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. $\square$

Comment #1099 by Filip Chindea on

It is obvious that $(f_j, x_i - g_i) \subset \ker(\alpha)$, but I don't know how to prove the reverse inclusion and I would be grateful for a hint. The proof in EGA is quite involved, and I would have liked to see an elementary alternative. Thank you.

Comment #1132 by on

For any ring $A$ and element $a \in A$ we have $A[x]/(x - a) = A$. Does this help? Or am I misunderstanding the question?

Comment #1158 by Filip Chindea on

Sorry for that, it was obvious: just consider the projection to $R[x_i, y_j]/(f_j, x_i - g_i)$ and see what happens after composition. Still the proof is amazingly short.

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

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