Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $L | K$ is a separable extension and $\sigma : L \rightarrow \bar K$ varies over the different $K$-embeddings of $L$ into an algebraic closure $\bar K$ of $K$, then how to prove that $$f_x(t) = \Pi (t - \sigma(x))?$$ $f_x(t)$ is the characteristic polynomial of the linear transformation $T_x:L \rightarrow L$ where $T_x(a)=xa$

share|improve this question
4  
Consider the degree and the number of automorphisms of L over K, along with the fact that all conjugates of a root are also roots. –  awllower Jan 20 '13 at 19:47
add comment

1 Answer 1

up vote 2 down vote accepted

First assume $L = K(x)$. By the Cayley-Hamilton Theorem, $f_x(x) = 0$, so $f_x$ is a multiple of the minimal polynomial of $x$ which is $\prod_\sigma (t-\sigma(x))$. Since both polynomials are monic and have the same degree, they are in fact equal.

For the general case, choose a basis $b_1,\ldots,b_r$ of $L$ over $K(x)$. Then, as $K$-vector spaces, $L = \bigoplus_{i=1}^r K(x)b_i$, and $T_x$ acts on the direct summands separately. Therefore, the characteristic polynomial of $T_x: L \to L$ is the product of the characteristic polynomials of the restricted maps $T_x: K(x)b_i \to K(x)b_i$. All those restricted maps have the same characteristic polynomial, namely the minimal polynomial $g$ of $x$. So the characteristic polynomial of $T_x: L\to L$ is $g^{[L:K(x)]}$. Since every embedding $\tilde\sigma: K(x) \to \overline K$ can be extended to an embedding $\sigma: L \to \overline K$ in exactly $[L:K(x)]$ ways, this equals $\prod_\sigma (t-\sigma(x))$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.