Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $D\in\mathbb{Z}$ be a squarefree integer and let $a\in\mathbb{Q}$ be a nonzero rational number. Please show that $\mathbb{Q}(\sqrt{a\sqrt{D}})$ cannot be a cyclic extension of degree 4 over $\mathbb{Q}$. Also show that if $\mathbb{Q}(\sqrt{a\sqrt{D}})$ is Galois over $\mathbb{Q}$ then $D=-1$.

I have the case $D=-1,|a|\neq2b^2$ for some $b\in\mathbb{Q}$ that the Galois group is $\mathbb{Z_2}^2$, the Klein four group (which is not cyclic). How to show in general that a group is not cyclic?

share|improve this question

1 Answer 1

Some ideas for you to mull and complete:

$$x=\sqrt{a\sqrt D}\implies x^2=a\sqrt D\implies x^4=a^2D$$

The above almost proves the minimal polynomial (over the rationals) of $\,\sqrt{a\sqrt D}\,$ is $\,p(x)=x^4-a^2D\in\Bbb Q[x]\,$ (can you complete this?)

We get that a basis for the linear space $\,\Bbb Q(\sqrt{a\sqrt D})_{\Bbb Q}\,$ can be

$$\{1\,,\,w\,,\,w^2\,,\,w^3\}\;,\;\;w:=\sqrt{a\sqrt D}$$

But we have that

$$x^4-w^4=(x^2-w^2)(x^2+w^2)\in\Bbb Q(w^2)[x]$$

and this means that no embedding of $\,\Bbb Q(w)\,$ in some algebraic closed field containing it (take $\,\Bbb C\,$ , for example) can map a root of $\,x^2-w^2\,$ to a root of $\,x^2+w^2\,$ and neither the other way around, which already shows the extension cannot be cyclic.

That the extension is Galois means it is normal, and this in turn means $\,Q(w)\,$ contains all the roots of $\,x^4-w^4\,$ , among which there are the ones of $\,x^2+w^2\,$ :

$$\pm w\,i=\pm\sqrt{a\sqrt D}\,i=\pm\sqrt{a\sqrt D(-1)}\;\ldots$$

share|improve this answer
    
Too much good work to go un-upvoted for an hour! +1 –  amWhy Apr 4 '13 at 4:00
    
I was able to show $D<0$, but how do you show $|D|=1$? Of course assuming the degree extension is 4. Thanks. –  user70520 Apr 4 '13 at 5:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.