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 $P = Q \cap R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.