spaces-more-morphisms-lemma-relative-pseudo-coherent-is-moot

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $E$ in $D_\QCoh(\mathcal{O}_X)$. If $f$ is flat and locally of finite presentation, then the following are equivalent \begin{enumerate} \item $E$ is pseudo-coherent relative to $Y$, and \item $E$ is pseudo-coherent on $X$. \end{enumerate}


spaces-more-morphisms-lemma-pseudo-coherent-finite-presentation

A pseudo-coherent morphism is locally of finite presentation.


spaces-more-morphisms-lemma-perfect-proper-perfect-direct-image

Let $S$ be a scheme. Let $Y$ be a Noetherian algebraic space over $S$. Let $f : X \to Y$ be a perfect proper morphism of algebraic spaces. Let $E \in D(\mathcal{O}_X)$ be perfect. Then $Rf_*E$ is a perfect object of $D(\mathcal{O}_Y)$.


spaces-more-morphisms-lemma-flat-base-change-lci

A flat base change of a local complete intersection morphism is a local complete intersection morphism.


spaces-more-morphisms-lemma-composition-lci

A composition of local complete intersection morphisms is a local complete intersection morphism.


spaces-more-morphisms-lemma-flat-lci

\begin{slogan} Syntomic equals flat plus lci (for algebraic spaces). \end{slogan} Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent \begin{enumerate} \item $f$ is flat and a local complete intersection morphism, and \item $f$ is syntomic. \end{enumerate}


spaces-more-morphisms-lemma-base-change-lci-fibres

Let $S$ be a scheme. Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces over $S$. Assume that both $p$ and $q$ are flat and locally of finite presentation. Then there exists an open subspace $U(f) \subset X$ such that $|U(f)| \subset |X|$ is the set of points where $f$ is Koszul. Moreover, for any morphism of algebraic spaces $Z' \to Z$, if $f' : X' \to Y'$ is the base change of $f$ by $Z' \to Z$, then $U(f')$ is the inverse image of $U(f)$ under the projection $X' \to X$.


spaces-more-morphisms-lemma-unramified-lci

Let $S$ be a scheme. Let $f : X \to Y$ be a local complete intersection morphism of algebraic spaces over $S$. Then $f$ is unramified if and only if $f$ is formally unramified and in this case the conormal sheaf $\mathcal{C}_{X/Y}$ is finite locally free on $X$.


spaces-more-morphisms-lemma-transitivity-conormal-lci

Let $S$ be a scheme. Let $Z \to Y \to X$ be formally unramified morphisms of algebraic spaces over $S$. Assume that $Z \to Y$ is a local complete intersection morphism. The exact sequence $$ 0 \to i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0 $$ of Lemma \ref{lemma-transitivity-conormal} is short exact.


spaces-more-morphisms-lemma-where-unramified

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that $p$ is locally of finite type and closed. Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is unramified.


spaces-more-morphisms-lemma-where-unramified-universally-injective

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is locally of finite type, \item $p$ is closed, and \item $p_2 : X \times_Y X \to Z$ is closed. \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is unramified and universally injective.


spaces-more-morphisms-lemma-where-closed-immersion

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is locally of finite type, \item $p$ is universally closed, and \item $q : Y \to Z$ is separated. \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is a closed immersion.


spaces-more-morphisms-lemma-where-flat

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is locally of finite presentation, \item $p$ is flat, \item $p$ is closed, and \item $q$ is locally of finite type. \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is flat.


spaces-more-morphisms-lemma-where-surjective-flat

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is locally of finite presentation, \item $p$ is flat, \item $p$ is closed, \item $q$ is locally of finite type, and \item $q$ is closed. \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is surjective and flat.


spaces-more-morphisms-lemma-where-isomorphism

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is locally of finite presentation, \item $p$ is flat, \item $p$ is universally closed, \item $q$ is locally of finite type, \item $q$ is closed, and \item $q$ is separated. \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is an isomorphism.


spaces-more-morphisms-lemma-where-lci

Consider a commutative diagram $$ \xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[ld]^q \\ & Z } $$ of algebraic spaces. Assume that \begin{enumerate} \item $p$ is flat and locally of finite presentation, \item $p$ is closed, and \item $q$ is flat and locally of finite presentation, \end{enumerate} Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is a local complete intersection morphism.


spaces-more-morphisms-lemma-case-of-tor-independence

Let $S$ be a scheme. Consider a commutative diagram of algebraic spaces $$ \xymatrix{ Z' \ar[d] \ar[r] & Y' \ar[d] \\ X' \ar[r] & B' } $$ over $S$. Let $B \to B'$ be a morphism. Denote by $X$ and $Y$ the base changes of $X'$ and $Y'$ to $B$. Assume $Y' \to B'$ and $Z' \to X'$ are flat. Then $X \times_B Y$ and $Z'$ are Tor independent over $X' \times_{B'} Y'$.


spaces-more-morphisms-lemma-derived-chow

Let $A$ be a ring. Let $X$ be a separated algebraic space of finite presentation over $A$. Let $x \in |X|$. Then there exist an $n \geq 0$, a closed subspace $Z \subset X \times_A \mathbf{P}^n_A$, a point $z \in |Z|$, an open $V \subset \mathbf{P}^n_A$, and an object $E$ in $D(\mathcal{O}_{X \times_A \mathbf{P}^n_A})$ such that \begin{enumerate} \item $Z \to X \times_A \mathbf{P}^n_A$ is of finite presentation, \item $c : Z \to \mathbf{P}^n_A$ is a closed immersion over $V$, set $W = c^{-1}(V)$, \item the restriction of $b : Z \to X$ to $W$ is \'etale, $z \in W$, and $b(z) = x$, \item $E|_{X \times_A V} \cong (b, c)_*\mathcal{O}_Z|_{X \times_A V}$, \item $E$ is pseudo-coherent and supported on $Z$. \end{enumerate}


spaces-more-morphisms-lemma-compute-Fourier-Mukai-for-derived-chow

Let $X/A$, $x \in |X|$, and $n, Z, z, V, E$ be as in Lemma \ref{lemma-derived-chow}. For any $K \in D_\QCoh(\mathcal{O}_X)$ we have $$ Rq_*(Lp^*K \otimes^\mathbf{L} E)|_V = R(W \to V)_*K|_W $$ where $p : X \times_A \mathbf{P}^n_A \to X$ and $q : X \times_A \mathbf{P}^n_A \to \mathbf{P}^n_A$ are the projections and where the morphism $W \to V$ is the finitely presented closed immersion $c|_W : W \to V$.


spaces-more-morphisms-lemma-characterize-pseudo-coherent

Let $A$ be a ring. Let $X$ be an algebraic space separated and of finite presentation over $A$. Let $K \in D_\QCoh(\mathcal{O}_X)$. If $R\Gamma(X, E \otimes^\mathbf{L} K)$ is pseudo-coherent in $D(A)$ for every pseudo-coherent $E$ in $D(\mathcal{O}_X)$, then $K$ is pseudo-coherent relative to $A$ (Definition \ref{definition-relative-pseudo-coherence}).


spaces-more-morphisms-lemma-characterize-pseudo-coh-improved

Let $A$ be a ring. Let $X$ be an algebraic space separated and of finite presentation over $A$. Let $K \in D_\QCoh(\mathcal{O}_X)$. If  $R \Gamma (X, E \otimes ^{\mathbf{L}} K)$ is pseudo-coherent in $D(A)$ for every perfect  $E \in D(\mathcal{O}_X)$, then $K$ is pseudo-coherent relative to $A$.


spaces-more-morphisms-lemma-affine-locally-rel-perfect

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. Let $E \in D_\QCoh(\mathcal{O}_X)$. The following are equivalent: \begin{enumerate} \item $E$ is $Y$-perfect, \item for every commutative diagram $$ \xymatrix{ U \ar[d] \ar[r]_g & V \ar[d] \\ X \ar[r]^f & Y } $$ where $U$, $V$ are schemes and the vertical arrows are \'etale, the complex $E|_U$ is $V$-perfect in the sense of Derived Categories of Schemes, Definition \ref{perfect-definition-relatively-perfect}, \item for some commutative diagram as in (2) with $U \to X$ surjective, the complex $E|_U$ is $V$-perfect in the sense of Derived Categories of Schemes, Definition \ref{perfect-definition-relatively-perfect}, \item for every commutative diagram as in (2) with $U$ and $V$ affine the complex $R\Gamma(U, E)$ is $\mathcal{O}_Y(V)$-perfect. \end{enumerate}


spaces-more-morphisms-lemma-perfect-relatively-perfect

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. A perfect object of $D(\mathcal{O}_X)$ is $Y$-perfect. If $K, M \in D(\mathcal{O}_X)$, then $K \otimes_{\mathcal{O}_X}^\mathbf{L} M$ is $Y$-perfect if $K$ is perfect and $M$ is $Y$-perfect.


spaces-more-morphisms-lemma-base-change-relatively-perfect

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. Let $g : Y' \to Y$ be a morphism of algebraic spaces over $S$. Set $X' = Y' \times_Y X$ and denote $g' : X' \to X$ the projection. If $K \in D(\mathcal{O}_X)$ is $Y$-perfect, then $L(g')^*K$ is $Y'$-perfect.


spaces-more-morphisms-lemma-relative-descend-homomorphisms

In Situation \ref{situation-relative-descent}. Let $K_0$ and $L_0$ be objects of $D(\mathcal{O}_{X_0})$. Set $K_i = Lf_{i0}^*K_0$ and $L_i = Lf_{i0}^*L_0$ for $i \geq 0$ and set $K = Lf_0^*K_0$ and $L = Lf_0^*L_0$. Then the map $$ \colim_{i \geq 0} \Hom_{D(\mathcal{O}_{X_i})}(K_i, L_i) \longrightarrow \Hom_{D(\mathcal{O}_X)}(K, L) $$ is an isomorphism if $K_0$ is pseudo-coherent and $L_0 \in D_\QCoh(\mathcal{O}_{X_0})$ has (locally) finite tor dimension as an object of $D((X_0 \to Y_0)^{-1}\mathcal{O}_{Y_0})$


spaces-more-morphisms-lemma-descend-relatively-perfect

In Situation \ref{situation-relative-descent} the category of $Y$-perfect objects of $D(\mathcal{O}_X)$ is the colimit of the categories of $Y_i$-perfect objects of $D(\mathcal{O}_{X_i})$.


spaces-more-morphisms-lemma-derived-pushforward-rel-perfect

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat, proper, and of finite presentation. Let $E \in D(\mathcal{O}_X)$ be $Y$-perfect. Then $Rf_*E$ is a perfect object of $D(\mathcal{O}_Y)$ and its formation commutes with arbitrary base change.


spaces-more-morphisms-lemma-compute-ext-rel-perfect

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $E, K \in D(\mathcal{O}_X)$. Assume \begin{enumerate} \item $Y$ is quasi-compact and quasi-separated, \item $f$ is proper, flat, and of finite presentation, \item $E$ is $Y$-perfect, \item $K$ is pseudo-coherent. \end{enumerate} Then there exists a pseudo-coherent $L \in D(\mathcal{O}_Y)$ such that $$ Rf_*R\SheafHom(K, E) = R\SheafHom(L, \mathcal{O}_Y) $$ and the same is true after arbitrary base change: given $$ \vcenter{ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ Y' \ar[r]^g & Y } } \quad\quad \begin{matrix} \text{cartesian, then we have } \\ Rf'_*R\SheafHom(L(g')^*K, L(g')^*E) \\ = R\SheafHom(Lg^*L, \mathcal{O}_{Y'}) \end{matrix} $$


spaces-more-morphisms-lemma-bounded-on-fibres

Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ such that the structure morphism $f : X \to S$ is flat and locally of finite presentation. Let $E$ be a pseudo-coherent object of $D(\mathcal{O}_X)$. The following are equivalent \begin{enumerate} \item $E$ is $S$-perfect, and \item $E$ is locally bounded below and for every point $s \in S$ the object $L(X_s \to X)^*E$ of $D(\mathcal{O}_{X_s})$ is locally bounded below. \end{enumerate}


spaces-more-morphisms-lemma-characterize-relatively-perfect

Let $A$ be a ring. Let $X$ be an algebraic space separated, of finite presentation, and flat over $A$. Let $K \in D_\QCoh(\mathcal{O}_X)$. If  $R \Gamma (X, E \otimes^\mathbf{L} K)$ is perfect in $D(A)$ for every perfect $E \in D(\mathcal{O}_X)$, then $K$ is $\Spec(A)$-perfect.


spaces-more-morphisms-lemma-diagonal-picard-flat-proper

Let $S$ be a scheme. Let $f : X \to Y$ be a flat, proper morphism of finite presentation of algebraic spaces over $S$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_X$-module. For a morphism $g : Y' \to Y$ consider the base change diagram $$ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^f \\ Y' \ar[r]^g & Y } $$ Assume $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_{X'}$ is an isomorphism for all $g : Y' \to Y$. Then there exists an immersion $j : Z \to Y$ of finite presentation such that a morphism $g : Y' \to Y$ factors through $Z$ if and only if there exists a finite locally free $\mathcal{O}_{Y'}$-module $\mathcal{N}$ with $(f')^*\mathcal{N} \cong (g')^*\mathcal{L}$.


