# extending ring homomorphism into fields

Let $$A$$ be a subring of $$B$$ such that $$B$$ is integral over $$A$$.

Show that every ring homomorphism $$f:A\rightarrow K$$ with $$K$$ an algebraically closed field can be extended to a ring homomorphism $$\tilde f:B\rightarrow K$$.

My attempt

Wlog assume that $$f$$ is injective (otherwise consider the restriction to $$A/\ker f$$, call it $$f'$$ and do the below for $$f'$$ and at the end extend to $$A$$ by mapping the rest to zero)

First assume that $$A$$ is an integral domain. Let $$F=\mathrm{Frac}(A)$$ denote the field of fractions of $$A$$ and let $$\overline F$$ denote its algebraic closure. Since $$B$$ is integral over $$A$$, we have an inclusion $$B\hookrightarrow\overline F$$.

Now $$\overline F$$ is the smallest algebraically closed field where there exists an inclusion $$A\hookrightarrow\overline F$$. In particular $$\overline F/K$$ is a field extension, thus $$A\subset\overline F\subset K$$ (here we use that $$f$$ is injective). Now define $$\tilde f:\overline F\rightarrow K,\quad \frac{a\cdot x}{a'\cdot x'}\mapsto \frac{f(a)x}{f(a')x'},$$ where $$a,a'\in A$$ and $$x,x'$$ have no factor in $$A$$. Then $$\tilde f$$ is a ring homomorphism $$\overline F\rightarrow K$$ with $$\tilde f|_A=f$$, hence $$f$$ extends to $$B\subset\overline F$$.

If my argument correct when $$A$$ is an integral domain?

Now I am having some trouble reducing to the latter case if $$A$$ is not an integral domain (a hint suggests to do this).

• So I figured that if $ab=0$, then $f(a)=0$ or $f(b)=0$, but not necessarily both. What do I do in that case? – Pink Panther Jun 1 at 11:18

I think it would be preferable to use Zorn's lemma to the inductive set $$\mathcal{F}=\{ (C,g)\mid ,A\subset C , g_{\vert C}=f\}$$ where $$(C_1,g_1)\leq (C_2,g_2)$$ if $$C_1\subset C_2$$ and $$g_2\vert_{C_1}=g_1$$. To prove that the maximal element of $$\mathcal{F}$$ is $$(B, g)$$ (that is that you can extend $$f$$ to the whole $$B$$), show that if there is some $$\alpha\in B\setminus C$$, then a map $$C\to K$$ may be extended to $$C[\alpha]\to K$$ (as you can imagine)