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.

I'm working my way though a classical geometry book by Hartshorne right now, but this problem popped up in a section I'm reading. It is Problem 13.10 from Hartshorne's Geometry: Euclid and Beyond if you're curious.

Anyway, the problem states:

If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$, then the square root is actually in $\mathbb{Z}[\sqrt{2}]$.

I'm not super familiar with algebra, so I'm having trouble interpreting the question, but I would like to know how to solve it.

I looked up $\mathbb{Q}(\sqrt{2})$ on wikipedia, and it seems that it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Q}\}$. I couldn't find $\mathbb{Z}[\sqrt{2}]$, but I assume it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Z}\}$.

So if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$ means there exists some $c+d\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ such that $$ (c+d\sqrt{2})^2=c^2+2d^2+2cd\sqrt{2}=a+b\sqrt{2}. $$ This implies (I think?) that $c^2+2d^2=a$ and $2cd=b$. If this is the correct path, is there then someway to conclude that $c$ and $d$ are in fact integers? Thanks.

By the way, is this exercise easily related to some aspect of classical geometry? It seems kind of out of the blue to me.

share|improve this question
1  
If you wish to construct the diagonal of the triangle with sides $a,b\in \mathbb{Z}$, then if the diagonal is a rational multiple of $\sqrt{2}$, then it actually is an integer multiple. That's the quieckest geometrical interpretation I can think of. I have to go now, but I'll think about the actual solution of the problem when I get back. –  Bruno Stonek Mar 6 '11 at 20:45
    
@Bruno, Thanks, that's a nice way to look at it. I'd be glad to see your solution later. –  yunone Mar 6 '11 at 21:21

2 Answers 2

up vote 3 down vote accepted

First $\rm\ 2\: c^2\: $ times $\rm\ c^2+2\ d^2 =\: a\ $ yields $\rm\ 2\: c^4 + b^2 =\: 2\: a\ c^2\ $ hence $\rm\ 2\: c\in \mathbb Z\ $ by the Rational Root Test.

Next $\:4\ $ times $\rm\ 2\ d^2 = \:a-c^2\: \ \to\ \ 8\ d^2 = \:4\ a\ - (2\:c)^2 \in \mathbb Z\ \:$ thus $\rm\: 2\: d\in \mathbb Z\ $

Finally $\rm\: 4\ a - 2\ (2\: d)^2 \:=\: (2\:c)^2\: \Rightarrow\ 2\:|\:(2\ c)^2\Rightarrow 2\: |\: 2\: c\ \Rightarrow\ c\in \mathbb Z\ \Rightarrow\ d\in \mathbb Z\quad\quad$ QED

share|improve this answer
    
Thanks Bill. To see if I'm applying this correctly, in the first case, by RRT, possible rationals roots are given by $c=\pm\frac{p}{1}$ or $c=\pm\frac{p}{2}$, for $p$ a factor of $b^2$? The first implies $c$ is an integer, and the second implies $2c$ is an integer, but how do you know that $c$ equals one of these possibilities? For aren't they only possibilities, and maybe the polynomial has no rational roots? –  yunone Mar 6 '11 at 22:44
    
@yun: Above shows $\rm\:c\:$ is a root of $\rm\:2\ x^4 - 2\:a\ x^2 + b^2\in \mathbb Z[x]\:.\ $ Since the Rational Root Test implies that the leading coefficient suffices as a denominator for every rational root, we infer $\rm\ 2\ c \in \mathbb Z\:.$ –  Bill Dubuque Mar 6 '11 at 23:02
    
Oh ok, silly of me, of course there is a rational root since $c$ is one. Thank you, this is a nice clean answer. –  yunone Mar 6 '11 at 23:07

First of all $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt{2}$, where $\mathbb{Z}[\sqrt{2}]$ is the smallest ring containing $\mathbb{Z}$ and $\sqrt{2}$. (That is square brackets mean ring, parentheses mean field).

From what you have, I am confident you can conclude that $c$ and $d$ are integers. Here's what i would do to find it directly:

Let $c=\frac{p}{q}, d=\frac{r}{s}$ with $gcd(p,q)=1$ and $gcd(r,s)=1$. Since $cd=\frac{pr}{qs}$ is an integer, this implies that $qs$ divides $pr$ and because of the conditions $gcd(p,q)=1$ and $gcd(r,s)=1$, $qm=r$ and $sn=p$, with $n$ and $m$ integers.

And we know $a=\frac{s^2n^2}{q^2}+2\frac{q^2m^2}{s^2}=\frac{s^4n^2+2q^4m^2}{q^2s^2}$ is an integer. Hence $q^2s^2$ divides $s^4n^2+2q^4m^2$. Since $q^2$ divides the second term and the whole thing, it must divide the first term, and the same with $s^2$. But since $q^2$ and $s^2n^2$ are relatively prime, this implies $q^2$ is 1, and the same follows for $s$.

I'm fairly certain there is a cleaner way to do this using the Gauss Lemma, but this is something you can work out directly, which is also nice.

share|improve this answer
    
Thanks for this answer Becca. One thing, how do you know $cd=\frac{pr}{qs}$ is an integer? I only see that $2cd$ is an integer, but what's to stop $cd$ from being something like $\frac{9}{2}$? –  yunone Mar 6 '11 at 20:57
    
Ah, yes, I missed the 2. Well, we have that either $q$ or $s$ is even, and divide whichever one it is by 2. So for instance, if it is $q$, then $qm/2=r$. Playing with the two cases where $q$ is even or $s$ is even, I think will get you there. –  Becca Winarski Mar 6 '11 at 22:34

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.