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 $R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m) \cong S$.

Comment #6759 by Reginald Anderson on

How does this depend on the map at all? It seems to be completely independent of the map, and only depend on the rings R and S

Comment #6760 by on

@#6759: Why do you say so? For example, suppose that $R = \mathbf{Z}[x_1, x_2, x_3, \ldots]$ and $S = \mathbf{Z}[y_1, y_2, y_3, \ldots]$. Then $R \to S$ is of finite type if $x_i \mapsto y_i$ but $R \to S$ is not of finite type if $x_i \mapsto y_{2i}$. OK?

Comment #8115 by Reginald Anderson on

The way this page is currently written reads is "a ring map R -> S is if finite type/presentation..." "...if certain properties of R and S hold, independent of the map given"

Comment #8223 by on

Dear Reginald Anderson, this is indeed a common practice in mathematics. Please only comment on this when you see a case of it where you are genuinly confused.

Comment #8275 by Luis Turcio on

Maybe what causes confution is that the surjection $R[x_1,\ldots,x_n]\to S$ in (1) and the isomorphism $R[x_1,\ldots,x_n]/(f_1,\ldots,f_m)\cong S$ in (2) both extends the given ring map $R\to S$. Am I right? If the above is the case I agree with Stacks Project when it says " this is indeed a common practice in mathematics".

Comment #8276 by Anon on

In both (1) and (2) the condition is that something is true as $R$-algebras. The $R$-algebra structure on $S$ that this refers to comes from the given map $R \to S$. So the conditions do refer to the given ring map.

There are also:

• 2 comment(s) on Section 10.6: Ring maps of finite type and of finite presentation

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