0
$\begingroup$

Let $ E $ be an algebraic extension $ F $ and $ x \in E $ and $ \sigma: E \to E $ be an automorphism of $ E $ fixing $ F. $ Prove that $ \sigma(x) $ and $ x $ are conjugate over $ F. $

I am starting to learn about extension field, automorphism and Galois theory now and there's a lot of stuffs that confuse me so any help for this question is really appreciated.

In order to prove that $ \sigma(x) $ and $ x $ are conjugate over $ F, $ I need to find an irreducible polynomial $ p(x) \in F[x] $ such that $ p(x) = p(\sigma(x)) = 0. $ If $ x \in F, $ then the proof completes, but I still stuck on the case when $ x\notin F. $

There is a theorem in my book which says that if $ F $ is a field and $ \alpha $ and $ \beta $ are algebraic over $ F $ with $ deg(\alpha, F) = n. $ The the map $$ \psi_{\alpha, \; \beta}:(c_{0} + c_{1}\alpha + \dots + c_{n - 1}\alpha^{n - 1}) = c_{0} + c_{1}\beta + \dots + c_{n - 1}\beta^{n - 1} $$ is an isomorphism of $ F(\alpha) $ onto $ F(\beta) $ if and only if $ \alpha $ and $ \beta $ are conjugate over $ F. $ I attempt this approach but fail to prove that $ \psi_{\sigma(x), \; x} $ is an isomorphism.

$\endgroup$
1
$\begingroup$

In fact any polynomial $p \in F[X]$ with $p(x)=0$ satisfies $p(\sigma (x))=\sigma(p(x))=\sigma(0)=0$, in particular the minimal polynomial of $x$ is the minimal polynomial of $\sigma(x)$.

$\endgroup$
  • $\begingroup$ How do you get $ p(\sigma(x)) = \sigma(p(x))? $ $\endgroup$ – user298251 May 11 '16 at 7:21
  • $\begingroup$ Just write down the polynomial expressions and use that $\sigma$ is a ring homomorphism, which fixes the coefficients of the polynomial. $\endgroup$ – MooS May 11 '16 at 7:23
0
$\begingroup$

Let $p(X)$ be an irreducible polynomial for $\alpha$ over $F$ (so that $p(X) \in F[X]$ and $p(\alpha)=0$). Suppose that $f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0$, then $a_i \in F$ and is fixed by $\sigma$, hence $$\sigma(p(x))=\sigma\left(\sum_{i=0}^n a_ix^i\right) = \sum_{i=0}^n \sigma(a_ix^i)=\sum_{i=0}^n a_i\sigma(x)^i = p(\sigma(x)).$$

Then $p(\sigma(x))=\sigma(p(x))= \sigma(0)=0$ and it follows that the degree of $\sigma(x)$ is smaller or equal to the degree of $x$. By a symmetric argument using the inverse of $\sigma$, we obtain that the degree of $\sigma(x)$ is smaller or equal to the degree of $x$ over $F$. Since $p(\sigma(x))=0$ and $p(X)$ is irreducible, it follows that $irr(\sigma(x),F)=p(X)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy