**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 m_{a/x} to be the minimum polynomial of a over the field X. So far i've argued as follows: Let m_{a/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 m_{a/K}=m_{b/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 m_{a/L} divides m_{a/K} and so m_{a/K}=g(x)m_{a/L}, assuming g(x)=/=1 otherwise we are done, we can deduce that m_{a/K} is reducible over L, otherwise we would have m_{a/K}=m_{a/L}. m_{a/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 X^{4}+3X^{2}+2=(X^{2}+2)(X^{2}+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 m_{a/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 |