The Stacks project

Proof. It is clear that if $R \to S$ is of finite type then $R' \to S'$ is of finite type. Assume that $R' \to S'$ is of finite type. Say $y_1, \ldots , y_ m$ generate $S'$ over $R'$. Write $y_ j = \sum _ i a_{ij} \otimes x_{ji}$ for some $a_{ij} \in R'$ and $x_{ji} \in S$. Let $A \subset S$ be the $R$-subalgebra generated by the $x_{ij}$. By flatness we have $A' := R' \otimes _ R A \subset S'$, and by construction $y_ j \in A'$. Hence $A' = S'$. By faithful flatness $A = S$. $\square$

Proof. It is clear that if $R \to S$ is of finite presentation then $R' \to S'$ is of finite presentation. Assume that $R' \to S'$ is of finite presentation. By Lemma 10.126.1 we see that $R \to S$ is of finite type. Write $S = R[x_1, \ldots , x_ n]/I$. By flatness $S' = R'[x_1, \ldots , x_ n]/R'\otimes I$. Say $g_1, \ldots , g_ m$ generate $R'\otimes I$ over $R'[x_1, \ldots , x_ n]$. Write $g_ j = \sum _ i a_{ij} \otimes f_{ji}$ for some $a_{ij} \in R'$ and $f_{ji} \in I$. Let $J \subset I$ be the ideal generated by the $f_{ij}$. By flatness we have $R' \otimes _ R J \subset R'\otimes _ R I$, and both are ideals over $R'[x_1, \ldots , x_ n]$. By construction $g_ j \in R' \otimes _ R J$. Hence $R' \otimes _ R J = R'\otimes _ R I$. By faithful flatness $J = I$. $\square$

Proof. Proof of (1). Choose a short exact sequence $0 \to K' \to (R')^{\oplus n} \to M' \to 0$. Let $K \subset R^{\oplus n}$ be the inverse image of $K'$ under the map $R^{\oplus n} \to (R')^{\oplus n}$. Then $M = R^{\oplus n}/K$ works.

Proof of (2). Choose a presentation $(R')^{\oplus m} \to (R')^{\oplus n} \to M' \to 0$. Suppose that the first map is given by the matrix $A' = (a'_{ij})$ and the second map is determined by generators $x'_ i \in M'$, $i = 1, \ldots , n$. As $R' = S^{-1}(R/I)$ we can choose $s \in S$ and a matrix $A = (a_{ij})$ with coefficients in $R$ such that $a'_{ij} = a_{ij} / s \bmod S^{-1}I$. Let $M$ be the finitely presented $R$-module with presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$ where the first map is given by the matrix $A$ and the second map is determined by generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $M \to M'$, $x_ i \mapsto x'_ i$ induces an isomorphism $S^{-1}(M/IM) \cong M'$. $\square$

Proof. Proof of (1). Let $x_1, \ldots , x_ n \in M$ be elements which generate $S^{-1}M$ as an $S^{-1}R$-module. Let $M'$ be the $R$-submodule of $M$ generated by $x_1, \ldots , x_ n$.

Proof of (2). Let $x_1, \ldots , x_ n \in M$ be elements which generate $S^{-1}M$ as an $S^{-1}R$-module. Let $K = \mathop{\mathrm{Ker}}(R^{\oplus n} \to M)$ where the map is given by the rule $(a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i$. By Lemma 10.5.3 we see that $S^{-1}K$ is a finite $S^{-1}R$-module. By (1) we can find a finite submodule $K' \subset K$ with $S^{-1}K' = S^{-1}K$. Take $M' = \mathop{\mathrm{Coker}}(K' \to R^{\oplus n})$. $\square$

Proof. This is a special case of Lemma 10.126.4 $\square$

Proof. Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Let $a_ i \in R_\mathfrak p$ be an element mapping to the image of $x_ i$ in $S_\mathfrak q$. Write $a_ i = b_ i/f$ for some $f \in R$, $f \not\in \mathfrak p$. After replacing $R$ by $R_ f$ and $x_ i$ by $x_ i - a_ i$ we may assume that $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ such that $x_ i$ maps to zero in $S_\mathfrak q$. Then if $c_ j$ denotes the constant term of $g_ j$ we conclude that $c_ j$ maps to zero in $R_\mathfrak p$. After another replacement of $R$ we may assume that the constant coefficients $c_ j$ of the $g_ j$ are zero. Thus we obtain an $R$-algebra map $S \to R$, $x_ i \mapsto 0$ whose kernel is the ideal $(x_1, \ldots , x_ n)$.

