Let be a subfield of . Let be an element of and let where denotes the real root of .

Prove thatif is any Galois extension of contained in then .

I have no idea how to approach this problem.

thanks everyone!

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- April 12th 2010, 03:52 PMmathman88Galois Extension
Let be a subfield of . Let be an element of and let where denotes the real root of .

Prove thatif is any Galois extension of contained in then .

I have no idea how to approach this problem.

thanks everyone! - April 12th 2010, 09:32 PMFancyMouse
First, you need to know that the minimal polynomial of over F is for some (not a hard lemma)

Then if L is Galois over F, we must have in order to throw all conjugates of into K. But clearly is not real unless . - April 13th 2010, 03:34 AMtonio
- April 13th 2010, 12:21 PMmathman88
- April 13th 2010, 03:36 PMFancyMouse
Sorry. I mixed up with something else. The ``lemma'' I stated before is not only unnecessary but also wrong. The correct version of the lemma needs the assumption that . The proof of the correct version needs Hilbert Theorem 90 using norms (or traces if n divides the characteristic of F), which is not hard but nontrivial to prove.

However, we don't even need this. is a root of x^n-a implies that all roots are of the form for some k (but not the converse, i.e. not all things of this form are the roots of x^n-a), since the minimal polynomial of must divide x^n-a. But use my idea before, that only two of the can be real. This implies that can only have at most two conjugates, hence the degree is <= 2 - April 13th 2010, 04:04 PMmathman88