algebra-lemma-local-syntomic

Let $R \to S$ be a ring map. Suppose we have $g_1, \ldots g_m \in S$ which generate the unit ideal such that each $R \to S_{g_i}$ is syntomic. Then $R \to S$ is syntomic.


algebra-lemma-huber

Let $S$ be a finitely presented $R$-algebra which has a presentation $S = R[x_1, \ldots, x_n]/I$ such that $I/I^2$ is free over $S$. Then $S$ has a presentation $S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_c)$ such that $(f_1, \ldots, f_c)/(f_1, \ldots, f_c)^2$ is free with basis given by the classes of $f_1, \ldots, f_c$.


algebra-lemma-adjoin-roots

Suppose that $A$ is a ring, and $P(x) = x^n + b_1 x^{n-1} + \ldots + b_n \in A[x]$ is a monic polynomial over $A$. Then there exists a syntomic, finite locally free, faithfully flat ring extension $A \subset A'$ such that $P(x) = \prod_{i = 1, \ldots, n} (x - \beta_i)$ for certain $\beta_i \in A'$.


algebra-lemma-base-change-relative-global-complete-intersection

Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ be a relative global complete intersection (Definition \ref{definition-relative-global-complete-intersection}) \begin{enumerate} \item For any $R \to R'$ the base change $R' \otimes_R S = R'[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ is a relative global complete intersection. \item For any $g \in S$ which is the image of $h \in R[x_1, \ldots, x_n]$ the ring $S_g = R[x_1, \ldots, x_n, x_{n + 1}]/(f_1, \ldots, f_c, hx_{n + 1} - 1)$ is a relative global complete intersection. \item If $R \to S$ factors as $R \to R_f \to S$ for some $f \in R$. Then the ring $S = R_f[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ is a relative global complete intersection over $R_f$. \end{enumerate}


algebra-lemma-localize-relative-complete-intersection

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. We will find $h \in R[x_1, \ldots, x_n]$ which maps to $g \in S$ such that $$ S_g = R[x_1, \ldots, x_n, x_{n + 1}]/(f_1, \ldots, f_c, hx_{n + 1} - 1) $$ is a relative global complete intersection with a presentation as in Definition \ref{definition-relative-global-complete-intersection} in each of the following cases: \begin{enumerate} \item Let $I \subset R$ be an ideal. If the fibres of $\Spec(S/IS) \to \Spec(R/I)$ have dimension $n - c$, then we can find $(h, g)$ as above such that $g$ maps to $1 \in S/IS$. \item Let $\mathfrak p \subset R$ be a prime. If $\dim(S \otimes_R \kappa(\mathfrak p)) = n - c$, then we can find $(h, g)$ as above such that $g$ maps to a unit of $S \otimes_R \kappa(\mathfrak p)$. \item Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. If $\dim_{\mathfrak q}(S/R) = n - c$, then we can find $(h, g)$ as above such that $g \not \in \mathfrak q$. \end{enumerate}


algebra-lemma-relative-global-complete-intersection-Noetherian

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ be a relative global complete intersection (Definition \ref{definition-relative-global-complete-intersection}). There exist a finite type $\mathbf{Z}$-subalgebra $R_0 \subset R$ such that $f_i \in R_0[x_1, \ldots, x_n]$ and such that $$ S_0 = R_0[x_1, \ldots, x_n]/(f_1, \ldots, f_c) $$ is a relative global complete intersection.


algebra-lemma-relative-global-complete-intersection-conormal

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ be a relative global complete intersection (Definition \ref{definition-relative-global-complete-intersection}). For every prime $\mathfrak q$ of $S$, let $\mathfrak q'$ denote the corresponding prime of $R[x_1, \ldots, x_n]$. Then \begin{enumerate} \item $f_1, \ldots, f_c$ is a regular sequence in the local ring $R[x_1, \ldots, x_n]_{\mathfrak q'}$, \item each of the rings $R[x_1, \ldots, x_n]_{\mathfrak q'}/(f_1, \ldots, f_i)$ is flat over $R$, and \item the $S$-module $(f_1, \ldots, f_c)/(f_1, \ldots, f_c)^2$ is free with basis given by the elements $f_i \bmod (f_1, \ldots, f_c)^2$. \end{enumerate}


algebra-lemma-relative-global-complete-intersection

A relative global complete intersection is syntomic, i.e., flat.


algebra-lemma-syntomic

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. The following are equivalent: \begin{enumerate} \item There exists an element $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is syntomic. \item There exists an element $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a relative global complete intersection over $R$. \item There exists an element $g \in S$, $g \not \in \mathfrak q$, such that $R \to S_g$ is of finite presentation, the local ring map $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat, and the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is a complete intersection ring over $\kappa(\mathfrak p)$ (see Definition \ref{definition-lci-local-ring}). \end{enumerate}


algebra-lemma-syntomic-presentation-ideal-mod-squares

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/I$ for some finitely generated ideal $I$. If $g \in S$ is such that $S_g$ is syntomic over $R$, then $(I/I^2)_g$ is a finite projective $S_g$-module.


algebra-lemma-composition-syntomic

Let $R \to S$, $S \to S'$ be ring maps. \begin{enumerate} \item If $R \to S$ and $S \to S'$ are syntomic, then $R \to S'$ is syntomic. \item If $R \to S$ and $S \to S'$ are relative global complete intersections, then $R \to S'$ is a relative global complete intersection. \end{enumerate}


algebra-lemma-lift-syntomic

Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a syntomic map. Then there exists elements $\overline{g}_i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_i} \cong S_i/IS_i$ for some relative global complete intersection $S_i$ over $R$.


algebra-lemma-base-change-smooth

\begin{slogan} Smoothness is preserved under base change \end{slogan} Let $R \to S$ be a smooth ring map. Let $R \to R'$ be any ring map. Then the base change $R' \to S' = R' \otimes_R S$ is smooth.


algebra-lemma-smooth-over-field

Let $k$ be a field. Let $S$ be a smooth $k$-algebra. Then $S$ is a local complete intersection.


algebra-lemma-standard-smooth

Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c) = R[x_1, \ldots, x_n]/I$ be a standard smooth algebra. Then \begin{enumerate} \item the ring map $R \to S$ is smooth, \item the $S$-module $\Omega_{S/R}$ is free on $\text{d}x_{c + 1}, \ldots, \text{d}x_n$, \item the $S$-module $I/I^2$ is free on the classes of $f_1, \ldots, f_c$, \item for any $g \in S$ the ring map $R \to S_g$ is standard smooth, \item for any ring map $R \to R'$ the base change $R' \to R'\otimes_R S$ is standard smooth, \item if $f \in R$ maps to an invertible element in $S$, then $R_f \to S$ is standard smooth, and \item the ring $S$ is a relative global complete intersection over $R$. \end{enumerate}


algebra-lemma-compose-standard-smooth

A composition of standard smooth ring maps is standard smooth.


algebra-lemma-smooth-syntomic

Let $R \to S$ be a smooth ring map. There exists an open covering of $\Spec(S)$ by standard opens $D(g)$ such that each $S_g$ is standard smooth over $R$. In particular $R \to S$ is syntomic.


algebra-lemma-smooth-at-point

Let $R \to S$ be of finite presentation. Let $\mathfrak q$ be a prime of $S$. The following are equivalent \begin{enumerate} \item $R \to S$ is smooth at $\mathfrak q$, \item $H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega_{S/R, \mathfrak q}$ is a finite free $S_\mathfrak q$-module, \item $H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega_{S/R, \mathfrak q}$ is a projective $S_\mathfrak q$-module, and \item $H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega_{S/R, \mathfrak q}$ is a flat $S_\mathfrak q$-module. \end{enumerate}


algebra-lemma-locally-smooth

\begin{slogan} A ring map is smooth if and only if it is smooth at all primes of the target \end{slogan} Let $R \to S$ be a ring map. Then $R \to S$ is smooth if and only if $R \to S$ is smooth at every prime $\mathfrak q$ of $S$.


algebra-lemma-compose-smooth

A composition of smooth ring maps is smooth.


algebra-lemma-relative-global-complete-intersection-smooth

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ be a relative global complete intersection. Let $\mathfrak q \subset S$ be a prime. Then $R \to S$ is smooth at $\mathfrak q$ if and only if there exists a subset $I \subset \{1, \ldots, n\}$ of cardinality $c$ such that the polynomial $$ g_I = \det (\partial f_j/\partial x_i)_{j = 1, \ldots, c, \ i \in I}. $$ does not map to an element of $\mathfrak q$.


algebra-lemma-flat-fibre-smooth

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume \begin{enumerate} \item there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_g$ is of finite presentation, \item the local ring homomorphism $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat, \item the fibre $S \otimes_R \kappa(\mathfrak p)$ is smooth over $\kappa(\mathfrak p)$ at the prime corresponding to $\mathfrak q$. \end{enumerate} Then $R \to S$ is smooth at $\mathfrak q$.


algebra-lemma-flat-base-change-locus-smooth

Let $R \to S$ be a ring map of finite presentation. Let $R \to R'$ be a flat ring map. Denote $S' = R' \otimes_R S$ the base change. Let $U \subset \Spec(S)$ be the set of primes at which $R \to S$ is smooth. Let $V \subset \Spec(S')$ the set of primes at which $R' \to S'$ is smooth. Then $V$ is the inverse image of $U$ under the map $f : \Spec(S') \to \Spec(S)$.


algebra-lemma-smooth-field-change-local

Let $k \subset K$ be a field extension. Let $S$ be a finite type algebra over $k$. Let $\mathfrak q_K$ be a prime of $S_K = K \otimes_k S$ and let $\mathfrak q$ be the corresponding prime of $S$. Then $S$ is smooth over $k$ at $\mathfrak q$ if and only if $S_K$ is smooth at $\mathfrak q_K$ over $K$.


algebra-lemma-lift-smooth

Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a smooth ring map. Then there exists elements $\overline{g}_i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_i} \cong S_i/IS_i$ for some (standard) smooth ring $S_i$ over $R$.


algebra-lemma-base-change-fs

Let $R \to S$ be a formally smooth ring map. Let $R \to R'$ be any ring map. Then the base change $S' = R' \otimes_R S$ is formally smooth over $R'$.


algebra-lemma-compose-formally-smooth

A composition of formally smooth ring maps is formally smooth.


algebra-lemma-polynomial-ring-formally-smooth

A polynomial ring over $R$ is formally smooth over $R$.


algebra-lemma-characterize-formally-smooth

Let $R \to S$ be a ring map. Let $P \to S$ be a surjective $R$-algebra map from a polynomial ring $P$ onto $S$. Denote $J \subset P$ the kernel. Then $R \to S$ is formally smooth if and only if there exists an $R$-algebra map $\sigma : S \to P/J^2$ which is a right inverse to the surjection $P/J^2 \to S$.


algebra-lemma-characterize-formally-smooth-again

Let $R \to S$ be a ring map. Let $P \to S$ be a surjective $R$-algebra map from a polynomial ring $P$ onto $S$. Denote $J \subset P$ the kernel. Then $R \to S$ is formally smooth if and only if the sequence $$ 0 \to J/J^2 \to \Omega_{P/R} \otimes_P S \to \Omega_{S/R} \to 0 $$ of Lemma \ref{lemma-differential-seq} is a split exact sequence.


algebra-lemma-ses-formally-smooth

Let $A \to B \to C$ be ring maps. Assume $B \to C$ is formally smooth. Then the sequence $$ 0 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to \Omega_{C/B} \to 0 $$ of Lemma \ref{lemma-exact-sequence-differentials} is a split short exact sequence.


algebra-lemma-differential-seq-formally-smooth

Let $A \to B \to C$ be ring maps with $A \to C$ formally smooth and $B \to C$ surjective with kernel $J \subset B$. Then the exact sequence $$ 0 \to J/J^2 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to 0 $$ of Lemma \ref{lemma-differential-seq} is split exact.


algebra-lemma-application-NL-formally-smooth

Let $A \to B \to C$ be ring maps. Assume $A \to C$ is surjective (so also $B \to C$ is) and $A \to B$ formally smooth. Denote $I = \Ker(A \to C)$ and $J = \Ker(B \to C)$. Then the sequence $$ 0 \to I/I^2 \to J/J^2 \to \Omega_{B/A} \otimes_B B/J \to 0 $$ of Lemma \ref{lemma-application-NL} is split exact.


algebra-lemma-lift-formal-smoothness

Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Assume \begin{enumerate} \item $I^2 = 0$, \item $R \to S$ is flat, and \item $R/I \to S/IS$ is formally smooth. \end{enumerate} Then $R \to S$ is formally smooth.


