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Let $A$ be a regular Noetherian local ring and $p \in \mathrm{Spec}(A)$. This implies that $A_p$ is also regular, i.e., the vector space $p/p^2 \otimes k(p)$ has dimension equal to $\dim A_p$.

Question:

Assuming that $A/p$ is also regular, is it true that $p/p^2$ is a free $A/p$-module of rank equal to $\dim A_p$?

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  • $\begingroup$ The answer is yes. You can find a proof in any commutative algebra text, say Matsumura. $\endgroup$
    – Mohan
    Sep 26 '20 at 21:08
  • $\begingroup$ can you give an exact reference please? I am somewhat familiar with Matsumura, but have not found the result. $\endgroup$
    – Hammerhead
    Sep 26 '20 at 21:49
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    $\begingroup$ The result follows easily from Bruns and Herzog, Proposition 2.2.4. $\endgroup$
    – user26857
    Oct 16 '20 at 21:55
  • $\begingroup$ I see. It really just boils down to the fact that the minimal generator sets have the same cardinality for modules over local rings. $\endgroup$
    – Hammerhead
    Oct 17 '20 at 3:09
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Since the precise reference has not come through in the comments, let me try to answer my own question.

Let $m$ be the maximal ideal of $A$ and $k = A/m$. Since $A/p$ is regular we get that the $k$-dimension of $m/(m^2 + p)$ is equal to $\dim A - \dim A_p$.

We have the exact sequence of $k$-vector spaces: $$0 \to p/(p \cap m^2) \to m/m^2 \to m/(p+m^2) \to 0.$$ Hence the $k$-dimension of $p/(p \cap m^2)$ is equal to $\dim A_p$. Since $pm = p \cap m^2$ (indeed, if $m_1m_2 \in p$ and $m_1,m_2 \in m$ then $m_1m_2 \in pm$, giving $p \cap m^2 \subset pm$. The other inclusion is trivial.) we get that the $k$-dimension of $p/pm$ is equal to $\dim A_p$. In particular, due to Nakayama, one can find $r_1,\ldots, r_{\dim A_p} \in p$ that are $A$-generators $p$.

Since $ht(r_1,\ldots, r_{\dim A_p})=\dim A_p$ and $A$ is Cohen-Macaulay, by Matsumura Thm. 17.4.iii we get that $r_1,\ldots, r_{\dim A_p}$ forms a maximal regular sequence inside $p$.

Now by Matusumura Thm 16.2 we get that $p/p^2$ is a a free $A/p$-module of rank equal to $\dim A_p$, as desired.

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    $\begingroup$ The elements of $m^2$ are not of the form $m_1m_2$. $\endgroup$
    – user26857
    Oct 16 '20 at 21:25
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    $\begingroup$ As a matter of fact $pm=p\cap m^2$, but from other reasons. $\endgroup$
    – user26857
    Oct 16 '20 at 21:54
  • $\begingroup$ Thanks for catching this. Could you elaborate why $pm = p\cap m^2$ (by possibly changing the above argument)? I have no idea how to do it. $\endgroup$
    – Hammerhead
    Oct 17 '20 at 2:29
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    $\begingroup$ As we learn from Bruns and Herzog, the ideal $p$ is generated by a part of a regular system of parameters. These generators form an $A/m$-basis for $p/pm$, so the dimension of $p/pm$ is height of $p$. On the other side the dimension of the $A/m$-vectorspace $p/p\cap m^2$ equals the dimension of $(m^2+p)/m^2$ which is $\dim A-\dim A/p$, that is, the height of $p$. $\endgroup$
    – user26857
    Oct 17 '20 at 9:12
  • $\begingroup$ I see, so it really follows from what we would like to prove, and not vice versa. $\endgroup$
    – Hammerhead
    Oct 17 '20 at 17:26

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