I was reading Going Down Theorem from Atiyah-MacDonald. There is a theorem in the Rings and Modules of Fractions chapter
Theorem 3.16: Let $A\to B$ be a ring homomorphism and let $\mathfrak{p}$ be a prime ideal of $A$. Then $\mathfrak{p}$ is the contraction of a prime ideal $B$ if and only if $\mathfrak{p}^{ec}=\mathfrak{p}$
Now in the prove of going down theorem we have $\mathfrak{p}_1\supseteq \mathfrak{p}_2$ prime ideals in $A$ and a prime ideal $\mathfrak{q}_1$ in $B$ such that $\mathfrak{p}_1=\mathfrak{q}_1\cap A$. We want to find a prime ideal $\mathfrak{q}_2\subseteq \mathfrak{q}_1$ in $B$ such that $\mathfrak{p}_2=\mathfrak{q}_2\cap A$.
They want to show that $\mathfrak{p}_2$ is the contraction of a prime ideal in the ring $B_{\mathfrak{q}_1}$ or by the Theorem 3.16 $$B_{\mathfrak{q}_1}\mathfrak{p}_2\cap A=\mathfrak{p}_2$$
Following the Theorem 3.16 it means that the extension of $\mathfrak{p}_2$ in $B_{\mathfrak{q_1}}$ is $B_{\mathfrak{q}_1}\mathfrak{p}_2$. But i cant understand why that will be the case. Extension of $\mathfrak{p}_2$ in $B$ is $B\mathfrak{p}_2$. Then the prime ideal corresponding to $B\mathfrak{p}_2$ in $B_{\mathfrak{q}_1}$ is $(B\mathfrak{p}_2)_{\mathfrak{q}_1}$. Then why $$(B\mathfrak{p}_2)_{\mathfrak{q}_1}=B_{\mathfrak{q}_1}\mathfrak{p}_2$$
Also why then the contraction of $B_{\mathfrak{q}_1}\mathfrak{p}_2$ is just intersection with $A$.