spaces-more-morphisms-lemma-pseudo-coherent-descends-fpqc

Let $S$ be a scheme. Let $\{f_i : X_i \to X\}$ be an fpqc covering of algebraic spaces over $S$. Let $E \in D_\QCoh(\mathcal{O}_X)$. Let $m \in \mathbf{Z}$. Then $E$ is $m$-pseudo-coherent if and only if each $Lf_i^*E$ is $m$-pseudo-coherent.


spaces-more-morphisms-lemma-tor-amplitude-descends-fppf

Let $S$ be a scheme. Let $\{g_i : Y_i \to Y\}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces and set $X_i = Y_i \times_Y X$ with projections $f_i : X_i \to Y_i$ and $g'_i : X_i \to X$. Let $E \in D_\QCoh(\mathcal{O}_X)$. Let $a, b \in \mathbf{Z}$. Then the following are equivalent \begin{enumerate} \item $E$ has tor amplitude in $[a, b]$ as an object of $D(f^{-1}\mathcal{O}_Y)$, and \item $L(g'_i)^*E$ has tor amplitude in $[a, b]$ as a object of $D(f_i^{-1}\mathcal{O}_{Y_i})$ for all $i$. \end{enumerate} Also true if ``tor amplitude in $[a, b]$'' is replaced by ``locally finite tor dimension''.


spaces-more-morphisms-lemma-thickening-pseudo-coherent

Let $S$ be a scheme. Let $i : X \to X'$ be a finite order thickening of algebraic spaces. Let $K' \in D(\mathcal{O}_{X'})$ be an object such that $K = Li^*K'$ is pseudo-coherent. Then $K'$ is pseudo-coherent.


spaces-more-morphisms-lemma-thickening-relatively-perfect

Let $S$ be a scheme. Consider a cartesian diagram $$ \xymatrix{ X \ar[r]_i \ar[d]_f & X' \ar[d]^{f'} \\ Y \ar[r]^j & Y' } $$ of algebraic spaces over $S$. Assume $X' \to Y'$ is flat and locally of finite presentation and $Y \to Y'$ is a finite order thickening. Let $E' \in D(\mathcal{O}_{X'})$. If $E = Li^*(E')$ is $Y$-perfect, then $E'$ is $Y'$-perfect.


spaces-more-morphisms-lemma-henselian-relatively-perfect

Let $(R, I)$ be a pair consisting of a ring and an ideal $I$ contained in the Jacobson radical. Set $S = \Spec(R)$ and $S_0 = \Spec(R/I)$. Let $X$ be an algebraic space over $R$ whose structure morphism $f : X \to S$ is proper, flat, and of finite presentation. Denote $X_0 = S_0 \times_S X$. Let $E \in D(\mathcal{O}_X)$ be pseudo-coherent. If the derived restriction $E_0$ of $E$ to $X_0$ is $S_0$-perfect, then $E$ is $S$-perfect.


spaces-more-morphisms-lemma-nodal-local

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent: \begin{enumerate} \item $f$ is at-worst-nodal of relative dimension $1$, \item for every scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times_Y X \to Z$ is at-worst-nodal of relative dimension $1$, \item for every affine scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times_Y X \to Z$ is at-worst-nodal of relative dimension $1$, \item there exists a scheme $V$ and a surjective \'etale morphism $V \to Y$ such that $V \times_Y X \to V$ is at-worst-nodal of relative dimension $1$, \item there exists a scheme $U$ and a surjective \'etale morphism $\varphi : U \to X$ such that the composition $f \circ \varphi$ is at-worst-nodal of relative dimension $1$, \item for every commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } $$ where $U$, $V$ are schemes and the vertical arrows are \'etale the top horizontal arrow is at-worst-nodal of relative dimension $1$, \item there exists a commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } $$ where $U$, $V$ are schemes, the vertical arrows are \'etale, and $U \to X$ is surjective such that the top horizontal arrow is at-worst-nodal of relative dimension $1$, and \item there exist Zariski coverings $Y = \bigcup_{i \in I} Y_i$, and $f^{-1}(Y_i) = \bigcup X_{ij}$ such that each morphism $X_{ij} \to Y_i$ is at-worst-nodal of relative dimension $1$. \end{enumerate}


spaces-more-morphisms-lemma-locus-where-nodal

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. Then there is a maximal open subspace $X' \subset X$ such that $f|_{X'} : X' \to Y$ is at-worst-nodal of relative dimension $1$. Moreover, formation of $X'$ commutes with arbitrary base change.


algebra-theorem-chevalley

Suppose that $R \to S$ is of finite presentation. The image of a constructible subset of $\Spec(S)$ in $\Spec(R)$ is constructible.


algebra-theorem-nullstellensatz

Let $k$ be a field. \begin{enumerate} \item \label{item-finite-kappa} For any maximal ideal $\mathfrak m \subset k[x_1, \ldots, x_n]$ the field extension $k \subset \kappa(\mathfrak m)$ is finite. \item \label{item-polynomial-ring-Jacobson} Any radical ideal $I \subset k[x_1, \ldots, x_n]$ is the intersection of maximal ideals containing it. \end{enumerate} The same is true in any finite type $k$-algebra.


algebra-theorem-uncountable-nullstellensatz

Let $k$ be a field. Let $S$ be a $k$-algebra generated over $k$ by the elements $\{x_i\}_{i \in I}$. Assume the cardinality of $I$ is smaller than the cardinality of $k$. Then \begin{enumerate} \item for all maximal ideals $\mathfrak m \subset S$ the field extension $k \subset \kappa(\mathfrak m)$ is algebraic, and \item $S$ is a Jacobson ring. \end{enumerate}


algebra-theorem-lazard

Let $M$ be an $R$-module.  Then $M$ is flat if and only if it is the colimit of a directed system of free finite $R$-modules.


algebra-theorem-universally-exact-criteria

Let $$ 0 \to M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \to 0 $$ be an exact sequence of $R$-modules.  The following are equivalent: \begin{enumerate} \item The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is universally exact. \item For every finitely presented $R$-module $Q$, the sequence $$ 0 \to M_1 \otimes_R Q \to M_2 \otimes_R Q \to M_3 \otimes_R Q \to 0 $$ is exact. \item Given elements $x_i \in M_1$ $(i = 1, \ldots, n)$, $y_j \in M_2$ $(j = 1, \ldots, m)$, and $a_{ij} \in R$ $(i = 1, \ldots, n, j = 1, \ldots, m)$ such that for all $i$ $$ f_1(x_i) = \sum\nolimits_j a_{ij} y_j, $$ there exists $z_j \in M_1$ $(j =1, \ldots, m)$ such that for all $i$, $$ x_i = \sum\nolimits_j a_{ij} z_j . $$ \item Given a commutative diagram of $R$-module maps $$ \xymatrix{ R^n \ar[r] \ar[d] &  R^m \ar[d] \\ M_1 \ar[r]^{f_1}        &  M_2 } $$ where $m$ and $n$ are integers, there exists a map $R^m \to M_1$ making the top triangle commute. \item For every finitely presented $R$-module $P$, the $R$-module map $\Hom_R(P, M_2) \to \Hom_R(P, M_3)$ is surjective. \item The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is the colimit of a directed system of split exact sequences of the form $$ 0 \to M_{1} \to M_{2, i} \to M_{3, i} \to 0 $$ where the $M_{3, i}$ are finitely presented. \end{enumerate}


algebra-theorem-kaplansky-direct-sum

Suppose $M$ is a direct sum of countably generated $R$-modules.  If $P$ is a direct summand of $M$, then $P$ is also a direct sum of countably generated $R$-modules.


algebra-theorem-projective-direct-sum

\begin{slogan} Any projective module is a direct sum of countably generated projective modules. \end{slogan} If $P$ is a projective $R$-module, then $P$ is a direct sum of countably generated projective $R$-modules.


algebra-theorem-projective-free-over-local-ring

\begin{slogan} Projective modules over local rings are free. \end{slogan} If $P$ is a projective module over a local ring $R$, then $P$ is free.


algebra-theorem-projectivity-characterization

Let $M$ be an $R$-module.  Then $M$ is projective if and only it satisfies: \begin{enumerate} \item $M$ is flat, \item $M$ is Mittag-Leffler, \item $M$ is a direct sum of countably generated $R$-modules. \end{enumerate}


algebra-theorem-ffdescent-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 projective, then $M$ is projective.


algebra-theorem-main-theorem

Let $R$ be a ring. Let $R \to S$ be a finite type $R$-algebra. Let $S' \subset S$ be the integral closure of $R$ in $S$. Let $\mathfrak q \subset S$ be a prime of $S$. If $R \to S$ is quasi-finite at $\mathfrak q$ then there exists a $g \in S'$, $g \not \in \mathfrak q$ such that $S'_g \cong S_g$.


algebra-theorem-openness-flatness

Let $R$ be a ring. Let $R \to S$ be a ring map of finite presentation. Let $M$ be a finitely presented $S$-module. The set $$ \{ \mathfrak q \in \Spec(S) \mid M_{\mathfrak q}\text{ is flat over }R\} $$ is open in $\Spec(S)$.


algebra-theorem-cohen-structure-theorem

Let $(R, \mathfrak m)$ be a complete local ring. \begin{enumerate} \item $R$ has a coefficient ring (see Definition \ref{definition-coefficient-ring}), \item if $\mathfrak m$ is a finitely generated ideal, then $R$ is isomorphic to a quotient $$ \Lambda[[x_1, \ldots, x_n]]/I $$ where $\Lambda$ is either a field or a Cohen ring. \end{enumerate}


algebra-lemma-snake

\begin{reference} \cite[III, Lemma 3.3]{Cartan-Eilenberg} \end{reference} Suppose given a commutative diagram $$ \xymatrix{ & X \ar[r] \ar[d]^\alpha & Y \ar[r] \ar[d]^\beta & Z \ar[r] \ar[d]^\gamma & 0 \\ 0 \ar[r] & U \ar[r] & V \ar[r] & W } $$ of abelian groups with exact rows, then there is a canonical exact sequence $$ \Ker(\alpha) \to \Ker(\beta) \to \Ker(\gamma) \to \Coker(\alpha) \to \Coker(\beta) \to \Coker(\gamma) $$ Moreover, if $X \to Y$ is injective, then the first map is injective, and if $V \to W$ is surjective, then the last map is surjective.


algebra-lemma-lift-map

Let $R$ be a ring. Let $\alpha : R^{\oplus n} \to M$ and $\beta : N \to M$ be module maps. If $\Im(\alpha) \subset \Im(\beta)$, then there exists an $R$-module map $\gamma : R^{\oplus n} \to N$ such that $\alpha = \beta \circ \gamma$.


algebra-lemma-extension

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. \begin{enumerate} \item If $M_1$ and $M_3$ are finite $R$-modules, then $M_2$ is a finite $R$-module. \item If $M_1$ and $M_3$ are finitely presented $R$-modules, then $M_2$ is a finitely presented $R$-module. \item If $M_2$ is a finite $R$-module, then $M_3$ is a finite $R$-module. \item If $M_2$ is a finitely presented $R$-module and $M_1$ is a finite $R$-module, then $M_3$ is a finitely presented $R$-module. \item If $M_3$ is a finitely presented $R$-module and $M_2$ is a finite $R$-module, then $M_1$ is a finite $R$-module. \end{enumerate}


algebra-lemma-trivial-filter-finite-module

\begin{slogan} Finite modules have filtrations such that successive quotients are cyclic modules. \end{slogan} Let $R$ be a 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 $R/I_i$ for some ideal $I_i$ of $R$.


algebra-lemma-finite-over-subring

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $M$ is finite as an $R$-module, then $M$ is finite as an $S$-module.


algebra-lemma-compose-finite-type

The notions finite type and finite presentation have the following permanence properties. \begin{enumerate} \item A composition of ring maps of finite type is of finite type. \item A composition of ring maps of finite presentation is of finite presentation. \item Given $R \to S' \to S$ with $R \to S$ of finite type, then $S' \to S$ is of finite type. \item Given $R \to S' \to S$, with $R \to S$ of finite presentation, and $R \to S'$ of finite type, then $S' \to S$ is of finite presentation. \end{enumerate}


algebra-lemma-finite-presentation-independent

Let $R \to S$ be a ring map of finite presentation. For any surjection $\alpha : R[x_1, \ldots, x_n] \to S$ the kernel of $\alpha$ is a finitely generated ideal in $R[x_1, \ldots, x_n]$.


algebra-lemma-finitely-presented-over-subring

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume $R \to S$ is of finite type and $M$ is finitely presented as an $R$-module. Then $M$ is finitely presented as an $S$-module.


algebra-lemma-finite-module-over-finite-extension

Let $R \to S$ be a finite ring map. Let $M$ be an $S$-module. Then $M$ is finite as an $R$-module if and only if $M$ is finite as an $S$-module.


algebra-lemma-finite-transitive

Suppose that $R \to S$ and $S \to T$ are finite ring maps. Then $R \to T$ is finite.


