Zeroes of f(x) in a splitting field $E $ have the same multiplicity

Question

Let $f(x)$ be an irreducible polynomial over a field $F$ and let $E$ be a splitting field of $f(x)$ over $F$. Then all the zeroes of $f(x)$ in $E$ have the same multiplicity.

The proof of this theorem involves the following steps

Let $a$ and $b$ be distinct zeroes of $f(x)$ in $E.$ If $a$ has multiplicity $m,$ then in $E[x]$, we may write $f(x)= (x-a)^m g(x)$. It follows from a lemma that there is a field isomorphism from E to itself that carries a to b and acts as the identity on $F.$

Thus, $f(x)= \Phi(f(x))= (x-b)^m \Phi(g(x))$

I am not able to understand the above step. I understand that $a$ is carried to b by an isomorphism. But why is $\Phi(x-a)= x-b $

EDIT: Suppose the isomorphism is extended to $E[x]$ such that $\Phi(x)=x$. Does this mean that the required result of the theorem is true only when $\Phi(x)=x$ and not for all other isomorphism mappings ?

Thank you for your help..

Answer 1

If we have a homomorphism $\varphi:F\to K$, where $F,K$ are fields (or rings), there is an induced homomorphism $\Phi:F[x]\to K[x]$, which is defined by $\Phi|_F=\varphi,\;\Phi(x)=x$. Obviously we get $\Phi(x-a)=\Phi(x)-\Phi(a)=x-b$.

Answer 2

In fact $\Phi$ is an isomorphism of $E$, it is extended to $E[x]$ by $\Phi(x)=x$, so $\Phi(x-a)=\Phi(x)-\Phi(a)=x-b$.