We have the isomorphisms $R_\mathfrak p \to S_\mathfrak q \to R_\mathfrak p$ and $S \to R$ sends $x_ i$ to zero. Thus we must have $S_\mathfrak q = R_\mathfrak p[x_1, \ldots , x_ n]/(x_1, \ldots , x_ n)$ and a fortiori $S_\mathfrak q = S_\mathfrak p/(x_1, \ldots , x_ n)S_\mathfrak p$. This means that the finitely generated ideal $(x_1, \ldots , x_ n)S_\mathfrak p$ is pure in $S_\mathfrak p$, see Definition 10.108.1. Hence $(x_1, \ldots , x_ n)S_\mathfrak p$ is generated by an idempotent $e$ in $S_\mathfrak p$ by Lemma 10.108.5. After replacing $R \to S$ by $R_ f \to S_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we can find an idempotent $e' \in S$ mapping to $e$. Then $e'S$ and $(x_1, \ldots , x_ n)S$ are finitely generated ideals which become equal in $S_\mathfrak p$. Hence after replacing $R \to S$ by $R_ f \to S_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we may assume $e'S = (x_1, \ldots , x_ n)S$. Setting $C = e'S$ finishes the proof. $\square$

Proof. Let $\psi : S_{\mathfrak q} \to S'_{\mathfrak q'}$ be the isomorphism of the hypothesis of the lemma. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ r)$ and $S' = R[y_1, \ldots , y_ m]/J$. For each $i = 1, \ldots , n$ choose a fraction $h_ i/g_ i$ with $h_ i, g_ i \in R[y_1, \ldots , y_ m]$ and $g_ i \bmod J$ not in $\mathfrak q'$ which represents the image of $x_ i$ under $\psi $. After replacing $S'$ by $S'_{g_1 \ldots g_ n}$ and $R[y_1, \ldots , y_ m, y_{m + 1}]$ (mapping $y_{m + 1}$ to $1/(g_1\ldots g_ n)$) we may assume that $\psi (x_ i)$ is the image of some $h_ i \in R[y_1, \ldots , y_ m]$. Consider the elements $f_ j(h_1, \ldots , h_ n) \in R[y_1, \ldots , y_ m]$. Since $\psi $ kills each $f_ j$ we see that there exists a $g \in R[y_1, \ldots , y_ m]$, $g \bmod J \not\in \mathfrak q'$ such that $g f_ j(h_1, \ldots , h_ n) \in J$ for each $j = 1, \ldots , r$. After replacing $S'$ by $S'_ g$ and $R[y_1, \ldots , y_ m, y_{m + 1}]$ as before we may assume that $f_ j(h_1, \ldots , h_ n) \in J$. Thus we obtain a ring map $S \to S'$, $x_ i \mapsto h_ i$ which induces $\psi $ on local rings. By Lemma 10.6.2 the map $S \to S'$ is of finite presentation. By Lemma 10.126.6 we may assume that $S' = S \times C$. Thus localizing $S'$ at the idempotent corresponding to the factor $C$ we obtain the result. $\square$

Proof. Choose $s_1, \ldots , s_ n \in S$ whose images in $S/IS$ generate $S/IS$ as an algebra over $R/I$. By Lemma 10.20.1 part (11) we see that the $R$-algebra map $R[x_1, \ldots , x_ n \to S$, $x_ i \mapsto s_ i$ is surjective and we conclude. $\square$

Proof. Write $S' = R[x_1, \ldots , x_ m]/K$ for some ideal $K$. By assumption there exist $g_ j = x_ j + \sum \delta _{j, J} x^ J \in R[x_1, \ldots , x_ n]$ with $\delta _{j, J} \in I$ and with $g_ j \bmod K \in \mathop{\mathrm{Im}}(S \to S')$. Hence it suffices to show that $g_1, \ldots , g_ m$ generate $R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be a finitely generated $\mathbf{Z}$-subalgebra of $R$ containing at least the $\delta _{j, J}$. Then $R_0 \cap I$ is a nilpotent ideal (by Lemma 10.32.5). It follows that $R_0[x_1, \ldots , x_ n]$ is generated by $g_1, \ldots , g_ m$ (because $x_ j \mapsto g_ j$ defines an automorphism of $R_0[x_1, \ldots , x_ m]$; details omitted). Since $R$ is the union of the subrings $R_0$ we win. $\square$

Proof. Let $J = \mathop{\mathrm{Ker}}(S \to S')$. By Lemma 10.6.2 $J$ is a finitely generated ideal. Since $S'_\mathfrak q$ is flat over $R$ we see that $J_\mathfrak q/IJ_\mathfrak q \subset S_\mathfrak q/IS_{\mathfrak q}$ (apply Lemma 10.39.12 to $0 \to J \to S \to S' \to 0$). By assumption (2) we see that $J_\mathfrak q/IJ_\mathfrak q$ is zero. By Nakayama's lemma (Lemma 10.20.1) we see that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $J_ g = 0$. Hence $S_ g \cong S'_ g$ as desired. $\square$

Proof. By Lemma 10.126.9 the map $S \to S'$ is surjective. As $I$ is locally nilpotent, so are the ideals $IS$ and $IS'$ (Lemma 10.32.3). Hence every prime ideal $\mathfrak q$ of $S$ contains $IS$ and (trivially) $S_\mathfrak q/IS_\mathfrak q \cong S'_\mathfrak q/IS'_\mathfrak q$. Thus Lemma 10.126.10 applies and we see that $S_\mathfrak q \to S'_\mathfrak q$ is an isomorphism for every prime $\mathfrak q \subset S$. It follows that $S \to S'$ is injective for example by Lemma 10.23.1. $\square$