I am trying to solve exercise 28 from chapter 14 of Dummit & Foote's Abstract Algebra, which says the following:
- Let $f(x)\in F[x]$ be an irreducible polynomial of degree $n$ over the field $F$, let $L$ be the splitting field of $f(x)$ over $F$ and let $\alpha$ be a root of $f(x)$ in $L$. If $K$ is any Galois extension of $F$ contained in $L$, show that the polynomial $f(x)$ splits into a product of $m$ irreducible polynomials each of degree $d$ over $K$, where $m=[F(\alpha)\cap K:F]$ and $d=[K(\alpha):K]$. [If $H$ is the subgroup of the Galois group of $L$ over $F$ corresponding to $K$ then the factors of $f(x)$ over $K$ correspond to the orbits of $H$ on the roots of $f(x)$. Then use Exercise 9 of Section 4.1.]
After some effort, I managed to write a proof of this result for the case where $f(x)$ is separable, since in this case, $L/F$ is a Galois extension and I can make use of the Galois correspondence. The case where $f(x)$ is inseparable remains. In this case, I don't know how to proceed. My first attempt was to try to use the fact that $L$ contains $K$ (which is Galois over $F$, hence separable) to deduce that $L/F$ is separable, but I couldn't prove it, which led me to the following question:
- Are there finite Galois extensions contained in inseparable extensions? If so, can you give me an example?
The hint given in the exercise assumes that the Galois group of $L$ over $F$ is well-defined, meaning that $L/F$ is Galois. So, I wonder if I should assume that $f(x)$ (and consequently, $L/F$) is indeed separable. My other questions are:
- Is this result true if $f(x)$ is not separable? If so, how to prove this result in that case? If not, what is a counterexample?
