Let $L/K$ be an algebraic extension, $\Omega$ an algebraically closed field and $\phi:K\to \Omega$ an injective field homomorphism. Prove that $\phi$ can be extended to $L$.

I've seen some solutions for this using Zorn's lemma (among other things), which I found surprisingly complicated. Before looking it up I had came up with this much simpler solution:

If $\Omega'$ is an algebraic closure for $L$, then it is also an algebraic closure for $K$ and we have the inclusions $K\hookrightarrow L\hookrightarrow\Omega'$. For every $a\in K$ and $\omega\in \Omega$, define $a\cdot \omega:=\phi(a)\omega$. That way, every polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_0\in K[x]$ has roots in $\Omega$, so $\Omega$ is also an algebraic closure for $K$. Then, there is a $K$-isomorphism between $\Omega$ and $\Omega'$ and the composition $L\hookrightarrow \Omega'\xrightarrow{\sim}\Omega$ is the desired extension.

Is something wrong here? What am I missing?

  • 1
    $\begingroup$ At a glance, you are taking for granted the fact that any two algebraic closures of $K$ are isomorphic as $K$-algebras. But that is more or less equivalent to the statement you quote. $\endgroup$ – tomasz May 15 '17 at 22:36
  • $\begingroup$ You meant that any polynomial of $\phi(K)[x]$ has some roots in $\Omega$. And if $L = K(\alpha_1,\ldots,\alpha_n)$ is a finite extension of $K$, then you don't need Zorn's lemma. $\endgroup$ – reuns May 15 '17 at 22:49
  • $\begingroup$ @user1952009 $K[x]$ was a bit of a abuse of language to make it clearer that $\Omega$ is an algebraic extension of $K$. But yes, you're right. $\endgroup$ – rmdmc89 May 15 '17 at 22:51
  • $\begingroup$ Zorn's lemma is there in the case $L$ is an infinite extension of $K$ (for example $L = K(\zeta_\infty)$), writing $L$ as the limit of a chain of finite extensions, and extending the result for those finite extensions to the limit. $\endgroup$ – reuns May 15 '17 at 22:54

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