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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).

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  • $\begingroup$ 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? $\endgroup$ – Pink Panther Jun 1 at 11:18
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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)

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