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.

Following my previous question the book then asks

Use this to determine when the field extension $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ over $\mathbb{Q}$ is biquadratic (where $a,b\in\mathbb{Q}$)

"this" means what proved in the linked post.

My thoughts: First, I don't know what to say when $a^2-b$ is not a square. When $a^2-b$ is a square I think that the extension is of degree $2$ if $a=b$, this leaves the case $a\neq b$ which I am also having problems with and I don't know what to do (I have some cases like if $m,n$ are different primes then the degree of the extension is $4$, but I don't have something that is general)

Can someone please help my do this exercice ?

share|improve this question
    
@MTurgeon - I don't think I do, and I don't think its mentioned in the book (I read everything up till this question) –  Belgi Sep 9 '12 at 21:57
    
Sorry, I had my terminology a bit mixed up. I thougt you asked about extensions of degree $4$ and not $2$ (this is what the question also asks so I'll edit) –  Belgi Sep 9 '12 at 22:05
    
I think it's enough for $m,n$ to be coprime (not necessarily integer, in the natural sense). It's still not complete, though. –  tomasz Sep 9 '12 at 22:58
    
$m, n$ coprime is not sufficient: $\sqrt{7+\sqrt{48}}=2+\sqrt{3}$ is a counterexample. It's not that hard to prove the extension is biquadratic if neither $b$ nor $a^2-b$ is square by the bye. –  anonymous Sep 10 '12 at 1:14

1 Answer 1

up vote 2 down vote accepted

Note firstly that you may assume $a$ and $b$ are integers without loss of generality. What we want is for $(x^2-a)^2-b$ to be irreducible and thereby the minimal polynomial of $\sqrt{a+\sqrt{b}}$. So let's factor it over $\mathbb{R}$: $$(x^2-a)^2-b=(x-\sqrt{a+\sqrt{b}})(x+\sqrt{a+\sqrt{b}})(x-\sqrt{a-\sqrt{b}})(x+\sqrt{a-\sqrt{b}})$$ So if we can show that the quadratics $(x-\sqrt{a+\sqrt{b}})(x+\sqrt{a+\sqrt{b}})$, $(x-\sqrt{a+\sqrt{b}})(x-\sqrt{a-\sqrt{b}})$, $(x-\sqrt{a+\sqrt{b}})(x+\sqrt{a-\sqrt{b}})$ and all 4 linear terms are not polynomials over $\mathbb{Q}$, then it will be irreducible and the field extension will be biquadratic.

Let's look firstly at $$(x-\sqrt{a+\sqrt{b}})(x+\sqrt{a+\sqrt{b}})=x^2-a-\sqrt{b}$$ which is a polynomial over $\mathbb{Q}$ if and only if $b$ is a square. So let's assume $b$ is a nonsquare from now on. Then in particular $\pm\sqrt{a\pm\sqrt{b}}$ will never be rational, so that also deals with potential linear terms.

Now we deal with the other two quadratics: $$(x-\sqrt{a+\sqrt{b}})(x\pm\sqrt{a-\sqrt{b}})=x^2-x*(\sqrt{a+\sqrt{b}}\pm\sqrt{a-\sqrt{b}})\pm\sqrt{a^2-b}$$ So for that to be a polynomial over $\mathbb{Q}$ it must be the case that $a^2-b$ is a square which tells us from your previous question that there exist $m,n\in\mathbb{N}$ such that $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$. Looking at Hagen's answer to the previous question, we also get $\sqrt{a-\sqrt{b}}=\sqrt{m}-\sqrt{n}$, so depending on the sign, the linear term in our quadratic above will be either $2\sqrt{m}$ or $2\sqrt{n}$. We'll assume it's $2\sqrt{m}$ since the situation is symmetrical in $m$ and $n$. So the relevant condition then becomes that $m$ is a square. We now note the following: $$2\sqrt{b}=(\sqrt{a+\sqrt{b}})^2-(\sqrt{a-\sqrt{b}})^2=(\sqrt{m}+\sqrt{n})^2-(\sqrt{m}-\sqrt{n})^2=4\sqrt{m}\sqrt{n}$$ $$b=4mn$$ This implies in particular that 4 divides $b$ and in fact we also have a sort of converse: if 4 divides $b$, then we can factorize $b=4u^2v$ in different ways (depending on the amount of square factors $b$ has) and each such factorization corresponds to a choice of $m=u^2$, $n=v$ by our last equation and then $a=m+n=u^2+v$ by the answer to the previous question. You can check that $a^2-b$ is again a square.

So the conclusion is that for that extension to fail to be biquadratic, it is required that either $b$ is a square or there exists a factorization $b=4u^2v$ such that $a=u^2+v$ (in which case in particular $a^2-b$ is a square).

share|improve this answer

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.