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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?

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Your question is coming from a more general fact in commutative algebra: Let $S,R$ be commutative rings with $S \subseteq R$ and $R$ integral over $S$. Then $\dim R = \dim S$. The proof if I remember just comes from going up.

Exercise: Use the result I mention above to prove that for affine varieties $X,Y$, the dimension of the product variety $X \times Y$ is equal to the sum of the dimensions of $X$ and $Y$. This requires knowing the isomorphism $A(X\times Y) \cong A(X) \otimes_k A(Y)$ which can be deduced from the Yoneda Lemma.

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  • $\begingroup$ Awesome +1...i was actually missing this very important fact that you mention. $\endgroup$ – Manos Jul 9 '13 at 19:47
  • $\begingroup$ @Manos I have added an exercise. $\endgroup$ – user38268 Jul 9 '13 at 19:49
  • $\begingroup$ Thanks for the exercise; i need though to first establish $\dim R = \dim S$, then go to the varieties :) $\endgroup$ – Manos Jul 9 '13 at 19:53

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