Field Theory

Degree of an Irreducible Factor of a Composition of Polynomials

Field theory problems and solution in abstract algebra

Problem 83

Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$.

Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.

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Hint.

Use the following fact.

Let $h(x)$ is an irreducible polynomial over a field $F$.
Let $\alpha$ be a root of $h(x)$.
(That is, $h(x)$ is the minimal polynomial of $\alpha$ over $F$.)
Then we have
\[\deg(h(x))=[F(\alpha):F].\]

Proof.

Let $h(x)$ be an irreducible factor of $f(g(x))$. Let $\beta$ be a root of $h(x)$.
Put $\alpha=g(\beta)$. Then $\alpha$ is a root of $f(x)$ since we have
\[f(\alpha)=f(g(\beta))=0.\]

Since $f(x)$ is irreducible, $[F(\alpha):F]=n$.
Since $g(\beta)\in F(\beta)$, the field $F(g(\beta))$ is a subfield of the field $F(\beta)$.

Thus we have
\begin{align*}
\deg(h(x))&=[F(\beta):F] \text{ (since } h(x) \text{ is irreducible})\\
&=[F(\beta): F(g(\beta))] [F(g(\beta)): F]\\
&=[F(\beta): F(g(\beta))] [F(\alpha): F]\\
&= [F(\beta): F(g(\beta))]\cdot n.
\end{align*}

Hence $n$ divides the degree of the irreducible factor $h(x)$ of the composite $f(g(x))$.

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