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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.

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  • $\begingroup$ 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? $\endgroup$ Jun 18, 2021 at 18:36

2 Answers 2

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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.

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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.

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