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Let $E$ be a field extension over $F$. Let $\alpha \in E$ s.t. for $p(x) \in F[x]$, we have that $p(\alpha) = 0$. Furthermore, suppose $\phi: E \rightarrow E$ is a field automorphism s.t. $\phi(a) = a$ for all $a \in F$.

Goal: Show that $p(\phi(\alpha)) = 0$.

Attempt:

  1. Let $p(x) = a_0 + a_1 x^1 + \ldots + a_n x^n$.

  2. Then $p(\alpha) = a_0 + a_1 \alpha^1 + \ldots + a_n \alpha^n = 0$.

  3. Then $\phi(p(\alpha)) = \phi(a_0 + a_1 \alpha^1 + \ldots + a_n \alpha^n) = a_0 + a_1 \phi(\alpha^1) + \ldots + a_n \phi(\alpha^n) = 0$.

  4. Then $\phi(\alpha) = \frac{1}{\alpha_1} [-a_0 - a_2 \phi(\alpha^2) - \ldots - a_n \phi(\alpha^n)]$.

And at this point I'm not sure how I could use the equality in (4) to my advantage.

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    $\begingroup$ show $\phi(p(\alpha))=p(\phi(\alpha))$ (you almost did it in step 3) $\endgroup$ – user8268 Sep 24 '13 at 16:56
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Per user8268's comment, the completed proof is as follows:

  1. Let $p(x) = a_0 + a_1 x^1 + \ldots + a_n x^n$.

  2. Then $p(\alpha) = a_0 + a_1 \alpha^1 + \ldots + a_n \alpha^n = 0$ by hypothesis.

  3. Then $\phi(0) = \phi(p(\alpha)) = \phi(a_0 + a_1 \alpha^1 + \ldots + a_n \alpha^n) = a_0 + a_1 \phi(\alpha^1) + \ldots + a_n \phi(\alpha^n) = 0$ via the fact that $\phi(a_0) = a_0$, $0 \in F \implies \phi(0) = 0$, and making use of the automorphism properties.

  4. But then $\phi(p(\alpha)) = a_0 + a_1 \phi(\alpha) + a_2 \phi(\alpha)^2 + \ldots + a_n \phi(\alpha)^n = p(\phi(\alpha)) = 0$ via the fact that $\phi(\alpha^k) = \phi(\alpha)^k$ since $\phi$ is an automorphism and combining facts obtained in (3).

  5. Then $p(\phi(\alpha)) = 0$ as desired.

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