Galois groups

**Krahl** · March 14th 2010, 04:19 PM

If L:K (field extension) is finite and seperable but not normal, then
|Gal(L:K)|<[L:K]. where Gal(L:K) is the group of k-Auts on L. [L:K] is the degree of the extension.

Since L:K is finite and seperable it is simple so i can say $L=K(\alpha)$ .
I think i need to show that m(x) the minimal polynomial of $\alpha$ cannot split over L.
All i can say is that since it's not seperable, there is an irreducible polynomial in K[x] which doesn't split over L. how can i show that m(x) cannot split also.
Any help is good.
Thanks

EDIT: also i know [L:K]=degree of m(x) and |Gal(L:K)|=the number of zeroes m(x) has in L

EDIT2:Hang on can't i use the definition of normality and say; it's not normal so there is an irreducible poly f(x) over K[x] which has a zero in $L=K(\alpha)$ but doesn't split over L. so $f(\alpha)=0$ and so m(x)|f(x). and m(x)=f(x). so m(x) doesn't split???? i'm a bit wary of the $f(\alpha)=0$ part

**aliceinwonderland** · March 14th 2010, 05:39 PM

Originally Posted by Krahl

If L:K (field extension) is finite and seperable but not normal, then
|Gal(L:K)|<[L:K]. where Gal(L:K) is the group of k-Auts on L. [L:K] is the degree of the extension.

Since L:K is finite and seperable it is simple so i can say $L=K(\alpha)$ .
I think i need to show that m(x) the minimal polynomial of $\alpha$ cannot split over L.
All i can say is that since it's not seperable, there is an irreducible polynomial in K[x] which doesn't split over L. how can i show that m(x) cannot split also.
Any help is good.
Thanks

EDIT: also i know [L:K]=degree of m(x) and |Gal(L:K)|=the number of zeroes m(x) has in L

Let $K_1$ be a fixed field of Gal(L:K). It follows that $K \subset K_1 \subset L$ . Then $[L:K]=|Gal(L:K)||K_1:K|$ . Thus, $|Gal(L:K)| < [L:K]$ (It never equals because L is not the normal extension over K).

For example, let $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt[3]{2})$ . Then L is the finite and separable extension over K but not the normal extension over K. You see that $[L:K]=3, |Gal(L:K)|=1, |K_1:K|=3$ .

**Krahl** · March 14th 2010, 05:49 PM

hi thanks for your reply. what does |K':K| mean with the vertical bars.
but surely if $<br /> [L:K]=3, |Gal(L:K)|=1, |K_1:K|=3<br />$ is true then $L=K_1$ since $[L:K]=[L:K_1][K_1:K]$
by the tower law.
also how can $K_1$ be a fixed field of a group.

**aliceinwonderland** · March 14th 2010, 05:54 PM

Originally Posted by Krahl

hi thanks for your reply. what does |K':K| mean with the vertical bars.
but surely if $<br /> [L:K]=3, |Gal(L:K)|=1, |K_1:K|=3<br />$ is true then $L=K_1$ since $[L:K]=[L:K_1][K_1:K]$
by the tower law.
also how can $K_1$ be a fixed field of a group.

A typo there. [K_1:K] instead of |K_1:K|.

**Krahl** · March 20th 2010, 11:30 AM

Well i managed to do it i think for anyone interested;

Proof

$L:K \; finite \; and \; separable \Longrightarrow it's \; simple \Longrightarrow L=K(\alpha)$

$L:K \; finite \Longrightarrow algebraic$

so let m(x) be the minimum polynomial of $\alpha$ over K.

if deg(m(x))=1 then $\alpha \in K$ contradiction since K:K is normal.

so deg(m(x))>1, say n.

suppose m(x) splits in L.
$m(x)=(x-\alpha_1)...(x-\alpha_n)$ and $\alpha_1=\alpha$

let M be a splittin field for m(x) over K such that M is a subset of L. then we must have that $\alpha_i \in M$ for all i. and M is a field so $K(\alpha) \subset M$

so L is a splitting field for m(x) over K. but then this means L:K is finite and normal. so m(x) cannot split in L (or any subset of L). So

$|Gal(L:K)|= no. \; of \; zeros \; of \; m(x) \; in \; L < deg(m(x))=[L:K]$

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