Let $k$ be a field and $R=k[y_1,\cdots,y_d]$ where $y_i$ are algebraically independent over $k$. Suppose that $S=k[x_1,\cdots,x_d]$ is a subring of $R$ such that $R$ is a finitely generated $S$-module. It is well known that the Krull dimension of $S$ is equal to the transcendence degree of $k(x_1,\cdots,x_d)$ over $k$.
Question: Is it true that $S$ and $R$ have the same Krull dimension? If yes, how can we see that?