# Question about dimension of regular local rings

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 $$\dim A_p$$?

• The answer is yes. You can find a proof in any commutative algebra text, say Matsumura. Sep 26 '20 at 21:08
• can you give an exact reference please? I am somewhat familiar with Matsumura, but have not found the result. Sep 26 '20 at 21:49
• The result follows easily from Bruns and Herzog, Proposition 2.2.4. Oct 16 '20 at 21:55
• I see. It really just boils down to the fact that the minimal generator sets have the same cardinality for modules over local rings. Oct 17 '20 at 3:09

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.

• The elements of $m^2$ are not of the form $m_1m_2$. Oct 16 '20 at 21:25
• As a matter of fact $pm=p\cap m^2$, but from other reasons. Oct 16 '20 at 21:54
• 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. Oct 17 '20 at 2:29
• 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 $p$. Oct 17 '20 at 9:12
• I see, so it really follows from what we would like to prove, and not vice versa. Oct 17 '20 at 17:26