# Injective field homomorphism from algebraic extension to algebraic closure exists

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 $$\alpha \in \overline{F}$$.

• 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 $\overline{F}$. – Arturo Magidin May 6 at 2:18
• 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 $\overline{F}$. – Arturo Magidin May 6 at 2:33
• (Also: you only defined a map from $F(\alpha)$ to $\overline{F}$, not from all of $E$ to $\overline{F}$). – Arturo Magidin May 6 at 2:33
• 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 $E$. – Arturo Magidin May 6 at 2:35
• The difficulty of algebraic closure is to identify $\Bbb{Q}(i+\sqrt{2}) = \Bbb{Q}[x]/(x^4-2x^2+9)$ with $\Bbb{Q}(i,\sqrt{2}) = \Bbb{Q}[y,z]/(y^2+1,z^2-2)$ – reuns May 6 at 2:40