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Let $f:(A,\mathfrak{m}) \to (B,\mathfrak{n})$ be a local homomorphism of local rings (i.e. the image of the maximal ideal $\mathfrak{m}$ is contained in $\mathfrak{n}$). Suppose that

  • $B$ is a local regular ring, i.e. $\dim B =\dim_{B/\mathfrak{n}} \mathfrak{n}/\mathfrak{n}^2$
  • $f$ is surjective, and
  • $\mathfrak{m}/\mathfrak{m}^2 =\mathfrak{n}/\mathfrak{n}^2$.

Does it necessarily follow that $f$ is an isomorphism?

I tried considering a system of parameters $x_1, \ldots, x_d$ of $B$, and then passing to dimensions: $$\dim B/(x_1, \ldots, x_d)= \dim B -d =\dim_{B/\mathfrak{n}} \mathfrak{n}/\mathfrak{n}^2 -d = \dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 -d.$$ I have no idea if this argument yields something useful, but its the only thing that made sense considering the assumptions of the problem.

mathfan24
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1 Answers1

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It's harmless to pass to the completion to suppose $A$ and $B$ are complete, and are thus quotients of the same regular local ring $(S,\mathfrak{l})$. We may thus assume we have ideals $I \subseteq J$ in $S$ with $A=S/I$, $B=S/J$, and $f:A \to B$ is the natural surjection. We may furthermore suppose the presentation of $A$ is minimal, so $I \subseteq \mathfrak{l}^2$ and $\mu_S(\mathfrak{l}/\mathfrak{l}^2)=\mu_S(\mathfrak{m}/\mathfrak{m}^2)$. As $\mu_S(\mathfrak{m}/\mathfrak{m}^2)=\mu_S(\mathfrak{n}/\mathfrak{n}^2)$, it thus follows that $J \subseteq \mathfrak{l}^2$. But $S/J$ is regular, so it can only be that $I=J=0$, proving the claim.

metalspringpro
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  • Could you please explain what it means for $A$ to have "minimal presentation" and why it is safe to assume that? Also, I am not familiar with the notation $\mu_S$. – mathfan24 Feb 05 '23 at 22:48
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    @StillALittleChild If $(S,\mathfrak{l})$ is a regular local ring and $x \in \mathfrak{l}-\mathfrak{l}^2$, then $S/xS$ is still a regular local ring. So if we have an $R=S/I$ and $I$ contains such an $x$, we could write $R=(S/x)/(I/x)$ and present $R$ as the quotient of a smaller regular local ring. Minimal thus means $I \subseteq \mathfrak{l}^2$ so we cannot do this. – metalspringpro Feb 05 '23 at 22:53
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    The notation $\mu_S(M)$ is the standard notation for the minimal number of generators of an $S$-module $M$. Equivalently, $\mu_S(M)=\dim_{S/\mathfrak{l}}(M/\mathfrak{l}M)$. – metalspringpro Feb 05 '23 at 22:53
  • So the assumption that $A$ be of minimal presentation is allowed since the process you describe terminates at some point? – mathfan24 Feb 05 '23 at 22:58
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    @StillALittleChild Yes, each iteration drops the dimension of $S$ by $1$. This process will terminate when $\dim(S)=\mu_S(\mathfrak{m})$, where $\mathfrak{m}$ is the maximal ideal of $R$. – metalspringpro Feb 05 '23 at 23:03
  • Is passing to completions harmless because it is exact in this case? – mathfan24 Feb 05 '23 at 23:11
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    @StillALittleChild Faithfully exact, yes. All of your hypotheses pass to the completion and the conclusion descends from the completion. – metalspringpro Feb 05 '23 at 23:16
  • I'm sorry, I still don't see where the passage to completions was used. Also, in the last line of the argument, I don't understand the implication "But $S/J$ is regular, so it can only be that $I=J=0$." – mathfan24 Feb 06 '23 at 12:09
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    Not every local ring is the quotient of a regular local ring, but Cohen's Structure Theorem implies every complete local ring is. If $(S,\mathfrak{l})$ is a regular local ring, then $S/J$ is regular if and only if $J$ is generated by a subset of a regular system of parameters, i.e., part of an s.o.p $x_1,\dots,x_r$ contained in $\mathfrak{l}-\mathfrak{l}^2$ (see Proposition 2.2.4 in "Cohen-Macaulay Rings" by Bruns and Herzog). Thus $S/J$ cannot be regular if $J \ne 0$ and $J \subseteq \mathfrak{l}^2$. – metalspringpro Feb 06 '23 at 13:45
  • A side remark: I think that the same argument still works if $A$ was assumed regular instead of $B$. Am I correct? – mathfan24 Feb 23 '23 at 15:43
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    @StillALittleChild No, it does not. Indeed, every complete local ring is the quotient of a regular local ring. So for instance, take $A=k[![x,y]!]$ and $B=k[![x,y]!]/(x^2)$. – metalspringpro Feb 23 '23 at 19:41