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