TITLE: Going up  theorem (basic question)
QUESTION [5 upvotes]: 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 $T_{0} \subset T_{1} \subset...$ ?

REPLY [5 votes]: 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".

REPLY [4 votes]: 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.