# Finite field extension and roots of a polynomial of degree 3

Let $$L/K$$ be a finite field extension with $$[L:K] = 2^k$$ and $$f \in K[X]$$ be a polynomial of degree 3 that has a root in $$L$$. Show that $$f$$ has a root in $$K$$.

I was thinking about proving this with induction but I have trouble with the base case:

$$k=1$$: Without loss of generality we can assume that $$f$$ is monic, so $$f = X^3 + aX^2 + bX + c$$. Then $$[L:K] = 2$$ and since $$2$$ is prime, there exists an $$\alpha \in L$$ such that $$L = K(\alpha)$$. Now the Minimal polynomial $$g$$ of $$\alpha$$ over $$K$$ is of degree 2 and since it is also monic, it must be a divider of $$f$$ (I'm not sure about this one). So $$f = g\cdot h$$ with $$\deg(h) = 1$$. Since $$h$$ is a linear polynomial over $$K$$ it must have a root in $$K$$.

I'm not sure if this was correct since I didn't really need that $$f$$ has a root in $$L$$? Any hints would be greatly appreciated.

• $[L:K(\alpha)][K(\alpha):K]=[L:K]$ Nov 30, 2020 at 16:34

If $$f$$ has no solution over $$K$$ and a root $$a$$ in $$L$$, then $$f$$ is irreducible in $$K$$ and $$L$$ contains a subfield $$K(a)$$ isomorphic to $$K[X]/(f)$$. Such a field has degree $$3$$ and cannot be contained in $$L$$, since $$3$$ doesn't divide $$[L:K]$$
• You say that $K(\alpha)$ has degree $3$ but why exactly is that? How can we be sure that $f$ is the minimal polynomial of $a$ over $K$? Nov 30, 2020 at 16:49
• Because it is irreducible and $f(a)=0$. Nov 30, 2020 at 17:18