Let $R$ be a Noetherian ring, free as $\mathbb{Z}$-module. Suppose $R \otimes K$ is Cohen-Macaulay for some field $K$ of characteristic $p > 0$. Show that $R \otimes L$ is Cohen-Macaulay for every field $L$ of characteristic $0$.
At first I thought I can draw some map from $L$ to $K$ so that I can reduce the case to $K$ but after some simple calculation I figure out that it is only the trivial map, which is of no use. Since $R \otimes K$ is Cohen-Macaulay, the depth and dimension of $R \otimes K$ are same. Since $R$ is Noetherian, I think by Hilbert's Basis Theorem we can say that $R \otimes K$ and $R \otimes L$ are also Noetherian. But I am stuck here. Any help will be appreciated.