algebra-lemma-finite-finite-type

Let $\varphi : R \to S$ be a ring map. \begin{enumerate} \item If $\varphi$ is finite, then $\varphi$ is of finite type. \item If $S$ is of finite presentation as an $R$-module, then $\varphi$ is of finite presentation. \end{enumerate}


algebra-lemma-colimit

Let $(M_i, \mu_{ij})$ be a system of $R$-modules over the preordered set $I$. The colimit of the system $(M_i, \mu_{ij})$ is the quotient $R$-module $(\bigoplus_{i\in I} M_i) /Q$ where $Q$ is the $R$-submodule generated by all elements $$ \iota_i(x_i) - \iota_j(\mu_{ij}(x_i)) $$ where $\iota_i : M_i \to \bigoplus_{i\in I} M_i$ is the natural inclusion. We denote the colimit $M = \colim_i M_i$. We denote $\pi : \bigoplus_{i\in I} M_i \to M$ the projection map and $\phi_i = \pi \circ \iota_i : M_i \to M$.


algebra-lemma-directed-colimit

Let $(M_i, \mu_{ij})$ be a system of $R$-modules over the preordered set $I$. Assume that $I$ is directed. The colimit of the system $(M_i, \mu_{ij})$ is canonically isomorphic to the module $M$ defined as follows: \begin{enumerate} \item as a set let $$ M = \left(\coprod\nolimits_{i \in I} M_i\right)/\sim $$ where for $m \in M_i$ and $m' \in M_{i'}$ we have $$ m \sim m' \Leftrightarrow \mu_{ij}(m) = \mu_{i'j}(m')\text{ for some }j \geq i, i' $$ \item as an abelian group for $m \in M_i$ and $m' \in M_{i'}$ we define the sum of the classes of $m$ and $m'$ in $M$ to be the class of $\mu_{ij}(m) + \mu_{i'j}(m')$ where $j \in I$ is any index with $i \leq j$ and $i' \leq j$, and \item as an $R$-module define for $m \in M_i$ and $x \in R$ the product of $x$ and the class of $m$ in $M$ to be the class of $xm$ in $M$. \end{enumerate} The canonical maps $\phi_i : M_i \to M$ are induced by the canonical maps $M_i \to \coprod_{i \in I} M_i$.


algebra-lemma-zero-directed-limit

Let $(M_i, \mu_{ij})$ be a directed system. Let $M = \colim M_i$ with $\mu_i : M_i \to M$. Then, $\mu_i(x_i) = 0$ for $x_i \in M_i$ if and only if there exists $j \geq i$ such that $\mu_{ij}(x_i) = 0$.


algebra-lemma-homomorphism-limit

Let $(M_i, \mu_{ij})$, $(N_i, \nu_{ij})$ be systems of $R$-modules over the same preordered set. A morphism of systems $\Phi = (\phi_i)$ from $(M_i, \mu_{ij})$ to $(N_i, \nu_{ij})$ induces a unique homomorphism $$ \colim \phi_i : \colim M_i \longrightarrow \colim N_i $$ such that $$ \xymatrix{ M_i \ar[r] \ar[d]_{\phi_i} & \colim M_i \ar[d]^{\colim \phi_i} \\ N_i \ar[r] & \colim N_i } $$ commutes for all $i \in I$.


algebra-lemma-directed-colimit-exact

\begin{slogan} Filtered colimits are exact. Directed colimits are exact. \end{slogan} Let $I$ be a directed set. Let $(L_i, \lambda_{ij})$, $(M_i, \mu_{ij})$, and $(N_i, \nu_{ij})$ be systems of $R$-modules over $I$. Let $\varphi_i : L_i \to M_i$ and $\psi_i : M_i \to N_i$ be morphisms of systems over $I$. Assume that for all $i \in I$ the sequence of $R$-modules $$ \xymatrix{ L_i \ar[r]^{\varphi_i} & M_i \ar[r]^{\psi_i} & N_i } $$ is a complex with homology $H_i$. Then the $R$-modules $H_i$ form a system over $I$, the sequence of $R$-modules $$ \xymatrix{ \colim_i L_i \ar[r]^\varphi & \colim_i M_i \ar[r]^\psi & \colim_i N_i } $$ is a complex as well, and denoting $H$ its homology we have $$ H = \colim_i H_i. $$


algebra-lemma-almost-directed-colimit-exact

Let $\mathcal{I}$ be an index category satisfying the assumptions of Categories, Lemma \ref{categories-lemma-split-into-directed}. Then taking colimits of diagrams of abelian groups over $\mathcal{I}$ is exact (i.e., the analogue of Lemma \ref{lemma-directed-colimit-exact} holds in this situation).


algebra-lemma-localization-zero

The localization $S^{-1}A$ is the zero ring if and only if $0\in S$.


algebra-lemma-localization-and-modules

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. The category of $S^{-1}R$-modules is equivalent to the category of $R$-modules $N$ with the property that every $s \in S$ acts as an automorphism on $N$.


algebra-lemma-universal-property-localization-module

Let $R$ be a ring. Let $S \subset R$ a multiplicative subset. Let $M$, $N$ be $R$-modules. Assume all the elements of $S$ act as automorphisms on $N$. Then the canonical map $$ \Hom_R(S^{-1}M, N) \longrightarrow \Hom_R(M, N) $$ induced by the localization map, is an isomorphism.


algebra-lemma-localization-colimit

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module. Then $$ S^{-1}M = \colim_{f \in S} M_f $$ where the preorder on $S$ is given by $f \geq f' \Leftrightarrow f = f'f''$ for some $f'' \in R$ in which case the map $M_{f'} \to M_f$ is given by $m/(f')^e \mapsto m(f'')^e/f^e$.


algebra-lemma-localize-quotient-modules

Localization respects quotients, i.e. if $N$ is a submodule of $M$, then $S^{-1}(M/N)\simeq (S^{-1}M)/(S^{-1}N)$.


algebra-lemma-submodule-localization

Any submodule $N'$ of $S^{-1}M$ is of the form $S^{-1}N$ for some $N\subset M$. Indeed one can take $N$ to be the inverse image of $N'$ in $M$.


algebra-lemma-hom-exact

Exactness and $\Hom_R$. Let $R$ be a ring. \begin{enumerate} \item Let $M_1 \to M_2 \to M_3 \to 0$ be a complex of $R$-modules. Then $M_1 \to M_2 \to M_3 \to 0$ is exact if and only if $0 \to \Hom_R(M_3, N) \to \Hom_R(M_2, N) \to \Hom_R(M_1, N)$ is exact for all $R$-modules $N$. \item  Let $0 \to M_1 \to M_2 \to M_3$ be a complex of $R$-modules. Then $0 \to M_1 \to M_2 \to M_3$ is exact if and only if $0 \to \Hom_R(N, M_1) \to \Hom_R(N, M_2) \to \Hom_R(N, M_3)$ is exact for all $R$-modules $N$. \end{enumerate}


algebra-lemma-hom-from-finitely-presented

Let $R$ be a ring. Let $M$ be a finitely presented $R$-module. Let $N$ be an $R$-module. \begin{enumerate} \item For $f \in R$ we have $\Hom_R(M, N)_f = \Hom_{R_f}(M_f, N_f) = \Hom_R(M_f, N_f)$, \item for a multiplicative subset $S$ of $R$ we have $$ S^{-1}\Hom_R(M, N) = \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N) = \Hom_R(S^{-1}M, S^{-1}N). $$ \end{enumerate}


algebra-lemma-characterize-finite-module-hom

Let $R$ be a ring. Let $N$ be an $R$-module. The following are equivalent \begin{enumerate} \item $N$ is a finite $R$-module, \item for any filtered colimit $M = \colim M_i$ of $R$-modules the map $\colim \Hom_R(N, M_i) \to \Hom_R(K, M)$ is injective. \end{enumerate}


algebra-lemma-module-colimit-fp

Let $R$ be a ring and let $M$ be an $R$-module. Then $M$ is the colimit of a directed system $(M_i, \mu_{ij})$ of $R$-modules with all $M_i$ finitely presented $R$-modules.


algebra-lemma-tensor-product

Let $M, N$ be $R$-modules. Then there exists a pair $(T, g)$ where $T$ is an $R$-module, and $g : M \times N \to T$ an $R$-bilinear mapping, with the following universal property: For any $R$-module $P$ and any $R$-bilinear mapping $f : M \times N \to P$, there exists a unique $R$-linear mapping $\tilde{f} : T \to P$ such that $f = \tilde{f} \circ g$. In other words, the following diagram commutes: $$ \xymatrix{ M \times N \ar[rr]^f \ar[dr]_g & & P\\ & T \ar[ur]_{\tilde f} } $$ Moreover, if $(T, g)$ and $(T', g')$ are two pairs with this property, then there exists a unique isomorphism $j : T \to T'$ such that $j\circ g = g'$.


algebra-lemma-flip-tensor-product

Let $M, N, P$ be $R$-modules, then the bilinear maps \begin{align*} (x, y) & \mapsto y \otimes x\\ (x + y, z) & \mapsto x \otimes z + y \otimes z\\ (r, x) & \mapsto rx \end{align*} induce unique isomorphisms \begin{align*} M \otimes_R N & \to N \otimes_R M, \\ (M\oplus N)\otimes_R P & \to (M \otimes_R P)\oplus(N \otimes_R P),  \\ R \otimes_R M & \to M \end{align*}


algebra-lemma-multilinear

Let $M_1, \ldots, M_r$ be $R$-modules. Then there exists a pair $(T, g)$ consisting of an $R$-module T and an $R$-multilinear mapping $g : M_1\times \ldots \times M_r \to T$ with the universal property: For any $R$-multilinear mapping $f : M_1\times \ldots \times M_r \to P$ there exists a unique $R$-module homomorphism $f' : T \to P$ such that $f'\circ g = f$. Such a module $T$ is unique up to unique isomorphism. We denote it $M_1\otimes_R \ldots \otimes_R M_r$ and we denote the universal multilinear map $(m_1, \ldots, m_r) \mapsto m_1 \otimes \ldots \otimes m_r$.


algebra-lemma-transitive

The homomorphisms $$ (M \otimes_R N)\otimes_R P \to M \otimes_R N \otimes_R P \to M \otimes_R (N \otimes_R P) $$ such that $f((x \otimes y)\otimes z) = x \otimes y \otimes z$ and $g(x \otimes y \otimes z) = x \otimes (y \otimes z)$, $x\in M, y\in N, z\in P$ are well-defined and are isomorphisms.


algebra-lemma-tensor-with-bimodule

For $A$-module $M$, $B$-module $P$ and $(A, B)$-bimodule $N$, the modules $(M \otimes_A N)\otimes_B P$ and $M \otimes_A(N \otimes_B P)$ can both be given $(A, B)$-bimodule structure, and moreover $$ (M \otimes_A N)\otimes_B P \cong M \otimes_A(N \otimes_B P). $$


algebra-lemma-hom-from-tensor-product

For any three $R$-modules $M, N, P$, $$ \Hom_R(M \otimes_R N, P) \cong \Hom_R(M, \Hom_R(N, P)) $$


algebra-lemma-tensor-products-commute-with-limits

Let $(M_i, \mu_{ij})$ be a system over the preordered set $I$. Let $N$ be an $R$-module. Then $$ \colim (M_i \otimes N) \cong (\colim M_i)\otimes N. $$ Moreover, the isomorphism is induced by the homomorphisms $\mu_i \otimes 1: M_i \otimes N \to M \otimes N$ where $M = \colim_i M_i$ with natural maps $\mu_i : M_i \to M$.


algebra-lemma-tensor-product-exact

Let \begin{align*} M_1\xrightarrow{f} M_2\xrightarrow{g} M_3 \to 0 \end{align*} be an exact sequence of $R$-modules and homomorphisms, and let $N$ be any $R$-module. Then the sequence \begin{equation} \label{equation-2ndex} M_1\otimes N\xrightarrow{f \otimes 1} M_2\otimes N \xrightarrow{g \otimes 1} M_3\otimes N \to 0 \end{equation} is exact. In other words, the functor $- \otimes_R N$ is {\it right exact}, in the sense that tensoring each term in the original right exact sequence preserves the exactness.


algebra-lemma-tensor-finiteness

Let $R$ be a ring. Let $M$ and $N$ be $R$-modules. \begin{enumerate} \item If $N$ and $M$ are finite, then so is $M \otimes_R N$. \item If $N$ and $M$ are finitely presented, then so is $M \otimes_R N$. \end{enumerate}


algebra-lemma-tensor-localization

Let $M$ be an $R$-module. Then the $S^{-1}R$-modules $S^{-1}M$ and $S^{-1}R \otimes_R M$ are canonically isomorphic, and the canonical isomorphism $f : S^{-1}R \otimes_R M \to S^{-1}M$ is given by $$ f((a/s) \otimes m) = am/s, \forall a \in R, m \in M, s \in S $$


algebra-lemma-tensor-product-localization

Let $M, N$ be $R$-modules, then there is a canonical $S^{-1}R$-module isomorphism $f : S^{-1}M \otimes_{S^{-1}R}S^{-1}N \to S^{-1}(M \otimes_R N)$, given by $$ f((m/s)\otimes(n/t)) = (m \otimes n)/st $$


algebra-lemma-free-tensor-algebra

Let $R$ be a ring. Let $M$ be an $R$-module. If $M$ is a free $R$-module, so is each symmetric and exterior power.


