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If $S \subset R$ are commutative rings with $1$ and $R$ is an integral extension of $S$ then they have the same dimension. Basically the proof uses the going up theorem.

But I have a question about a part of the proof:

Let $P_{0} \subset P_{1} \subset P_{2}$ ... be an ascending chain of prime ideals of $S$. Then by the going up theorem, we can find $T_{i} \in Spec(R)$ such that $T_{i} \cap S = P_{i}$.

Question: how we know that also $T_{0} \subset T_{1} \subset ....$? I.e, why are inclusions preserved? All we know is that $T_{i} \cap S \subset T_{j} \cap S$, yes? Why do we have $T_{0} \subset T_{1} \subset...$ ?

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How about using the going-up theorem? Or is this theorem before the going-up? – Gobi Jun 2 '11 at 2:35
up vote 4 down vote accepted

You should argue inductively. Namely, first choose $T_0$ lying over $P_0$. Now consider the integral extension $S/P_0 \hookrightarrow R/T_0$. Apply the lying over theorem to the prime ideal $P_1/P_0$ of $S/P_0$, to obtain a prime ideal of $R/T_0$ lying over $P_1/P_0$; this prime ideal is of the form $T_1/T_0$, for some prime ideal $T_1$ of $R$ which lies over $P_1$, and contains $T_0$. Now continue in the same way, applying lying over to the prime ideal $P_2/P_1$ in the extension $S/P_1 \hookrightarrow R/T_1$, and so on.

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thanks! – user6495 Jun 2 '11 at 2:41
Can you please explain why we know that $T_{1}$ lies over $P_{1}$, we know that $T_{1}/T_{0} \cap S/P_{0} = P_{1}/P_{0}$, why from this it follows that $T_{1} \cap S = P_{1}$? – user6495 Jun 2 '11 at 20:52
@user6495: By construction the map $S/P_1 = (S/P_0)/(P_1/P_0) \to R/T_1 = (R/T_0)/(T_1/T_0)$ is injective. Regards, – Matt E Jun 2 '11 at 22:37

Prof Emerton has answered your question, but I wanted to add here (since I can't comment) that the result you want is often called "going up".

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thank you, yes, my mistake. – user6495 Jun 2 '11 at 2:38

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