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 $S = S'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m, h_1 - \bar y_1, \ldots , h_ a - \bar y_ a)$. $\square$

## Comments (3)

Comment #584 by Wei Xu on

A in last sentence: "Then it is clear that $S' = S[x_1, \ldots, x_n]/(f_1, \ldots, f_m, h_1 - \bar y_1, \ldots, h_m - \bar y_m)$.", $S'$ should be $S$ and $S$ should be $S'$.

Comment #586 by Anfang on

There is also an index error. It should be $S = S'[x_1, \ldots, x_n]/(f_1, \ldots, f_m, h_1 - \bar y_1, \ldots, h_a - \bar y_a)$.

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