Lemma 10.126.1. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is of finite type if and only if $R' \to S'$ is of finite type.

## 10.126 Algebras and modules of finite presentation

In this section we discuss some standard results where the key feature is that the assumption involves a finite type or finite presentation assumption.

**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 $A = S$.
$\square$

Lemma 10.126.2. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is of finite presentation if and only if $R' \to S'$ is of finite presentation.

**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 $J = I$.
$\square$

Lemma 10.126.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \subset R$ be a multiplicative subset. Set $R' = S^{-1}(R/I) = S^{-1}R/S^{-1}I$.

For any finite $R'$-module $M'$ there exists a finite $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$.

For any finitely presented $R'$-module $M'$ there exists a finitely presented $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$.

**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 $S^{-1}(M/IM) \cong M'$. $\square$

Lemma 10.126.4. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module.

If $S^{-1}M$ is a finite $S^{-1}R$-module then there exists a finite $R$-module $M'$ and a map $M' \to M$ which induces an isomorphism $S^{-1}M' \to S^{-1}M$.

If $S^{-1}M$ is a finitely presented $S^{-1}R$-module then there exists an $R$-module $M'$ of finite presentation and a map $M' \to M$ which induces an isomorphism $S^{-1}M' \to S^{-1}M$.

**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 $M' = \mathop{\mathrm{Coker}}(K' \to R^{\oplus n})$. $\square$

Lemma 10.126.5. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $M$ be an $R$-module.

If $M_{\mathfrak p}$ is a finite $R_{\mathfrak p}$-module then there exists a finite $R$-module $M'$ and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$.

If $M_{\mathfrak p}$ is a finitely presented $R_{\mathfrak p}$-module then there exists an $R$-module $M'$ of finite presentation and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$.

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

Lemma 10.126.6. Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Assume

$S$ is of finite presentation over $R$,

$\varphi $ induces an isomorphism $R_\mathfrak p \cong S_\mathfrak q$.

Then there exist $f \in R$, $f \not\in \mathfrak p$ and an $R_ f$-algebra $C$ such that $S_ f \cong R_ f \times C$ as $R_ f$-algebras.

**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 $(x_1, \ldots , x_ n)$.

Note that $\mathfrak q = \mathfrak pS + (x_1, \ldots , x_ n)$. Write $g_ j = \sum a_{ji}x_ i + h.o.t.$. Since $S_\mathfrak q = R_\mathfrak p$ we have $\mathfrak p \otimes \kappa (\mathfrak p) = \mathfrak q \otimes \kappa (\mathfrak q)$. It follows that $m \times n$ matrix $A = (a_{ji})$ defines a surjective map $\kappa (\mathfrak p)^{\oplus m} \to \kappa (\mathfrak p)^{\oplus n}$. Thus after inverting some element of $R$ not in $\mathfrak p$ we may assume there are $b_{ij} \in R$ such that $\sum b_{ij} g_ j = x_ i + h.o.t.$. We conclude that $(x_1, \ldots , x_ n) = (x_1, \ldots , x_ n)^2$ in $S$. It follows from Lemma 10.21.5 that $(x_1, \ldots , x_ n)$ is generated by an idempotent $e$. Setting $C = eS$ finishes the proof. $\square$

Lemma 10.126.7. Let $R$ be a ring. Let $S$, $S'$ be of finite presentation over $R$. Let $\mathfrak q \subset S$ and $\mathfrak q' \subset S'$ be primes. If $S_{\mathfrak q} \cong S'_{\mathfrak q'}$ as $R$-algebras, then there exist $g \in S$, $g \not\in \mathfrak q$ and $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S_ g \cong S'_{g'}$ as $R$-algebras.

**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.
$\square$

Lemma 10.126.8. Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $S$ be an $R$-algebra such that $R/I \to S/IS$ is of finite type. Then $R \to S$ is of finite type.

**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.
$\square$

Lemma 10.126.9. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $S \to S'$ be an $R$-algebra map such that $S \to S'/IS'$ is surjective and such that $S'$ is of finite type over $R$. Then $S \to S'$ is surjective.

**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.
$\square$

Lemma 10.126.10. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Let $IS \subset \mathfrak q \subset S$ be a prime ideal. Assume that

$S \to S'$ is surjective,

$S_\mathfrak q/IS_\mathfrak q \to S'_\mathfrak q/IS'_\mathfrak q$ is an isomorphism,

$S$ is of finite type over $R$,

$S'$ of finite presentation over $R$, and

$S'_\mathfrak q$ is flat over $R$.

Then $S_ g \to S'_ g$ is an isomorphism for some $g \in S$, $g \not\in \mathfrak q$.

**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.
$\square$

Lemma 10.126.11. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Assume that

$I$ is locally nilpotent,

$S/IS \to S'/IS'$ is an isomorphism,

$S$ is of finite type over $R$,

$S'$ of finite presentation over $R$, and

$S'$ is flat over $R$.

Then $S \to S'$ is an isomorphism.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)