5
$\begingroup$

I'm trying to solve the last part of an exercise in Dummit and Foote.

Let $K/F$ be any finite separable extension, and let $\alpha\in K$. Let $L$ be a Galois extension of $F$ containing $K$ and let $H\leq \text{Gal}(L/F)$ be the subgroup corresponding to $K$. Define the norm of $\alpha$ from $K$ to $F$ to be $$ N_{K/F}(\alpha)=\prod_\sigma\sigma(\alpha), $$ where the product is taken over all the embeddings of $K$ into an algebraic closure of $F$, (so over a set of coset representatives for $H$ in $\text{Gal}(L/F)$ by the Fundamental Theorem of Galois Theory.

The part I'm stuck on is

(d) Let $m_\alpha(x)=x^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0\in F[x]$ be the minimal polynomial for $\alpha\in K$ over $F$. Let $n=[K:F]$. Prove that $d$ divides $n$, that there are $d$ distinct Galois conjugates of $\alpha$ which are all repeated $n/d$ times in the product above and conlcude that $N_{K/F}(\alpha)=(-1)^na_0^{n/d}$.

I see that $d$ divides $n$, since $d=[F(\alpha):F]$, and this divides $[K:F]$ by multiplicativity of the degree. I also see that there are $d$ distinct Galois conjugates of $\alpha$ since the minimal polynomial is irreducible and has no repeated roots since $K/F$ is a separable extension. But why are they repeated $n/d$ times? After that, I think I can make the final conclusion.

I let $H'$ be the subgroup corresponding to $F(\alpha)$, so $H\leq H'\leq\text{Gal}(L/F)$. I calculate that $[H':H]=n/d$, but does this imply the result in some way?

$\endgroup$

1 Answer 1

6
$\begingroup$

$H = \text{Gal}(L/F)^K \subset H' = \text{Gal}(L/F)^{F(x)} \subset \text{Gal}(L/F)$

$n = [K:F] = [ \text{Gal}(L/F) : H]$, and $d = [K(x):F] = [\text{Gal}(L/F) : H']$, so $n/d = [H' : H]$

$N_{K/F}(x) = \prod_{\sigma H \in \text{Gal}(L/F) / H} \sigma(x)$ So you need to find out when two $H$-cosets induce the same image for $\sigma(x)$.

The image $\sigma(x)$ only depends on the $H'$-coset containing $\sigma$, since $x$ is fixed by $H'$. Now, since $H \subset H'$, an $H'$-coset is a reunion of $H$-cosets : $H' = \bigcup_{\tau H \in H'/H} \tau H$, so for every $\sigma H' \in \text{Gal}(L/F) / H'$, $\sigma H' = \bigcup_{\tau H \in H'/H} \sigma \tau H$

Every $H'$-coset contains exactly $n/d$ $H$-cosets, so when you group the factors accordingly,

$N_{K/F}(x) = \prod_{\sigma H' \in \text{Gal}(L/F) / H'} \prod_{\tau H \in H'/ H}\sigma\tau(x) = \prod_{\sigma H' \in \text{Gal}(L/F) / H'} (\sigma(x))^{n/d}$

which is precisely what you need to show.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .