# Polynomial of degree 5 that has no root in a normal extension, then it is irreducible over that extension

This is a problem from my Qual exams

"Let $$L/K$$ be a Galois extension of fields of characteristic 0. Suppose $$f(x)$$ is an irreducible polynomial of degree 5 in $$K[x]$$ and has no root in $$L$$. Prove that $$f$$ is irreducible in $$L[x]$$."

I usually only deal with normal extension of finite degree. So in this problem it is really hard to understand $$L$$. Anyway, suppose $$f$$ is reducible, then it must be a product of a quadratic and a cubic. How does this lead to contradiction?

You have one possible case where $$f=g_2g_3$$ where $$g_2$$ and $$g_3$$ are irreducible over $$L$$ of degrees $$2$$ and $$3$$. We can assume $$g_2$$ and $$g_3$$ are monic. We can assume that $$L$$ is the Galois closure of the extension of $$K$$ generated by the coefficients of the $$g_j$$, so is a finite extension of $$K$$. Let $$G$$ be the Galois group of $$L/K$$. Then $$f=g_2^\sigma g_3^\sigma$$ for any $$\sigma\in G$$. That's also an irreducible factorisation of $$f$$ so must be the same as $$f=g_2g_3$$. Therefore $$g_2^\sigma=g_2$$ etc. (this is where we need $$g_2$$ and $$g_3$$ to have different degrees). So the coefficients of $$g_2$$ are fixed by $$G$$ so lie in $$K$$, contradicting $$f$$ being irreducible over $$K$$.