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 $T_{0} \subset T_{1} \subset...$ ?