algebra-lemma-finite-presentation-fs-Noetherian

Let $R \to S$ be a smooth ring map. Then there exists a subring $R_0 \subset R$ of finite type over $\mathbf{Z}$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes_{R_0} S_0$.


algebra-lemma-descent-formally-smooth

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


algebra-lemma-smooth-strong-lift

Let $R \to S$ be a smooth ring map. Given a commutative solid diagram $$ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } $$ where $I \subset A$ is a locally nilpotent ideal, a dotted arrow exists which makes the diagram commute.


algebra-lemma-triangle-differentials-smooth

Given ring maps $A \to B \to C$ with $B \to C$ smooth, then the sequence $$ 0 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0 $$ of Lemma \ref{lemma-exact-sequence-differentials} is exact.


algebra-lemma-differential-seq-smooth

Let $A \to B \to C$ be ring maps with $A \to C$ smooth and $B \to C$ surjective with kernel $J \subset B$. Then the exact sequence $$ 0 \to J/J^2 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to 0 $$ of Lemma \ref{lemma-differential-seq} is split exact.


algebra-lemma-application-NL-smooth

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


algebra-lemma-rank-omega

Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. Then $$ \dim_{\kappa(\mathfrak m)} \Omega_{S/k} \otimes_S \kappa(\mathfrak m) = \dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2. $$


algebra-lemma-characterize-smooth-kbar

Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. The following are equivalent: \begin{enumerate} \item The ring $S_{\mathfrak m}$ is a regular local ring. \item We have $\dim_{\kappa(\mathfrak m)} \Omega_{S/k} \otimes_S \kappa(\mathfrak m) \leq \dim(S_{\mathfrak m})$. \item We have $\dim_{\kappa(\mathfrak m)} \Omega_{S/k} \otimes_S \kappa(\mathfrak m) = \dim(S_{\mathfrak m})$. \item There exists a $g \in S$, $g \not \in \mathfrak m$ such that $S_g$ is smooth over $k$. In other words $S/k$ is smooth at $\mathfrak m$. \end{enumerate}


algebra-lemma-characterize-smooth-over-field

Let $k$ be any field. Let $S$ be a finite type $k$-algebra. Let $X = \Spec(S)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$. The following are equivalent: \begin{enumerate} \item The $k$-algebra $S$ is smooth at $\mathfrak q$ over $k$. \item We have $\dim_{\kappa(\mathfrak q)} \Omega_{S/k} \otimes_S \kappa(\mathfrak q) \leq \dim_x X$. \item We have $\dim_{\kappa(\mathfrak q)} \Omega_{S/k} \otimes_S \kappa(\mathfrak q) = \dim_x X$. \end{enumerate} Moreover, in this case the local ring $S_{\mathfrak q}$ is regular.


algebra-lemma-computation-differential

Let $k$ be a field. Let $R$ be a Noetherian local ring containing $k$. Assume that the residue field $\kappa = R/\mathfrak m$ is a finitely generated separable extension of $k$. Then the map $$ \text{d} : \mathfrak m/\mathfrak m^2 \longrightarrow \Omega_{R/k} \otimes_R \kappa(\mathfrak m) $$ is injective.


algebra-lemma-separable-smooth

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset S$ be a prime. Assume $\kappa(\mathfrak q)$ is separable over $k$. The following are equivalent: \begin{enumerate} \item The algebra $S$ is smooth at $\mathfrak q$ over $k$. \item The ring $S_{\mathfrak q}$ is regular. \end{enumerate}


algebra-lemma-characteristic-zero

Let $R \to S$ be a $\mathbf{Q}$-algebra map. Let $f \in S$ be such that $\Omega_{S/R} = S \text{d}f \oplus C$ for some $S$-submodule $C$. Then \begin{enumerate} \item $f$ is not nilpotent, and \item if $S$ is a Noetherian local ring, then $f$ is a nonzerodivisor in $S$. \end{enumerate}


algebra-lemma-characteristic-zero-local-smooth

Let $k$ be a field of characteristic $0$. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset S$ be a prime. The following are equivalent: \begin{enumerate} \item The algebra $S$ is smooth at $\mathfrak q$ over $k$. \item The $S_{\mathfrak q}$-module $\Omega_{S/k, \mathfrak q}$ is (finite) free. \item The ring $S_{\mathfrak q}$ is regular. \end{enumerate}


algebra-lemma-smooth-at-generic-point

Let $R \to S$ be an injective finite type ring map with $R$ and $S$ domains. Then $R \to S$ is smooth at $\mathfrak q = (0)$ if and only if the induced extension $L/K$ of fraction fields is separable.


algebra-lemma-smooth-test-artinian

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime ideal of $S$ lying over $\mathfrak p \subset R$. Assume $R$ is Noetherian and $R \to S$ of finite type. The following are equivalent: \begin{enumerate} \item $R \to S$ is smooth at $\mathfrak q$, \item for every surjection of local $R$-algebras $(B', \mathfrak m') \to (B, \mathfrak m)$ with $\Ker(B' \to B)$ having square zero and every solid commutative diagram $$ \xymatrix{ S \ar[r] \ar@{-->}[rd] & B \\ R \ar[r] \ar[u] & B' \ar[u] } $$ such that $\mathfrak q = S \cap \mathfrak m$ there exists a dotted arrow making the diagram commute, \item same as in (2) but with $B' \to B$ ranging over small extensions, and \item same as in (2) but with $B' \to B$ ranging over small extensions such that in addition $S \to B$ induces an isomorphism $\kappa(\mathfrak q) \cong \kappa(\mathfrak m)$. \end{enumerate}


algebra-lemma-etale-standard-smooth

Any \'etale ring map is standard smooth. More precisely, if $R \to S$ is \'etale, then there exists a presentation $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_n)$ such that the image of $\det(\partial f_j/\partial x_i)$ is invertible in $S$.


algebra-lemma-etale

Results on \'etale ring maps. \begin{enumerate} \item The ring map $R \to R_f$ is \'etale for any ring $R$ and any $f \in R$. \item Compositions of \'etale ring maps are \'etale. \item A base change of an \'etale ring map is \'etale. \item The property of being \'etale is local: Given a ring map $R \to S$ and elements $g_1, \ldots, g_m \in S$ which generate the unit ideal such that $R \to S_{g_j}$ is \'etale for $j = 1, \ldots, m$ then $R \to S$ is \'etale. \item Given $R \to S$ of finite presentation, and a flat ring map $R \to R'$, set $S' = R' \otimes_R S$. The set of primes where $R' \to S'$ is \'etale is the inverse image via $\Spec(S') \to \Spec(S)$ of the set of primes where $R \to S$ is \'etale. \item An \'etale ring map is syntomic, in particular flat. \item If $S$ is finite type over a field $k$, then $S$ is \'etale over $k$ if and only if $\Omega_{S/k} = 0$. \item Any \'etale ring map $R \to S$ is the base change of an \'etale ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$. \item Let $A = \colim A_i$ be a filtered colimit of rings. Let $A \to B$ be an \'etale ring map. Then there exists an \'etale ring map $A_i \to B_i$ for some $i$ such that $B \cong A \otimes_{A_i} B_i$. \item Let $A$ be a ring. Let $S$ be a multiplicative subset of $A$. Let $S^{-1}A \to B'$ be \'etale. Then there exists an \'etale ring map $A \to B$ such that $B' \cong S^{-1}B$. \end{enumerate}


algebra-lemma-etale-over-field

Let $k$ be a field. A ring map $k \to S$ is \'etale if and only if $S$ is isomorphic as a $k$-algebra to a finite product of finite separable extensions of $k$.


algebra-lemma-etale-at-prime

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$ in $R$. If $S/R$ is \'etale at $\mathfrak q$ then \begin{enumerate} \item we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite separable. \end{enumerate}


algebra-lemma-etale-quasi-finite

An \'etale ring map is quasi-finite.


algebra-lemma-characterize-etale

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. If \begin{enumerate} \item $R \to S$ is of finite presentation, \item $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat \item $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite separable, \end{enumerate} then $R \to S$ is \'etale at $\mathfrak q$.


algebra-lemma-map-between-etale

Let $R \to S$ and $R \to S'$ be \'etale. Then any $R$-algebra map $S' \to S$ is \'etale.


algebra-lemma-surjective-flat-finitely-presented

Let $\varphi : R \to S$ be a ring map. If $R \to S$ is surjective, flat and finitely presented then there exist an idempotent $e \in R$ such that $S = R_e$.


algebra-lemma-lift-etale

\begin{slogan} \'Etale ring maps lift along surjections of rings \end{slogan} Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be an \'etale ring map. Then there exists an \'etale ring map $R \to S$ such that $\overline{S} \cong S/IS$ as $R/I$-algebras.


algebra-lemma-lift-etale-infinitesimal

Consider a commutative diagram $$ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } $$ with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps whose kernels are ideals of square zero. If $A \to B$ is \'etale, and $J = I \otimes_A B$, then $A' \to B'$ is \'etale.


algebra-lemma-factor-mod-lift-etale

Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $\mathfrak p$ be a prime of $R$. Let $f \bmod \mathfrak p = \overline{g} \overline{h}$ be a factorization of the image of $f$ in $\kappa(\mathfrak p)[x]$. If $\gcd(\overline{g}, \overline{h}) = 1$, then there exist \begin{enumerate} \item an \'etale ring map $R \to R'$, \item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, and \item a factorization $f = g h$ in $R'[x]$ \end{enumerate} such that \begin{enumerate} \item $\kappa(\mathfrak p) = \kappa(\mathfrak p')$, \item $\overline{g} = g \bmod \mathfrak p'$, $\overline{h} = h \bmod \mathfrak p'$, and \item the polynomials $g, h$ generate the unit ideal in $R'[x]$. \end{enumerate}


algebra-lemma-standard-etale

Let $R \to R[x]_g/(f)$ be standard \'etale. \begin{enumerate} \item The ring map $R \to R[x]_g/(f)$ is \'etale. \item For any ring map $R \to R'$ the base change $R' \to R'[x]_g/(f)$ of the standard \'etale ring map $R \to R[x]_g/(f)$ is standard \'etale. \item Any principal localization of $R[x]_g/(f)$ is standard \'etale over $R$. \item A composition of standard \'etale maps is {\bf not} standard \'etale in general. \end{enumerate}


algebra-lemma-make-etale-map-prescribed-residue-field

Let $R$ be a ring. Let $\mathfrak p$ be a prime of $R$. Let $\kappa(\mathfrak p) \subset L$ be a finite separable field extension. There exists an \'etale ring map $R \to R'$ together with a prime $\mathfrak p'$ lying over $\mathfrak p$ such that the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is isomorphic to $\kappa(\mathfrak p) \subset L$.


algebra-lemma-standard-etale-finite-flat-Zariski

Let $R \to S$ be a standard \'etale morphism. There exists a ring map $R \to S'$ with the following properties \begin{enumerate} \item $R \to S'$ is finite, finitely presented, and flat (in other words $S'$ is finite projective as an $R$-module), \item $\Spec(S') \to \Spec(R)$ is surjective, \item for every prime $\mathfrak q \subset S$, lying over $\mathfrak p \subset R$ and every prime $\mathfrak q' \subset S'$ lying over $\mathfrak p$ there exists a $g' \in S'$, $g' \not \in \mathfrak q'$ such that the ring map $R \to S'_{g'}$ factors through a map $\varphi : S \to S'_{g'}$ with $\varphi^{-1}(\mathfrak q'S'_{g'}) = \mathfrak q$. \end{enumerate}


algebra-lemma-etale-finite-flat-zariski

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is \'etale, and \item $\Spec(S) \to \Spec(R)$ is surjective. \end{enumerate} Then there exists a ring map $R \to S'$ such that \begin{enumerate} \item $R \to S'$ is finite, finitely presented, and flat (in other words it is finite projective as an $R$-module), \item $\Spec(S') \to \Spec(R)$ is surjective, \item for every prime $\mathfrak q' \subset S'$ there exists a $g' \in S'$, $g' \not \in \mathfrak q'$ such that the ring map $R \to S'_{g'}$ factors as $R \to S \to S'_{g'}$. \end{enumerate}


algebra-lemma-produce-finite

Let $R \to S' \to S$ be ring maps. Let $\mathfrak p \subset R$ be a prime. Let $g \in S'$ be an element. Assume \begin{enumerate} \item $R \to S'$ is integral, \item $R \to S$ is finite type, \item $S'_g \cong S_g$, and \item $g$ invertible in $S' \otimes_R \kappa(\mathfrak p)$. \end{enumerate} Then there exists a $f \in R$, $f \not \in \mathfrak p$ such that $R_f \to S_f$ is finite.


