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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?

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    Yes, they are root of the irreducible polynomial $f\in F[x]$. @arctictern – 2017-01-02
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    How are what defined? – 2017-01-02
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    The monomorphism $\sigma_a$ and $\sigma_b$. Let $y\in F[x]/\langle f(x)\rangle$. To what is $\sigma_a(y)$ or $\sigma_b(y)$ equal? @ZelosMalum – 2017-01-02
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    So you want them explicitly constructed? – 2017-01-02
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    So, do we not need it? How can we construct then the mapping $\sigma$ ? @ZelosMalum – 2017-01-02
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    No we do not, we only need to prove the existence of such a homomorphism, sometimes it is done by explicitly constructing it, sometimes it is not. – 2017-01-02
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    Ah ok... What I have done so far with $\sigma_a$ and $\sigma_b$ is correct? If yes, how could we continue to show the existence of $\sigma$ ? @ZelosMalum – 2017-01-02
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    From the theorem that I cited in my question, do we maybe have that $\sigma_a$ can be extended to $\sigma_a:C\rightarrow C$ ? @ZelosMalum – 2017-01-02

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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.

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    Could you explain to me further the last sentence? – 2017-01-20
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    What do you mean by "such that it commutes" ? – 2017-01-23
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    It means if you draw diagrams of the morphisms then the choosen path does not matter – 2017-01-24
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    So, we have this diagram https://i.stack.imgur.com/TnivR.png . Do you mean that if we get from $F[x]/(f(x))$ to $C$ (right) we get the same result when we get from $F[x]/(f(x))$ to $C$ (left) and then from $C$ to $C$ ? – 2017-01-24
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    Also, we get that $C\rightarrow C$ is an isomorphism. How do we get that it is a $F$-isomorphism? – 2017-01-24
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    I edited a little my initial question. Could you take a look at it? – 2017-01-24