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

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?

• Why is it that $f'(bm) = c \cdot f(m)$? Perhaps you are defining $f'(b) = c$ and then extending linearly? Commented Apr 6, 2014 at 15:05
• Also, how does showing $f'(b) \neq f'(m)$ give that $f'$ is well-defined? One needs to worry that extending linearly won't give two different places to send a single element. Commented Apr 6, 2014 at 15:06
• "I let f'(b) = c and f'(m) = f(m) for every m in M". Since f(m) is already understood to be well-defined, the only place where f' could not be well-defined is if f'(b) = f'(m) for some m, because that would give two distinct elements of M U {b} with the same image under f'. In the same way, the only way it wouldn't be a homomorphism is if f'(b) = c didn't "behave" with the already-existing f(m) for all m in M. Right? Commented Apr 6, 2014 at 17:39
• It might be that I got something confused after repeatedly looking at what "well-defined" means and did things backwards on that, though. Commented Apr 6, 2014 at 17:40
• Yes, you are showing that $f'$ is not one-to-one (injective). You need to show that each element of $\langle M, b\rangle$ (not the union!) is sent to precisely one element of $F$. Commented Apr 6, 2014 at 17:57

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.