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I am having trouble proving this theorem:

For a field $k$, let $f \in k[x]$ be irreducible, and $g \in k[x]$ be a polynomial. Let $h\in k[x]$ be an irreducible factor of $f\circ g$. Show that $\deg f \mid \deg h$.

My attempt so far was to prove this:

Let $\mathrm{Sq}(n,k)$ be the set of square $n\times n$ matrices with entries in $k$. Let $A\in > \mathrm{Sq}(n,k)$ which has an irreducible characteristic polynomial $f\in k[x]$. Let $g \in k[x]$. Then there exists a $B\in \mathrm{Sq}(n,k)$ satisfying $A=g(B)$ iff $f\circ g$ has a factor in $k[x]$ of degree $n$.

And here's my proof.

Proof. ($\Rightarrow$) Suppose $B$ exists and let $p$ be the minimal polynomial of $B$. Since $k[B]$ contains $k[A]$, $\deg p = n$. Since $f(g(B)) = f(A) = > 0$, we have $p \mid f\circ g$.

($\Leftarrow$) Let $h\mid f\circ g$ and $\deg h = n$. Let $C$ be the companion matrix of $h$. Then $f(g(C)) = 0$, and since $f$ is irreducible with $\deg f = n$, $g(C)$ is similar to the companion matrix of $f$. Therefore $g(C)$ is similar to $A$. As such, $A = T^{-1} g(C) T$ for some $T\in\mathrm{GL}(n,k)$. Then $g(T^{-1}CT)=A$, so $B=T^{-1}CT$. $\blacksquare$

I'm not sure where to go from here. Is this even relevant? Is there a better way to go about this? Hints would be helpful, a proof even more appreciated.

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1 Answer 1

up vote 4 down vote accepted

Suppose $h$ is an irreducible factor of $f \circ g$. Let $L = k(\alpha)$ for some root $\alpha$ of $h$ then $\dim_k (L)=\deg(h)$. Notice that $h(x) \mid f\circ g(x)$ so $\alpha$ is also a root of $f\circ g$ and so $g(\alpha)$ is a root of $f$.

Since $\beta=g(\alpha) \in k(\alpha)$ then $F=k(\beta)\subseteq L$. So to finish the proof $\deg(h)=[L:k]=[L:F][F:k]=[L:F] \deg(f)$.

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I'm an idiot. Thanks! –  Quadrescence Mar 7 '11 at 20:33
+1, but the fourth (and last) line should start with $deg(h)=...$ instead of $deg(g)=...$. –  Georges Elencwajg Mar 12 '11 at 1:36
@elgeorges : fixed it, thanks –  Prometheus Mar 13 '11 at 11:22

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