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My friend and I were working through old preliminary exams. This problem is the fifth one on this list, but we were unable to come up with a solution.

Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi:A\to B$ is a map such that $\phi(\mathfrak{m})\subset\mathfrak{n}$, and suppose that

  1. $A/\mathfrak{m}\to B/\mathfrak{n}$ is an isomorphism.
  2. $\mathfrak{m}\to\mathfrak{n}/\mathfrak{n}^2$ is surjective.
  3. $B$ is finitely generated as an $A$-module.

Prove that $\phi$ is surjective.

Does anyone have an idea on how to proceed, or possibly a solution to this?

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Samuel Cavazos
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1 Answers1

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This follows from Nakayama's lemma. By (1) it is enough to show that $\mathfrak n\subset\operatorname{im}(\phi)$:

First off, notice that by (3) and the hypothesis that $A$ is Noetherian, $\mathfrak n$ is a finitely generated $A$-module; a fortiori, it is a finitely generated ideal of $B$. Pick any finite set $S\subset\mathfrak n$ such that its projection $\bar S$ is a basis of $\mathfrak n/\mathfrak n^2$ over $B/\mathfrak n$. View it as a basis over $A/\mathfrak m$ using (1). Then, by Nakayama's lemma, $S$ is a generating set for $\mathfrak n$ over $A$. By (2), we can pick $x_1,\dots,x_r\in A$ such that $\overline{\phi(x_1)},\dots,\overline{\phi(x_r)}$ generates $\mathfrak n/\mathfrak n^2$. Setting $S=\{\phi(x_1),\dots,\phi(x_r)\}$, we see that $\phi$ surjects onto $\mathfrak n$.

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lol wut
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  • $B$ is also a noetherian ring, so $\mathfrak n$ is finitely generated. – user26857 May 30 '14 at 08:32
  • @user26857 The explicit assumption that $B$ is Noetherian ring is extraneous, so I decided to not use it. It is more important that $\mathfrak n$ is a finitely-generated $A$-module anyway. – lol wut May 30 '14 at 08:34
  • Yeah, it follows from 3. if $A$ is supposed noetherian. – user26857 May 30 '14 at 08:37
  • @user26857 Notice that I said exactly that... – lol wut May 30 '14 at 08:43
  • Yeah, however not explicitly. – user26857 May 30 '14 at 08:46
  • Ah I like this one. I wasn't sure how to use property (2) or how it came up. Thanks for your help. – Samuel Cavazos May 31 '14 at 01:17
  • @SamuelCavazos Yeah, it is not the easiest problem. I was initially confused by condition (2) as well, and at first I wanted to prove that a corresponding map on the completions was surjective. Then, I finally realized Nakayama's lemma would do the trick--it's funny how Nakayama's lemma is never the first thing I think of. :) – lol wut May 31 '14 at 11:08
  • To get a generator of $\mathfrak{n}$ as an $A$-module, don't we need to lift generators of $\mathfrak{n}/\mathfrak{m}\mathfrak{n}$, not $\mathfrak{n}/\mathfrak{n}^2$? I don't see how to do this... – nrkm May 08 '20 at 22:06