Let $K$ be a field, $f(x)\in K[x]$ monic and irreducible polynomial over $K[x]$, $E$ the splitting field of $f(x)$ over $K$ and $a,b\in E\ $ two roots of $f(x)$.
We would like to prove that there exists a field automorphism $\eta: E\longrightarrow E$ such that $\eta(a)=b.$
My attempt. Let's take the field automorphism $Id_K:K\longrightarrow K$.
We observe that
- $f(x)\in K[x]$ is monic, irreducible over $K$, and $f(a)=0_K$. So, $f(x)=Irr_{(K,a)}(x)$.
- $f(x)\in K[x]$ is monic, irreducible over $K$, and $f(b)=0_K$. So, $f(x)=Irr_{(K,b)}(x)$.
So, $f(x)=Irr_{(K,a)}(x)=Irr_{(K,b)}(x)\in K[x].$ Also, since $E$ is the splitting field of $f(x)=Irr_{(K,a)}(x)\in K[x]$ over $K$ and $E$ is the splitting field of $Id_K (f(x))=f(x)=Irr_{(K,b)}(x)\in K[x]$ over $K$ there exists an automorphism $$\eta : E \longrightarrow E$$ that extends $Id_K$.
- Is this attempt in the right way?
- How do we prove that $\eta(a)=b$?
Thank you.