algebra-lemma-etale-makes-quasi-finite-finite-one-prime

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p \subset R$. Assume $R \to S$ finite type and quasi-finite at $\mathfrak q$. Then there exists \begin{enumerate} \item an \'etale ring map $R \to R'$, \item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, \item a product decomposition $$ R' \otimes_R S = A \times B $$ \end{enumerate} with the following properties \begin{enumerate} \item $\kappa(\mathfrak p) = \kappa(\mathfrak p')$, \item $R' \to A$ is finite, \item $A$ has exactly one prime $\mathfrak r$ lying over $\mathfrak p'$, and \item $\mathfrak r$ lies over $\mathfrak q$. \end{enumerate}


algebra-lemma-etale-makes-quasi-finite-finite

Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume $R \to S$ finite type. Then there exists \begin{enumerate} \item an \'etale ring map $R \to R'$, \item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, \item a product decomposition $$ R' \otimes_R S = A_1 \times \ldots \times A_n \times B $$ \end{enumerate} with the following properties \begin{enumerate} \item we have $\kappa(\mathfrak p) = \kappa(\mathfrak p')$, \item each $A_i$ is finite over $R'$, \item each $A_i$ has exactly one prime $\mathfrak r_i$ lying over $\mathfrak p'$, and \item $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$. \end{enumerate}


algebra-lemma-etale-makes-quasi-finite-finite-variant

Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume $R \to S$ finite type. Then there exists \begin{enumerate} \item an \'etale ring map $R \to R'$, \item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, \item a product decomposition $$ R' \otimes_R S = A_1 \times \ldots \times A_n \times B $$ \end{enumerate} with the following properties \begin{enumerate} \item each $A_i$ is finite over $R'$, \item each $A_i$ has exactly one prime $\mathfrak r_i$ lying over $\mathfrak p'$, \item the finite field extensions $\kappa(\mathfrak p') \subset \kappa(\mathfrak r_i)$ are purely inseparable, and \item $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$. \end{enumerate}


algebra-lemma-trick

Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $R \to B$ be a ring map. If $h \in B[x]/(f)$ is integral over $R$, then the element $f' h$ can be written as $f'h = \sum_i b_i x^i$ with $b_i \in B$ integral over $R$.


algebra-lemma-integral-closure-commutes-etale

Let $R \to S$ be an \'etale ring map. Let $R \to B$ be any ring map. Let $A \subset B$ be the integral closure of $R$ in $B$. Let $A' \subset S \otimes_R B$ be the integral closure of $S$ in $S \otimes_R B$. Then the canonical map $S \otimes_R A \to A'$ is an isomorphism.


algebra-lemma-integral-closure-commutes-smooth

Let $R \to S$ be a smooth ring map. Let $R \to B$ be any ring map. Let $A \subset B$ be the integral closure of $R$ in $B$. Let $A' \subset S \otimes_R B$ be the integral closure of $S$ in $S \otimes_R B$. Then the canonical map $S \otimes_R A \to A'$ is an isomorphism.


algebra-lemma-integral-closure-commutes-colim-smooth

Let $R \to S$ and $R \to B$ be ring maps. Let $A \subset B$ be the integral closure of $R$ in $B$. Let $A' \subset S \otimes_R B$ be the integral closure of $S$ in $S \otimes_R B$. If $S$ is a filtered colimit of smooth $R$-algebras, then the canonical map $S \otimes_R A \to A'$ is an isomorphism.


algebra-lemma-characterize-formally-unramified

Let $R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item $R \to S$ is formally unramified, \item the module of differentials $\Omega_{S/R}$ is zero. \end{enumerate}


algebra-lemma-formally-unramified-local

Let $R \to S$ be a ring map. The following are equivalent: \begin{enumerate} \item $R \to S$ is formally unramified, \item $R \to S_{\mathfrak q}$ is formally unramified for all primes $\mathfrak q$ of $S$, and \item $R_{\mathfrak p} \to S_{\mathfrak q}$ is formally unramified for all primes $\mathfrak q$ of $S$ with $\mathfrak p = R \cap \mathfrak q$. \end{enumerate}


algebra-lemma-formally-unramified-localize

Let $A \to B$ be a formally unramified ring map. \begin{enumerate} \item For $S \subset A$ a multiplicative subset, $S^{-1}A \to S^{-1}B$ is formally unramified. \item For $S \subset B$ a multiplicative subset, $A \to S^{-1}B$ is formally unramified. \end{enumerate}


algebra-lemma-universal-thickening

Let $R \to S$ be a formally unramified ring map. There exists a surjection of $R$-algebras $S' \to S$ whose kernel is an ideal of square zero with the following universal property: Given any commutative diagram $$ \xymatrix{ S \ar[r]_a & A/I \\ R \ar[r]^b \ar[u] & A \ar[u] } $$ where $I \subset A$ is an ideal of square zero, there is a unique $R$-algebra map $a' : S' \to A$ such that $S' \to A \to A/I$ is equal to $S' \to S \to A/I$.


algebra-lemma-universal-thickening-quotient

Let $I \subset R$ be an ideal of a ring. The universal first order thickening of $R/I$ over $R$ is the surjection $R/I^2 \to R/I$. The conormal module of $R/I$ over $R$ is $C_{(R/I)/R} = I/I^2$.


algebra-lemma-universal-thickening-localize

Let $A \to B$ be a formally unramified ring map. Let $\varphi : B' \to B$ be the universal first order thickening of $B$ over $A$. \begin{enumerate} \item Let $S \subset A$ be a multiplicative subset. Then $S^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $S^{-1}A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/S^{-1}A}$. \item Let $S \subset B$ be a multiplicative subset. Then $S' = \varphi^{-1}(S)$ is a multiplicative subset in $B'$ and $(S')^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/A}$. \end{enumerate} Note that the lemma makes sense by Lemma \ref{lemma-formally-unramified-localize}.


algebra-lemma-differentials-universal-thickening

Let $R \to A  \to B$ be ring maps. Assume $A \to B$ formally unramified. Let $B' \to B$ be the universal first order thickening of $B$ over $A$. Then $B'$ is formally unramified over $A$, and the canonical map $\Omega_{A/R} \otimes_A B \to \Omega_{B'/R} \otimes_{B'} B$ is an isomorphism.


algebra-lemma-formally-etale-etale

Let $R \to S$ be a ring map of finite presentation. The following are equivalent: \begin{enumerate} \item $R \to S$ is formally \'etale, \item $R \to S$ is \'etale. \end{enumerate}


algebra-lemma-colimit-formally-etale

Let $R$ be a ring. Let $I$ be a directed set. Let $(S_i, \varphi_{ii'})$ be a system of $R$-algebras over $I$. If each $R \to S_i$ is formally \'etale, then $S = \colim_{i \in I} S_i$ is formally \'etale over $R$


algebra-lemma-localization-formally-etale

Let $R$ be a ring. Let $S \subset R$ be any multiplicative subset. Then the ring map $R \to S^{-1}R$ is formally \'etale.


algebra-lemma-formally-unramified-unramified

Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is formally unramified and of finite type, and \item $R \to S$ is unramified. \end{enumerate} Moreover, also the following are equivalent \begin{enumerate} \item $R \to S$ is formally unramified and of finite presentation, and \item $R \to S$ is G-unramified. \end{enumerate}


algebra-lemma-unramified

Properties of unramified and G-unramified ring maps. \begin{enumerate} \item The base change of an unramified ring map is unramified. The base change of a G-unramified ring map is G-unramified. \item The composition of unramified ring maps is unramified. The composition of G-unramified ring maps is G-unramified. \item Any principal localization $R \to R_f$ is G-unramified and unramified. \item If $I \subset R$ is an ideal, then $R \to R/I$ is unramified. If $I \subset R$ is a finitely generated ideal, then $R \to R/I$ is G-unramified. \item An \'etale ring map is G-unramified and unramified. \item If $R \to S$ is of finite type (resp.\ finite presentation), $\mathfrak q \subset S$ is a prime and $(\Omega_{S/R})_{\mathfrak q} = 0$, then $R \to S$ is unramified (resp.\ G-unramified) at $\mathfrak q$. \item If $R \to S$ is of finite type (resp.\ finite presentation), $\mathfrak q \subset S$ is a prime and $\Omega_{S/R} \otimes_S \kappa(\mathfrak q) = 0$, then $R \to S$ is unramified (resp.\ G-unramified) at $\mathfrak q$. \item If $R \to S$ is of finite type (resp.\ finite presentation), $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ and $(\Omega_{S \otimes_R \kappa(\mathfrak p)/\kappa(\mathfrak p)})_{\mathfrak q} = 0$, then $R \to S$ is unramified (resp.\ G-unramified) at $\mathfrak q$. \item If $R \to S$ is of finite type (resp.\ presentation), $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ and $(\Omega_{S \otimes_R \kappa(\mathfrak p)/\kappa(\mathfrak p)}) \otimes_{S \otimes_R \kappa(\mathfrak p)} \kappa(\mathfrak q) = 0$, then $R \to S$ is unramified (resp.\ G-unramified) at $\mathfrak q$. \item If $R \to S$ is a ring map, $g_1, \ldots, g_m \in S$ generate the unit ideal and $R \to S_{g_j}$ is unramified (resp.\ G-unramified) for $j = 1, \ldots, m$, then $R \to S$ is unramified (resp.\ G-unramified). \item If $R \to S$ is a ring map which is unramified (resp.\ G-unramified) at every prime of $S$, then $R \to S$ is unramified (resp.\ G-unramified). \item If $R \to S$ is G-unramified, then there exists a finite type $\mathbf{Z}$-algebra $R_0$ and a G-unramified ring map $R_0 \to S_0$ and a ring map $R_0 \to R$ such that $S = R \otimes_{R_0} S_0$. \item If $R \to S$ is unramified, then there exists a finite type $\mathbf{Z}$-algebra $R_0$ and an unramified ring map $R_0 \to S_0$ and a ring map $R_0 \to R$ such that $S$ is a quotient of $R \otimes_{R_0} S_0$. \end{enumerate}


algebra-lemma-diagonal-unramified

Let $R \to S$ be a ring map. If $R \to S$ is unramified, then there exists an idempotent $e \in S \otimes_R S$ such that $S \otimes_R S \to S$ is isomorphic to $S \otimes_R S \to (S \otimes_R S)_e$.


algebra-lemma-unramified-at-prime

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$ in $R$. If $S/R$ is unramified at $\mathfrak q$ then \begin{enumerate} \item we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite separable. \end{enumerate}


algebra-lemma-unramified-quasi-finite

Let $R \to S$ be a finite type ring map. Let $\mathfrak q$ be a prime of $S$. If $R \to S$ is unramified at $\mathfrak q$ then $R \to S$ is quasi-finite at $\mathfrak q$. In particular, an unramified ring map is quasi-finite.


algebra-lemma-characterize-unramified

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. If \begin{enumerate} \item $R \to S$ is of finite type, \item $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite separable, \end{enumerate} then $R \to S$ is unramified at $\mathfrak q$.


algebra-lemma-etale-flat-unramified-finite-presentation

Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is \'etale, \item $R \to S$ is flat and G-unramified, and \item $R \to S$ is flat, unramified, and of finite presentation. \end{enumerate}


algebra-lemma-characterize-etale-over-polynomial-ring

Let $k$ be a field. Let $$ \varphi : k[x_1, \ldots, x_n] \to A, \quad x_i \longmapsto a_i $$ be a finite type ring map. Then $\varphi$ is \'etale if and only if we have the following two conditions: (a) the local rings of $A$ at maximal ideals have dimension $n$, and (b) the elements $\text{d}(a_1), \ldots, \text{d}(a_n)$ generate $\Omega_{A/k}$ as an $A$-module.


algebra-lemma-etale-makes-unramified-closed

\begin{slogan} In an unramified ring map, one can separate the points in a fiber by passing to an \'etale neighbourhood. \end{slogan} Let $R \to S$ be a ring map. Let $\mathfrak p$ be a prime of $R$. If $R \to S$ is unramified then there exist \begin{enumerate} \item an \'etale ring map $R \to R'$, \item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$. \item a product decomposition $$ R' \otimes_R S = A_1 \times \ldots \times A_n \times B $$ \end{enumerate} with the following properties \begin{enumerate} \item $R' \to A_i$ is surjective, \item $\mathfrak p'A_i$ is a prime of $A_i$ lying over $\mathfrak p'$, and \item there is no prime of $B$ lying over $\mathfrak p'$. \end{enumerate}


algebra-lemma-uniqueness

Let $(R, \mathfrak m, \kappa)$ be a local ring. Let $f \in R[T]$. Let $a, b \in R$ such that $f(a) = f(b) = 0$, $a = b \bmod \mathfrak m$, and $f'(a) \not \in \mathfrak m$. Then $a = b$.


