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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 $T_x(a)=xa$

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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
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 $\prod_\sigma (t-\sigma(x))$.

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