I am reading Atiyah-MacDonald's Introduction to Commutative Algebra. Theorem 5.10 states the following:
Let $A\subset B$ be rings, $B$ integral over $A$, and let $\mathfrak{p}$ be a prime ideal of $A$. Then there exists a prime ideal $\mathfrak{q}$ of $B$ such that $\mathfrak{q}\cap A=\mathfrak{p}$.
Then, the "Going-up" theorem is presented, which states the following:
Let $A \subset B$ be rings, $B$ integral over $A$; let $\mathfrak{p}_1 \subset \dotsm \subset \mathfrak{p}_n$ be a chain of prime ideals of $A$ and $\mathfrak{q}_1 \subset \dotsm \subset \mathfrak{q}_m$ $(m < n)$ a chain of prime ideals of $B$ such that $\mathfrak{q}_i \cap A = \mathfrak{p}_i$ ($1 \leq i \leq m$). Then the chain $\mathfrak{q}_1 \subset \dotsm \subset \mathfrak{q}_m$ can be extended to a chain $\mathfrak{q}_1 \subset \dotsm \subset \mathfrak{q}_n$ such that $\mathfrak{q}_i \cap A = \mathfrak{p}_i$ for $1 \leq i \leq n$.
Now; my problem is that I see the "Going-up" theorem as a trivial corollary of Theorem 5.10, and I think this must be because I am misunderstanding something. As I see it, if you have a prime ideal $\mathfrak{p}_i$ of $A$, then Theorem 5.10 grants the existance of a prime ideal $\mathfrak{q}_i$ of $B$ such that $\mathfrak{p}_i=A\cap \mathfrak{q}_i$. Moreover, doing this must preserve the inclusions, so you can easily find the chain of ideals $\mathfrak{q}_1 \subset \dotsm \subset \mathfrak{q}_n$. This is not, however, how Atiyah proves the result. Instead, he takes quotients of the form $A/\mathfrak{p_i}$ to find the ideal $\overline{\mathfrak{q}_{i+1}}$ of $B/\mathfrak{q}_i$ and then lift it back to $A$. I don't see why taking quotients is necessary.
From the mere fact that $q\cap A=p$ and $q'\cap A=p'$ and $p\subseteq p'$,
it does not logically follow that $q\subseteq q'$.