algebra-lemma-characterize-henselian

\begin{slogan} Characterizations of henselian local rings \end{slogan} Let $(R, \mathfrak m, \kappa)$ be a local ring. The following are equivalent \begin{enumerate} \item $R$ is henselian, \item for every $f \in R[T]$ and every root $a_0 \in \kappa$ of $\overline{f}$ such that $\overline{f'}(a_0) \not = 0$ there exists an $a \in R$ such that $f(a) = 0$ and $a_0 = \overline{a}$, \item for any monic $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$, \item for any monic $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and moreover $\deg_T(g) = \deg_T(g_0)$, \item for any $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$, \item for any $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and moreover $\deg_T(g) = \deg_T(g_0)$, \item for any \'etale ring map $R \to S$ and prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$ there exists a section $\tau : S \to R$ of $R \to S$, \item for any \'etale ring map $R \to S$ and prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$ there exists a section $\tau : S \to R$ of $R \to S$ with $\mathfrak q = \tau^{-1}(\mathfrak m)$, \item any finite $R$-algebra is a product of local rings, \item any finite $R$-algebra is a finite product of local rings, \item any finite type $R$-algebra $S$ can be written as $A \times B$ with $R \to A$ finite and $R \to B$ not quasi-finite at any prime lying over $\mathfrak m$, \item any finite type $R$-algebra $S$ can be written as $A \times B$ with $R \to A$ finite such that each irreducible component of $\Spec(B \otimes_R \kappa)$ has dimension $\geq 1$, and \item any quasi-finite $R$-algebra $S$ can be written as $S = A \times B$ with $R \to A$ finite such that $B \otimes_R \kappa = 0$. \end{enumerate}


algebra-lemma-finite-over-henselian

Let $(R, \mathfrak m, \kappa)$ be a henselian local ring. \begin{enumerate} \item If $R \subset S$ is a finite ring extension then $S$ is a finite product of henselian local rings. \item If $R \subset S$ is a finite local homomorphism of local rings, then $S$ is a henselian local ring. \item If $R \to S$ is a finite type ring map, and $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$ at which $R \to S$ is quasi-finite, then $S_{\mathfrak q}$ is henselian. \item If $R \to S$ is quasi-finite then $S_{\mathfrak q}$ is henselian for every prime $\mathfrak q$ lying over $\mathfrak m$. \end{enumerate}


algebra-lemma-mop-up

Let $(R, \mathfrak m, \kappa)$ be a henselian local ring. Any finite type $R$-algebra $S$ can be written as $S = A_1 \times \ldots \times A_n \times B$ with $A_i$ local and finite over $R$ and $R \to B$ not quasi-finite at any prime of $B$ lying over $\mathfrak m$.


algebra-lemma-mop-up-strictly-henselian

Let $(R, \mathfrak m, \kappa)$ be a strictly henselian local ring. Any finite type $R$-algebra $S$ can be written as $S = A_1 \times \ldots \times A_n \times B$ with $A_i$ local and finite over $R$ and $\kappa \subset \kappa(\mathfrak m_{A_i})$ finite purely inseparable and $R \to B$ not quasi-finite at any prime of $B$ lying over $\mathfrak m$.


algebra-lemma-henselian-cat-finite-etale

Let $(R, \mathfrak m, \kappa)$ be a henselian local ring. The category of finite \'etale ring extensions $R \to S$ is equivalent to the category of finite \'etale algebras $\kappa \to \overline{S}$ via the functor $S \mapsto S/\mathfrak mS$.


algebra-lemma-complete-henselian

\begin{slogan} Complete local rings are Henselian by Newton's method \end{slogan} Let $(R, \mathfrak m, \kappa)$ be a complete local ring, see Definition \ref{definition-complete-local-ring}. Then $R$ is henselian.


algebra-lemma-local-dimension-zero-henselian

\begin{slogan} Local rings of dimension zero are henselian. \end{slogan} Let $(R, \mathfrak m)$ be a local ring of dimension $0$. Then $R$ is henselian.


algebra-lemma-map-into-henselian

Let $R \to S$ be a ring map with $S$ henselian local. Given \begin{enumerate} \item an \'etale ring map $R \to A$, \item a prime $\mathfrak q$ of $A$ lying over $\mathfrak p = R \cap \mathfrak m_S$, \item a $\kappa(\mathfrak p)$-algebra map $\tau : \kappa(\mathfrak q) \to S/\mathfrak m_S$, \end{enumerate} then there exists a unique homomorphism of $R$-algebras $f : A \to S$ such that $\mathfrak q = f^{-1}(\mathfrak m_S)$ and $f \bmod \mathfrak q = \tau$.


algebra-lemma-strictly-henselian-solutions

Let $\varphi : R \to S$ be a local homomorphism of strictly henselian local rings. Let $P_1, \ldots, P_n \in R[x_1, \ldots, x_n]$ be polynomials such that $R[x_1, \ldots, x_n]/(P_1, \ldots, P_n)$ is \'etale over $R$. Then the map $$ R^n \longrightarrow S^n, \quad (h_1, \ldots, h_n) \longmapsto (\varphi(h_1), \ldots, \varphi(h_n)) $$ induces a bijection between $$ \{ (r_1, \ldots, r_n) \in R^n \mid P_i(r_1, \ldots, r_n) = 0, \ i = 1, \ldots, n \} $$ and $$ \{ (s_1, \ldots, s_n) \in S^n \mid P'_i(s_1, \ldots, s_n) = 0, \ i = 1, \ldots, n \} $$ where $P'_i \in S[x_1, \ldots, x_n]$ are the images of the $P_i$ under $\varphi$.


algebra-lemma-split-ML-henselian

Let $R$ be a henselian local ring. Any countably generated Mittag-Leffler module over $R$ is a direct sum of finitely presented $R$-modules.


algebra-lemma-base-change-colimit-etale

Let $R \to A$ and $R \to R'$ be ring maps. If $A$ is a filtered colimit of \'etale ring maps, then so is $R' \to R' \otimes_R A$.


algebra-lemma-composition-colimit-etale

Let $A \to B \to C$ be ring maps. If $A \to B$ is a filtered colimit of \'etale ring maps and $B \to C$ is a filtered colimit of \'etale ring maps, then $A \to C$ is a filtered colimit of \'etale ring maps.


algebra-lemma-colimit-colimit-etale

Let $R$ be a ring. Let $A = \colim A_i$ be a filtered colimit of $R$-algebras such that each $A_i$ is a filtered colimit of \'etale $R$-algebras. Then $A$ is a filtered colimit of \'etale $R$-algebras.


algebra-lemma-colimits-of-etale

Let $R$ be a ring. Let $A \to B$ be an $R$-algebra homomorphism. If $A$ and $B$ are filtered colimits of \'etale $R$-algebras, then $B$ is a filtered colimit of \'etale $A$-algebras.


algebra-lemma-map-into-henselian-colimit

Let $R \to S$ be a ring map with $S$ henselian local. Given \begin{enumerate} \item an $R$-algebra $A$ which is a filtered colimit of \'etale $R$-algebras, \item a prime $\mathfrak q$ of $A$ lying over $\mathfrak p = R \cap \mathfrak m_S$, \item a $\kappa(\mathfrak p)$-algebra map $\tau : \kappa(\mathfrak q) \to S/\mathfrak m_S$, \end{enumerate} then there exists a unique homomorphism of $R$-algebras $f : A \to S$ such that $\mathfrak q = f^{-1}(\mathfrak m_S)$ and $f \bmod \mathfrak q = \tau$.


algebra-lemma-uniqueness-henselian

Let $R$ be a ring. Given a commutative diagram of ring maps $$ \xymatrix{ S \ar[r] & K \\ R \ar[u] \ar[r] & S' \ar[u] } $$ where $S$, $S'$ are henselian local, $S$, $S'$ are filtered colimits of \'etale $R$-algebras, $K$ is a field and the arrows $S \to K$ and  $S' \to K$ identify $K$ with the residue field of both $S$ and $S'$. Then there exists an unique $R$-algebra isomorphism $S \to S'$ compatible with the maps to $K$.


algebra-lemma-colimit-henselian

A filtered colimit of henselian local rings along local homomorphisms is henselian.


algebra-lemma-henselization

Let $(R, \mathfrak m, \kappa)$ be a local ring. There exists a local ring map $R \to R^h$ with the following properties \begin{enumerate} \item $R^h$ is henselian, \item $R^h$ is a filtered colimit of \'etale $R$-algebras, \item $\mathfrak m R^h$ is the maximal ideal of $R^h$, and \item $\kappa = R^h/\mathfrak m R^h$. \end{enumerate}


algebra-lemma-strict-henselization

Let $(R, \mathfrak m, \kappa)$ be a local ring. Let $\kappa \subset \kappa^{sep}$ be a separable algebraic closure. There exists a commutative diagram $$ \xymatrix{ \kappa \ar[r] & \kappa \ar[r] & \kappa^{sep} \\ R \ar[r] \ar[u] & R^h \ar[r] \ar[u] & R^{sh} \ar[u] } $$ with the following properties \begin{enumerate} \item the map $R^h \to R^{sh}$ is local \item $R^{sh}$ is strictly henselian, \item $R^{sh}$ is a filtered colimit of \'etale $R$-algebras, \item $\mathfrak m R^{sh}$ is the maximal ideal of $R^{sh}$, and \item $\kappa^{sep} = R^{sh}/\mathfrak m R^{sh}$. \end{enumerate}


algebra-lemma-henselian-functorial

Let $R \to S$ be a local map of local rings. Let $R \to R^h$ and $S \to S^h$ be the henselizations. There exists a unique local ring map $R^h \to S^h$ fitting into the commutative diagram $$ \xymatrix{ R^h \ar[r]_f & S^h \\ R \ar[u] \ar[r] & S \ar[u] } $$


algebra-lemma-henselization-different

Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is \'etale and $\mathfrak q$ is a prime lying over $\mathfrak p$ such that $\kappa(\mathfrak p) = \kappa(\mathfrak q)$. This category is filtered and $$ (R_{\mathfrak p})^h = \colim_{(S, \mathfrak q)} S = \colim_{(S, \mathfrak q)} S_{\mathfrak q} $$ canonically.


algebra-lemma-henselian-functorial-improve

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $R \to R^h$ and $S \to S^h$ be the henselizations of $R_\mathfrak p$ and $S_\mathfrak q$. The local ring map $R^h \to S^h$ of Lemma \ref{lemma-henselian-functorial} identifies $S^h$ with the henselization of $R^h \otimes_R S$ at the unique prime lying over $\mathfrak m^h$ and $\mathfrak q$.


algebra-lemma-quasi-finite-henselization

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram $$ \xymatrix{ R_{\mathfrak p}^h \ar[r] & S_{\mathfrak q}^h \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] } $$ of Lemma \ref{lemma-henselian-functorial} identifies $S_{\mathfrak q}^h$ with the localization of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ at the prime generated by $\mathfrak q$.


algebra-lemma-quotient-henselization

\begin{slogan} Henselization is compatible with quotients. \end{slogan} Let $R$ be a local ring with henselization $R^h$. Let $I \subset \mathfrak m_R$. Then $R^h/IR^h$ is the henselization of $R/I$.


algebra-lemma-strictly-henselian-functorial-prepare

Let $\varphi : R \to S$ be a local map of local rings. Let $S/\mathfrak m_S \subset \kappa^{sep}$ be a separable algebraic closure. Let $S \to S^{sh}$ be the strict henselization of $S$ with respect to $S/\mathfrak m_S \subset \kappa^{sep}$. Let $R \to A$ be an \'etale ring map and let $\mathfrak q$ be a prime of $A$ lying over $\mathfrak m_R$. Given any commutative diagram $$ \xymatrix{ \kappa(\mathfrak q) \ar[r]_{\phi} & \kappa^{sep} \\ R/\mathfrak m_R \ar[r]^{\varphi} \ar[u] & S/\mathfrak m_S \ar[u] } $$ there exists a unique morphism of rings $f : A \to S^{sh}$ fitting into the commutative diagram $$ \xymatrix{ A \ar[r]_f & S^{sh} \\ R \ar[u] \ar[r]^{\varphi} & S \ar[u] } $$ such that $f^{-1}(\mathfrak m_{S^h}) = \mathfrak q$ and the induced map $\kappa(\mathfrak q) \to \kappa^{sep}$ is the given one.


