
## 15.73 Relatively finitely presented modules

Let $R$ be a ring. Let $A \to B$ be a finite map of finite type $R$-algebras. Let $M$ be a finite $B$-module. In this case it is not true that

$M\text{ of finite presentation over }B \Leftrightarrow M\text{ of finite presentation over }A$

A counter example is $R = k[x_1, x_2, x_3, \ldots ]$, $A = R$, $B = R/(x_ i)$, and $M = B$. To “fix” this we introduce a relative notion of finite presentation.

Lemma 15.73.1. Let $R \to A$ be a ring map of finite type. Let $M$ be an $A$-module. The following are equivalent

1. for some presentation $\alpha : R[x_1, \ldots , x_ n] \to A$ the module $M$ is a finitely presented $R[x_1, \ldots , x_ n]$-module,

2. for all presentations $\alpha : R[x_1, \ldots , x_ n] \to A$ the module $M$ is a finitely presented $R[x_1, \ldots , x_ n]$-module, and

3. for any surjection $A' \to A$ where $A'$ is a finitely presented $R$-algebra, the module $M$ is finitely presented as $A'$-module.

In this case $M$ is a finitely presented $A$-module.

Proof. If $\alpha : R[x_1, \ldots , x_ n] \to A$ and $\beta : R[y_1, \ldots , y_ m] \to A$ are presentations. Choose $f_ j \in R[x_1, \ldots , x_ n]$ with $\alpha (f_ j) = \beta (y_ j)$ and $g_ i \in R[y_1, \ldots , y_ m]$ with $\beta (g_ i) = \alpha (x_ i)$. Then we get a commutative diagram

$\xymatrix{ R[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \ar[d]^{x_ i \mapsto g_ i} \ar[rr]_-{y_ j \mapsto f_ j} & & R[x_1, \ldots , x_ n] \ar[d] \\ R[y_1, \ldots , y_ m] \ar[rr] & & A }$

Hence the equivalence of (1) and (2) follows by applying Algebra, Lemmas 10.6.4 and 10.35.23. The equivalence of (2) and (3) follows by choosing a presentation $A' = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and using Algebra, Lemma 10.35.23 to show that $M$ is finitely presented as $A'$-module if and only if $M$ is finitely presented as a $R[x_1, \ldots , x_ n]$-module. $\square$

Definition 15.73.2. Let $R \to A$ be a finite type ring map. Let $M$ be an $A$-module. We say $M$ is an $A$-module finitely presented relative to $R$ if the equivalent conditions of Lemma 15.73.1 hold.

Note that if $R \to A$ is of finite presentation, then $M$ is an $A$-module finitely presented relative to $R$ if and only if $M$ is a finitely presented $A$-module. It is equally clear that $A$ as an $A$-module is finitely presented relative to $R$ if and only if $A$ is of finite presentation over $R$. If $R$ is Noetherian the notion is uninteresting. Now we can formulate the result we were looking for.

Lemma 15.73.3. Let $R$ be a ring. Let $A \to B$ be a finite map of finite type $R$-algebras. Let $M$ be a $B$-module. Then $M$ is an $A$-module finitely presented relative to $R$ if and only if $M$ is a $B$-module finitely presented relative to $R$.

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose $y_1, \ldots , y_ m \in B$ which generate $B$ over $A$. As $A \to B$ is finite each $y_ i$ satisfies a monic equation with coefficients in $A$. Hence we can find monic polynomials $P_ j(T) \in R[x_1, \ldots , x_ n][T]$ such that $P_ j(y_ j) = 0$ in $B$. Then we get a commutative diagram

$\xymatrix{ R[x_1, \ldots , x_ n] \ar[d] \ar[r] & R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/(P_ j(y_ j)) \ar[d] \\ A \ar[r] & B }$

Since the top arrow is a finite and finitely presented ring map we conclude by Algebra, Lemma 10.35.23 and the definition. $\square$

With this result in hand we see that the relative notion makes sense and behaves well with regards to finite maps of rings of finite type over $R$. It is also stable under localization, stable under base change, and "glues" well.

Lemma 15.73.4. Let $R$ be a ring, $f \in R$ an element, $R_ f \to A$ is a finite type ring map, $g \in A$, and $M$ an $A$-module. If $M$ of finite presentation relative to $R_ f$, then $M_ g$ is an $A_ g$-module of finite presentation relative to $R$.

Proof. Choose a presentation $R_ f[x_1, \ldots , x_ n] \to A$. We write $R_ f = R[x_0]/(fx_0 - 1)$. Consider the presentation $R[x_0, x_1, \ldots , x_ n, x_{n + 1}] \to A_ g$ which extends the given map, maps $x_0$ to the image of $1/f$, and maps $x_{n + 1}$ to $1/g$. Choose $g' \in R[x_0, x_1, \ldots , x_ n]$ which maps to $g$ (this is possible). Suppose that

$R_ f[x_1, \ldots , x_ n]^{\oplus s} \to R_ f[x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

is a presentation of $M$ given by a matrix $(h_{ij})$. Pick $h'_{ij} \in R[x_0, x_1, \ldots , x_ n]$ which map to $h_{ij}$. Then

$R[x_0, x_1, \ldots , x_ n, x_{n + 1}]^{\oplus s + 2t} \to R[x_0, x_1, \ldots , x_ n, x_{n + 1}]^{\oplus t} \to M_ g \to 0$

is a presentation of $M_ f$. Here the $t \times (s + 2t)$ matrix defining the map has a first $t \times s$ block consisting of the matrix $h'_{ij}$, a second $t \times t$ block which is $(x_0f - )I_ t$, and a third block which is $(x_{n + 1}g' - 1)I_ t$. $\square$

Lemma 15.73.5. Let $R \to A$ be a finite type ring map. Let $M$ be an $A$-module finitely presented relative to $R$. For any ring map $R \to R'$ the $A \otimes _ R R'$-module

$M \otimes _ A A' = M \otimes _ R R'$

is finitely presented relative to $R'$.

