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Given a number fields extension $F\subseteq K$ I'm trying to prove that the characteristic polynomial of an element $\alpha\in K$ - which is defined to be the characteristic polynomial of the $F$-endomorphism of $K$ given by the multiplication $x\mapsto \alpha x$ - is a power of its minimal polynomial.

I know that if $F\subseteq K \subseteq E$ is the Galois closure of $K/F$ then the minimal polynomial of $\alpha$ is $$ f^{min}_\alpha (x)=\prod_{\sigma\in Gal(E/F)}(x-\sigma(\alpha))$$ while if $\text{Hom}_F(K,\mathbb{C})=\{\sigma_1, \dots, \sigma_n\}$ its characteristic polynomial is given by

$$ f^{char}_\alpha (x)=\prod_{i=1}^n (x-\sigma_i(\alpha))$$

The result is the following $$ f^{char}_\alpha (x)=f^{min}_\alpha(x)^{[K/F(\alpha)]}$$

and the textbook proves it as follows. The multiplicity of the factor $(x-\sigma_i(\alpha))$ in the characteristic polynomial is given by the number of $\sigma_j$ s.t. $\sigma_j(\alpha)=\sigma_i(\alpha)$ which is the number of $F$-homomorphism $K\to E$ extending $\sigma_i\big|_{F(\alpha)}$.

How do I prove that number is given by $[K/F(\alpha)]$? If $K/F$ was Galois I would use the fundamental theorem, but in this general case?

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3 Answers 3

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Let $\beta$ be a primitive element for $K/F(\alpha)$. Then with $p_\beta(x)\in F(\alpha)[x]$ the minimal polynomial for $\beta$ with $\deg p_{\beta}(x)=k$, we have

$$K\cong F(\alpha)[x]/(p_\beta(x))$$

And each choice of root gives at most one different embedding of $F(\alpha)(\beta)$ into the Galois closure, so the number of embeddings is bounded by $[K:F(\alpha)]$, i.e. there are at most that many. However, each root gives a different embedding as--if we number the roots as $\beta=\beta_1,\ldots,\beta_k$ then writing the isomorphisms

$$\phi_i: F(\alpha)(\beta_i)=K_i\to F(\alpha)[x]/(p_{\beta}(x))$$

where the equivalence class $[x]\in F(\alpha)[x]/(p_{\beta}(x))$ is mapped to $\beta_i$ as usual, then $\phi_i\circ \phi_j^{-1}$, $i\ne j$, is a non-identity--$\beta_i\mapsto\beta_j$ and $p_\beta(x)$ is irreducible--isomorphism between $K_i$ and $K_j$ as subfields of the Galois closure, hence the number of embeddings is at least $[K:F(\alpha)]$ proving equality.


Note the first inequality holds in general, the second one relies on the fact that an irreducible polynomial in characteristic $0$ is separable, i.e. all roots are distinct. This is critical for the equality, as the general case only has that the number of embeddings is at most $[K:F(\alpha)]$.

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  • $\begingroup$ Thanks a lot! And what happens if the extension is not simple? I know that number fields extensions are simple, but is it true in general for an arbitrary separable and finite extension? $\endgroup$
    – Oromis
    Commented Jan 24, 2017 at 16:40
  • $\begingroup$ @Oromis this is exactly the content of the primitive element theorem. (All finite, separable extensions are simple). $\endgroup$ Commented Jan 24, 2017 at 16:42
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I think you want that if $E/K$ is a finite extension and $F$ is an intermediate field, and if $x \in F$ then

$$\chi_{E/K}(x)=(\chi_{F/K}(x))^{[E:F]}$$

This is proved as follows: Let $\alpha_1, \ldots , \alpha_n$ be a basis for $F/K$ and $\beta _1, \ldots , \beta_m$ a basis for $E/F$. Then $\{\alpha_i\beta_j\}$ is a basis for $E/K$ and with respect to this basis the matrix for $T_{x}$ is the block diagonal $\begin{pmatrix} M&0&0\\ 0&M&0\\ 0&0&M\\ \end{pmatrix}$ where $M$ is the matrix for $T_{F/K}$.

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For clarity, let us first recall your definitions and notations: $K/F$ is a finite extension of number fields of degree $n$ (I guess that any finite separable extension would do), $\alpha\in K$ is given, $f_m (X)$ and $f_c (X)$ are resp. the minimal and characteristic polynomials of $\alpha$, and you want to show that $f_c= (f_m)^r$ , where $r=[K:F(\alpha)]$. Proceed in two steps:

1) Suppose $K=F(\alpha)$ , i.e. $r=1$. Denote by $\mu$ the multiplication by $\alpha$ in $K$ and consider the matrix $M$ of this endomorphism (in the sense of linear algebra) in the basis {$1, \alpha, ... , \alpha^{n-1}$}. I don't know how to type a matrix (!), but it is immediate that $M$ is the so called "companion matrix" of $f_m (X)=X^n + a_{n-1}X^{n-1} + ... +a_0$ (just write it down) and it is a classical exercise to show that $det (X.Id - M)$ is nothing but $f_m (X)$. Thus $f_c= f_m$ as wanted.

2) In the general case, let $(y_i)$ be a basis of $F(\alpha)/F$ and $(z_j)$ a basis of $K/F(\alpha)$. Then $(y_i z_j)$ is a basis of $K/F$, and the matrix of $\mu$ in this basis is a "diagonal block matrix" with $r$ blocks $M$ on the diagonal (the same $M$ as before, just replacing $n$ by $\frac nr$) and $0$ elsewhere, as is displayed in the beautiful typing of @Rene Schipperus. Then $f_c= (f_m)^r$ as desired.

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