I have found this question on the Papantonopoulou's Algebra book:
Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$ with deg$ f(x) = 15$ and deg$ g(x) = 14$. Let $\alpha$ be a zero of $f(x)$ in some extension field of $F$. Show that $g(x)$ is still irreducible over $F(\alpha)$.
I have some ideas about how to prove it, but none of then use the degrees 15 and 14, which makes me think they will fail...
For instance, I know that a polynomial $p(x)$ is irreducible over a field $K$ iff $\frac{K[x]}{\langle p(x) \rangle}$ is a field. So, I could try to find a isomorphism between $\frac{F(a)[x]}{\langle g(x) \rangle}$ and any field...
I know that the conditions given in the exercise imply in $[F(\alpha) : F ] = 15$, so, maybe I could generate a contradiction by supposing that $g(x)$ is reducible over $F(\alpha)$ and finding basis for $F(\alpha)$ with less than $15$ elements...
Well, anybody has an idea about how to solve this exercise?
Thank you very much.
As you noted, $\;[F(\alpha):F]=\deg f=15\;$, so if $\;g(x)\;$ is reducible in $\;F(\alpha)[x]\;$, say $\;g(x)=h(x)k(x)\;,\;\;h,k\in F(\alpha)[x]\;$ and say $\;\beta\;$ is a root of $\;h(x)\;$ , then
$$\;\begin{cases}I\;\;[F(\alpha)(\beta):F(\alpha)]=m<14\\{}\\II\;\;F(\beta)(\alpha):F(\beta)]=m'\le15\end{cases}\;,\;\;\text{yet}\;\;[F(\beta):F]=14\;,\;\;[F(\alpha):F]=15\;,\;\;\text{so}$$
$$[F(\alpha,\beta):F]=\begin{cases}[F(\alpha)(\beta):F(\alpha)][F(\alpha):F]=15m\\{}\\ [F(\beta)(\alpha):F(\beta)][F(\beta):F]=14m'\end{cases}$$
Observe that it must be $\;m=14\cdot x\;,\;\;x\in\Bbb N\;$ , yet this would contradict $I$ above