# Extending a homomorphism of a ring to an algebraically closed field [duplicate]

Let $A$ be a subring of a ring $B$. Suppose $B$ is integral over $A$. Let $\Omega$ be an algebraically closed field. Then every homomorphism $\psi\colon A \rightarrow \Omega$ can be extended to a homomorphism $\phi\colon B \rightarrow \Omega$?

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## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 11 '15 at 18:43

The question is not a duplicate. The question is about homomorphism extensions over integral ring extensions, not over finite ring extensions. A finite ring extension is an integral ring extension, but the converse is not necessarily true. – Makoto Kato 2 days ago

We know that $\mathfrak{p} = \ker \psi$ is a prime ideal of $A$ since the image of $\psi$ is an subring of $\Omega$, in particular an integral domain. The lying-over theorem implies that there exists a prime ideal $\mathfrak{q}$ of $B$ lying over $\mathfrak{p}$ since $B$ is integral over $A$. Now consider the compositum $A \to B \to B/\mathfrak{q}$. The kernel of this compositum is precisely $\mathfrak{p}$ and so we have an injective homomorphism

$$f : A/\mathfrak{p} \to B/\mathfrak{q}$$

such that $B/\mathfrak{q}$ is integral over $A/\mathfrak{p}$. Taking fraction fields, we have that $\textrm{Frac}(B/\mathfrak{q})$ is an algebraic extension of $\textrm{Frac}(A/\mathfrak{p})$. The following steps will now complete the problem:

1. The universal property of quotients gives us a unique ring homomorphism $\overline{\psi} : A/\mathfrak{p} \to \Omega$ that is injective.

2. The universal property of the fraction field now gives a unique ring homomorphism $\Psi: \textrm{Frac}(A/\mathfrak{p} ) \to \Omega$ that extends $\psi$.

3. Use Theorem 1. of Keith Conrad's notes here to give us a homomorphism $\Phi : \textrm{Frac}(B/\mathfrak{q}) \to \Omega$.

4. The desired homomorphism from $B$ to $\Omega$ is obtained by doing $$B \to B/\mathfrak{q} \to \textrm{Frac}(B/\mathfrak{q}) \stackrel{\Phi}{\longrightarrow} \Omega.$$

We can put this all in two diagrams:

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As a complement to BenjaLim's perfect answer, let me give an example of an integral extension $A\subset B$ and of a morphism $\psi:A\to \Omega$ with infinitely many extensions $B\to \Omega$ .

Take $A=\mathbb F_p$ and fix an algebraic closure $B=\Omega=\bar {\mathbb F_p}$ of $\mathbb F_p$.
The inclusion $\psi: \mathbb F_p\to \Omega$ extends to $\phi: B=\Omega\to \Omega:x\to x^p$, the Frobenius automorphism .
But there are many other extensions, the simplest being the infinitely many powers of that Frobenius automorphism, namely the $\phi^n: B=\Omega\to \Omega:x\to x^{p^n}$ where $n\in \mathbb Z$ is arbitrary.

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Hint: 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?

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