algebra-lemma-presentation-sym-exterior

Let $R$ be a ring. Let $M_2 \to M_1 \to M \to 0$ be an exact sequence of $R$-modules. There are exact sequences $$ M_2 \otimes_R \text{Sym}^{n - 1}(M_1) \to \text{Sym}^n(M_1) \to \text{Sym}^n(M) \to 0 $$ and similarly $$ M_2 \otimes_R \wedge^{n - 1}(M_1) \to \wedge^n(M_1) \to \wedge^n(M) \to 0 $$


algebra-lemma-present-sym-wedge

Let $R$ be a ring. Let $M$ be an $R$-module. Let $x_i$, $i \in I$ be a given system of generators of $M$ as an $R$-module. Let $n \geq 2$. There exists a canonical exact sequence $$ \bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i_1, i_2 \in I} \text{T}^{n - 2}(M) \oplus \bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i \in I} \text{T}^{n - 2}(M) \to \text{T}^n(M) \to \wedge^n(M) \to 0 $$ where the pure tensor $m_1 \otimes \ldots \otimes m_{n - 2}$ in the first summand maps to \begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ + \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*} and $m_1 \otimes \ldots \otimes m_{n - 2}$ in the second summand maps to $$ \underbrace{ m_1 \otimes \ldots \otimes x_i \otimes \ldots \otimes x_i \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i} \text{ and } x_{i} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} $$ There is also a canonical exact sequence $$ \bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i_1, i_2 \in I} \text{T}^{n - 2}(M) \to \text{T}^n(M) \to \text{Sym}^n(M) \to 0 $$ where the pure tensor $m_1 \otimes \ldots \otimes m_{n - 2}$ maps to \begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ - \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*}


algebra-lemma-colimit-tensor-algebra

\begin{slogan} Taking tensor algebras commutes with filtered colimits. \end{slogan} Let $R$ be a ring. Let $M_i$ be a directed system of $R$-modules. Then $\colim_i \text{T}(M_i) = \text{T}(\colim_i M_i)$ and similarly for the symmetric and exterior algebras.


algebra-lemma-base-change-finiteness

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $R \to R'$ be a ring map and let $S' = S \otimes_R R'$ and $M' = M \otimes_R R'$ be the base changes. \begin{enumerate} \item If $M$ is a finite $S$-module, then the base change $M'$ is a finite $S'$-module. \item If $M$ is an $S$-module finite presentation, then the base change $M'$ is an $S'$-module of finite presentation. \item If $R \to S$ is of finite type, then the base change $R' \to S'$ is of finite type. \item If $R \to S$ is of finite presentation, then the base change $R' \to S'$ is of finite presentation. \end{enumerate}


algebra-lemma-adjoint-tensor-restrict

Let $R \to S$ be a ring map. The functors $\text{Mod}_S \to \text{Mod}_R$, $N \mapsto N_R$ (restriction) and $\text{Mod}_R \to \text{Mod}_S$, $M \mapsto M \otimes_R S$ (base change) are adjoint functors. In a formula $$ \Hom_R(M, N_R) = \Hom_S(M \otimes_R S, N) $$


algebra-lemma-adjoint-hom-restrict

Let $R \to S$ be a ring map. The functors $\text{Mod}_S \to \text{Mod}_R$, $N \mapsto N_R$ (restriction) and $\text{Mod}_R \to \text{Mod}_S$, $M \mapsto \Hom_R(S, M)$ are adjoint functors. In a formula $$ \Hom_R(N_R, M) = \Hom_S(N, \Hom_R(S, M)) $$


algebra-lemma-hom-from-tensor-product-variant

Let $R \to S$ be a ring map. Given $S$-modules $M, N$ and an $R$-module $P$ we have $$ \Hom_R(M \otimes_S N, P) = \Hom_S(M, \Hom_R(N, P)) $$


algebra-lemma-product-ideals-in-prime

Let $R$ be a ring, $I$ and $J$ two ideals and $\mathfrak p$ a prime ideal containing the product $IJ$. Then $\mathfrak{p}$ contains $I$ or $J$.


algebra-lemma-silly

\begin{slogan} 1. In an affine scheme if a finite number of points are contained in an open subset then they are contained in a smaller principal open subset. 2. Affine opens are cofinal among the neighborhoods of a given finite set of an affine scheme \end{slogan} Let $R$ be a ring. Let $I_i \subset R$, $i = 1, \ldots, r$, and $J \subset R$ be ideals. Assume \begin{enumerate} \item $J \not\subset I_i$ for $i = 1, \ldots, r$, and \item all but two of $I_i$ are prime ideals. \end{enumerate} Then there exists an $x \in J$, $x\not\in I_i$ for all $i$.


algebra-lemma-silly-silly

Let $R$ be a ring. Let $x \in R$, $I \subset R$ an ideal, and $\mathfrak p_i$, $i = 1, \ldots, r$ be prime ideals. Suppose that $x + I \not \subset \mathfrak p_i$ for $i = 1, \ldots, r$. Then there exists an $y \in I$ such that $x + y \not \in \mathfrak p_i$ for all $i$.


algebra-lemma-chinese-remainder

Let $R$ be a ring. \begin{enumerate} \item If $I_1, \ldots, I_r$ are ideals such that $I_a + I_b = R$ when $a \not = b$, then $I_1 \cap \ldots \cap I_r = I_1I_2\ldots I_r$ and $R/(I_1I_2\ldots I_r) \cong R/I_1 \times \ldots \times R/I_r$. \item If $\mathfrak m_1, \ldots, \mathfrak m_r$ are pairwise distinct maximal ideals then $\mathfrak m_a + \mathfrak m_b = R$ for $a \not = b$ and the above applies. \end{enumerate}


algebra-lemma-matrix-left-inverse

Let $R$ be a ring. Let $n \geq m$. Let $A$ be an $n \times m$ matrix with coefficients in $R$. Let $J \subset R$ be the ideal generated by the $m \times m$ minors of $A$. \begin{enumerate} \item For any $f \in J$ there exists a $m \times n$ matrix $B$ such that $BA = f 1_{m \times m}$. \item If $f \in R$ and $BA = f 1_{m \times m}$ for some $m \times n$ matrix $B$, then $f^m \in J$. \end{enumerate}


algebra-lemma-matrix-right-inverse

Let $R$ be a ring. Let $n \geq m$. Let $A = (a_{ij})$ be an $n \times m$ matrix with coefficients in $R$, written in block form as $$ A = \left( \begin{matrix} A_1 \\ A_2 \end{matrix} \right) $$ where $A_1$ has size $m \times m$. Let $B$ be the adjugate (transpose of cofactor) matrix to $A_1$. Then $$ AB =  \left( \begin{matrix} f 1_{m \times m} \\ C \end{matrix} \right) $$ where $f = \det(A_1)$ and $c_{ij}$ is (up to sign) the determinant of the $m \times m$ minor of $A$ corresponding to the rows $1, \ldots, \hat j, \ldots, m, i$.


algebra-lemma-map-cannot-be-injective

\begin{slogan} A map of finite free modules cannot be injective if the source has rank bigger than the target. \end{slogan} Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a nonzero kernel.


algebra-lemma-rank

\begin{slogan} The rank of a finite free module is well defined. \end{slogan} Let $R$ be a nonzero ring. Let $n, m \geq 0$ be integers. If $R^{\oplus n}$ is isomorphic to $R^{\oplus m}$ as $R$-modules, then $n = m$.


algebra-lemma-charpoly

Let $R$ be a ring. Let $A = (a_{ij})$ be an $n \times n$ matrix with coefficients in $R$. Let $P(x) \in R[x]$ be the characteristic polynomial of $A$ (defined as $\det(x\text{id}_{n \times n} - A)$). Then $P(A) = 0$ in $\text{Mat}(n \times n, R)$.


algebra-lemma-charpoly-module

Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be an endomorphism. Then there exists a monic polynomial $P \in R[T]$ such that $P(\varphi) = 0$ as an endomorphism of $M$.


algebra-lemma-charpoly-module-ideal

Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be an endomorphism such that $\varphi(M) \subset IM$. Then there exists a monic polynomial $P = t^n + a_1 t^{n - 1} + \ldots + a_n \in R[T]$ such that $a_j \in I^j$ and $P(\varphi) = 0$ as an endomorphism of $M$.


algebra-lemma-fun

Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be a surjective $R$-module map. Then $\varphi$ is an isomorphism.


algebra-lemma-Zariski-topology

Let $R$ be a ring. \begin{enumerate} \item The spectrum of a ring $R$ is empty if and only if $R$ is the zero ring. \item Every nonzero ring has a maximal ideal. \item Every nonzero ring has a minimal prime ideal. \item Given an ideal $I \subset R$ and a prime ideal $I \subset \mathfrak p$ there exists a prime $I \subset \mathfrak q \subset \mathfrak p$ such that $\mathfrak q$ is minimal over $I$. \item If $T \subset R$, and if $(T)$ is the ideal generated by $T$ in $R$, then $V((T)) = V(T)$. \item If $I$ is an ideal and $\sqrt{I}$ is its radical, see basic notion (\ref{item-radical-ideal}), then $V(I) = V(\sqrt{I})$. \item Given an ideal $I$ of $R$ we have $\sqrt{I} = \bigcap_{I \subset \mathfrak p} \mathfrak p$. \item If $I$ is an ideal then $V(I) = \emptyset$ if and only if $I$ is the unit ideal. \item If $I$, $J$ are ideals of $R$ then $V(I) \cup V(J) = V(I \cap J)$. \item If $(I_a)_{a\in A}$ is a set of ideals of $R$ then $\cap_{a\in A} V(I_a) = V(\cup_{a\in A} I_a)$. \item If $f \in R$, then $D(f) \amalg V(f) = \Spec(R)$. \item If $f \in R$ then $D(f) = \emptyset$ if and only if $f$ is nilpotent. \item If $f = u f'$ for some unit $u \in R$, then $D(f) = D(f')$. \item If $I \subset R$ is an ideal, and $\mathfrak p$ is a prime of $R$ with $\mathfrak p \not\in V(I)$, then there exists an $f \in R$ such that $\mathfrak p \in D(f)$, and $D(f) \cap V(I) = \emptyset$. \item If $f, g \in R$, then $D(fg) = D(f) \cap D(g)$. \item If $f_i \in R$ for $i \in I$, then $\bigcup_{i\in I} D(f_i)$ is the complement of $V(\{f_i \}_{i\in I})$ in $\Spec(R)$. \item If $f \in R$ and $D(f) = \Spec(R)$, then $f$ is a unit. \end{enumerate}


algebra-lemma-spec-functorial

\begin{slogan} Functoriality of the spectrum \end{slogan} Suppose that $\varphi : R \to R'$ is a ring homomorphism. The induced map $$ \Spec(\varphi) : \Spec(R') \longrightarrow \Spec(R), \quad \mathfrak p' \longmapsto \varphi^{-1}(\mathfrak p') $$ is continuous for the Zariski topologies. In fact, for any element $f \in R$ we have $\Spec(\varphi)^{-1}(D(f)) = D(\varphi(f))$.


algebra-lemma-spec-localization

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. The map $R \to S^{-1}R$ induces via the functoriality of $\Spec$ a homeomorphism $$ \Spec(S^{-1}R) \longrightarrow \{\mathfrak p \in \Spec(R) \mid S \cap \mathfrak p = \emptyset \} $$ where the topology on the right hand side is that induced from the Zariski topology on $\Spec(R)$. The inverse map is given by $\mathfrak p \mapsto S^{-1}\mathfrak p$.


algebra-lemma-standard-open

Let $R$ be a ring. Let $f \in R$. The map $R \to R_f$ induces via the functoriality of $\Spec$ a homeomorphism $$ \Spec(R_f) \longrightarrow D(f) \subset \Spec(R). $$ The inverse is given by $\mathfrak p \mapsto \mathfrak p \cdot R_f$.


algebra-lemma-spec-closed

Let $R$ be a ring. Let $I \subset R$ be an ideal. The map $R \to R/I$ induces via the functoriality of $\Spec$ a homeomorphism $$ \Spec(R/I) \longrightarrow V(I) \subset \Spec(R). $$ The inverse is given by $\mathfrak p \mapsto \mathfrak p / I$.


algebra-lemma-in-image

Let $\varphi : R \to S$ be a ring map. Let $\mathfrak p$ be a prime of $R$. The following are equivalent \begin{enumerate} \item $\mathfrak p$ is in the image of $\Spec(S) \to \Spec(R)$, \item $S \otimes_R \kappa(\mathfrak p) \not = 0$, \item $S_{\mathfrak p}/\mathfrak p S_{\mathfrak p} \not = 0$, \item $(S/\mathfrak pS)_{\mathfrak p} \not = 0$, and \item $\mathfrak p = \varphi^{-1}(\mathfrak pS)$. \end{enumerate}


algebra-lemma-quasi-compact

\begin{slogan} The spectrum of a ring is quasi-compact \end{slogan} Let $R$ be a ring. The space $\Spec(R)$ is quasi-compact.