algebra-lemma-strictly-henselian-functorial

Let $R \to S$ be a local map of local rings. Choose separable algebraic closures $R/\mathfrak m_R \subset \kappa_1^{sep}$ and $S/\mathfrak m_S \subset \kappa_2^{sep}$. Let $R \to R^{sh}$ and $S \to S^{sh}$ be the corresponding strict henselizations. Given any commutative diagram $$ \xymatrix{ \kappa_1^{sep} \ar[r]_{\phi} & \kappa_2^{sep} \\ R/\mathfrak m_R \ar[r]^{\varphi} \ar[u] & S/\mathfrak m_S \ar[u] } $$ There exists a unique local ring map $R^{sh} \to S^{sh}$ fitting into the commutative diagram $$ \xymatrix{ R^{sh} \ar[r]_f & S^{sh} \\ R \ar[u] \ar[r] & S \ar[u] } $$ and inducing $\phi$ on the residue fields of $R^{sh}$ and $S^{sh}$.


algebra-lemma-strict-henselization-different

Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $\kappa(\mathfrak p) \subset \kappa^{sep}$ be a separable algebraic closure. Consider the category of triples $(S, \mathfrak q, \phi)$ where $R \to S$ is \'etale, $\mathfrak q$ is a prime lying over $\mathfrak p$, and $\phi : \kappa(\mathfrak q) \to \kappa^{sep}$ is a $\kappa(\mathfrak p)$-algebra map. This category is filtered and $$ (R_{\mathfrak p})^{sh} = \colim_{(S, \mathfrak q, \phi)} S = \colim_{(S, \mathfrak q, \phi)} S_{\mathfrak q} $$ canonically.


algebra-lemma-strictly-henselian-functorial-improve

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Choose separable algebraic closures $\kappa(\mathfrak p) \subset \kappa_1^{sep}$ and $\kappa(\mathfrak q) \subset \kappa_2^{sep}$. Let $R^{sh}$ and $S^{sh}$ be the corresponding strict henselizations of $R_\mathfrak p$ and $S_\mathfrak q$. Given any commutative diagram $$ \xymatrix{ \kappa_1^{sep} \ar[r]_{\phi} & \kappa_2^{sep} \\ \kappa(\mathfrak p) \ar[r]^{\varphi} \ar[u] & \kappa(\mathfrak q) \ar[u] } $$ The local ring map $R^{sh} \to S^{sh}$ of Lemma \ref{lemma-strictly-henselian-functorial} identifies $S^{sh}$ with the strict henselization of $R^{sh} \otimes_R S$ at a prime lying over $\mathfrak q$ and the maximal ideal $\mathfrak m^{sh} \subset R^{sh}$.


algebra-lemma-quasi-finite-strict-henselization

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Let $\kappa(\mathfrak q) \subset \kappa^{sep}$ be a separable algebraic closure. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram $$ \xymatrix{ R_{\mathfrak p}^{sh} \ar[r] & S_{\mathfrak q}^{sh} \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] } $$ of Lemma \ref{lemma-strictly-henselian-functorial} identifies $S_{\mathfrak q}^{sh}$ with a localization of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$.


algebra-lemma-quotient-strict-henselization

Let $R$ be a local ring with strict henselization $R^{sh}$. Let $I \subset \mathfrak m_R$. Then $R^{sh}/IR^{sh}$ is a strict henselization of $R/I$.


algebra-lemma-sh-from-h-map

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$ such that $\kappa(\mathfrak p) \to \kappa(\mathfrak q)$ is an isomorphism. Choose a separable algebraic closure $\kappa^{sep}$ of $\kappa(\mathfrak p) = \kappa(\mathfrak q)$. Then $$ (S_\mathfrak q)^{sh} = (S_\mathfrak q)^h \otimes_{(R_\mathfrak p)^h} (R_\mathfrak p)^{sh} $$


algebra-lemma-criterion-no-embedded-primes

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following are equivalent: \begin{enumerate} \item $M$ has no embedded associated prime, and \item $M$ has property $(S_1)$. \end{enumerate}


algebra-lemma-criterion-reduced

\begin{slogan} Reduced equals R0 plus S1. \end{slogan} Let $R$ be a Noetherian ring. The following are equivalent: \begin{enumerate} \item $R$ is reduced, and \item $R$ has properties $(R_0)$ and $(S_1)$. \end{enumerate}


algebra-lemma-criterion-normal

\begin{reference} \cite[IV, Theorem 5.8.6]{EGA} \end{reference} \begin{slogan} Normal equals R1 plus S2. \end{slogan} Let $R$ be a Noetherian ring. The following are equivalent: \begin{enumerate} \item $R$ is a normal ring, and \item $R$ has properties $(R_1)$ and $(S_2)$. \end{enumerate}


algebra-lemma-regular-normal

A regular ring is normal.


algebra-lemma-normal-domain-intersection-localizations-height-1

Let $R$ be a Noetherian normal domain with fraction field $K$. Then \begin{enumerate} \item for any nonzero $a \in R$ the quotient $R/aR$ has no embedded primes, and all its associated primes have height $1$ \item $$ R = \bigcap\nolimits_{\text{height}(\mathfrak p) = 1} R_{\mathfrak p} $$ \item For any nonzero $x \in K$ the quotient $R/(R \cap xR)$ has no embedded primes, and all its associates primes have height $1$. \end{enumerate}


algebra-lemma-characterize-separable-algebraic-field-extensions

Let $k \subset K$ be a finitely generated field extension. The following are equivalent \begin{enumerate} \item $K$ is a finite separable field extension of $k$, \item $\Omega_{K/k} = 0$, \item $K$ is formally unramified over $k$, \item $K$ is unramified over $k$, \item $K$ is formally \'etale over $k$, \item $K$ is \'etale over $k$. \end{enumerate}


algebra-lemma-derivative-zero-pth-power

Let $k$ be a perfect field of characteristic $p > 0$. Let $K/k$ be an extension. Let $a \in K$. Then $\text{d}a = 0$ in $\Omega_{K/k}$ if and only if $a$ is a $p$th power.


algebra-lemma-size-extension-pth-roots

Let $k$ be a field of characteristic $p > 0$. Let $a_1, \ldots, a_n \in k$ be elements such that $\text{d}a_1, \ldots, \text{d}a_n$ are linearly independent in $\Omega_{k/\mathbf{F}_p}$. Then the field extension $k(a_1^{1/p}, \ldots, a_n^{1/p})$ has degree $p^n$ over $k$.


algebra-lemma-separable-differentials

Let $k$ be a field of characteristic $p > 0$. The following are equivalent: \begin{enumerate} \item the field extension $K/k$ is separable (see Definition \ref{definition-separable-field-extension}), and \item the map $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K/\mathbf{F}_p}$ is injective. \end{enumerate}


algebra-lemma-formally-smooth-implies-separable

Let $k \subset K$ be an extension of fields. If $K$ is formally smooth over $k$, then $K$ is a separable extension of $k$.


algebra-lemma-characterize-formally-smooth-field-extension

Let $k \subset K$ be an extension of fields. Then $K$ is formally smooth over $k$ if and only if $H_1(L_{K/k}) = 0$.


algebra-lemma-formally-smooth-extensions-easy

Let $k \subset K$ be an extension of fields. \begin{enumerate} \item If $K$ is purely transcendental over $k$, then $K$ is formally smooth over $k$. \item If $K$ is separable algebraic over $k$, then $K$ is formally smooth over $k$. \item If $K$ is separable over $k$, then $K$ is formally smooth over $k$. \end{enumerate}


algebra-lemma-fields-are-formally-smooth

\begin{slogan} Formally smooth equals separable for field extensions. \end{slogan} Let $k$ be a field. \begin{enumerate} \item If the characteristic of $k$ is zero, then any extension field of $k$ is formally smooth over $k$. \item If the characteristic of $k$ is $p > 0$, then $k \subset K$ is formally smooth if and only if it is a separable field extension. \end{enumerate}


algebra-lemma-localization-smooth-separable

Let $k \subset K$ be a finitely generated field extension. Then $K$ is separable over $k$ if and only if $K$ is the localization of a smooth $k$-algebra.


algebra-lemma-colimit-syntomic

Let $k \subset K$ be a field extension. Then $K$ is a filtered colimit of global complete intersection algebras over $k$. If $K/k$ is separable, then $K$ is a filtered colimit of smooth algebras over $k$.


algebra-lemma-flat-local-given-residue-field

Let $(R, \mathfrak m, k)$ be a local ring. Let $k \subset K$ be a field extension. There exists a local ring $(R', \mathfrak m', k')$, a flat local ring map $R \to R'$ such that $\mathfrak m' = \mathfrak mR'$ and such that $k \subset k'$ is isomorphic to $k \subset K$.


algebra-lemma-colimit-finite-etale-given-residue-field

Let $(R, \mathfrak m, k)$ be a local ring. If $k \subset K$ is a separable algebraic extension, then there exists a directed set $I$ and a system of finite \'etale extensions $R \subset R_i$, $i \in I$ of local rings such that $R' = \colim R_i$ has residue field $K$ (as extension of $k$).


algebra-lemma-finite-free-given-residue-field-extension

Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime and let $\kappa(\mathfrak p) \subset L$ be a finite extension of fields. Then there exists a finite free ring map $R \to S$ such that $\mathfrak q = \mathfrak pS$ is prime and $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is isomorphic to the given extension $\kappa(\mathfrak p) \subset L$.


algebra-lemma-quotient-complete-local

Let $R$ be a Noetherian complete local ring. Any quotient of $R$ is also a Noetherian complete local ring. Given a finite ring map $R \to S$, then $S$ is a product of Noetherian complete local rings.


algebra-lemma-complete-local-ring-Noetherian

Let $(R, \mathfrak m)$ be a complete local ring. If $\mathfrak m$ is a finitely generated ideal then $R$ is Noetherian.


algebra-lemma-cohen-rings-exist

Let $p$ be a prime number. Let $k$ be a field of characteristic $p$. There exists a Cohen ring $\Lambda$ with $\Lambda/p\Lambda \cong k$.


algebra-lemma-cohen-ring-formally-smooth

Let $p > 0$ be a prime. Let $\Lambda$ be a Cohen ring with residue field of characteristic $p$. For every $n \geq 1$ the ring map $$ \mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda $$ is formally smooth.


algebra-lemma-regular-complete-containing-coefficient-field

Let $(R, \mathfrak m)$ be a Noetherian complete local ring. Assume $R$ is regular. \begin{enumerate} \item If $R$ contains either $\mathbf{F}_p$ or $\mathbf{Q}$, then $R$ is isomorphic to a power series ring over its residue field. \item If $k$ is a field and $k \to R$ is a ring map inducing an isomorphism $k \to R/\mathfrak m$, then $R$ is isomorphic as a $k$-algebra to a power series ring over $k$. \end{enumerate}


algebra-lemma-complete-local-Noetherian-domain-finite-over-regular

Let $(R, \mathfrak m)$ be a Noetherian complete local domain. Then there exists a $R_0 \subset R$ with the following properties \begin{enumerate} \item $R_0$ is a regular complete local ring, \item $R_0 \subset R$ is finite and induces an isomorphism on residue fields, \item $R_0$ is either isomorphic to $k[[X_1, \ldots, X_d]]$ where $k$ is a field or $\Lambda[[X_1, \ldots, X_d]]$ where $\Lambda$ is a Cohen ring. \end{enumerate}


algebra-lemma-localize-N

Let $R$ be a domain. If $R$ is N-1 then so is any localization of $R$. Same for N-2.


algebra-lemma-Japanese-local

Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$. Same for N-2.


algebra-lemma-quasi-finite-over-Noetherian-japanese

Let $R$ be a domain. Let $R \subset S$ be a quasi-finite extension of domains (for example finite). Assume $R$ is N-2 and Noetherian. Then $S$ is N-2.


algebra-lemma-Laurent-ring-N-1

Let $R$ be a Noetherian domain. If $R[z, z^{-1}]$ is N-1, then so is $R$.


algebra-lemma-finite-extension-N-2

Let $R$ be a Noetherian domain, and let $R \subset S$ be a finite extension of domains. If $S$ is N-1, then so is $R$. If $S$ is N-2, then so is $R$.


