# Minimal polynomial for some galois extension

The following problem is an exercise in Dummit & Foote's textbook - Abstract Algebra(3rd) in section 14.4:

Suppose that $$K/F$$ is Galois with Galois group $$G$$, and $$\theta$$ is a primitive element for $$K$$, i.e., $$K=F(\theta)$$. For any subgroup $$H$$ of $$G$$, let $$f(x):=\prod_{\sigma\in H}\big(x-\sigma(\theta)\big).$$

1. Show that $$f(x)\in E[x]$$ where $$E$$ is the fixed field of $$H$$ in $$K$$, and that $$f(x)$$ is the minimal polynomial for $$\theta$$ over $$E$$.

2. Prove that the coefficients of $$f(x)$$ generate $$E$$ over $$F$$.

In view of this problem, I wonder if it can be applied to solve the following problem:

Let $$K=\mathbb{Q}(\alpha)$$ be a simple extension over $$\mathbb{Q}$$(possibly non-Galois over $$\mathbb{Q}$$) where $$\alpha$$ is algebraic over $$\mathbb{Q}$$, and let $$F$$ be a subfield of $$K$$. Suppose that the minimal polynomial for $$\alpha$$ over $$F$$ given by $$\textrm{irr}(\alpha,F):=x^{r}+a_{1}x^{r-1}+\cdots+a_{r-1}x+a_{r},$$ where $$a_{1},a_{2},\ldots,a_{r}\in F$$. Show that $$F=\mathbb{Q}(a_{1},a_{2},\ldots,a_{n})$$.

When I think as follows, I thought that the first and second problems have the same meaning:

If we consider the Galois closure $$L$$ of $$K$$ over $$F$$, and take the fixed field $$L_{H}$$ of $$H:=\textrm{Gal}(L/F)$$ in $$L$$, then we can prove that $$f(x)=\prod_{\sigma\in H}\big(x-\sigma(\alpha)\big)$$ is exactly the minimal polynomial for $$\alpha$$ over $$L_{H}$$, and which derive $$F=\mathbb{Q}(a_{1},a_{2},\ldots,a_{n})$$.

It seems possible, but it is difficult to fill in the details.

Can anyone help me a little? Thank you.

• What difficulty do you have in filling in the details? It seems your first statement implies the second with no work at all. Or are you asking how to prove the first statement?
– user208649
Apr 28, 2020 at 5:49

I will give a proof for your first problem, which will also prove the second statement. The point is that the proof of the second part of first problem doesn't require $$K/F$$ being Galois (only simple extension is enough) so it can be copied to the second problem.

Proof of first part. Since $$E\subset K$$ is fixed field of subgroup $$H\subset \text{Aut}(K)$$ so $$K/E$$ is a Galois extension with Galois group $$H$$. This follows $$|H|=[K:E]$$.

Let $$f\in E[x]$$ be minimal polynomial of $$\theta$$ over $$E$$ then as $$K=F(\theta)=E(\theta)$$ so $$\text{deg}(f)=[K:E]=|H|$$.

On the other hand, as $$H=\text{Aut}(K/E)$$ so $$\sigma(\theta)$$ is root of $$f$$ for all $$\sigma\in H$$. This follows $$f(x)=\prod_{\sigma\in H} (x-\sigma(\theta))$$.

Proof of second part. Let $$f(x)=x^r+a_{r-1}x^{r-1}+\cdots+a_1x_1+a_0$$ then $$a_i\in E$$ so $$F(a_0,\ldots, a_{r-1})\subset E\subset K$$. This inclusion implies two things:

One is that $$[K:F(a_0,\ldots, a_{r-1})]\ge [K:E]$$.

Second, the minimal polynomial $$g$$ of $$\theta$$ over $$F(a_0,a_1,\ldots,a_{r-1})$$ must divide $$f$$, implying $$\deg g\le \deg f=[K:E]$$. On the other hand, since $$K=F(\theta)=F(a_0,\ldots, a_{r-1})(\theta)$$ is simple so $$\deg g=[K:F(a_0,\ldots, a_{r-1})]$$.

Thus, $$[K:F(a_0,\ldots, a_{r-1})]=[K:E]$$ so $$E=F(a_0,\ldots, a_{r-1})$$, as desired.