The Stacks project

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

Comments (6)

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 and . Then is of finite type if but is not of finite type if . 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 in (1) and the isomorphism in (2) both extends the given ring map . 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 -algebras. The -algebra structure on that this refers to comes from the given map . 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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