We know the fact that if $R$ is a regular local ring then $R_{P}$ the localization of $R$ at $P$, $P\in\mathrm{Spec}(R)$ is a regular local ring.
So, I wonder the converse is true or not? My counter-example is taking $R=\mathbb Z$.
-If $P=(0)$ then $R_{P}$ is a field so $R_{P}$ is a regular local ring
-If $P=(p)$, $p$ is a prime number then $\dim R_{p}=1=v(PR_{P})$ (generate by $\frac{p}{1}$) so it is also a regular local ring.
However, $\mathbb Z$ is not a regular local ring. Can you check that to me please?
Let $I$ be a prime ideal of $R_P$, then $I^c$ is a prime ideal of $\mathbb{Z}$, so $I^c=q\mathbb{Z}$ for some prime number $q$.
We have $I=I^{ce}=(q\mathbb{Z})^e=(\frac{q}{1})$. However, $q\mathbb{Z}$ is maximal so $I$ is also maximal. So an arbitrary nonzero prime ideal of $R_P$ is maximal, which means $\dim R_P=1=v(PR_P)$