algebra-lemma-topology-spec

Let $R$ be a ring. The topology on $X = \Spec(R)$ has the following properties: \begin{enumerate} \item $X$ is quasi-compact, \item $X$ has a basis for the topology consisting of quasi-compact opens, and \item the intersection of any two quasi-compact opens is quasi-compact. \end{enumerate}


algebra-lemma-characterize-local-ring

Let $R$ be a ring. The following are equivalent: \begin{enumerate} \item $R$ is a local ring, \item $\Spec(R)$ has exactly one closed point, \item $R$ has a maximal ideal $\mathfrak m$ and every element of $R \setminus \mathfrak m$ is a unit, and \item $R$ is not the zero ring and for every $x \in R$ either $x$ or $1 - x$ is invertible or both. \end{enumerate}


algebra-lemma-contained-in-radical

Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $I \subset R$ be an ideal. The following are equivalent \begin{enumerate} \item $I \subset \text{rad}(R)$, and \item every element of $1 + I$ is a unit in $R$. \end{enumerate} In this case every element of $R$ which maps to a unit of $R/I$ is a unit.


algebra-lemma-NAK

\begin{reference} \cite[1.M Lemma (NAK) page 11]{MatCA} \end{reference} \begin{history} We quote from \cite{MatCA}: ``This simple but important lemma is due to T.~Nakayama, G.~Azumaya and W.~Krull. Priority is obscure, and although it is usually called the Lemma of Nakayama, late Prof.~Nakayama did not like the name.'' \end{history} Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $M$ be an $R$-module. Let $I \subset R$ be an ideal. \begin{enumerate} \item \label{item-nakayama} If $IM = M$ and $M$ is finite, then there exists a $f \in 1 + I$ such that $fM = 0$. \item If $IM = M$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M = 0$. \item If $N, N' \subset M$, $M = N + IN'$, and $N'$ is finite, then there exists a $f \in 1 + I$ such that $fM \subset N$ and $M_f = N_f$. \item If $N, N' \subset M$, $M = N + IN'$, $N'$ is finite, and $I \subset \text{rad}(R)$, then $M = N$. \item If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, and $M$ is finite, then there exists a $f \in 1 + I$ such that $N_f \to M_f$ is surjective. \item If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, $M$ is finite, and $I \subset \text{rad}(R)$, then $N \to M$ is surjective. \item If $x_1, \ldots, x_n \in M$ generate $M/IM$ and $M$ is finite, then there exists an $f \in 1 + I$ such that $x_1, \ldots, x_n$ generate $M_f$ over $R_f$. \item If $x_1, \ldots, x_n \in M$ generate $M/IM$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M$ is generated by $x_1, \ldots, x_n$. \item If $IM = M$, $I$ is nilpotent, then $M = 0$. \item If $N, N' \subset M$, $M = N + IN'$, and $I$ is nilpotent then $M = N$. \item If $N \to M$ is a module map, $I$ is nilpotent, and $N/IN \to M/IM$ is surjective, then $N \to M$ is surjective. \item If $\{x_\alpha\}_{\alpha \in A}$ is a set of elements of $M$ which generate $M/IM$ and $I$ is nilpotent, then $M$ is generated by the $x_\alpha$. \end{enumerate}


algebra-lemma-when-surjective-local

Let $A \to B$ be a local homomorphism of local rings. Assume \begin{enumerate} \item $B$ is finite as an $A$-module, \item $\mathfrak m_B$ is a finitely generated ideal, \item $A \to B$ induces an isomorphism on residue fields, and \item $\mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2$ is surjective. \end{enumerate} Then $A \to B$ is surjective.


algebra-lemma-idempotent-spec

Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case $$ \Spec(R) = D(e) \amalg D(1-e). $$


algebra-lemma-spec-product

Let $R_1$ and $R_2$ be rings. Let $R = R_1 \times R_2$. The maps $R \to R_1$, $(x, y) \mapsto x$ and $R \to R_2$, $(x, y) \mapsto y$ induce continuous maps $\Spec(R_1) \to \Spec(R)$ and $\Spec(R_2) \to \Spec(R)$. The induced map $$ \Spec(R_1) \amalg \Spec(R_2) \longrightarrow \Spec(R) $$ is a homeomorphism. In other words, the spectrum of $R = R_1\times R_2$ is the disjoint union of the spectrum of $R_1$ and the spectrum of $R_2$.


algebra-lemma-disjoint-decomposition

Let $R$ be a ring. For each $U \subset \Spec(R)$ which is open and closed there exists a unique idempotent $e \in R$ such that $U = D(e)$. This induces a 1-1 correspondence between open and closed subsets $U \subset \Spec(R)$ and idempotents $e \in R$.


algebra-lemma-characterize-spec-connected

Let $R$ be a nonzero ring. Then $\Spec(R)$ is connected if and only if $R$ has no nontrivial idempotents.


algebra-lemma-ideal-is-squared-union-connected

Let $R$ be a ring. Let $I$ be a finitely generated ideal. Assume that $I = I^2$. Then $V(I)$ is open and closed in $\Spec(R)$, and $R/I \cong R_e$ for some idempotent $e \in R$.


algebra-lemma-connected-component

Let $R$ be a ring. A connected component of $\Spec(R)$ is of the form $V(I)$, where $I$ is an ideal generated by idempotents such that every idempotent of $R$ either maps to $0$ or $1$ in $R/I$.


algebra-lemma-characterize-zero-local

Let $R$ be a ring. \begin{enumerate} \item For an element $x$ of an $R$-module $M$ the following are equivalent \begin{enumerate} \item $x = 0$, \item $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \in \Spec(R)$, \item $x$ maps to zero in $M_{\mathfrak m}$ for all maximal ideals $\mathfrak m$ of $R$. \end{enumerate} In other words, the map $M \to \prod_{\mathfrak m} M_{\mathfrak m}$ is injective. \item Given an $R$-module $M$ the following are equivalent \begin{enumerate} \item $M$ is zero, \item $M_{\mathfrak p}$ is zero for all $\mathfrak p \in \Spec(R)$, \item $M_{\mathfrak m}$ is zero for all maximal ideals $\mathfrak m$ of $R$. \end{enumerate} \item Given a complex $M_1 \to M_2 \to M_3$ of $R$-modules the following are equivalent \begin{enumerate} \item $M_1 \to M_2 \to M_3$ is exact, \item for every prime $\mathfrak p$ of $R$ the localization $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$ is exact, \item for every maximal ideal $\mathfrak m$ of $R$ the localization $M_{1, \mathfrak m} \to M_{2, \mathfrak m} \to M_{3, \mathfrak m}$ is exact. \end{enumerate} \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent \begin{enumerate} \item $f$ is injective, \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is injective for all primes $\mathfrak p$ of $R$, \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is injective for all maximal ideals $\mathfrak m$ of $R$. \end{enumerate} \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent \begin{enumerate} \item $f$ is surjective, \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is surjective for all primes $\mathfrak p$ of $R$, \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is surjective for all maximal ideals $\mathfrak m$ of $R$. \end{enumerate} \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent \begin{enumerate} \item $f$ is bijective, \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is bijective for all primes $\mathfrak p$ of $R$, \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is bijective for all maximal ideals $\mathfrak m$ of $R$. \end{enumerate} \end{enumerate}


algebra-lemma-cover

\begin{slogan} Zariski-local properties of modules and algebras \end{slogan} Let $R$ be a ring. Let $M$ be an $R$-module. Let $S$ be an $R$-algebra. Suppose that $f_1, \ldots, f_n$ is a finite list of elements of $R$ such that $\bigcup D(f_i) = \Spec(R)$, in other words $(f_1, \ldots, f_n) = R$. \begin{enumerate} \item If each $M_{f_i} = 0$ then $M = 0$. \item If each $M_{f_i}$ is a finite $R_{f_i}$-module, then $M$ is a finite $R$-module. \item If each $M_{f_i}$ is a finitely presented $R_{f_i}$-module, then $M$ is a finitely presented $R$-module. \item Let $M \to N$ be a map of $R$-modules. If $M_{f_i} \to N_{f_i}$ is an isomorphism for each $i$ then $M \to N$ is an isomorphism. \item Let $0 \to M'' \to M \to M' \to 0$ be a complex of $R$-modules. If $0 \to M''_{f_i} \to M_{f_i} \to M'_{f_i} \to 0$ is exact for each $i$, then $0 \to M'' \to M \to M' \to 0$ is exact. \item If each $R_{f_i}$ is Noetherian, then $R$ is Noetherian. \item If each $S_{f_i}$ is a finite type $R$-algebra, so is $S$. \item If each $S_{f_i}$ is of finite presentation over $R$, so is $S$. \end{enumerate}


algebra-lemma-cover-upstairs

Let $R \to S$ be a ring map. Suppose that $g_1, \ldots, g_n$ is a finite list of elements of $S$ such that $\bigcup D(g_i) = \Spec(S)$ in other words $(g_1, \ldots, g_n) = S$. \begin{enumerate} \item If each $S_{g_i}$ is of finite type over $R$, then $S$ is of finite type over $R$. \item If each $S_{g_i}$ is of finite presentation over $R$, then $S$ is of finite presentation over $R$. \end{enumerate}


algebra-lemma-cover-module

Let $R$ be a ring. Let $f_1, \ldots, f_n$ be elements of $R$ generating the unit ideal. Let $M$ be an $R$-module. The sequence $$ 0 \to M \xrightarrow{\alpha} \bigoplus\nolimits_{i = 1}^n M_{f_i} \xrightarrow{\beta} \bigoplus\nolimits_{i, j = 1}^n M_{f_i f_j} $$ is exact, where $\alpha(m) = (m/1, \ldots, m/1)$ and $\beta(m_1/f_1^{e_1}, \ldots, m_n/f_n^{e_n}) = (m_i/f_i^{e_i} - m_j/f_j^{e_j})_{(i, j)}$.


algebra-lemma-standard-covering

Let $R$ be a ring, and let $f_1, f_2, \ldots f_n\in R$ generate the unit ideal in $R$. Then the following sequence is exact: $$ 0 \longrightarrow R \longrightarrow \bigoplus\nolimits_i R_{f_i} \longrightarrow \bigoplus\nolimits_{i, j}R_{f_if_j} $$ where the maps $\alpha : R \longrightarrow \bigoplus_i R_{f_i}$ and $\beta : \bigoplus_i R_{f_i} \longrightarrow \bigoplus_{i, j} R_{f_if_j}$ are defined as $$ \alpha(x) = \left(\frac{x}{1}, \ldots, \frac{x}{1}\right) \text{ and } \beta\left(\frac{x_1}{f_1^{r_1}}, \ldots, \frac{x_n}{f_n^{r_n}}\right) = \left(\frac{x_i}{f_i^{r_i}}-\frac{x_j}{f_j^{r_j}}~\text{in}~R_{f_if_j}\right). $$


algebra-lemma-disjoint-implies-product

Let $R$ be a ring. If $\Spec(R) = U \amalg V$ with both $U$ and $V$ open then $R \cong R_1 \times R_2$ with $U \cong \Spec(R_1)$ and $V \cong \Spec(R_2)$ via the maps in Lemma \ref{lemma-spec-product}. Moreover, both $R_1$ and $R_2$ are localizations as well as quotients of the ring $R$.


algebra-lemma-when-injective-covering

Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$. Let $M$ be an $R$-module. Then $M \to \bigoplus M_{f_i}$ is injective if and only if $$ M \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} M, \quad m \longmapsto (f_1m, \ldots, f_nm) $$ is injective.


algebra-lemma-glue-modules

Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$. Suppose we are given the following data: \begin{enumerate} \item For each $i$ an $R_{f_i}$-module $M_i$. \item For each pair $i, j$ an $R_{f_if_j}$-module isomorphism $\psi_{ij} : (M_i)_{f_j} \to (M_j)_{f_i}$. \end{enumerate} which satisfy the ``cocycle condition'' that all the diagrams $$ \xymatrix{ (M_i)_{f_jf_k} \ar[rd]_{\psi_{ij}} \ar[rr]^{\psi_{ik}} & & (M_k)_{f_if_j} \\ & (M_j)_{f_if_k} \ar[ru]_{\psi_{jk}} } $$ commute (for all triples $i, j, k$). Given this data define $$ M = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_i \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_i)_{f_j} \right) $$ where $(m_1, \ldots, m_n)$ maps to the element whose $(i, j)$th entry is $m_i/1 - \psi_{ji}(m_j/1)$. Then the natural map $M \to M_i$ induces an isorphism $M_{f_i} \to M_i$. Moreover $\psi_{ij}(m/1) = m/1$ for all $m \in M$ (with obvious notation).


algebra-lemma-minimal-prime-reduced-ring

Let $\mathfrak p$ be a minimal prime of a ring $R$. Every element of the maximal ideal of $R_{\mathfrak p}$ is nilpotent. If $R$ is reduced then $R_{\mathfrak p}$ is a field.


algebra-lemma-reduced-ring-sub-product-fields

Let $R$ be a reduced ring. Then \begin{enumerate} \item $R$ is a subring of a product of fields, \item $R \to \prod_{\mathfrak p\text{ minimal}} R_{\mathfrak p}$ is an embedding into a product of fields, \item $\bigcup_{\mathfrak p\text{ minimal}} \mathfrak p$ is the set of zerodivisors of $R$. \end{enumerate}


algebra-lemma-total-ring-fractions

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset consisting of nonzerodivisors. Then $Q(R) \cong Q(S^{-1}R)$. In particular $Q(R) \cong Q(Q(R))$.


