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