## 10.6 Ring maps of finite type and of finite presentation

Definition 10.6.1. Let $R \to S$ be a ring map.

1. We say $R \to S$ is of finite type, or that $S$ is a finite type $R$-algebra if there exist an $n \in \mathbf{N}$ and an surjection of $R$-algebras $R[x_1, \ldots , x_ n] \to S$.

2. We say $R \to S$ is of finite presentation if there exist integers $n, m \in \mathbf{N}$ and polynomials $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$ and an isomorphism of $R$-algebras $R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m) \cong S$.

Informally, $R \to S$ is of finite presentation if and only if $S$ is finitely generated as an $R$-algebra and the ideal of relations among the generators is finitely generated. A choice of a surjection $R[x_1, \ldots , x_ n] \to S$ as in the definition is sometimes called a presentation of $S$.

Lemma 10.6.2. The notions finite type and finite presentation have the following permanence properties.

1. A composition of ring maps of finite type is of finite type.

2. A composition of ring maps of finite presentation is of finite presentation.

3. Given $R \to S' \to S$ with $R \to S$ of finite type, then $S' \to S$ is of finite type.

4. Given $R \to S' \to S$, with $R \to S$ of finite presentation, and $R \to S'$ of finite type, then $S' \to S$ is of finite presentation.

Proof. We only prove the last assertion. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and $S' = R[y_1, \ldots , y_ a]/I$. Say that the class $\bar y_ i$ of $y_ i$ maps to $h_ i \bmod (f_1, \ldots , f_ m)$ in $S$. Then it is clear that $S = S'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m, h_1 - \bar y_1, \ldots , h_ a - \bar y_ a)$. $\square$

Lemma 10.6.3. Let $R \to S$ be a ring map of finite presentation. For any surjection $\alpha : R[x_1, \ldots , x_ n] \to S$ the kernel of $\alpha$ is a finitely generated ideal in $R[x_1, \ldots , x_ n]$.

Proof. Write $S = R[y_1, \ldots , y_ m]/(f_1, \ldots , f_ k)$. Choose $g_ i \in R[y_1, \ldots , y_ m]$ which are lifts of $\alpha (x_ i)$. Then we see that $S = R[x_ i, y_ j]/(f_ l, x_ i - g_ i)$. Choose $h_ j \in R[x_1, \ldots , x_ n]$ such that $\alpha (h_ j)$ corresponds to $y_ j \bmod (f_1, \ldots , f_ k)$. Consider the map $\psi : R[x_ i, y_ j] \to R[x_ i]$, $x_ i \mapsto x_ i$, $y_ j \mapsto h_ j$. Then the kernel of $\alpha$ is the image of $(f_ l, x_ i - g_ i)$ under $\psi$ and we win. $\square$

Lemma 10.6.4. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume $R \to S$ is of finite type and $M$ is finitely presented as an $R$-module. Then $M$ is finitely presented as an $S$-module.

Proof. This is similar to the proof of part (4) of Lemma 10.6.2. We may assume $S = R[x_1, \ldots , x_ n]/J$. Choose $y_1, \ldots , y_ m \in M$ which generate $M$ as an $R$-module and choose relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ which generate the kernel of $R^{\oplus m} \to M$. For any $i = 1, \ldots , n$ and $j = 1, \ldots , m$ write

$x_ i y_ j = \sum a_{ijk} y_ k$

for some $a_{ijk} \in R$. Consider the $S$-module $N$ generated by $y_1, \ldots , y_ m$ subject to the relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ and $x_ i y_ j = \sum a_{ijk} y_ k$, $i = 1, \ldots , n$ and $j = 1, \ldots , m$. Then $N$ has a presentation

$S^{\oplus nm + t} \longrightarrow S^{\oplus m} \longrightarrow N \longrightarrow 0$

By construction there is a surjective map $\varphi : N \to M$. To finish the proof we show $\varphi$ is injective. Suppose $z = \sum b_ j y_ j \in N$ for some $b_ j \in S$. We may think of $b_ j$ as a polynomial in $x_1, \ldots , x_ n$ with coefficients in $R$. By applying the relations of the form $x_ i y_ j = \sum a_{ijk} y_ k$ we can inductively lower the degree of the polynomials. Hence we see that $z = \sum c_ j y_ j$ for some $c_ j \in R$. Hence if $\varphi (z) = 0$ then the vector $(c_1, \ldots , c_ m)$ is an $R$-linear combination of the vectors $(a_{i1}, \ldots , a_{im})$ and we conclude that $z = 0$ as desired. $\square$

Comment #5686 by Albrecht on

In the proof to Lemma 10.6.2 it would be better to write $S=S′[x_1,\dots,x_n,y_1,\dots,y_a]/(f_1,…,f_m,h_1−y_1,\dots,h_a−y_a)$.

Comment #5688 by Albrecht on

Sorry, I was wrong!

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