algebra-lemma-total-ring-fractions-no-embedded-points

Let $R$ be a ring. Assume that $R$ has finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, and that $\mathfrak q_1 \cup \ldots \cup \mathfrak q_t$ is the set of zerodivisors of $R$. Then the total ring of fractions $Q(R)$ is equal to $R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}$.


algebra-lemma-irreducible

Let $R$ be a ring. \begin{enumerate} \item For a prime $\mathfrak p \subset R$ the closure of $\{\mathfrak p\}$ in the Zariski topology is $V(\mathfrak p)$. In a formula $\overline{\{\mathfrak p\}} = V(\mathfrak p)$. \item The irreducible closed subsets of $\Spec(R)$ are exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a prime. \item The irreducible components (see Topology, Definition \ref{topology-definition-irreducible-components}) of $\Spec(R)$ are  exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a minimal prime. \end{enumerate}


algebra-lemma-spec-spectral

The spectrum of a ring is a spectral space, see Topology, Definition \ref{topology-definition-spectral-space}.


algebra-lemma-irreducible-components-containing-x

Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime. \begin{enumerate} \item the set of irreducible closed subsets of $\Spec(R)$ passing through $\mathfrak p$ is in one-to-one correspondence with primes $\mathfrak q \subset R_{\mathfrak p}$. \item The set of irreducible components of $\Spec(R)$ passing through $\mathfrak p$ is in one-to-one correspondence with minimal primes $\mathfrak q \subset R_{\mathfrak p}$. \end{enumerate}


algebra-lemma-standard-open-containing-maximal-point

Let $R$ be a ring. Let $\mathfrak p$ be a minimal prime of $R$. Let $W \subset \Spec(R)$ be a quasi-compact open not containing the point $\mathfrak p$. Then there exists an $f \in R$, $f \not \in \mathfrak p$ such that $D(f) \cap W = \emptyset$.


algebra-lemma-ring-with-only-minimal-primes

Let $R$ be a ring. Let $X = \Spec(R)$ as a topological space. The following are equivalent \begin{enumerate} \item $X$ is profinite, \item $X$ is Hausdorff, \item $X$ is totally disconnected. \item every quasi-compact open of $X$ is closed, \item there are no nontrivial inclusions between its prime ideals, \item every prime ideal is a maximal ideal, \item every prime ideal is minimal, \item every standard open $D(f) \subset X$ is closed, and \item add more here. \end{enumerate}


algebra-lemma-colon

Let $R$ be a ring. For a principal ideal $J \subset R$, and for any ideal $I \subset J$ we have $I = J (I : J)$.


algebra-lemma-simple

Let $R$ be a ring. Let $S$ be a multiplicative subset of $R$. An ideal $I \subset R$ which is maximal with respect to the property that $I \cap S = \emptyset$ is prime.


algebra-lemma-cohen

Let $R$ be a ring. \begin{enumerate} \item An ideal $I \subset R$ maximal with respect to not being finitely generated is prime. \item If every prime ideal of $R$ is finitely generated, then every ideal of $R$ is finitely generated\footnote{Later we will say that $R$ is Noetherian.}. \end{enumerate}


algebra-lemma-qc-open

Let $U \subset \Spec(R)$ be open. The following are equivalent: \begin{enumerate} \item $U$ is retrocompact in $\Spec(R)$, \item $U$ is quasi-compact, \item $U$ is a finite union of standard opens, and \item there exists a finitely generated ideal $I \subset R$ such that $X \setminus V(I) = U$. \end{enumerate}


algebra-lemma-affine-map-quasi-compact

Let $\varphi : R \to S$ be a ring map. The induced continuous map $f : \Spec(S) \to \Spec(R)$ is quasi-compact. For any constructible set $E \subset \Spec(R)$ the inverse image $f^{-1}(E)$ is constructible in $\Spec(S)$.


algebra-lemma-constructible

Let $R$ be a ring. A subset of $\Spec(R)$ is constructible if and only if it can be written as a finite union of subsets of the form $D(f) \cap V(g_1, \ldots, g_m)$ for $f, g_1, \ldots, g_m \in R$.


algebra-lemma-constructible-is-image

Let $R$ be a ring and let $T \subset \Spec(R)$ be constructible. Then there exists a ring map $R \to S$ of finite presentation such that $T$ is the image of $\Spec(S)$ in $\Spec(R)$.


algebra-lemma-open-fp

Let $R$ be a ring. Let $f$ be an element of $R$. Let $S = R_f$. Then the image of a constructible subset of $\Spec(S)$ is constructible in $\Spec(R)$.


algebra-lemma-closed-fp

Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $S = R/I$. Then the image of a constructible of $\Spec(S)$ is constructible in $\Spec(R)$.


algebra-lemma-affineline-open

Let $R$ be a ring. The map $\Spec(R[x]) \to \Spec(R)$ is open, and the image of any standard open is a quasi-compact open.


algebra-lemma-characteristic-polynomial-prime

Let $R \to A$ be a ring homomorphism. Assume $A \cong R^{\oplus n}$ as an $R$-module. Let $f \in A$. The multiplication map $m_f: A \to A$ is $R$-linear and hence has a characteristic polynomial $P(T) = T^n + r_{n-1}T^{n-1} + \ldots + r_0 \in R[T]$. For any prime $\mathfrak{p} \in \Spec(R)$, $f$ acts nilpotently on $A \otimes_R \kappa(\mathfrak{p})$ if and only if $\mathfrak p \in V(r_0, \ldots, r_{n-1})$.


algebra-lemma-affineline-special

Let $R$ be a ring. Let $f, g \in R[x]$ be polynomials. Assume the leading coefficient of $g$ is a unit of $R$. There exists elements $r_i\in R$, $i = 1\ldots, n$ such that the image of $D(f) \cap V(g)$ in $\Spec(R)$ is $\bigcup_{i = 1, \ldots, n} D(r_i)$.


algebra-lemma-generic-finite-presentation

Let $R \subset S$ be an inclusion of domains. Assume that $R \to S$ is of finite type. There exists a nonzero $f \in R$, and a nonzero $g \in S$ such that $R_f \to S_{fg}$ is of finite presentation.


algebra-lemma-characterize-image-finite-type

Let $R \to S$ be a finite type ring map. Denote $X = \Spec(R)$ and $Y = \Spec(S)$. Write $f : Y \to X$ the induced map of spectra. Let $E \subset Y = \Spec(S)$ be a constructible set. If a point $\xi \in X$ is in $f(E)$, then $\overline{\{\xi\}} \cap f(E)$ contains an open dense subset of $\overline{\{\xi\}}$.


algebra-lemma-surjective-spec-radical-ideal

Let $\varphi : R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item The map $\Spec(S) \to \Spec(R)$ is surjective. \item For any ideal $I \subset R$ the inverse image of $\sqrt{IS}$ in $R$ is equal to $\sqrt{I}$. \item For any radical ideal $I \subset R$ the inverse image of $IS$ in $R$ is equal to $I$. \item For every prime $\mathfrak p$ of $R$ the inverse image of $\mathfrak p S$ in $R$ is $\mathfrak p$. \end{enumerate} In this case the same is true after any base change: Given a ring map $R \to R'$ the ring map $R' \to R' \otimes_R S$ has the equivalent properties (1), (2), (3) as well.


algebra-lemma-domain-image-dense-set-points-generic-point

Let $R$ be a domain. Let $\varphi : R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item The ring map $R \to S$ is injective. \item The image $\Spec(S) \to \Spec(R)$ contains a dense set of points. \item There exists a prime ideal $\mathfrak q \subset S$ whose inverse image in $R$ is $(0)$. \end{enumerate}


algebra-lemma-injective-minimal-primes-in-image

Let $R \subset S$ be an injective ring map. Then $\Spec(S) \to \Spec(R)$ hits all the minimal primes.


algebra-lemma-image-dense-generic-points

Let $R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item The kernel of $R \to S$ consists of nilpotent elements. \item The minimal primes of $R$ are in the image of $\Spec(S) \to \Spec(R)$. \item The image of $\Spec(S) \to \Spec(R)$ is dense in $\Spec(R)$. \end{enumerate}


algebra-lemma-minimal-prime-image-minimal-prime

Let $R \to S$ be a ring map. If a minimal prime $\mathfrak p \subset R$ is in the image of $\Spec(S) \to \Spec(R)$, then it is the image of a minimal prime.


algebra-lemma-Noetherian-permanence

\begin{slogan} Noetherian property is stable by passage to finite type extension and localization. \end{slogan} Any finitely generated ring over a Noetherian ring is Noetherian. Any localization of a Noetherian ring is Noetherian.


algebra-lemma-Noetherian-power-series

If $R$ is a Noetherian ring, then so is the formal power series ring $R[[x_1, \ldots, x_n]]$.


algebra-lemma-obvious-Noetherian

Any finite type algebra over a field is Noetherian. Any finite type algebra over $\mathbf{Z}$ is Noetherian.


algebra-lemma-Noetherian-finite-type-is-finite-presentation

Let $R$ be a Noetherian ring. \begin{enumerate} \item Any finite $R$-module is of finite presentation. \item Any finite type $R$-algebra is of finite presentation over $R$. \end{enumerate}


algebra-lemma-Noetherian-topology

If $R$ is a Noetherian ring then $\Spec(R)$ is a Noetherian topological space, see Topology, Definition \ref{topology-definition-noetherian}.


algebra-lemma-Noetherian-irreducible-components

\begin{slogan} A Noetherian affine scheme has finitely many generic points. \end{slogan} If $R$ is a Noetherian ring then $\Spec(R)$ has finitely many irreducible components. In other words $R$ has finitely many minimal primes.


algebra-lemma-Noetherian-base-change-finite-type

Let $R \to S$ be a ring map. Let $R \to R'$ be of finite type. If $S$ is Noetherian, then the base change $S' = R' \otimes_R S$ is Noetherian.


algebra-lemma-Noetherian-field-extension

Let $k$ be a field and let $R$ be a Noetherian $k$-algebra. If $K/k$ is a finitely generated field extension then $K \otimes_k R$ is Noetherian.


algebra-lemma-subring-of-local-ring

Let $R$ be a ring and $\mathfrak p \subset R$ be a prime. There exists an $f \in R$, $f \not \in \mathfrak p$ such that $R_f \to R_\mathfrak p$ is injective in each of the following cases \begin{enumerate} \item $R$ is a domain, \item $R$ is Noetherian, or \item $R$ is reduced and has finitely many minimal primes. \end{enumerate}


algebra-lemma-surjective-endo-noetherian-ring-is-iso

Any surjective endomorphism of a Noetherian ring is an isomorphism.


algebra-lemma-locally-nilpotent

Let $R \to R'$ be a ring map and let $I \subset R$ be a locally nilpotent ideal. Then $IR'$ is a locally nilpotent ideal of $R'$.


algebra-lemma-locally-nilpotent-unit

Let $R$ be a ring and let $I \subset R$ be a locally nilpotent ideal. An element $x$ of $R$ is a unit if and only if the image of $x$ in $R/I$ is a unit.


algebra-lemma-Noetherian-power

\begin{slogan} An ideal in a Noetherian ring is nilpotent if each element of the ideal is nilpotent. \end{slogan} Let $R$ be a Noetherian ring. Let $I, J$ be ideals of $R$. Suppose $J \subset \sqrt{I}$. Then $J^n \subset I$ for some $n$. In particular, in a Noetherian ring the notions of ``locally nilpotent ideal'' and ``nilpotent ideal'' coincide.


algebra-lemma-lift-idempotents

Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Then $R \to R/I$ induces a bijection on idempotents.


algebra-lemma-lift-idempotents-noncommutative

Let $A$ be a possibly noncommutative algebra. Let $e \in A$ be an element such that $x = e^2 - e$ is nilpotent. Then there exists an idempotent of the form $e' = e + x(\sum a_{i, j}e^ix^j) \in A$ with $a_{i, j} \in \mathbf{Z}$.


algebra-lemma-lift-nth-roots

Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $n \geq 1$ be an integer which is invertible in $R/I$. Then \begin{enumerate} \item the $n$th power map $1 + I \to 1 + I$, $1 + x \mapsto (1 + x)^n$ is a bijection, \item a unit of $R$ is a $n$th power if and only if its image in $R/I$ is an $n$th power. \end{enumerate}


algebra-lemma-invert-closed-quotient

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Assume the image of the map $\Spec(S^{-1}R) \to \Spec(R)$ is closed. Then $S^{-1}R \cong R/I$ for some ideal $I \subset R$.


algebra-lemma-field-finite-type-over-domain

Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is of finite type over $R$, then there exists an $f \in R$ such that $R_f$ is a field, and $R_f \subset K$ is a finite field extension.


algebra-lemma-finite-type-field-Jacobson

Any algebra of finite type over a field is Jacobson.


algebra-lemma-jacobson

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


algebra-lemma-characterize-jacobson

Let $R$ be a ring. If $R$ is not Jacobson there exist a prime $\mathfrak p \subset R$, an element $f \in R$ such that the following hold \begin{enumerate} \item $\mathfrak p$ is not a maximal ideal, \item $f \not \in \mathfrak p$, \item $V(\mathfrak p) \cap D(f) = \{\mathfrak p\}$, and \item $(R/\mathfrak p)_f$ is a field. \end{enumerate} On the other hand, if $R$ is Jacobson, then for any pair $(\mathfrak p, f)$ such that (1) and (2) hold the set $V(\mathfrak p) \cap D(f)$ is infinite.


