# Do quotienting and contraction commute?

$$\DeclareMathOperator{\Spec}{Spec}$$

Let $$A \subset B$$ be an integral extension of rings. Consider the map $$f:\Spec B \to \Spec A$$ by $$f(Q) = Q^c = Q \cap A$$. Show that $$f$$ is a closed map.

My idea was to consider a typical closed set $$V(I)$$ of $$\Spec B$$ and show that $$f[V(I)] = V(I^c)$$.
The inclusion $$(\subset)$$ was easy to show. It is the reverse direction that I am struggling with.

Here's what I have done: Consider $$P \in V(I^c)$$. Thus, $$P \in \Spec A$$ and $$P \supset I^c$$.
Note that the extension $$A/I^c \hookrightarrow B/I$$ is integral and $$P/I^c$$ is a prime in $$A/I^c$$.
Then, by the lying over theorem, it follows that there exists a prime $$Q/I \in \Spec(B/I)$$ contracting to $$P/I^c$$.

Thus, we have gotten hold of a prime ideal $$Q \in V(I).$$ I am now not sure how to prove that $$Q \cap A = P$$. (Is this even true?)

An alternate approach could have been to directly use the lying over theorem for $$P$$ to conclude that there exists a prime $$Q \in \Spec B$$ lying over $$P$$. But then it wasn't clear why $$Q \supset I$$.

Recall that the inclusion $$i : R/I^c \hookrightarrow S/I$$ is given as $$i(r + I^c) = r + I$$. Regarding $$R/I^c$$ as a subset of $$S/I$$ under this map, we show that $$(Q/I) \cap (R/I^c) = (Q \cap R)/I^c.$$ (You already know that the left set is $$P/I^c$$.) Note that $$Q \cap R$$ indeed contains $$I^c = I \cap R.$$

$$(\subset)$$ Let $$x \in (Q/I) \cap (R/I^c)$$. Then $$x = r + I^c = q + I$$ for some $$r \in R$$ and $$q \in Q$$. However, we identify $$r + I^c$$ with $$r + I$$ in $$S/I$$. Thus, we see that $$r + I = q + I$$ in $$S/I$$ and thus, $$r - q \in I \subset Q$$.
Thus, we see that $$r \in q + Q = Q$$. Since $$r \in R$$ to begin with, we conclude that $$r \in Q \cap R$$ and thus, $$x = r + I^c \in (Q \cap R)/I^c$$.

$$(\supset)$$ Let $$x \in (Q \cap R)/I^c$$.
Then, $$x = y + I^c$$ for some $$y \in Q \cap R$$. Under the identification, we have $$x = y + I^c = y + I.$$ The second expression shows that $$x \in R/I^c$$ and the third that $$x \in Q/I$$.

Conclusion. The main issue here is getting around the (abuse of) notation and then concluding that $$P/I^c = (Q/I) \cap (R/I^c) = (Q \cap R)/I^c.$$

Now, $$P$$ and $$Q \cap R$$ are ideals containing $$I^c$$. By the ideal correspondence, we see that $$P = Q \cap R$$.