algebra-lemma-Noetherian-normal-domain-finite-separable-extension

Let $R$ be a Noetherian normal domain with fraction field $K$. Let $K \subset L$ be a finite separable field extension. Then the integral closure of $R$ in $L$ is finite over $R$.


algebra-lemma-Noetherian-normal-domain-insep-extension

Let $R$ be a Noetherian normal domain with fraction field $K$ of characteristic $p > 0$. Let $a \in K$ be an element such that there exists a derivation $D : R \to R$ with $D(a) \not = 0$. Then the integral closure of $R$ in $L = K[x]/(x^p - a)$ is finite over $R$.


algebra-lemma-domain-char-zero-N-1-2

A Noetherian domain whose fraction field has characteristic zero is N-1 if and only if it is N-2 (i.e., Japanese).


algebra-lemma-domain-char-p-N-1-2

Let $R$ be a Noetherian domain with fraction field $K$ of characteristic $p > 0$. Then $R$ is N-2 if and only if for every finite purely inseparable extension $K \subset L$ the integral closure of $R$ in $L$ is finite over $R$.


algebra-lemma-polynomial-ring-N-2

Let $R$ be a Noetherian domain. If $R$ is N-1 then $R[x]$ is N-1. If $R$ is N-2 then $R[x]$ is N-2.


algebra-lemma-openness-normal-locus

Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_f$ is normal then $$ U = \{\mathfrak p \in \Spec(R) \mid R_{\mathfrak p} \text{ is normal}\} $$ is open in $\Spec(R)$.


algebra-lemma-characterize-N-1

Let $R$ be a Noetherian domain. Then $R$ is N-1 if and only if the following two conditions hold \begin{enumerate} \item there exists a nonzero $f \in R$ such that $R_f$ is normal, and \item for every maximal ideal $\mathfrak m \subset R$ the local ring $R_{\mathfrak m}$ is N-1. \end{enumerate}


algebra-lemma-tate-japanese

\begin{reference} \cite[Theorem 23.1.3]{EGA} \end{reference} Let $R$ be a ring. Let $x \in R$. Assume \begin{enumerate} \item $R$ is a normal Noetherian domain, \item $R/xR$ is a domain and N-2, \item $R \cong \lim_n R/x^nR$ is complete with respect to $x$. \end{enumerate} Then $R$ is N-2.


algebra-lemma-power-series-over-N-2

Let $R$ be a ring. If $R$ is Noetherian, a domain, and N-2, then so is $R[[x]]$.


algebra-lemma-nagata-in-reduced-finite-type-finite-integral-closure

Let $R$ be a Nagata ring. Let $R \to S$ be essentially of finite type with $S$ reduced. Then the integral closure of $R$ in $S$ is finite over $R$.


algebra-lemma-check-universally-japanese

Let $R$ be a ring. To check that $R$ is universally Japanese it suffices to show: If $R \to S$ is of finite type, and $S$ a domain then $S$ is N-1.


algebra-lemma-universally-japanese

If $R$ is universally Japanese then any algebra essentially of finite type over $R$ is universally Japanese.


algebra-lemma-quasi-finite-over-nagata

Let $R$ be a Nagata ring. If $R \to S$ is a quasi-finite ring map (for example finite) then $S$ is a Nagata ring also.


algebra-lemma-nagata-localize

A localization of a Nagata ring is a Nagata ring.


algebra-lemma-nagata-local

Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. \begin{enumerate} \item  If each $R_{f_i}$ is universally Japanese then so is $R$. \item  If each $R_{f_i}$ is Nagata then so is $R$. \end{enumerate}


algebra-lemma-Noetherian-complete-local-Nagata

A Noetherian complete local ring is a Nagata ring.


algebra-lemma-analytically-unramified-easy

Let $(R, \mathfrak m)$ be a Noetherian local ring. \begin{enumerate} \item If $R$ is analytically unramified, then $R$ is reduced. \item If $R$ is analytically unramified, then each minimal prime of $R$ is analytically unramified. \item If $R$ is reduced with minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, and each $\mathfrak q_i$ is analytically unramified, then $R$ is analytically unramified. \item If $R$ is analytically unramified, then the integral closure of $R$ in its total ring of fractions $Q(R)$ is finite over $R$. \item If $R$ is a domain and analytically unramified, then $R$ is N-1. \end{enumerate}


algebra-lemma-codimension-1-analytically-unramified

Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Assume \begin{enumerate} \item $R_{\mathfrak p}$ is a discrete valuation ring, and \item $\mathfrak p$ is analytically unramified. \end{enumerate} Then for any associated prime $\mathfrak q$ of $R^\wedge/\mathfrak pR^\wedge$ the local ring $(R^\wedge)_{\mathfrak q}$ is a discrete valuation ring.


algebra-lemma-criterion-analytically-unramified

Let $(R, \mathfrak m)$ be a Noetherian local domain. Let $x \in \mathfrak m$. Assume \begin{enumerate} \item $x \not = 0$, \item $R/xR$ has no embedded primes, and \item for each associated prime $\mathfrak p \subset R$ of $R/xR$ we have \begin{enumerate} \item the local ring $R_{\mathfrak p}$ is regular, and \item $\mathfrak p$ is analytically unramified. \end{enumerate} \end{enumerate} Then $R$ is analytically unramified.


algebra-lemma-local-nagata-domain-analytically-unramified

Let $(R, \mathfrak m)$ be a local ring. If $R$ is Noetherian, a domain, and Nagata, then $R$ is analytically unramified.


algebra-lemma-local-nagata-and-analytically-unramified

Let $(R, \mathfrak m)$ be a Noetherian local ring. The following are equivalent \begin{enumerate} \item $R$ is Nagata, \item for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$ maximal the local ring $S_{\mathfrak m'}$ is analytically unramified, \item for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite local homomorphism with $S$ a domain, then $S$ is analytically unramified. \end{enumerate}


algebra-lemma-nagata-pth-roots

Let $(A, \mathfrak m)$ be a Noetherian local domain which is Nagata and has fraction field of characteristic $p$. If $a \in A$ has a $p$th root in $A^\wedge$, then $a$ has a $p$th root in $A$.


algebra-lemma-apply-grothendieck-module

\begin{reference} \cite[IV, Proposition 6.3.1]{EGA} \end{reference} We have $$ \text{depth}(M \otimes_R N) = \text{depth}(M) + \text{depth}(N/\mathfrak m_RN) $$ where $R \to S$ is a local homomorphism of local Noetherian rings, $M$ is a finite $R$-module, and $N$ is a finite $S$-module flat over $R$.


algebra-lemma-apply-grothendieck

Suppose that $R \to S$ is a flat and local ring homomorphism of Noetherian local rings. Then $$ \text{depth}(S) = \text{depth}(R) + \text{depth}(S/\mathfrak m_RS). $$


algebra-lemma-CM-goes-up

Let $R \to S$ be a flat local homomorphism of local Noetherian rings. Then the following are equivalent \begin{enumerate} \item $S$ is Cohen-Macaulay, and \item $R$ and $S/\mathfrak m_RS$ are Cohen-Macaulay. \end{enumerate}


algebra-lemma-Sk-goes-up

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $R$ is Noetherian, \item $S$ is Noetherian, \item $\varphi$ is flat, \item the fibre rings $S \otimes_R \kappa(\mathfrak p)$ are $(S_k)$, and \item $R$ has property $(S_k)$. \end{enumerate} Then $S$ has property $(S_k)$.


algebra-lemma-Rk-goes-up

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $R$ is Noetherian, \item $S$ is Noetherian \item $\varphi$ is flat, \item the fibre rings $S \otimes_R \kappa(\mathfrak p)$ have property $(R_k)$, and \item $R$ has property $(R_k)$. \end{enumerate} Then $S$ has property $(R_k)$.


algebra-lemma-reduced-goes-up-noetherian

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $R$ is Noetherian, \item $S$ is Noetherian \item $\varphi$ is flat, \item the fibre rings $S \otimes_R \kappa(\mathfrak p)$ are reduced, \item $R$ is reduced. \end{enumerate} Then $S$ is reduced.


algebra-lemma-reduced-goes-up

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ is smooth, \item $R$ is reduced. \end{enumerate} Then $S$ is reduced.


algebra-lemma-normal-goes-up-noetherian

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $R$ is Noetherian, \item $S$ is Noetherian, \item $\varphi$ is flat, \item the fibre rings $S \otimes_R \kappa(\mathfrak p)$ are normal, and \item $R$ is normal. \end{enumerate} Then $S$ is normal.


algebra-lemma-normal-goes-up

Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ is smooth, \item $R$ is normal. \end{enumerate} Then $S$ is normal.


algebra-lemma-regular-goes-up

\begin{slogan} Regularity ascends along smooth maps of rings. \end{slogan} Let $\varphi : R \to S$ be a ring map. Assume \begin{enumerate} \item $\varphi$ is smooth, \item $R$ is a regular ring. \end{enumerate} Then $S$ is regular.


algebra-lemma-descent-Noetherian

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is Noetherian. \end{enumerate} Then $R$ is Noetherian.


algebra-lemma-descent-reduced

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is reduced. \end{enumerate} Then $R$ is reduced.


algebra-lemma-descent-normal

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is a normal ring. \end{enumerate} Then $R$ is a normal ring.


algebra-lemma-descent-regular

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is a regular ring. \end{enumerate} Then $R$ is a regular ring.


algebra-lemma-descent-Sk

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is Noetherian and has property $(S_k)$. \end{enumerate} Then $R$ is Noetherian and has property $(S_k)$.


algebra-lemma-descent-Rk

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is faithfully flat, and \item $S$ is Noetherian and has property $(R_k)$. \end{enumerate} Then $R$ is Noetherian and has property $(R_k)$.


algebra-lemma-descent-nagata

Let $R \to S$ be a ring map. Assume that \begin{enumerate} \item $R \to S$ is smooth and surjective on spectra, and \item $S$ is a Nagata ring. \end{enumerate} Then $R$ is a Nagata ring.


algebra-lemma-geometrically-normal

Let $k$ be a field. Let $A$ be a $k$-algebra. The following properties of $A$ are equivalent: \begin{enumerate} \item $k' \otimes_k A$ is a normal ring for every field extension $k'/k$, \item $k' \otimes_k A$ is a normal ring for every finitely generated field extension $k'/k$, \item $k' \otimes_k A$ is a normal ring for every finite purely inseparable extension $k'/k$, \item $k^{perf} \otimes_k A$ is a normal ring. \end{enumerate} Here normal ring is defined in Definition \ref{definition-ring-normal}.


algebra-lemma-localization-geometrically-normal-algebra

\begin{slogan} Localization preserves geometric normality. \end{slogan} Let $k$ be a field. A localization of a geometrically normal $k$-algebra is geometrically normal.


algebra-lemma-separable-field-extension-geometrically-normal

Let $k$ be a field. Let $K/k$ be a separable field extension. Then $K$ is geometrically normal over $k$.


algebra-lemma-geometrically-normal-tensor-normal

Let $k$ be a field. Let $A, B$ be $k$-algebras. Assume $A$ is geometrically normal over $k$ and $B$ is a normal ring. Then $A \otimes_k B$ is a normal ring.


algebra-lemma-geometrically-normal-over-separable-algebraic

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


algebra-lemma-geometrically-regular

Let $k$ be a field. Let $A$ be a $k$-algebra. Assume $A$ is Noetherian. The following properties of $A$ are equivalent: \begin{enumerate} \item $k' \otimes_k A$ is regular for every finitely generated field extension $k \subset k'$, and \item $k' \otimes_k A$ is regular for every finite purely inseparable extension $k \subset k'$. \end{enumerate} Here regular ring is as in Definition \ref{definition-regular}.


algebra-lemma-geometrically-regular-descent

\begin{slogan} Geometric regularity descends through faithfully flat maps of algebras \end{slogan} Let $k$ be a field. Let $A \to B$ be a faithfully flat $k$-algebra map. If $B$ is geometrically regular over $k$, so is $A$.


algebra-lemma-geometrically-regular-goes-up

Let $k$ be a field. Let $A \to B$ be a smooth ring map of $k$-algebras. If $A$ is geometrically regular over $k$, then $B$ is geometrically regular over $k$.


algebra-lemma-geometrically-regular-over-subfields

Let $k$ be a field. Let $A$ be an algebra over $k$. Let $k = \colim k_i$ be a directed colimit of subfields. If $A$ is geometrically regular over each $k_i$, then $A$ is geometrically regular over $k$.