algebra-lemma-pid-jacobson

The ring $\mathbf{Z}$ is a Jacobson ring. More generally, let $R$ be a ring such that \begin{enumerate} \item $R$ is a domain, \item $R$ is Noetherian, \item any nonzero prime ideal is a maximal ideal, and \item $R$ has infinitely many maximal ideals. \end{enumerate} Then $R$ is a Jacobson ring.


algebra-lemma-finite-residue-extension-closed

Let $R \to S$ be a ring map. Let $\mathfrak m \subset R$ be a maximal ideal. Let $\mathfrak q \subset S$ be a prime ideal lying over $\mathfrak m$ such that $\kappa(\mathfrak m) \subset \kappa(\mathfrak q)$ is an algebraic field extension. Then $\mathfrak q$ is a maximal ideal of $S$.


algebra-lemma-dimension

Suppose that $k$ is a field and suppose that $V$ is a nonzero vector space over $k$. Assume the dimension of $V$ (which is a cardinal number) is smaller than the cardinality of $k$. Then for any linear operator $T : V \to V$ there exists some monic polynomial $P(t) \in k[t]$ such that $P(T)$ is not invertible.


algebra-lemma-base-change-Jacobson

Let $k$ be a field. Let $S$ be a $k$-algebra. For any field extension $k \subset K$ whose cardinality is larger than the cardinality of $S$ we have \begin{enumerate} \item for every maximal ideal $\mathfrak m$ of $S_K$ the field $\kappa(\mathfrak m)$ is algebraic over $K$, and \item $S_K$ is a Jacobson ring. \end{enumerate}


algebra-lemma-Jacobson-invert-element

Let $R$ be a Jacobson ring. Let $f \in R$. The ring $R_f$ is Jacobson and maximal ideals of $R_f$ correspond to maximal ideals of $R$ not containing $f$.


algebra-lemma-Jacobson-mod-ideal

Let $R$ be a Jacobson ring. Let $I \subset R$ be an ideal. The ring $R/I$ is Jacobson and maximal ideals of $R/I$ correspond to maximal ideals of $R$ containing $I$.


algebra-lemma-silly-jacobson

Let $R$ be a Jacobson ring. Let $K$ be a field. Let $R \subset K$ and $K$ is of finite type over $R$. Then $R$ is a field and $K/R$ is a finite field extension.


algebra-lemma-image-finite-type-map-Jacobson-rings

Let $R \to S$ be a finite type ring map of Jacobson rings. Denote $X = \Spec(R)$ and $Y = \Spec(S)$. Write $f : Y \to X$ the induced map of spectra. Let $E \subset Y = \Spec(S)$ be a constructible set. Denote with a subscript ${}_0$ the set of closed points of a topological space. \begin{enumerate} \item We have $f(E)_0 = f(E_0) = X_0 \cap f(E)$. \item A point $\xi \in X$ is in $f(E)$ if and only if $\overline{\{\xi\}} \cap f(E_0)$ is dense in $\overline{\{\xi\}}$. \end{enumerate}


algebra-lemma-characterize-integral-element

Let $\varphi : R \to S$ be a ring map. Let $y \in S$. If there exists a finite $R$-submodule $M$ of $S$ such that $1 \in M$ and $yM \subset M$, then $y$ is integral over $R$.


algebra-lemma-finite-is-integral

A finite ring extension is integral.


algebra-lemma-characterize-integral

Let $\varphi : R \to S$ be a ring map. Let $s_1, \ldots, s_n$ be a finite set of elements of $S$. In this case $s_i$ is integral over $R$ for all $i = 1, \ldots, n$ if and only if there exists an $R$-subalgebra $S' \subset S$ finite over $R$ containing all of the $s_i$.


algebra-lemma-characterize-finite-in-terms-of-integral

Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is finite, \item $R \to S$ is integral and of finite type, and \item there exist $x_1, \ldots, x_n \in S$ which generate $S$ as an algebra over $R$ such that each $x_i$ is integral over $R$. \end{enumerate}


algebra-lemma-integral-transitive

\begin{slogan} A composition of integral ring maps is integral \end{slogan} Suppose that $R \to S$ and $S \to T$ are integral ring maps. Then $R \to T$ is integral.


algebra-lemma-integral-closure-is-ring

Let $R \to S$ be a ring homomorphism. The set $$ S' = \{s \in S \mid s\text{ is integral over }R\} $$ is an $R$-subalgebra of $S$.


algebra-lemma-finite-product-integral

Let $R_i\to S_i$ be ring maps $i = 1, \ldots, n$. Let $R$ and $S$ denote the product of the $R_i$ and $S_i$ respectively. Then an element $s = (s_1, \ldots, s_n) \in S$ is integral over $R$ if and only if each $s_i$ is integral over $R_i$.


algebra-lemma-finite-product-integral-closure

Let $R_i\to S_i$ be ring maps $i = 1, \ldots, n$. Denote the integral closure of $R_i$ in $S_i$ by $S'_i$. Further let $R$ and $S$ denote the product of the $R_i$ and $S_i$ respectively. Then the integral closure of $R$ in $S$ is the product of the $S'_i$. In particular $R \to S$ is integrally closed if and only if each $R_i \to S_i$ is integrally closed.


algebra-lemma-integral-closure-localize

Integral closure commutes with localization: If $A \to B$ is a ring map, and $S \subset A$ is a multiplicative subset, then the integral closure of $S^{-1}A$ in $S^{-1}B$ is $S^{-1}B'$, where $B' \subset B$ is the integral closure of $A$ in $B$.


algebra-lemma-integral-closure-stalks

\begin{slogan} An element of an algebra over a ring is integral over the ring if and only if it is locally integral at every prime ideal of the ring. \end{slogan} Let $\varphi : R \to S$ be a ring map. Let $x \in S$. The following are equivalent: \begin{enumerate} \item $x$ is integral over $R$, and \item for every prime ideal $\mathfrak p \subset R$ the element $x \in S_{\mathfrak p}$ is integral over $R_{\mathfrak p}$. \end{enumerate}


algebra-lemma-base-change-integral

\begin{slogan} Integrality and finiteness are preserved under base change. \end{slogan} Let $R \to S$ and $R \to R'$ be ring maps. Set $S' = R' \otimes_R S$. \begin{enumerate} \item If $R \to S$ is integral so is $R' \to S'$. \item If $R \to S$ is finite so is $R' \to S'$. \end{enumerate}


algebra-lemma-integral-local

Let $R \to S$ be a ring map. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. \begin{enumerate} \item If each $R_{f_i} \to S_{f_i}$ is integral, so is $R \to S$. \item If each $R_{f_i} \to S_{f_i}$ is finite, so is $R \to S$. \end{enumerate}


algebra-lemma-integral-permanence

Let $A \to B \to C$ be ring maps. \begin{enumerate} \item If $A \to C$ is integral so is $B \to C$. \item If $A \to C$ is finite so is $B \to C$. \end{enumerate}


algebra-lemma-integral-closure-transitive

Let $A \to B \to C$ be ring maps. Let $B'$ be the integral closure of $A$ in $B$, let $C'$ be the integral closure of $B'$ in $C$. Then $C'$ is the integral closure of $A$ in $C$.


algebra-lemma-integral-overring-surjective

Suppose that $R \to S$ is an integral ring extension with $R \subset S$. Then $\varphi : \Spec(S) \to \Spec(R)$ is surjective.


algebra-lemma-integral-under-field

Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is integral over $R$, then $R$ is a field and $K$ is an algebraic extension. If $R \subset K$ and $K$ is finite over $R$, then $R$ is a field and $K$ is a finite algebraic extension.


algebra-lemma-integral-over-field

Let $k$ be a field. Let $S$ be a $k$-algebra over $k$. \begin{enumerate} \item If $S$ is a domain and finite dimensional over $k$, then $S$ is a field. \item If $S$ is integral over $k$ and a domain, then $S$ is a field. \item If $S$ is integral over $k$ then every prime of $S$ is a maximal ideal (see Lemma \ref{lemma-ring-with-only-minimal-primes} for more consequences). \end{enumerate}


algebra-lemma-integral-no-inclusion

Suppose $R \to S$ is integral. Let $\mathfrak q, \mathfrak q' \in \Spec(S)$ be distinct primes having the same image in $\Spec(R)$. Then neither $\mathfrak q \subset \mathfrak q'$ nor $\mathfrak q' \subset \mathfrak q$.


algebra-lemma-finite-finite-fibres

Suppose $R \to S$ is finite. Then the fibres of $\Spec(S) \to \Spec(R)$ are finite.


algebra-lemma-integral-going-up

Let $R \to S$ be a ring map such that $S$ is integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q$ be a prime of $S$ mapping to $\mathfrak p$. Then there exists a prime $\mathfrak q'$ with $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p'$.


algebra-lemma-finite-finitely-presented-extension

Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. Then $M$ is finitely presented as an $R$-module if and only if $M$ is finitely presented as an $S$-module.


algebra-lemma-integral-closure-in-normal

Let $R \to S$ be a ring map. If $S$ is a normal domain, then the integral closure of $R$ in $S$ is a normal domain.


algebra-lemma-almost-integral

Let $R$ be a domain with fraction field $K$. If $u, v \in K$ are almost integral over $R$, then so are $u + v$ and $uv$. Any element $g \in K$ which is integral over $R$ is almost integral over $R$. If $R$ is Noetherian then the converse holds as well.


algebra-lemma-localize-normal-domain

Any localization of a normal domain is normal.


algebra-lemma-PID-normal

A principal ideal domain is normal.


algebra-lemma-prepare-polynomial-ring-normal

Let $R$ be a domain with fraction field $K$. Suppose $f = \sum \alpha_i x^i$ is an element of $K[x]$. \begin{enumerate} \item If $f$ is integral over $R[x]$ then all $\alpha_i$ are integral over $R$, and \item If $f$ is almost integral over $R[x]$ then all $\alpha_i$ are almost integral over $R$. \end{enumerate}


algebra-lemma-polynomial-domain-normal

Let $R$ be a normal domain. Then $R[x]$ is a normal domain.


algebra-lemma-power-series-over-Noetherian-normal-domain

Let $R$ be a Noetherian normal domain. Then $R[[x]]$ is a Noetherian normal domain.


algebra-lemma-normality-is-local

Let $R$ be a domain. The following are equivalent: \begin{enumerate} \item The domain $R$ is a normal domain, \item for every prime $\mathfrak p \subset R$ the local ring $R_{\mathfrak p}$ is a normal domain, and \item for every maximal ideal $\mathfrak m$ the ring $R_{\mathfrak m}$ is a normal domain. \end{enumerate}


algebra-lemma-normal-ring-integrally-closed

A normal ring is integrally closed in its total ring of fractions.


algebra-lemma-localization-normal-ring

A localization of a normal ring is a normal ring.


algebra-lemma-polynomial-ring-normal

Let $R$ be a normal ring. Then $R[x]$ is a normal ring.


algebra-lemma-finite-product-normal

A finite product of normal rings is normal.


algebra-lemma-characterize-reduced-ring-normal

Let $R$ be a ring. Assume $R$ is reduced and has finitely many minimal primes. Then the following are equivalent: \begin{enumerate} \item $R$ is a normal ring, \item $R$ is integrally closed in its total ring of fractions, and \item $R$ is a finite product of normal domains. \end{enumerate}


algebra-lemma-colimit-normal-ring

Let $(R_i, \varphi_{ii'})$ be a directed system (Categories, Definition \ref{definition-directed-system}) of rings. If each $R_i$ is a normal ring so is $R = \colim_i R_i$.


algebra-lemma-characterize-integral-ideal

Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $A = \sum I^nt^n \subset R[t]$ be the subring of the polynomial ring generated by $R \oplus It \subset R[t]$. An element $s \in S$ is integral over $I$ if and only if the element $st \in S[t]$ is integral over $A$.


algebra-lemma-integral-over-ideal-is-submodule

Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. The set of elements of $S$ which are integral over $I$ form a $R$-submodule of $S$. Furthermore, if $s \in S$ is integral over $R$, and $s'$ is integral over $I$, then $ss'$ is integral over $I$.


algebra-lemma-integral-integral-over-ideal

Suppose $\varphi : R \to S$ is integral. Suppose $I \subset R$ is an ideal. Then every element of $IS$ is integral over $I$.


algebra-lemma-polynomials-divide

Let $K$ be a field. Let $n, m \in \mathbf{N}$ and $a_0, \ldots, a_{n - 1}, b_0, \ldots, b_{m - 1} \in K$. If the polynomial $x^n + a_{n - 1}x^{n - 1} + \ldots + a_0$ divides the polynomial $x^m + b_{m - 1} x^{m - 1} + \ldots + b_0$ in $K[x]$ then \begin{enumerate} \item $a_0, \ldots, a_{n - 1}$ are integral over any subring $R_0$ of $K$ containing the elements $b_0, \ldots, b_{m - 1}$, and \item each $a_i$ lies in $\sqrt{(b_0, \ldots, b_{m-1})R}$ for any subring $R \subset K$ containing the elements $a_0, \ldots, a_{n - 1}, b_0, \ldots, b_{m - 1}$. \end{enumerate}


algebra-lemma-minimal-polynomial-normal-domain

Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal. Let $g \in S$ be integral over $R$. Then the minimal polynomial of $g$ has coefficients in $R$.


