Let $K$ be a finite field extension of $F$. Let $\alpha\in K\setminus F$. Then multiplication by $\alpha$ is an F-linear transformation form $K\to K$.

Let linear transformation be called $T_{\alpha}$. Hence $T_{\alpha}:K\to K$ such that $s\to \alpha s$.

Prove that the minimal polynomial of $T_\alpha$ is the same as the minimal polynomial of $\alpha$ over $F$.

To generate the minimal polynomial of $T_{\alpha}$, I will have to determie its matrix representation. What will that look like? Like this $$\begin{pmatrix} \alpha & 0\\ 0 & 0\end{pmatrix}$$?

If this is true, then I will also have to represent all members of the field $K$ as $\begin{pmatrix} k\\0\end{pmatrix}$, in order to satisfy $\begin{pmatrix} \alpha & 0\\ 0 & 0\end{pmatrix}\begin{pmatrix} k\\0\end{pmatrix}=\alpha \begin{pmatrix} k\\0\end{pmatrix}$. I find all this to be a little weird. Am I on the right track?


First prove that $\alpha\rightarrow T_{\alpha}$ is a field isomorphism (from $K$ to whatever its image is under that map, with the appropriate definition of division of course). Then everything follow instantly.

  • $\begingroup$ +1 A nice conceptualization! Actually proving that it is a homomorphism of rings suffices, because a ring homomorphism from a field always has a trivial kernel for lack of alternative ideals. $\endgroup$ – Jyrki Lahtonen Jun 27 '14 at 6:41

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