Separable field extensions Log Out | Topics | Search Moderators | Register | Edit Profile

 Ask NRICH » Archive 2010-2011 » Higher Dimension » Separable field extensions « Previous Next »

Author Message
George
Veteran poster

Post Number: 466
 Posted on Tuesday, 22 March, 2011 - 01:35 pm:

Question: Let M/L/K be a tower of field extenstions. Prove that if L/K is separable and a belonging to M is separable over M, then a is separable over K.

My thoughts:

I'll use ma/x to be the minimum polynomial of a over the field X.

So far i've argued as follows:

Let ma/K be the minimum polynomial of a over K, then it is monic, irreducible and has a as a root. Assuming it is not linear, otherwise trivial case, it has other roots. Due to it being monic and irreducible ma/K=mb/K for all other roots b. Now if one of these roots belongs to L, then as L/K is separable we know that there exists a splitting field extension for which it has distinct roots.

So i'm now trying to show that it must have a root belonging to L.

Now ma/L divides ma/K and so ma/K=g(x)ma/L, assuming g(x)=/=1 otherwise we are done, we can deduce that ma/K is reducible over L, otherwise we would have ma/K=ma/L. ma/L is irreducible so we can split g(x) into irreducible factors. The fact it's reducible over L doesn't necessarily mean it's got roots in L X4+3X2+2=(X2+2)(X2+1) has no roots in the rationals but is reducible.

I can't see how to deduce that g has a root in L. ie why it can't have all of its roots in M. Part of me thinks that if all of its roots are in M then that part would be included in ma/L, but then it might be needed to bring the polynomial back into the ring K[x].

Any help would be appreciated, i think my whole approach just might not be the right way to go about it. I know we require characteristic >0 for separability to even be an issue, so maybe results from that might be the way forward.

Thanks