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 $\dim A_p$?

 A: 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.
