Let $E = F(\alpha)$ be a simple field extension of a finite field F by an algebraic element $\alpha$. Thinking of $E$ as an $F$-vector space, define a linear transformation $$T:E\rightarrow E\;\;\;\;\;\;\;x\mapsto \alpha \cdot x$$ Show that this transformation has no non-trivial invariant subspaces.
Suppose $U$ is a proper $T$-invariant subspace of $E$. If we define $S$ to be $T|_U$, I know that an important step in the solution is the fact that the minimal polynomial of $S$ divides the minimal polynomial of $\alpha$.
The thing is, I don't know why this is true. What does the minimal polynomial of $\alpha$ have to do with the minimal polynomials of transformations on $E=F(\alpha)$?
I would appreciate any help. Thanks.
Let $a$ have degree $n$, and let $b\ne0$ be in the invariant subspace $U$. Then $ab,a^2b,\dots,a^{n-1}b$ are also in the subspace. No linear combination of the $n$ elements $b,ab,\dots,a^{n-1}b$ can vanish; if it did, you'd have a polynomial of degree $n-1$ vanishing at $a$, contrary to degree of $a$ being $n$. So the dimension of $U$ is $n$, so $U$ is not a proper subspace.