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 .
I think i need to show that m(x) the minimal polynomial of 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 but doesn't split over L. so and so m(x)|f(x). and m(x)=f(x). so m(x) doesn't split???? i'm a bit wary of the part
Well i managed to do it i think for anyone interested;
Proof
so let m(x) be the minimum polynomial of over K.
if deg(m(x))=1 then contradiction since K:K is normal.
so deg(m(x))>1, say n.
suppose m(x) splits in L.
and
let M be a splittin field for m(x) over K such that M is a subset of L. then we must have that for all i. and M is a field so
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