Let $C$ be an algebraic closure of $F$, let $f\in F[x]$ be irreduccible and $a,b\in C$ the roots of $f$.
We have the following theorem:
If $E$ is an algebraic extension of $F$, $C$ is an algebraic closure of $F$, and $i$ is an embedding (that is, a monomorphism) of $F$ into $C$, then $i$ can be extended to an embedding of $E$ into $C$.
I want to show that there is a $F$-monomorhism $\sigma: C\hookrightarrow C$, that maps $a$ to $b$.
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I have done the following:
We have that $C$ is an algebraic closure of $F$ and $a,b\in C$. So, $C$ is also an algebraic closure of $F[a]$ and $F[b]$, right?
Then we have the algebraic extension $C /F[a]$ and $C/ F[b]$.
Since $F[a]$ and $F[b]$ are subfields of $C$, by definition there are the field monomorphisms $\sigma_a: F[a]\rightarrow C$ and $\sigma_b:F[b]\rightarrow C$, with $\sigma_a:(x)=a$ and $\sigma_b(x)=b$. Is this correct?
We have that $f\in F[x]$ is irreducible, then there is a field that contains $F$, in that $f(x)$ has a root. This field is $K=F[x]/\langle f(x)\rangle$.
From a theorem we have that $K\cong F[a]$ and $K\cong F[b]$.
Therefore, we have that $F[x]/\langle f(x)\rangle\cong F[a]$ and $F[x]/\langle f(x)\rangle\cong F[b]$.
So, we have the homomorphisms $\sigma_a:F[x]/\langle f(x)\rangle \rightarrow C$ with $\sigma_a(x)=a$ and $\sigma_b:F[x]/\langle f(x)\rangle \rightarrow C$ with $\sigma_b(x)=b$, right?
From the above cited theorem we have that there is an homomorphism $\tau : C\rightarrow C$ with $\tau \circ \sigma_a=\sigma_b$. So, $\tau (\sigma_a(y))=\sigma_b(y), y\in F[x]/\langle f(x)\rangle$, and so $\tau (a)=b$.
Since the field extensions are injective homomorphisms, we have the monomorphism $\tau : C\rightarrow C$ with $\tau (a)=b$.
Is everything correct so far?
How do we get that $\tau$ is a $F$-monomorphism?
As you say you have that $$F[\alpha]\cong F[x]/\langle f\rangle \cong F[\beta]$$ By your own theorem we have that $\imath:F\to C$ can be extended to $\imath^\ast:F[x]/\langle f\rangle\to C$ which also means that for $\imath_\alpha^\ast:F[\alpha] \to C$ and $\imath_\beta^\ast:F[\beta] \to C$ that $\text{Im }\imath^\ast_\alpha \cong \text{Im }\imath^\ast_\beta$. By how homomorphisms between field extensions we have that there must exist an automorphism such that it commutes.