algebra-lemma-flat-intersect-ideals

Let $R$ be a ring. Let $I, J \subset R$ be ideals. Let $M$ be a flat $R$-module. Then $IM \cap JM = (I \cap J)M$.


algebra-lemma-colimit-flat

Let $R$ be a ring. Let $\{M_i, \varphi_{ii'}\}$ be a directed system of flat $R$-modules. Then $\colim_i M_i$ is a flat $R$-module.


algebra-lemma-composition-flat

A composition of (faithfully) flat ring maps is (faithfully) flat. If $R \to R'$ is (faithfully) flat, and $M'$ is a (faithfully) flat $R'$-module, then $M'$ is a (faithfully) flat $R$-module.


algebra-lemma-flat

Let $M$ be an $R$-module. The following are equivalent: \begin{enumerate} \item \label{item-flat} $M$ is flat over $R$. \item \label{item-injective} for every injection of $R$-modules $N \subset N'$ the map $N \otimes_R M \to N'\otimes_R M$ is injective. \item \label{item-f-ideal} for every ideal $I \subset R$ the map $I \otimes_R M \to R \otimes_R M = M$ is injective. \item \label{item-ffg-ideal} for every finitely generated ideal $I \subset R$ the map $I \otimes_R M \to R \otimes_R M = M$ is injective. \end{enumerate}


algebra-lemma-colimit-rings-flat

Let $\{R_i, \varphi_{ii'}\}$ be a system of rings over the directed set $I$. Let $R = \colim_i R_i$. Let $M$ be an $R$-module such that $M$ is flat as an $R_i$-module for all $i$. Then $M$ is flat as an $R$-module.


algebra-lemma-flat-base-change

Suppose that $M$ is (faithfully) flat over $R$, and that $R \to R'$ is a ring map. Then $M \otimes_R R'$ is (faithfully) flat over $R'$.


algebra-lemma-flatness-descends

Let $R \to R'$ be a faithfully flat ring map. Let $M$ be a module over $R$, and set $M' = R' \otimes_R M$. Then $M$ is flat over $R$ if and only if $M'$ is flat over $R'$.


algebra-lemma-flatness-descends-more-general

Let $R$ be a ring. Let $S \to S'$ be a flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes_S M$. \begin{enumerate} \item If $M$ is flat over $R$, then $M'$ is flat over $R$. \item If $S \to S'$ is faithfully flat, then $M$ is flat over $R$ if and only if $M'$ is flat over $R$. \end{enumerate}


algebra-lemma-flat-permanence

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $M$ is flat as an $R$-module and faithfully flat as an $S$-module, then $R \to S$ is flat.


algebra-lemma-flat-eq

A module $M$ over $R$ is flat if and only if every relation in $M$ is trivial.


algebra-lemma-flat-tor-zero

Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$ a short exact sequence, and $N$ an $R$-module. If $M$ is flat then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the sequence $$ 0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0 $$ is a short exact sequence.


algebra-lemma-flat-ses

Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of $R$-modules. If $M'$ and $M''$ are flat so is $M$. If $M$ and $M''$ are flat so is $M'$.


algebra-lemma-easy-ff

Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent \begin{enumerate} \item $M$ is faithfully flat, and \item $M$ is flat and for all $R$-module homomorphisms $\alpha : N \to N'$ we have $\alpha = 0$ if and only if $\alpha \otimes \text{id}_M = 0$. \end{enumerate}


algebra-lemma-ff

\begin{slogan} A flat module is faithfully flat if and only if it has nonzero fibers. \end{slogan} Let $M$ be a flat $R$-module. The following are equivalent: \begin{enumerate} \item $M$ is faithfully flat, \item for every nonzero $R$-module $N$, then tensor product $M \otimes_R N$ is nonzero, \item for all $\mathfrak p \in \Spec(R)$ the tensor product $M \otimes_R \kappa(\mathfrak p)$ is nonzero, and \item for all maximal ideals $\mathfrak m$ of $R$ the tensor product $M \otimes_R \kappa(\mathfrak m) = M/{\mathfrak m}M$ is nonzero. \end{enumerate}


algebra-lemma-ff-rings

Let $R \to S$ be a flat ring map. The following are equivalent: \begin{enumerate} \item $R \to S$ is faithfully flat, \item the induced map on $\Spec$ is surjective, and \item any closed point $x \in \Spec(R)$ is in the image of the map $\Spec(S) \to \Spec(R)$. \end{enumerate}


algebra-lemma-local-flat-ff

A flat local ring homomorphism of local rings is faithfully flat.


algebra-lemma-flat-localization

Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. \begin{enumerate} \item The localization $S^{-1}R$ is a flat $R$-algebra. \item If $M$ is an $S^{-1}R$-module, then $M$ is a flat $R$-module if and only if $M$ is a flat $S^{-1}R$-module. \item Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak p}$ is a flat $R_{\mathfrak p}$-module for all primes $\mathfrak p$ of $R$. \item Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak m}$ is a flat $R_{\mathfrak m}$-module for all maximal ideals $\mathfrak m$ of $R$. \item Suppose $R \to A$ is a ring map, $M$ is an $A$-module, and $g_1, \ldots, g_m \in A$ are elements generating the unit ideal of $A$. Then $M$ is flat over $R$ if and only if each localization $M_{g_i}$ is flat over $R$. \item Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$) for all primes $\mathfrak q$ of $A$. \item Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p = R \cap \mathfrak m$) for all maximal ideals $\mathfrak m$ of $A$. \end{enumerate}


algebra-lemma-flat-going-down

Let $R \to S$ be flat. Let $\mathfrak p \subset \mathfrak p'$ be primes of $R$. Let $\mathfrak q' \subset S$ be a prime of $S$ mapping to $\mathfrak p'$. Then there exists a prime $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p$.


algebra-lemma-colimit-faithfully-flat

Let $R$ be a ring. Let $\{S_i, \varphi_{ii'}\}$ be a directed system of faithfully flat $R$-algebras. Then $S = \colim_i S_i$ is a faithfully flat $R$-algebra.


algebra-lemma-support-zero

\begin{slogan} A module over a ring has empty support if and only if it is the trivial module. \end{slogan} Let $R$ be a ring. Let $M$ be an $R$-module. Then $$ M = (0) \Leftrightarrow \text{Supp}(M) = \emptyset. $$


algebra-lemma-annihilator-flat-base-change

Let $R \to S$ be a flat ring map. Let $M$ be an $R$-module and $m \in M$. Then $\text{Ann}_R(m) S = \text{Ann}_S(m \otimes 1)$. If $M$ is a finite $R$-module, then $\text{Ann}_R(M) S = \text{Ann}_S(M \otimes_R S)$.


algebra-lemma-support-closed

Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is finite, then $\text{Supp}(M)$ is closed. More precisely, if $I = \text{Ann}(M)$ is the annihilator of $M$, then $V(I) = \text{Supp}(M)$.


algebra-lemma-support-base-change

Let $R \to R'$ be a ring map and let $M$ be a finite $R$-module. Then $\text{Supp}(M \otimes_R R')$ is the inverse image of $\text{Supp}(M)$.


algebra-lemma-support-element

Let $R$ be a ring, let $M$ be an $R$-module, and let $m \in M$. Then $\mathfrak p \in V(\text{Ann}(m))$ if and only if $m$ does not map to zero in $M_\mathfrak p$.


algebra-lemma-support-finite-presentation-constructible

Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is a finitely presented $R$-module, then $\text{Supp}(M)$ is a closed subset of $\Spec(R)$ whose complement is quasi-compact.


algebra-lemma-support-quotient

Let $R$ be a ring and let $M$ be an $R$-module. \begin{enumerate} \item If $M$ is finite then the support of $M/IM$ is $\text{Supp}(M) \cap V(I)$. \item If $N \subset M$, then $\text{Supp}(N) \subset \text{Supp}(M)$. \item If $Q$ is a quotient module of $M$ then $\text{Supp}(Q) \subset \text{Supp}(M)$. \item If $0 \to N \to M \to Q \to 0$ is a short exact sequence then $\text{Supp}(M) = \text{Supp}(Q) \cup \text{Supp}(N)$. \end{enumerate}


algebra-lemma-open-going-down

Let $R \to S$ be a ring map. If the induced map $\varphi : \Spec(S) \to \Spec(R)$ is open, then $R \to S$ satisfies going down.


algebra-lemma-going-up-down-specialization

Let $R \to S$ be a ring map. \begin{enumerate} \item $R \to S$ satisfies going down if and only if generalizations lift along the map $\Spec(S) \to \Spec(R)$, see Topology, Definition \ref{topology-definition-lift-specializations}. \item $R \to S$ satisfies going up if and only if specializations lift along the map $\Spec(S) \to \Spec(R)$, see Topology, Definition \ref{topology-definition-lift-specializations}. \end{enumerate}


algebra-lemma-going-up-down-composition

Suppose $R \to S$ and $S \to T$ are ring maps satisfying going down. Then so does $R \to T$. Similarly for going up.


algebra-lemma-image-stable-specialization-closed

Let $R \to S$ be a ring map. Let $T \subset \Spec(R)$ be the image of $\Spec(S)$. If $T$ is stable under specialization, then $T$ is closed.


algebra-lemma-going-up-closed

Let $R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item Going up holds for $R \to S$, and \item the map $\Spec(S) \to \Spec(R)$ is closed. \end{enumerate}


algebra-lemma-constructible-stable-specialization-closed

Let $R$ be a ring. Let $E \subset \Spec(R)$ be a constructible subset. \begin{enumerate} \item If $E$ is stable under specialization, then $E$ is closed. \item If $E$ is stable under generalization, then $E$ is open. \end{enumerate}


algebra-lemma-same-image

Let $k$ be a field, and let $R$, $S$ be $k$-algebras. Let $S' \subset S$ be a sub $k$-algebra, and let $f \in S' \otimes_k R$. In the commutative diagram $$ \xymatrix{ \Spec((S \otimes_k R)_f) \ar[rd] \ar[rr] & & \Spec((S' \otimes_k R)_f) \ar[ld] \\ & \Spec(R) & } $$ the images of the diagonal arrows are the same.


algebra-lemma-map-into-tensor-algebra-open

Let $k$ be a field. Let $R$ and $S$ be $k$-algebras. The map $\Spec(S \otimes_k R) \to \Spec(R)$ is open.


algebra-lemma-unique-prime-over-localize-below

Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume that \begin{enumerate} \item there exists a unique prime $\mathfrak q \subset S$ lying over $\mathfrak p$, and \item either \begin{enumerate} \item going up holds for $R \to S$, or \item going down holds for $R \to S$ and there is at most one prime of $S$ above every prime of $R$. \end{enumerate} \end{enumerate} Then $S_{\mathfrak p} = S_{\mathfrak q}$.


algebra-lemma-going-down-flat-module

Let $R \to S$ be a ring map. Let $N$ be a finite $S$-module flat over $R$. Endow $\text{Supp}(N) \subset \Spec(S)$ with the induced topology. Then generalizations lift along $\text{Supp}(N) \to \Spec(R)$.


algebra-lemma-subextensions-are-separable

Let $k \subset K$ be a separable field extension. For any subextension $k \subset K' \subset K$ the field extension $k \subset K'$ is separable.


algebra-lemma-generating-finitely-generated-separable-field-extensions

Let $k \subset K$ be a separably generated, and finitely generated field extension. Set $r = \text{trdeg}_k(K)$. Then there exist elements $x_1, \ldots, x_{r + 1}$ of $K$ such that \begin{enumerate} \item $x_1, \ldots, x_r$ is a transcendence basis of $K$ over $k$, \item $K = k(x_1, \ldots, x_{r + 1})$, and \item $x_{r + 1}$ is separable over $k(x_1, \ldots, x_r)$. \end{enumerate}


algebra-lemma-make-separably-generated

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 separably generated field extension.


algebra-lemma-subalgebra-separable

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


algebra-lemma-geometrically-reduced-permanence

Let $k$ be a field. If $R$ is geometrically reduced over $k$, and $S \subset R$ is a multiplicative subset, then the localization $S^{-1}R$ is geometrically reduced over $k$. If $R$ is geometrically reduced over $k$, then $R[x]$ is geometrically reduced over $k$.


algebra-lemma-limit-argument

Let $k$ be a field. Let $R$, $S$ be $k$-algebras. \begin{enumerate} \item If $R \otimes_k S$ is nonreduced, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes_k S'$ is not reduced. \item If $R \otimes_k S$ contains a nonzero zerodivisor, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes_k S'$ contains a nonzero zerodivisor. \item If $R \otimes_k S$ contains a nontrivial idempotent, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes_k S'$ contains a nontrivial idempotent. \end{enumerate}


algebra-lemma-geometrically-reduced-any-reduced-base-change

Let $k$ be a field. Let $S$ be a geometrically reduced $k$-algebra. Let $R$ be any reduced $k$-algebra. Then $R \otimes_k S$ is reduced.


algebra-lemma-separable-extension-preserves-reducedness

Let $k$ be a field. Let $S$ be a reduced $k$-algebra. Let $k \subset K$ be either a separable field extension, or a separably generated field extension. Then $K \otimes_k S$ is reduced.