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Consider a morphism surjective of rings $\varphi: R[T_1, \dots, T_m] \to R[T_1, \dots, T_n]/(f_1, \dots, f_k)$ such that $\varphi|_R = id_R$ (that is, a morphism of $R$-algebras). Is there an easy way to see that $\ker \varphi$ is finitely generated as an $R[T_1, \dots, T_m]$ module? I am using Bosch's book on algebraic geometry, and I was wondering if there is a less complicated proof than the one provided by the book. Can anyone shed some light?

user480840
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  • Is $\varphi$ any morphism between the two or is it a specific one (eg. the projection)? Also, are there any conditions on $R$? – Daniel Aug 16 '21 at 20:47
  • Oh, I forgot to mention that the morphism is surjective – user480840 Aug 16 '21 at 20:48
  • For context, would you mind adding where this comes from in Bosch? – Daniel Aug 16 '21 at 21:05
  • Lemma 8.3.2, I was paraphrasing the lemma in a simpler way. Maybe I did something wrong, but I am almost certain that the lemma is indeed false. – user480840 Aug 16 '21 at 21:07
  • @DanielApsley I think the point is that Bosch specifies that $\varphi$ is a morphism of $R$-algebras, not just a ring morphism, in the definition immediately preceeding the Lemma – Atticus Stonestrom Aug 16 '21 at 21:15
  • @AtticusStonestrom Right, and I think an important hypothesis here is that the quotient is finitely presented, as in Definition 1. – Daniel Aug 16 '21 at 21:20
  • @DanielApsley indeed! :) – Atticus Stonestrom Aug 16 '21 at 21:22
  • @user480840 No, the $\varphi$ in Atticus' counterexample is not a morphism of $R$-algebras. After all, there is only one $R-$algebra morphism $R \to R$. – Daniel Aug 16 '21 at 21:33
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    I think you're getting your diagrams mixed up. An algebra morphism should form a commutative triangle. If $\varphi$ commuted with Id$_R$, then $\varphi$ would be forced to be the identity on $R$! – Daniel Aug 16 '21 at 22:10

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As written, this will not be true in general. For example, suppose $m=n=k=0$, and let $R$ be the polynomial ring $F[x_k:k\in\mathbb{N}]$ in infinitely many variables over your favorite field $F$. Now consider the unique $F$-algebra morphism $\varphi:R\to R$ that sends each $x_{2k}$ to $0$ and each $x_{2k+1}$ to $x_k$. Then $\varphi$ is surjective (why?), but $\ker\varphi$ is equal to $\langle x_{2k}:k\in\mathbb{N}\rangle$, which is not finitely generated as an $R$-module.

Atticus Stonestrom
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    Well, I guess this means that the book was lying and hiding it behind an almost unreadable proof. Thanks. – user480840 Aug 16 '21 at 21:04
  • @user480840 happy it helped! :) but in fact, I believe Bosch's Lemma 8.3.2 is correct as written – in definition 8.3.1, note that $\Phi$ is taken to be a surjection of $R$-algebras... the map $\varphi$ that I give in my answer is a ring morphism, but it is not an $R$-algebra morphism (why not?) – Atticus Stonestrom Aug 16 '21 at 21:11
  • (in particular, the $R$-algebra structure on $R$ required to make my map $\varphi$ into an $R$-algebra morphism will not make $R$ into a finitely presented $R$-algebra, so the hypotheses of Bosch's lemma do not apply in this case.) – Atticus Stonestrom Aug 16 '21 at 21:20
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    Now I am really confused. No one pointed me to any proof (gotta remember the burden of proof), maybe Bosch's proof is correct, but it is borderline unreadable. I guess I have to let this ferment in my mind. I mean, it makes sense intuitively that the property of being finitely presented is independent of the morphism. – user480840 Aug 16 '21 at 21:31
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    dear @user480840, yes, I realize that this answer does not help with the question you set out to ask, for which I apologize :) ... unfortunately I do not know of any nicer proofs of this lemma, but hopefully someone will have some insight for you! (I will add however that the statement of this lemma is a good reminder to keep the distinction between $R$-algebra morphisms and ring morphisms close in mind when dealing with these kinds of results! there's definitely a certain amount of subtlety to it...) – Atticus Stonestrom Aug 16 '21 at 21:42
  • Thank you so much for your time and disposition, and I am sorry for my confrontational tone. – user480840 Aug 16 '21 at 21:53
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    (@user480840 oh truly no worries at all, I didn't find your tone confrontational in the slightest!! :) ) – Atticus Stonestrom Aug 16 '21 at 21:55
  • what is $m=n=k=0$? Your notation is rather ambiguous... – user480840 Aug 18 '21 at 17:06
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    dear @user480840, in your question statement, you write that $\varphi$ is a map from $R[T_1, \dots, T_m]$ to $R[T_1, \dots, T_n]/(f_1, \dots, f_k)$; by taking $m=n=k=0$, I am restricting attention to a map from $R$ to $R$. for a different example, if we had $m=n=k=1$, then we would be considering a map from $R[T_1]$ to $R[T_1]/(f_1)$ – Atticus Stonestrom Aug 18 '21 at 17:31
  • Ok, I see my mistake now. I wasn't interpreting things correctly, and I am very ashamed of my confusion. – user480840 Aug 27 '21 at 07:28
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    dear @user480840 you should absolutely not be ashamed of your confusion!!! it is very natural, and happens to all of us :) ... I am glad you have found clarity on the problem!! – Atticus Stonestrom Aug 27 '21 at 15:07