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).
Best Answer
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)