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Question

Let $K/k$ be a finite, separable field extension with $[K:k]=n$. Let

$$\rho,\rho':K\to M_n(k)$$

be two $k$-algebra homomorphisms. Show that there exists an invertible matrix $A$ in $M_n(k)$ such that

$$\rho'(x)=A\cdot \rho(x)\cdot A^{-1}$$

for all $x\in K$.

confused
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1 Answers1

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As $K/k$ is finite and separable, $K=k(\alpha)$ for some separable element $\alpha$. Then a $k$-basis for $K$ as a vector space is $(1,\alpha,\dots,\alpha^{n-1})$, and so since $\rho,\rho'$ are $k$-algebra homomorphisms, we see that they are uniquely determined by $\rho(\alpha)$ and $\rho'(\alpha)$, respectively. Thus, it is sufficient to show that $\rho'(\alpha)=A\cdot \rho(\alpha)\cdot A^{-1}$ for some $A\in M_n(k)$, i.e. that these matrices are similar.

Note that the minimal polynomial of $\alpha$ will also be the minimal polynomial for $\rho(\alpha)$ and $\rho'(\alpha)$, as it is a polynomial over $k$, thus preserved by $k$-algebra homomorphisms. Also, it must remain minimal, as the homomorphisms are injective. As the extension is separable, there are no multiple roots, and it is a linear algebra result (see below*) that this means the minimal polynomial is also the characteristic polynomial. It is easy to show that conjugation preserves the characteristic polynomial, but in our case, we even have the converse: If two matrices have the same characteristic polynomial, they are similar. This can be shown by using the Frobenius normal form, see here** for example.

This completes the proof.

(*): When are minimal and characteristic polynomials the same?

(**): https://en.wikipedia.org/wiki/Frobenius_normal_form#General_case_and_theory

SomeCallMeTim
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  • As I understand both $\rho(\alpha)$ and $\rho'(\alpha)$ have the same invariant factors (actually only one invariant factor which is the minimal polynomial). Therefore, they have the same rational canonical form. Thanks a lot! – confused Jul 07 '23 at 13:59
  • Also, can't we argue directly that minimal and characteristic polynomials are the same by looking at the degree of the polynomials? – confused Jul 07 '23 at 14:01
  • @confused sure, to both coments. About the latter, how would you show this equality of degrees without kind of doing the same argument? Also, I'm not sure this step is even necessary, by what you said in the first comment. – SomeCallMeTim Jul 07 '23 at 14:04
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    Since the matrix $\rho(\alpha)$ is an $n\times n$ matrix, its characteristic polynomial must be of degree $n$. And since the extension degree of $K/k$ is $n$, the degree of the minimal polynomial is also $n$. – confused Jul 07 '23 at 14:17