Localization of the module of Kahler differential of a regular ring

Question

Let $R$ be a regular local ring, and also an algebra of e.f.t over an algebraically closed field $k$ of characteristic zero. The module of Kahler differentials $\Omega_{R/k}$ is free of rank $d=\dim R$. Let $p$ be a prime ideal of height one: $\dim R_p =1$. The ring $R_p$ is regular, therefore the module of the Kahler differentials $\Omega_{R_p/k}$ is free has rank one $\operatorname{rk} \Omega_{R_p/k}=\dim R_p=1$. From the other hand, Kahler differentials commute with the localization $$ \Omega_{R_p/k} \cong (\Omega_{R/k})_p. $$ But $(\Omega_{R/k})_p$ is a free $R_p$-module of rank $d$. Where is my mistake in this sophism?

Answer 1

Consider $k\subset k(t)=R$. Then $R$ satisfies your eft hypothesis, regular and dimension is zero, but $\Omega^1_{R/k}$ is of rank one and not zero. So, be careful.