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

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.

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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

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

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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.

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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