# Minimal polynomial of $\alpha$ over a field extension

I have this problem:

Let $$K \subseteq L$$ be a Galois extension with Galois group $$G$$, let $$L \subseteq T$$ be an algebraic extension of $$L$$ and let $$\alpha \in T$$. Let $$f(x) \in L[x]$$ the minimal polynomial of $$\alpha$$ over $$L$$. Show that the minimal polynomial of $$\alpha$$ over $$K$$ is the product $$\prod_{h(x) \in H} h(x)$$ where $$H$$ is the set of different polynomials of the form $$\sigma f(x)$$ with $$\sigma \in G$$.

I got no idea how to begin with. I only know that if $$h(x) \in H$$, then $$h(x) = \sigma f(x)$$, so $$\prod_{h(x) \in H} h(x) = \prod_{\sigma \in G} \sigma f(x).$$ But I don't have any significant progress from this.

Any hint or answer is very welcome. Thanks in advance.

• To show that a polynomial $p \in K[x]$ is the minimal polynomial of $\alpha$, you need to do two things: first, show that $p(\alpha) = 0$, then show that $\operatorname{deg}(p) = [K(\alpha) : K]$. Before either of these, though, you need to make sure that the polynomial you have is actually an element of $K[x]$! How would you try to prove this? Jun 18, 2021 at 18:36

## 2 Answers

Let $$g(x)=\prod\limits_{h(x)\in H} h(x)$$. First of all, it is not hard to check that we have $$\tau g(x)=g(x)$$ for all $$\tau\in G$$. This means the coefficients of $$g$$ belong to the fixed field of $$L/K$$, which is $$K$$. So indeed $$g(x)\in K[x]$$.

Next, we clearly have $$g(\alpha)=0$$, because $$f(x)$$ itself is one of the factors in the product which defines $$g$$. Thus, if $$m(x)\in K[x]$$ is the minimal polynomial of $$\alpha$$ over $$K$$ then $$m|g$$. So if we show that also $$g|m$$ then we are done. In order to do this, first note that clearly $$f|m$$ over $$L$$, because $$f$$ is the minimal polynomial of $$\alpha$$ over $$L$$. It follows that for all $$\sigma\in G$$ we have $$\sigma f|\sigma m$$. But the coefficients of $$m$$ belong to $$K$$, and so $$\sigma m=m$$ for all $$\sigma\in G$$. Thus $$\sigma f|m$$ over $$L$$ for all $$\sigma\in G$$. In particular, for all $$h\in H$$ we have $$h|m$$ over $$L$$. Since the elements of $$H$$ are distinct irreducible polynomials in $$L[x]$$, they are pairwise coprime, and so their product (which is $$g$$) divides $$m$$ as well.

Let $$\operatorname {deg} (f(x)) =n$$ and let $$[L:K] =m$$. Now we have $$[L(\alpha) :L] =n$$ and hence $$[L(\alpha) :K] =mn$$. And therefore $$\alpha$$ is algebraic over $$K$$ and of degree $$mn$$ over $$K$$.

Thus if $$g(x)\in K[x]$$ is any monic polynomial of degree $$mn$$ with $$g(\alpha) =0$$ then $$g(x)$$ is the minimal polynomial for $$\alpha$$ over $$K$$.

We show that $$g(x) =\prod_{\sigma \in G} \sigma(f(x))$$ is monic, of degree $$mn$$ and has coefficients in $$K$$ and $$g(\alpha) =0$$.

The part $$g(\alpha) =0$$ follows because there is one identity map $$\sigma\in G$$ for which $$\sigma(f(x)) =f(x)$$ and $$f(\alpha) =0$$.

Further since $$f(x)$$ is monic $$\sigma(f(x))$$ is also monic for every $$\sigma\in G$$ (why? Because $$\sigma$$ maps the leading coefficient $$1$$ of $$f$$ to $$1$$). And thus $$g(x)$$ is also monic. And since $$|G|=[L:K] =m$$ we have $$\operatorname {deg} (g(x)) =mn$$.

It remains to show that $$g(x) \in K[x]$$. Well, this is where the $$G$$ being Galois group comes into picture. Let $$\tau\in G$$ and then $$\tau(g(x)) =\tau\prod_{\sigma\in G}\sigma(f(x)) =\prod_{\sigma\in G} \tau(\sigma(f(x)))$$ As $$\sigma$$ runs through all members of $$G$$ so does $$\sigma'=\tau\sigma$$ and hence we have $$\tau(g(x)) =\prod_{\sigma'\in G} \sigma '(f(x)) =g(x)$$ And this holds for every $$\tau\in G$$. Thus coefficients of $$g(x)$$ lie in the fixed field of $$G$$ which is $$K$$. Then $$g(x) \in K[x]$$ and the proof is now complete.