The Stacks project

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

@#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?

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"

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.

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".

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.