With B integral over subring A, homomorphism from A to algebraically closed field F can be extended to B.

Question

Here's the problem I am working on:

Let $A$ be a subring of $B$ such that $B$ is integral over $A$, and let $f: A \rightarrow F$ be a homomorphism of $A$ into an algebraically closed field $F$. Show that $f$ can be extended to a homomorphism of $B$ into $F$.

Now, I know this problem has been asked about before here, but the answers on that question don't quite do the job for me. The top answer uses a theorem from some paper somewhere that I'm sure I can't use for this class, and whose proof looks too complicated to include inside of my proof of this statement. So instead I tried another answer from that question, specifically "Using Zorn's lemma, there is a maximal ring $M\subset B$ to which $\psi$ can be extended. Assume $b\in B\setminus M$. What can you conclude?"

So I did exactly that. Letting $M$ be a maximal subring of $B$ so that $f$ extends to $M$, I suppose that $B$ is not equal to $M$, i.e., that $B\setminus M$ is not empty, and let $b$ be in $B\setminus M$. Since $B$ is integral over $A$, there is some polynomial $p(x)$ in $A[x]$ (with coefficients $a_i$) so that $p(b) = 0$; since $F$ is integrally closed, if I take a polynomial in $F[x]$ whose coefficients are $f(a_i)$ in the same order, there is some $c$ in $F$ so that putting $c$ into $x$ in that polynomial makes it $0$ in $F$. So I let $f'(b) = c$ and $f'(m) = f(m)$ for every $m \in M$. If I can show that this is a well-defined homomorphism, then I am done. And in fact, it's fairly easy if I knew it was well-defined to show that it's a homomorphism, since f is already a homomorphism and the only difference between $f$ and $f'$ is $b$, and $f'(bm) = c \cdot f(m) = f'(b)\cdot f'(m)$, $f'(b+m)=c+f(m)=f'(b)+f'(m).$

The problem comes in trying to show that assuming $f'(b) = f'(m)$ leads to some kind of contradiction. Since $f'(m) = f'(b)$, $f'(p(b)) = f'(p(m)) = f(p(m)) = 0$, which only says that $p(m)$ is in the kernel of $f$. I've tried doing something with the kernel of $f$ in $M$, but I can't come up with any meaningful statement from there since $B$ is integral over $A$, and $M$ is not necessarily just $A$. I know that $m$ is in $B$, so there's some polynomial $q(m) = 0$ with coefficients also in $A$, but I'm not sure if that's useful here, either.

If this proof won't work, what will? The textbook gives a hint to use the theorem stating that if $B$ is integral over the subring $A$ and $P$ is some prime ideal of $A$, then there exists a prime ideal $Q$ of $B$ so that $Q \cap A = P$, which fact the answer I mentioned I couldn't really use used. Is there a more "sensible" way to apply this theorem to the problem?

Answer 1

It turns out this problem needs a theorem we hadn't learned quite yet, but have now. The theorem is as follows: If K is a field and F is an integrally closed field, there is a set $\sum$ of pairs (A, f) where A is a subring of K and f is a homomorphism mapping A into F, with the ordering relation $\leq$ defined by $(A, f) \leq (B, g)$ if and only if A is a subring of B and g restricted to A is f. For any element (A, f) of this set, there exists a maximal element (B, g) so that $(A, f) \leq (B, g)$, and futhermore, (B, g) is such a maximal element if and only if B is a valuation ring of K.

Given that theorem, the proof goes like this: Note that the image under f of A is an integral domain (as a subring of field F), so the kernel P of f is a prime ideal of A. Since B is integral over A, there exists a prime ideal Q of B so that $Q \cap A = P$. Now we take K to be the field of fractions of B/Q, which is an integral domain, and observe that (A/P, f) must be in the $\sum$ as defined above for this K and the F given by the problem. After that, it is a matter of taking the maximal element (C, h) so that $(A/P, f) \leq (C, h)$ and using the fact that C is a valuation ring of K = Frac(B/Q) to show that B/Q is a subring of C, and h restricted to B/Q is a homomorphism from B/Q to F. Composing that with the natural homomorphism from B to B/Q gives the desired extension of f.