Let $p<q$ be primes, $k$ be a field and $f(x), g(x) \in k[x]$ be polynomials of degree $p$ and $q$ respectively.
Given :
- $f(x)$ is irreducible.
- $L$ is the splitting field of $f(x)$ over $k$.
- $\exists$ $\alpha \in L$ such that $g(\alpha) = 0$.
Question : Is $g(x)$ reducible in $k[x]$?
Attempt : I tried applying the division algorithm of $f$ and $g$ but I can't really figure out how to proceed.
Consider the minimal polynomial $\rho(x) \in k[x]$ of $\alpha \in L$ over $k$. This minimal polynomial divides $g$ in $k[x]$. In case this statement is not familiar, I prove this now.
If you anticipate the division algorithm, it is here that the division algorithm comes into play. Divide $g$ by $p$. $$ g(x) = \rho(x)q(x) + r(x). $$ The remainder $r$ has degree strictly smaller than that of $\rho$. Now evaluate both sides of the equality at $x = \alpha$ to obtain $0 = r(\alpha)$. This contradicts the minimality of $\rho$ unless $r = 0$. Thus, the divisibility of $g$ by $\rho$ is verified.
I return to the reducibility of $g$. $$ k \subseteq k(\alpha) \subseteq L. $$ Since $[L:k]$ divides $p!$, so does the degree of $\rho$. In particular, the degree of $\rho$ does not share any prime factors with the degree $q$ of $g$, for all prime factors of $\operatorname{deg} \rho$ are consequently known to be at most $p$. Therefore, $g$ is reducible.