algebra-lemma-geometrically-regular-over-separable-algebraic

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


algebra-lemma-tensor-fields-CM

Let $k$ be a field and let $k \subset K$ and $k \subset L$ be two field extensions such that one of them is a field extension of finite type. Then $K \otimes_k L$ is a Noetherian Cohen-Macaulay ring.


algebra-lemma-CM-geometrically-CM

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


algebra-lemma-flat-finite-presentation-limit-flat

Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume that \begin{enumerate} \item $R \to S$ is of finite presentation, \item $M$ is a finitely presented $S$-module, and \item $M$ is flat over $R$. \end{enumerate} In this case we have the following: \begin{enumerate} \item There exists a finite type $\mathbf{Z}$-algebra $R_0$ and a finite type ring map $R_0 \to S_0$ and a finite $S_0$-module $M_0$ such that $M_0$ is flat over $R_0$, together with a ring maps $R_0 \to R$ and $S_0 \to S$ and an $S_0$-module map $M_0 \to M$ such that $S \cong R \otimes_{R_0} S_0$ and $M = S \otimes_{S_0} M_0$. \item If $R = \colim_{\lambda \in \Lambda} R_\lambda$ is written as a directed colimit, then there exists a $\lambda$ and a ring map $R_\lambda \to S_\lambda$ of finite presentation, and an $S_\lambda$-module $M_\lambda$ of finite presentation such that $M_\lambda$ is flat over $R_\lambda$ and such that $S = R \otimes_{R_\lambda} S_\lambda$ and $M = S \otimes_{S_{\lambda}} M_\lambda$. \item If $$ (R \to S, M) = \colim_{\lambda \in \Lambda} (R_\lambda \to S_\lambda, M_\lambda) $$ is written as a directed colimit such that \begin{enumerate} \item $R_\mu \otimes_{R_\lambda} S_\lambda \to S_\mu$ and $S_\mu \otimes_{S_\lambda} M_\lambda \to M_\mu$ are isomorphisms for $\mu \geq \lambda$, \item $R_\lambda \to S_\lambda$ is of finite presentation, \item $M_\lambda$ is a finitely presented $S_\lambda$-module, \end{enumerate} then for all sufficiently large $\lambda$ the module $M_\lambda$ is flat over $R_\lambda$. \end{enumerate}


algebra-lemma-descend-faithfully-flat-finite-presentation

Let $R \to A \to B$ be ring maps. Assume $A \to B$ faithfully flat of finite presentation. Then there exists a commutative diagram $$ \xymatrix{ R \ar[r] \ar@{=}[d] & A_0 \ar[d] \ar[r] & B_0 \ar[d] \\ R \ar[r] & A \ar[r] & B } $$ with $R \to A_0$ of finite presentation, $A_0 \to B_0$ faithfully flat of finite presentation and $B = A \otimes_{A_0} B_0$.


algebra-lemma-colimit-finite

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is finite, \item $C_0$ is of finite type over $B_0$. \end{enumerate} Then there exists an $i \geq 0$ such that the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is finite.


algebra-lemma-colimit-surjective

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is surjective, \item $C_0$ is of finite type over $B_0$. \end{enumerate} Then for some $i \geq 0$ the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is surjective.


algebra-lemma-colimit-unramified

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is unramified, \item $C_0$ is of finite type over $B_0$. \end{enumerate} Then for some $i \geq 0$ the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is unramified.


algebra-lemma-colimit-isomorphism

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is an isomorphism, \item $B_0 \to C_0$ is of finite presentation. \end{enumerate} Then for some $i \geq 0$ the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is an isomorphism.


algebra-lemma-colimit-etale

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is \'etale, \item $B_0 \to C_0$ is of finite presentation. \end{enumerate} Then for some $i \geq 0$ the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is \'etale.


algebra-lemma-colimit-smooth

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is smooth, \item $B_0 \to C_0$ is of finite presentation. \end{enumerate} Then for some $i \geq 0$ the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is smooth.


algebra-lemma-colimit-lci

Let $A = \colim_{i \in I} A_i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi_0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume \begin{enumerate} \item $A \otimes_{A_0} B_0 \to A \otimes_{A_0} C_0$ is syntomic (resp.\ a relative global complete intersection), \item $C_0$ is of finite presentation over $B_0$. \end{enumerate} Then there exists an $i \geq 0$ such that the map $A_i \otimes_{A_0} B_0 \to A_i \otimes_{A_0} C_0$ is syntomic (resp.\ a relative global complete intersection).


algebra-lemma-fppf-fpqf

Let $R \to S$ be a faithfully flat ring map of finite presentation. Then there exists a commutative diagram $$ \xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] } $$ where $R \to S'$ is quasi-finite, faithfully flat and of finite presentation.


algebra-proposition-universal-property-localization

Let $f : A \to B$ be a ring map that sends every element in $S$ to a unit of $B$. Then there is a unique homomorphism $g : S^{-1}A \to B$ such that the following diagram commutes. $$ \xymatrix{ A \ar[rr]^{f} \ar[dr] & & B \\ & S^{-1}A \ar[ur]_g } $$


algebra-proposition-localize-twice

Let $\overline{S}$ be the image of $S$ in $S'^{-1}A$, then $(SS')^{-1}A$ is isomorphic to $\overline{S}^{-1}(S'^{-1}A)$.


algebra-proposition-localize-twice-module

View $S'^{-1}M$ as an $A$-module, then $S^{-1}(S'^{-1}M)$ is isomorphic to $(SS')^{-1}M$.


algebra-proposition-localization-exact

\begin{slogan} Localization is exact. \end{slogan} Let $L\xrightarrow{u} M\xrightarrow{v} N$ be an exact sequence of $R$-modules. Then $S^{-1}L \to S^{-1}M \to S^{-1}N$ is also exact.


algebra-proposition-localize-quotient

Let $I$ be an ideal of $A$, $S$ a multiplicative set of $A$. Then $S^{-1}I$ is an ideal of $S^{-1}A$ and $\overline{S}^{-1}(A/I)$ is isomorphic to $S^{-1}A/S^{-1}I$, where $\overline{S}$ is the image of $S$ in $A/I$.


algebra-proposition-oka

If $\mathcal{F}$ is an Oka family of ideals, then any maximal element of the complement of $\mathcal{F}$ is prime.


algebra-proposition-Jacobson-permanence

Let $R$ be a Jacobson ring. Let $R \to S$ be a ring map of finite type. Then \begin{enumerate} \item The ring $S$ is Jacobson. \item The map $\Spec(S) \to \Spec(R)$ transforms closed points to closed points. \item For $\mathfrak m' \subset S$ maximal lying over $\mathfrak m \subset R$ the field extension $\kappa(\mathfrak m')/\kappa(\mathfrak m)$ is finite. \end{enumerate}


algebra-proposition-going-down-normal-integral

Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal and $S$ integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q'$ be a prime of $S$ with $\mathfrak p' = R \cap \mathfrak q'$. Then there exists a prime $\mathfrak q$ with $\mathfrak q \subset \mathfrak q'$ such that $\mathfrak p = R \cap \mathfrak q$. In other words: the going down property holds for $R \to S$, see Definition \ref{definition-going-up-down}.


algebra-proposition-fppf-open

Let $R \to S$ be flat and of finite presentation. Then $\Spec(S) \to \Spec(R)$ is open. More generally this holds for any ring map $R \to S$ of finite presentation which satisfies going down.


algebra-proposition-graded-hilbert-polynomial

Suppose that $S$ is a Noetherian graded ring and $M$ a finite graded $S$-module. Consider the function $$ \mathbf{Z} \longrightarrow K'_0(S_0), \quad n \longmapsto [M_n] $$ see Lemma \ref{lemma-graded-module-fg}. If $S_{+}$ is generated by elements of degree $1$, then this function is a numerical polynomial.


algebra-proposition-hilbert-function-polynomial

Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Let $I \subset R$ be an ideal of definition. The Hilbert function $\varphi_{I, M}$ and the function $\chi_{I, M}$ are numerical polynomials.


algebra-proposition-dimension-zero-ring

Let $R$ be a ring. The following are equivalent: \begin{enumerate} \item $R$ is Artinian, \item $R$ is Noetherian and $\dim(R) = 0$, \item $R$ has finite length as a module over itself, \item $R$ is a finite product of Artinian local rings, \item $R$ is Noetherian and $\Spec(R)$ is a finite discrete topological space, \item $R$ is a finite product of Noetherian local rings of dimension $0$, \item $R$ is a finite product of Noetherian local rings $R_i$ with $d(R_i) = 0$, \item $R$ is a finite product of Noetherian local rings $R_i$ whose maximal ideals are nilpotent, \item $R$ is Noetherian, has finitely many maximal ideals and its Jacobson radical ideal is nilpotent, and \item $R$ is Noetherian and there are no strict inclusions among its primes. \end{enumerate}


algebra-proposition-dimension

Let $R$ be a local Noetherian ring. Let $d \geq 0$ be an integer. The following are equivalent: \begin{enumerate} \item \label{item-dim-d} $\dim(R) = d$, \item \label{item-d-d} $d(R) = d$, \item \label{item-ideal-d} there exists an ideal of definition generated by $d$ elements, and no ideal of definition is generated by fewer than $d$ elements. \end{enumerate}


algebra-proposition-minimal-primes-associated-primes

Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following sets of primes are the same: \begin{enumerate} \item The minimal primes in the support of $M$. \item The minimal primes in $\text{Ass}(M)$. \item For any filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_{n-1} \subset M_n = M$ with $M_i/M_{i-1} \cong R/\mathfrak p_i$ the minimal primes of the set $\{\mathfrak p_i\}$. \end{enumerate}


algebra-proposition-ffdescent-finite-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 finite projective, then $M$ is finite projective.


algebra-proposition-ML-characterization

Let $M$ be an $R$-module.  Let $(M_i, f_{ij})$ be a directed system of finitely presented $R$-modules, indexed by $I$, such that $M = \colim M_i$.  Let $f_i: M_i \to M$ be the canonical map.  The following are equivalent: \begin{enumerate} \item For every finitely presented $R$-module $P$ and module map $f: P \to M$, there exists a finitely presented $R$-module $Q$ and a module map $g: P \to Q$ such that $g$ and $f$ dominate each other, i.e., $\Ker(f \otimes_R \text{id}_N) = \Ker(g \otimes_R \text{id}_N)$ for every $R$-module $N$. \item For each $i \in I$, there exists $j \geq i$ such that $f_{ij}: M_i \to M_j$ dominates $f_i: M_i \to M$. \item For each $i \in I$, there exists $j \geq i$ such that $f_{ij}: M_i \to M_j$ factors through $f_{ik}: M_i \to M_k$ for all $k \geq i$. \item For every $R$-module $N$, the inverse system $(\Hom_R(M_i, N), \Hom_R(f_{ij}, N))$ is Mittag-Leffler. \item For $N = \prod_{s \in I} M_s$, the inverse system $(\Hom_R(M_i, N), \Hom_R(f_{ij}, N))$ is Mittag-Leffler. \end{enumerate}


algebra-proposition-fg-tensor

Let $M$ be an $R$-module.  The following are equivalent: \begin{enumerate} \item $M$ is finitely generated. \item For every family $(Q_{\alpha})_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes_R \left( \prod_{\alpha} Q_{\alpha} \right) \to \prod_{\alpha} (M \otimes_R Q_{\alpha})$ is surjective. \item For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes_R Q^{A} \to (M \otimes_R Q)^{A}$ is surjective. \item For every set $A$, the canonical map $M \otimes_R R^{A} \to M^{A}$ is surjective. \end{enumerate}


algebra-proposition-fp-tensor

Let $M$ be an $R$-module.  The following are equivalent: \begin{enumerate} \item $M$ is finitely presented. \item For every family $(Q_{\alpha})_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes_R \left( \prod_{\alpha} Q_{\alpha} \right) \to \prod_{\alpha} (M \otimes_R Q_{\alpha})$ is bijective. \item For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes_R Q^{A} \to (M \otimes_R Q)^{A}$ is bijective. \item For every set $A$, the canonical map $M \otimes_R R^{A} \to M^{A}$ is bijective. \end{enumerate}


algebra-proposition-ML-tensor

Let $M$ be an $R$-module.  The following are equivalent: \begin{enumerate} \item $M$ is Mittag-Leffler. \item For every family $(Q_{\alpha})_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes_R \left( \prod_{\alpha} Q_{\alpha} \right) \to \prod_{\alpha} (M \otimes_R Q_{\alpha})$ is injective. \end{enumerate}


algebra-proposition-characterize-coherent

\begin{reference} This is \cite[Theorem 2.1]{Chase}. \end{reference} Let $R$ be a ring. The following are equivalent \begin{enumerate} \item $R$ is coherent, \item any product of flat $R$-modules is flat, and \item for every set $A$ the module $R^A$ is flat. \end{enumerate}


algebra-proposition-what-exact

\begin{reference} \cite[Corollary 1]{WhatExact} \end{reference} In Situation \ref{situation-complex}, suppose $R$ is a local Noetherian ring. The following are equivalent \begin{enumerate} \item $0 \to R^{n_e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_e}, \ldots, R^{n_1}$, and \item for all $i$, $1 \leq i \leq e$ the following two conditions are satisfied: \begin{enumerate} \item $\text{rank}(\varphi_i) = r_i$ where $r_i = n_i - n_{i + 1} + \ldots + (-1)^{e-i-1} n_{e-1} + (-1)^{e-i} n_e$, \item $I(\varphi_i) = R$, or $I(\varphi_i)$ contains a regular sequence of length $i$. \end{enumerate} \end{enumerate}


algebra-proposition-CM-module

Let $R$ be a Noetherian local ring, with maximal ideal $\mathfrak m$. Let $M$ be a Cohen-Macaulay module over $R$ whose support has dimension $d$. Suppose that $g_1, \ldots, g_c$ are elements of $\mathfrak m$ such that $\dim(\text{Supp}(M/(g_1, \ldots, g_c)M)) = d - c$. Then $g_1, \ldots, g_c$ is an $M$-regular sequence, and can be extended to a maximal $M$-regular sequence.


