Proving an extension of the conjugate root theorem In our analysis class, we just proved the complex conjugate theorem - that for any polynomial with real coefficients and a root $z$, $\overline{z}$ is also a root of that polynomial. I searched online to try to find a generalization of this fact, and ended up piecing together something like the following:

Let $L/K$ be a field extension, let $p\in K[x]$ and $z\in L$ such that $p(z)=0$. If $\sigma\colon L\rightarrow L$ is a ring homomorphism such that $\sigma$ fixes the elements of $K$, then $\sigma(z)$ is a root of $p$.

This would certainly be nice if true, but coming from an intro to analysis class I don't have the right tools to prove it and can't find a proof online. It doesn't stem easily from the (purely algebraic) proof of the complex conjugate theorem we were shown, either. How can I prove this statement?
 A: Hint: Write the expression for $p(z)=0$ and apply $\sigma$ to both sides; use that $\sigma$ is a ring homomorphism that fixes the elements of $K$.
A: Key Idea $\rm\:\sigma:\ w\,\mapsto\, \overline w\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products},\,$ and $\rm\:\color{#0a0}{fixes\ elements}\in\color{#0a0}{  K},\:$ therefore by induction, it preserves polynomials $\rm\  \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{K}\:\![x],\ $ having all $\,\color{#0a0}{{\rm coef's\in K}},\,$  since such polynomials are compositions of said basic operations. $ $ More explicitly
$$ \begin{eqnarray}
\rm \overline{f(w)}\:
&=&\rm\ \  \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\
&=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in L\\
&=&\rm\,\  \overline a_n\,  \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in L \\
&=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \  \forall\ \color{#0a0}a\in \color{#0a0}{K}\\
&=&\rm\  f(\overline w)\\
\rm\!\! So\ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w),\ \ i.e.\ \ w\ root\ of\ f\,\Rightarrow\, \overline w\ root\ of\ f\quad {\bf QED}
\end{eqnarray}\qquad$$
Remark $ $ It's just the obvious generalization of the proof of the  complex Conjugate Root Theorem (case $L/K=\Bbb C/\Bbb R)$.
More conceptually, see the notion of algebraically indistinguishable.
