Assume $F$ is a field, $E_1$ and $E_2$ are extensions of F, $\sigma:E_1 \rightarrow E_2$ is a field homomorphism, $f(x) \in F[x]$ is an irreducible polynomial. For any $a \in F$, $\sigma(a)=a$, i.e. $E_1$ and $E_2$ are isomorphic over $F$.
Prove that: if $u \in E_1$ is a root of $f(x)$ over $E_1$, then $\sigma(u)$ is a root of $f(x)$ over $E_2$
Thought: For the root $u \in F \subseteq E_1$ it is obvious because $f(\sigma(u))=f(u)=0$, but how to extend to those roots $u \notin F$? Moreover, how to use the condition $f(x)$ is irreducible?
Let
$f(x) = \displaystyle \sum_0^{\deg f} f_i x^i \in F[x]; \tag 1$
then
$f_i \in F, \; 0 \le i \le \deg f; \tag 2$
it follows that
$\sigma(f_i) = f_i, \; 0 \le i \le n; \tag 3$
then if
$f(u) = \displaystyle \sum_0^{\deg f} f_i u^i = 0, \tag 4$
we have
$f(\sigma(u)) = \displaystyle \sum_0^{\deg f} f_i (\sigma(u))^i = \sum_0^{\deg f} \sigma(f_i) (\sigma(u))^i = \sum_0^{\deg f} \sigma(f_i) \sigma(u^i)$ $= \displaystyle \sum_0^{\deg f} \sigma(f_iu^i) = \sigma \left (\sum_0^{\deg f} f_iu^i \right ) = \sigma(f(u)) = \sigma(0) = 0, \tag 5$
that is, $\sigma(u)$ is also a root of $f(x)$.
Note that, as pointed out by Jyrki Lahtonen in his comment to the question itself, it is not necessary to assume $f(x)$ is irreducible to achieve this result.