Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $K$ be a Galois extension of $F$, and let $a\in K$. Let $n=[K:F]$, $r=[F(a):F]$, and $H=\mathrm{Gal}(K/F(a))$. Let $z_1, z_2,\ldots,z_r$ be left coset representatives of $H$ in $G$. Show that $$\min(F,a)=\prod_{i=1}^r\left( x - z_i(a)\right).$$

In this product $\min(F,a)$ is of degree $r$. That is true from fundamental theorem.

And if one of $z_i$ is the identity, then $a$ satisfies the polynomial.

My question is:

What is the guarantee that $z_i(a)$ belongs to $F$ for all $i$?

What if I choose a representative which is not the identity of $\mathrm{Gal}(K/F)$?

share|improve this question
    
$z_i(a)$ is not in $F$, unless $a$ is already in $F$. –  Dylan Moreland Feb 16 '12 at 20:51
add comment

1 Answer

  1. There is no guarantee that $z_i(a)\in F$ for all $a$; in fact, it will never be the case unless $a\in F$. But you don't need each to be in $F$, you need the coefficients of the product to be in $F$.

    In order to show that the coefficients of the product are in $F$, it suffices to show that for every $\sigma\in G$ we have $$\sigma\left(\prod_{i=1}^r(x-z_i(a))\right) = \prod_{i=1}^r\left(x - \sigma(z_i(a))\right) = \prod_{i=1}^r(x-z_i(a)),$$ because the coefficients lie in $F$ if and only if the coefficients are invariant under the action of the Galois group.

  2. If, say $z_iH = \mathrm{id}_GH$, then that means that $z_i\in H$. But $H$ is precisely the subgroup of elements that fix $F(a)$. So what is $z_i(a)$ then?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.