Transitivity of the notion of regularity for local rings Let $(R,\mathfrak{m})$ be a Noetherian catenary local ring and $\mathfrak{p}$ a prime in $R$.

Question: Is it true that if $(R/\mathfrak{p}, \mathfrak{m})$ and $(R_\mathfrak{p}, \mathfrak{p})$ are both regular local rings then $(R,\mathfrak{m})$ is also regular?

As I am assuming $R$ to be catenary we have $$\dim R_\mathfrak{p} + \dim R/\mathfrak{p}=\dim R$$
Hence, it would be enough to prove that
$$\dim_{k(\mathfrak{p})} \mathfrak{p}/\mathfrak{p}^2 + \dim_{k(\mathfrak{m})} \mathfrak{m}/(\mathfrak{p}+\mathfrak{m}^2)=\dim_{k(\mathfrak{m})} \mathfrak{m}/\mathfrak{m}^2$$
I am interested in this because of its translation to algebraic geometry: If $X$ is an algebraic variety, $Y\subseteq X$ is an irreducible subvariety whose local ring is regular and $x\in Y$ is a regular point in $Y$, then $x$ is regular in $X$.
 A: No. Let $R=k[[x,y]]/(xy)$; then $R$ is a dimension $1$ reduced complete local ring.  In particular, the localization at any minimal prime is regular.  For example, such a minimal prime is $\mathfrak{p}=(x)$, and concretely, $R_{\mathfrak{p}} \cong \operatorname{Frac}(k[[y]])$.  Furthermore, $R/xR \cong k[[y]]$ is also regular, but $R$ is evidently singular.
Update: For a normal counterexample, let $R=k[[x,y,z]]/(xz-y^2)$, which is isomorphic to the second Veronese subring $k[[x^2,xy,y^2]]$ of $k[[x,y]]$.  In particular, $R$ is a normal domain and is singular. Let $\mathfrak{p}=(x,y) \in R$.  Then $R/\mathfrak{p} \cong k[[z]]$ which is regular, and, as $R$ is normal, it satisfies Serre's condition $R_1$, so automatically $R_{\mathfrak{p}}$ is regular.  However, we can also check this by hand: localizing at $(x,y)$ makes $z$ a unit, so the equation $xz-y^2$ becomes a linear form. A similar example can be taken by starting on the graded side and localizing at the irrelevant maximal ideal. 