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose a presentation

$R[x_1, \ldots , x_ n]^{\oplus s} \to R[x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

Then

$R'[x_1, \ldots , x_ n]^{\oplus s} \to R'[x_1, \ldots , x_ n]^{\oplus t} \to M \otimes _ R R' \to 0$

is a presentation of the base change and we win. $\square$

Lemma 15.73.6. Let $R \to A$ be a finite type ring map. Let $M$ be an $A$-module finitely presented relative to $R$. Let $A \to A'$ be a ring map of finite presentation. The $A'$-module $M \otimes _ A A'$ is finitely presented relative to $R$.

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose a presentation $A' = A[y_1, \ldots , y_ m]/(g_1, \ldots , g_ l)$. Pick $g'_ i \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ mapping to $g_ i$. Say

$R[x_1, \ldots , x_ n]^{\oplus s} \to R[x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

is a presentation of $M$ given by a matrix $(h_{ij})$. Then

$R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]^{\oplus s + tl} \to R[x_0, x_1, \ldots , x_ n, y_1, \ldots , y_ m]^{\oplus t} \to M \otimes _ A A' \to 0$

is a presentation of $M \otimes _ A A'$. Here the $t \times (s + lt)$ matrix defining the map has a first $t \times s$ block consisting of the matrix $h_{ij}$, followed by $l$ blocks of size $t \times t$ which are $g'_ iI_ t$. $\square$

Lemma 15.73.7. Let $R \to A \to B$ be finite type ring maps. Let $M$ be a $B$-module. If $M$ is finitely presented relative to $A$ and $A$ is of finite presentation over $R$, then $M$ is finitely presented relative to $R$.

Proof. Choose a surjection $A[x_1, \ldots , x_ n] \to B$. Choose a presentation

$A[x_1, \ldots , x_ n]^{\oplus s} \to A[x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

given by a matrix $(h_{ij})$. Choose a presentation

$A = R[y_1, \ldots , y_ m]/(g_1, \ldots , g_ u).$

Choose $h'_{ij} \in R[y_1, \ldots , y_ m, x_1, \ldots , x_ n]$ mapping to $h_{ij}$. Then we obtain the presentation

$R[y_1, \ldots , y_ m, x_1, \ldots , x_ n]^{\oplus s + tu} \to R[y_1, \ldots , y_ m, x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

where the $t \times (s + tu)$-matrix is given by a first $t \times s$ block consisting of $h'_{ij}$ followed by $u$ blocks of size $t \times t$ given by $g_ iI_ t$, $i = 1, \ldots , u$. $\square$

Lemma 15.73.8. Let $R \to A$ be a finite type ring map. Let $M$ be an $A$-module. Let $f_1, \ldots , f_ r \in A$ generate the unit ideal. The following are equivalent

1. each $M_{f_ i}$ is finitely presented relative to $R$, and

2. $M$ is finitely presented relative to $R$.

Proof. The implication (2) $\Rightarrow$ (1) is in Lemma 15.73.4. Assume (1). Write $1 = \sum f_ ig_ i$ in $A$. Choose a surjection $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r] \to A$. such that $y_ i$ maps to $f_ i$ and $z_ i$ maps to $g_ i$. Then we see that there exists a surjection

$P = R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r]/(\sum y_ iz_ i - 1) \longrightarrow A.$

By Lemma 15.73.1 we see that $M_{f_ i}$ is a finitely presented $A_{f_ i}$-module, hence by Algebra, Lemma 10.22.2 we see that $M$ is a finitely presented $A$-module. Hence $M$ is a finite $P$-module (with $P$ as above). Choose a surjection $P^{\oplus t} \to M$. We have to show that the kernel $K$ of this map is a finite $P$-module. Since $P_{y_ i}$ surjects onto $A_{f_ i}$ we see by Lemma 15.73.1 and Algebra, Lemma 10.5.3 that the localization $K_{y_ i}$ is a finitely generated $P_{y_ i}$-module. Choose elements $k_{i, j} \in K$, $i = 1, \ldots , r$, $j = 1, \ldots , s_ i$ such that the images of $k_{i, j}$ in $K_{y_ i}$ generate. Set $K' \subset K$ equal to the $P$-module generated by the elements $k_{i, j}$. Then $K/K'$ is a module whose localization at $y_ i$ is zero for all $i$. Since $(y_1, \ldots , y_ r) = P$ we see that $K/K' = 0$ as desired. $\square$

Lemma 15.73.9. Let $R \to A$ be a finite type ring map. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $A$-modules.

1. If $M', M''$ are finitely presented relative to $R$, then so is $M$.

2. If $M'$ is a finite type $A$-module and $M$ is finitely presented relative to $R$, then $M''$ is finitely presented relative to $R$.

Proof. Follows immediately from Algebra, Lemma 10.5.3. $\square$

Lemma 15.73.10. Let $R \to A$ be a finite type ring map. Let $M, M'$ be $A$-modules. If $M \oplus M'$ is finitely presented relative to $R$, then so are $M$ and $M'$.

Proof. Omitted. $\square$

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).