Galois Extension

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April 12th 2010, 02:52 PM
mathman88

Galois Extension

Let $F$ be a subfield of $\mathbb{R}$ . Let $a$ be an element of $F$ and let $K=F(\sqrt[n]{a})$ where $\sqrt[n]{a}$ denotes the real $n^{th}$ root of $a$ .

Prove thatif $L$ is any Galois extension of $F$ contained in $K$ then $[L:F]\leq2$ .

I have no idea how to approach this problem.

thanks everyone!
April 12th 2010, 08:32 PM
FancyMouse

First, you need to know that the minimal polynomial of $\sqrt[n]{a}$ over F is $x^d-b$ for some $d|n,b\in F$ (not a hard lemma)
Then if L is Galois over F, we must have $\zeta_d\in F$ in order to throw all conjugates of $\sqrt[d]{b}$ into K. But clearly $\zeta_d$ is not real unless $d\leq2$ .
April 13th 2010, 02:34 AM
tonio

Quote:

Originally Posted by FancyMouse

First, you need to know that the minimal polynomial of $\sqrt[n]{a}$ over F is $x^d-b$ for some $d|n,b\in F$ (not a hard lemma)
Then if L is Galois over F, we must have $\zeta_d\in F$ in order to throw all conjugates of $\sqrt[d]{b}$ into K. But clearly $\zeta_d$ is not real unless $d\leq2$ .

Remark: By $\zeta_d$ Fancymouse most probably meant a primitive d-th root of unity in $\mathbb{C}$

Tonio
April 13th 2010, 11:21 AM
mathman88

Quote:

Originally Posted by FancyMouse

First, you need to know that the minimal polynomial of $\sqrt[n]{a}$ over F is $x^d-b$ for some $d|n,b\in F$ (not a hard lemma)
Then if L is Galois over F, we must have $\zeta_d\in F$ in order to throw all conjugates of $\sqrt[d]{b}$ into K. But clearly $\zeta_d$ is not real unless $d\leq2$ .

How is that lemma proven?

Thanks!
April 13th 2010, 02:36 PM
FancyMouse

Quote:

Originally Posted by mathman88

How is that lemma proven?

Thanks!

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 $\zeta_n\in F$ . 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. $\alpha$ is a root of x^n-a implies that all roots are of the form $\zeta_n^k\sqrt[n]{a}$ 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 $\sqrt[n]{a}$ must divide x^n-a. But use my idea before, that only two of the $\zeta_n^k$ can be real. This implies that $\sqrt[n]{a}$ can only have at most two conjugates, hence the degree is <= 2
April 13th 2010, 03:04 PM
mathman88

Quote:

Originally Posted by FancyMouse

But use my idea before, that only two of the $\zeta_n^k$ can be real. This implies that $\sqrt[n]{a}$ can only have at most two conjugates, hence the degree is <= 2

Sorry, but I don't see why this is always true. I see why only up to 2 $\zeta_n^k$ can be real, but I don't see why this implies that $\sqrt[n]{a}$ can only have at most two conjugates, hence the degree is <= 2.