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Let $F$ be a field and $\overline{F}$ its algebraic closure. If $E \geq F$ is algebraic over $F$ then there exists an injective field homomorphism from $E$ into $\overline{F}$ which is the identity when restricted to $F$.

My attempt: Let $\alpha \in E$. Since $E$ is algebraic over $F$, there is a nonzero polynomial $f(x) \in F[x]$ such that $\alpha$ is a root of $f(x)$. Since $\overline{F}$ is an algebraic closure of $F$, $f(x)$ must split completely into linear factors of the form $(x - r)$ in $\overline{F}$. Thus $\alpha$ must also be in $\overline{F}$ since it is a root of a polynomial in $f$. We can define the injective homomorphism by $\phi : E \to \overline{F}$ defined by $\alpha_E \mapsto \alpha_{\overline{F}}$. It's easy to see that $\phi$ as defined is injective. Restricted to $F$, $\phi$ is of course the identity.

I'm wondering if my logic is correct in saying that $\alpha \in \overline{F}$.

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  • $\begingroup$ It’s not right, because $E$ could technically not be contained in $\overline{F}$ at all. That’s why your desired conclusion is just that there is an embedding of $E$ into $\overline{F}$, rather than that $E$ is contained in $\overline{F}$. $\endgroup$ – Arturo Magidin May 6 at 2:18
  • $\begingroup$ Formally, $E$ doesn’t have to be contained in $\overline{F}$. For example, take two objects that are distinct; call them $i$ and $j$. Construct two algebraic closures of $\mathbb{R}$, one of them by adding $i$ subject to $i^2=-1$; the other by adding $j$ subject to $j^2=-1$. Then $\mathbb{R}[i]\neq\mathbb{R}[j]$, and neither is contained in the other (since $i\neq j$); but of course there is an isomorphism between them. They are different, but isomorphic. Your $E$ need not be contained in $\overline{F}$, it just has to be isomorphic to a subfield of $\overline{F}$. $\endgroup$ – Arturo Magidin May 6 at 2:33
  • $\begingroup$ (Also: you only defined a map from $F(\alpha)$ to $\overline{F}$, not from all of $E$ to $\overline{F}$). $\endgroup$ – Arturo Magidin May 6 at 2:33
  • $\begingroup$ The key here is to use your idea to show that if $\alpha$ is a single algebraic element over $F$, then there is an embedding of $F(\alpha)$ into $\overline{F}$ that is the identity on $F$. Then use that to show that if $E$ is an algebraic extension, and you already have an embedding $j\colon E\to \overline{F}$ such that $j|_F$ is the identity, and $\alpha$ is an element that is algebraic over $F$, then you can extend $j$ to an embedding of $E(\alpha)$ into $\overline{F}$. And finally, use Zorn’s Lemma to show you can embedd an arbitrary $E$. $\endgroup$ – Arturo Magidin May 6 at 2:35
  • $\begingroup$ The difficulty of algebraic closure is to identify $\Bbb{Q}(i+\sqrt{2}) = \Bbb{Q}[x]/(x^4-2x^2+9)$ with $\Bbb{Q}(i,\sqrt{2}) = \Bbb{Q}[y,z]/(y^2+1,z^2-2)$ $\endgroup$ – reuns May 6 at 2:40

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