Question: let $f,g\in F[x]$ be irreducible polynomials of degree $m$ and $n$, respectively, for $(m,n)=1$. Show that if $\alpha$ is a root of $f$ in some extension of $F$, then $g$ is ireducible in $F(\alpha)[x]$.
Thoughts: First, suppose $\beta$ is a root of $g$ in some extension of $F$. First, we claim that $\beta\notin F(\alpha)$. To show this, we can assume that $\beta\in F(\alpha)$. Then, $F\subseteq F(\beta)\subseteq F(\alpha)$, thus $[F(\alpha):F]=[F(\alpha):F(\beta)][F(\beta):F]\implies m=[F(\alpha):F(\beta)]n\implies n|m$, which is a contradiction as $m$ and $n$ are coprime. Now, back to the proof of the question. We want to show that $g(x)$ is irreducible in $F(\alpha)[x]$. Suppose, by contradiction, that $g(x)$ is reducible in $F(\alpha)[x]$. Then, we can write $g(x)$ as a product of irreducible factors, $g_1\cdots g_k$, $k\in\mathbb{Z}$, where the $g_i$'s are in $F(\alpha)[x]$, and at least one of the $g_i$'s equal $0$ when evaluated at $\beta$. Suppose $g_1(\beta)=0$. Then $[F(\alpha,\beta):F(\alpha)]=deg(g_1)<n$… but I can't seem to get to a contradiction to show the result. Any help is greatly appreciated! Thank you.
Best Answer
Let $\beta$ be a root of $g$. It is not necessarily the case that $F(\beta)$ is contained in $F(\alpha)$, so we should instead look at the extension $F(\alpha, \beta)$ of $F$, which contains both $F(\alpha)$ and $F(\beta)$. We can write the degree of $F(\alpha,\beta)$ over $F$ in two ways:
$$ [F(\alpha,\beta) : F(\alpha)][F(\alpha) : F] = [F(\alpha, \beta) : F(\beta)][F(\beta) : F]$$
or in other words,
$$ m[F(\alpha,\beta) : F(\alpha)] = n[F(\alpha,\beta) : F(\beta)]$$
Clearly, $n$ divides the product of $m$ (to which it is relatively prime) and $[F(\alpha,\beta) : F(\alpha)]$. If $g$, whose degree is $n$, is not irreducible over $F(\alpha)$, then the extension $[F(\alpha,\beta) : F(\alpha)]$ must be an integer which is strictly less than $n$. See if you can get a contradiction from here.