\mathfrak q}$ is flat over $R$ for every $\mathfrak q \subset S$ such that $M \otimes_S \kappa(\mathfrak q)$ is nonzero.