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Suppose $S$ is a finitely presented $R$-algebra. If $g:R[x_1, \ldots, x_n] \to S$ is surjective, then $\ker(g)$ is finitely generated.

We can write $S$ as $R[y_1, \ldots, y_m]/(f_1,\ldots,f_t)$ and write $g$ as $g:R[x_1,\ldots,x_n] \to R[y_1, \ldots, y_m]/(f_1,\ldots,f_t)$, but I don't know how to find the kernel.

user26857
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afzd
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1 Answers1

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This is stated and proved at http://stacks.math.columbia.edu/tag/00R2.

Keenan Kidwell
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  • Hi, I am confused about the last step in the proof. It says that the kernel is the image of $(f_j, x_i - g_i)$ under $\psi$. It's obvious that the image of $(f_j, x_i - g_i)$ under $\psi$ is in the kernel, but why is the inverse direction correct ? – afzd May 26 '14 at 15:01
  • Oh, I know the reason now. Thanks very much. – afzd May 26 '14 at 15:19
  • It's unfortunate that these little bits get deleted, since I personally got stuck on the same thing: why is the kernel generated only by those elements? I'd appreciate a clear answer, left here for everyone that comes after me. Thank you in advance. – Filip Chindea Oct 19 '14 at 09:30