How could we show the equality?

Question

We have that $C$ is an algebraic closure of $F$ and $a,b\in C$.

We have the $F$-monomorphism $\tau: C\hookrightarrow C$ with $\tau (a)=b$.

From the mapping we have that $\tau (C)\subseteq C$.

Since $C$ is an algebraic closure of $F$, we have that $F\subseteq C$, and therefore $\tau (F)\subseteq \tau (C)$.

Since $\tau$ is a $F$-monomorphism, it holds that $\tau (F)=F$.

So, we have that $F\subseteq \tau (C)\subseteq C$.

How could we show that $\tau (C)=C$ ?

Answer 1

Hint: Prove that $\tau(C)$ is an algebraic closure of $F$. Then, since $C$ cannot contain a proper subset that is also an algebraic closure, $\tau(C)$ must equal $C$.

Sketch: Suppose that $f\in F[x]$ is a polynomial with coefficients in $F$. Let $c\in C$ be a root of $f$. Since $f(c)=0$, apply $\tau$ to both sides to show that $f(\tau(c))=0$ so $\tau(C)$ also contains a root of $f$.