Since is a field,
I hope I didn't make mistakes, but I've found and
Therefore, if then and
(In fact we even have ).
Say we have a field F and such that Q is a subfield of F and additionally F contains the element
Show that F contains the element
By considering the fact that F must contain we can find that F contains , and hence , and various other results like this. Can we show using various combinations of these results that F contains y? If so, then I'll be able to use this to answer a question about splitting fields.
(I want to find [Q(x,y):Q], and clearly this will just be [Q(x),Q] if y is in Q(x) which will simplify the problem).
and therefore and
Proof: the only non-trivial part is to show that which follows from Eisenstein's criterion because
Proof: as a vector space, is generated by so if then we'll have for some which is clearly impossible.
therefore and hence:
Proof: by the last two parts of we only need to show that so suppose that that is then since
we'll have: call this (1). now dividing both sides of (1) by will give us which is impossible by unless i.e.
but then (1) will give us which is again impossible by .
Remark: since we have you may use this to find another solution to your problem.