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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 $\mathrm{Gal}(K/F)$?

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$z_i(a)$ is not in $F$, unless $a$ is already in $F$. –  Dylan Moreland Feb 16 '12 at 20:51
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  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?

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