algebra-proposition-regular-finite-gl-dim

Let $R$ be a regular local ring of dimension $d$. Every finite $R$-module $M$ of depth $e$ has a finite free resolution $$ 0 \to F_{d-e} \to \ldots \to F_0 \to M \to 0. $$ In particular a regular local ring has global dimension $\leq d$.


algebra-proposition-finite-gl-dim-regular

A Noetherian local ring whose residue field has finite projective dimension is a regular local ring. In particular a Noetherian local ring of finite global dimension is a regular local ring.


algebra-proposition-Auslander-Buchsbaum

Let $R$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module which has finite projective dimension $\text{pd}_R(M)$. Then we have $$ \text{depth}(R) = \text{pd}_R(M) + \text{depth}(M) $$


algebra-proposition-finite-gl-dim-polynomial-ring

A polynomial algebra in $n$ variables over a field is a regular ring. It has global dimension $n$. All localizations at maximal ideals are regular local rings of dimension $n$.


algebra-proposition-characterize-formally-smooth

Let $R \to S$ be a ring map. Consider a formally smooth $R$-algebra $P$ and a surjection $P \to S$ with kernel $J$. The following are equivalent \begin{enumerate} \item $S$ is formally smooth over $R$, \item for some $P \to S$ as above there exists a section to $P/J^2 \to S$, \item for all $P \to S$ as above there exists a section to $P/J^2 \to S$, \item for some $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega_{P/R} \otimes S \to \Omega_{S/R} \to 0$ is split exact, \item for all $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega_{P/R} \otimes S \to \Omega_{S/R} \to 0$ is split exact, and \item the naive cotangent complex $\NL_{S/R}$ is quasi-isomorphic to a projective $S$-module placed in degree $0$. \end{enumerate}


algebra-proposition-smooth-formally-smooth

Let $R \to S$ be a ring map. The following are equivalent \begin{enumerate} \item $R \to S$ is of finite presentation and formally smooth, \item $R \to S$ is smooth. \end{enumerate}


algebra-proposition-etale-locally-standard

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime. If $R \to S$ is \'etale at $\mathfrak q$, then there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is standard \'etale.


algebra-proposition-unramified-locally-standard

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime. If $R \to S$ is unramified at $\mathfrak q$, then there exist \begin{enumerate} \item a $g \in S$, $g \not \in \mathfrak q$, \item a standard \'etale ring map $R \to S'$, and \item a surjective $R$-algebra map $S' \to S_g$. \end{enumerate}


algebra-proposition-characterize-separable-field-extensions

Let $k \subset K$ be a field extension. If the characteristic of $k$ is zero then \begin{enumerate} \item $K$ is separable over $k$, \item $K$ is geometrically reduced over $k$, \item $K$ is formally smooth over $k$, \item $H_1(L_{K/k}) = 0$, and \item the map $K \otimes_k \Omega_{k/\mathbf{Z}} \to \Omega_{K/\mathbf{Z}}$ is injective. \end{enumerate} If the characteristic of $k$ is $p > 0$, then the following are equivalent: \begin{enumerate} \item $K$ is separable over $k$, \item the ring $K \otimes_k k^{1/p}$ is reduced, \item $K$ is geometrically reduced over $k$, \item the map $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K/\mathbf{F}_p}$ is injective, \item $H_1(L_{K/k}) = 0$, and \item $K$ is formally smooth over $k$. \end{enumerate}


algebra-proposition-nagata-universally-japanese

Let $R$ be a ring. The following are equivalent: \begin{enumerate} \item $R$ is a Nagata ring, \item any finite type $R$-algebra is Nagata, and \item $R$ is universally Japanese and Noetherian. \end{enumerate}


algebra-proposition-ubiquity-nagata

The following types of rings are Nagata and in particular universally Japanese: \begin{enumerate} \item fields, \item Noetherian complete local rings, \item $\mathbf{Z}$, \item Dedekind domains with fraction field of characteristic zero, \item finite type ring extensions of any of the above. \end{enumerate}


moduli-curves-theorem-stable-reduction

\begin{reference} \cite[Corollary 2.7]{DM} \end{reference} Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_C) = K$ and genus $g \geq 2$. Then \begin{enumerate} \item there exists an extension of discrete valuation rings $R \subset R'$ inducing a finite separable extension of fraction fields $K'/K$ and a stable family of curves $Y \to \Spec(R')$ of genus $g$ with $Y_{K'} \cong C_{K'}$ over $K'$, and \item there exists a finite separable extension $L/K$ and a stable family of curves $Y \to \Spec(A)$ of genus $g$ where $A \subset L$ is the integral closure of $R$ in $L$ such that $Y_L \cong C_L$ over $L$. \end{enumerate}


moduli-curves-lemma-polarized-curves-in-polarized

The morphism $\textit{PolarizedCurves} \to \Polarizedstack$ is an open and closed immersion.


moduli-curves-lemma-polarized-curves-over-curves

The morphism $\textit{PolarizedCurves} \to \Curvesstack$ is smooth and surjective.


moduli-curves-lemma-etale-locally-scheme

Let $X \to S$ be a family of curves. Then there exists an \'etale covering $\{S_i \to S\}$ such that $X_i = X \times_S S_i$ is a scheme. We may even assume $X_i$ is H-projective over $S_i$.


moduli-curves-lemma-curves-diagonal-separated-fp

The diagonal of $\Curvesstack$ is separated and of finite presentation.


moduli-curves-lemma-curves-qs-lfp

The morphism $\Curvesstack \to \Spec(\mathbf{Z})$ is quasi-separated and locally of finite presentation.


moduli-curves-lemma-DM-curves

There exist an open substack $\Curvesstack^{DM} \subset \Curvesstack$ with the following properties \begin{enumerate} \item $\Curvesstack^{DM} \subset \Curvesstack$ is the maximal open substack which is DM, \item given a family of curves $X \to S$ the following are equivalent \begin{enumerate} \item the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack^{DM}$, \item the group algebraic space $\mathit{Aut}_S(X)$ is unramified over $S$, \end{enumerate} \item given $X$ a proper scheme over a field $k$ of dimension $\leq 1$ the following are equivalent \begin{enumerate} \item the classifying morphism $\Spec(k) \to \Curvesstack$ factors through $\Curvesstack^{DM}$, \item $\mathit{Aut}(X)$ is geometrically reduced over $k$ and has dimension $0$, \item $\mathit{Aut}(X) \to \Spec(k)$ is unramified. \end{enumerate} \end{enumerate}


moduli-curves-lemma-in-DM-locus-vector-fields

Let $X$ be a proper scheme over a field $k$ of dimension $\leq 1$. Then properties (3)(a), (b), (c) are also equivalent to $\text{Der}_k(\mathcal{O}_X, \mathcal{O}_X) = 0$.


moduli-curves-lemma-CM-curves

There exist an open substack $\Curvesstack^{CM} \subset \Curvesstack$ such that \begin{enumerate} \item given a family of curves $X \to S$ the following are equivalent \begin{enumerate} \item the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack^{CM}$, \item the morphism $X \to S$ is Cohen-Macaulay, \end{enumerate} \item given a scheme $X$ proper over a field $k$ with $\dim(X) \leq 1$ the following are equivalent \begin{enumerate} \item the classifying morphism $\Spec(k) \to \Curvesstack$ factors through $\Curvesstack^{CM}$, \item $X$ is Cohen-Macaulay. \end{enumerate} \end{enumerate}


moduli-curves-lemma-CM-1-curves

There exist an open substack $\Curvesstack^{CM, 1} \subset \Curvesstack$ such that \begin{enumerate} \item given a family of curves $X \to S$ the following are equivalent \begin{enumerate} \item the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack^{CM, 1}$, \item the morphism $X \to S$ is Cohen-Macaulay and has relative dimension $1$ (Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-relative-dimension}), \end{enumerate} \item given a scheme $X$ proper over a field $k$ with $\dim(X) \leq 1$ the following are equivalent \begin{enumerate} \item the classifying morphism $\Spec(k) \to \Curvesstack$ factors through $\Curvesstack^{CM, 1}$, \item $X$ is Cohen-Macaulay and $X$ is equidimensional of dimension $1$. \end{enumerate} \end{enumerate}


moduli-curves-lemma-pre-genus-curves

There exist an open substack $\Curvesstack^{h0, 1} \subset \Curvesstack$ such that \begin{enumerate} \item given a family of curves $f : X \to S$ the following are equivalent \begin{enumerate} \item the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack^{h0, 1}$, \item $f_*\mathcal{O}_X = \mathcal{O}_S$, this holds after arbitrary base change, and the fibres of $f$ have dimension $1$, \end{enumerate} \item given a scheme $X$ proper over a field $k$ with $\dim(X) \leq 1$ the following are equivalent \begin{enumerate} \item the classifying morphism $\Spec(k) \to \Curvesstack$ factors through $\Curvesstack^{h0, 1}$, \item $H^0(X, \mathcal{O}_X) = k$ and $\dim(X) = 1$. \end{enumerate} \end{enumerate}


moduli-curves-lemma-genus

Let $f : X \to S$ be a family of curves such that $\kappa(s) = H^0(X_s, \mathcal{O}_{X_s})$ for all $s \in S$, i.e., the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack^{h0, 1}$ (Lemma \ref{lemma-pre-genus-curves}). Then \begin{enumerate} \item $f_*\mathcal{O}_X = \mathcal{O}_S$ and this holds universally, \item $R^1f_*\mathcal{O}_X$ is a finite locally free $\mathcal{O}_S$-module, \item for any morphism $h : S' \to S$ if $f' : X' \to S'$ is the base change, then $h^*(R^1f_*\mathcal{O}_X) = R^1f'_*\mathcal{O}_{X'}$. \end{enumerate}


moduli-curves-lemma-pre-genus-one-piece-per-genus

There is a decomposition into open and closed substacks $$ \Curvesstack^{h0, 1} = \coprod\nolimits_{g \geq 0} \Curvesstack_g $$ where each $\Curvesstack_g$ is characterized as follows: \begin{enumerate} \item given a family of curves $f : X \to S$ the following are equivalent \begin{enumerate} \item the classifying morphism $S \to \Curvesstack$ factors through $\Curvesstack_g$, \item $f_*\mathcal{O}_X = \mathcal{O}_S$, this holds after arbitrary base change, the fibres of $f$ have dimension $1$, and $R^1f_*\mathcal{O}_X$ is a locally free $\mathcal{O}_S$-module of rank $g$, \end{enumerate} \item given a scheme $X$ proper over a field $k$ with $\dim(X) \leq 1$ the following are equivalent \begin{enumerate} \item the classifying morphism $\Spec(k) \to \Curvesstack$ factors through $\Curvesstack_g$, \item $\dim(X) = 1$, $k = H^0(X, \mathcal{O}_X)$, and the genus of $X$ is $g$. \end{enumerate} \end{enumerate}