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Finding the norm of an element in a field, say, $\mathbb{Z}[\sqrt{19}]$ is rather easy, it just requires the computation of $(a+b\sqrt{19})(a-b\sqrt{19})$. However, given a value, say, 2, how do we find elements in $\mathbb{Z}[\sqrt{19}]$ that have this norm? I guess it boils down to finding integer solutions of the equation $2=a^2-19b^2$, how do we solve this equation?

For some context to my question I'm trying to prove that $\mathbb{Z}[\sqrt{19}]$ is a UFD, and I'm showing that the two $\mathbb{Z}$-primes less than $\sqrt{19}$, 2 and 3, factor into principal ideals, therefore I'm checking that one of ±2 and one of ±3 is a norm.

Thanks in advance!

DurhamManiac
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  • $2$ is not a quadratic residue modulo $19$, so the equation $2 =a^2 - 19 b^2$ can't have a solution. The same happen with $3 = a^2 - 19 b^2$. – themaker Apr 23 '17 at 17:25
  • Okay, but for the case where $-2=a^2-19b^2$ has solution $a=13, b=3$ which I got through trial and error, I'm just wondering if there's a way to find these solutions without using trial and error? – DurhamManiac Apr 23 '17 at 17:30
  • Or maybe there's a way of generating more solutions from the first solution you find with trial and error? I don't know, any thoughts would be appreciated! :-) – DurhamManiac Apr 23 '17 at 17:33
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    yes, if you have one solution, you can generate more in the following way: let a, b your particular solution, and let c, d a solution of the pell equation $c^2 - 19 d^2 = 1$, then the number $(a + \sqrt{19}b)(c + \sqrt{19} d)$ is a solution as well. – themaker Apr 23 '17 at 17:39
  • Ah fantastic, that's very useful thank you! I'll leave this question open to see if there's a concrete way of finding an element without trial and error but using Pell's equation is a great idea if you have an element already! – DurhamManiac Apr 23 '17 at 17:42

1 Answers1

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This is quite difficult in general, but in a quadratic field, knowing the fundamental unit can be useful.

Let $\varepsilon>1$ be the fundamental unit in your quadratic field. If you seek $\alpha$ with $N(\alpha)=m>0$ then if there is such an $\alpha$ there will be one with $1\le\alpha<\varepsilon$. Its conjugate $\alpha^*$ satisfies $m\ge\alpha^*>m/\varepsilon$. (If $m>0$ we need slightly different inequalities). This gives bounds on $\alpha\pm\alpha^*$ which must be an integer or $\sqrt d$ times an integer (where your field is $\Bbb Q(\sqrt d)$). This gives a finite (maybe large) set of possibilities to check.

Angina Seng
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  • So I've tried this with a simple example, $\mathbb{Z}[\sqrt{3}]$, looking for an element with norm 7. The fundamental unit is $2+\sqrt{3}$. Therefore we get $1\leq\alpha<2+\sqrt{3}$ and $7\geq\alpha^>14-7\sqrt{3}$ and therefore $15-7\sqrt{3}<\alpha+\alpha^<9+\sqrt{3}$. I'm struggling as to what to do now though. – DurhamManiac Apr 23 '17 at 17:48
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    If $\alpha = a+b\sqrt 3$ that means that $2a$ is between $15-7\sqrt3$ and $9+\sqrt3$. That's not many possibilities for $a$. – Angina Seng Apr 23 '17 at 17:53
  • Oh of course, so we get $a\in{2, 3, 4, 5}$. Then substituting these into the equation $a^2-3b^2=7$ gives us no solutions. Fantastic, thanks very much for your help! I might just get a degree this year, who knows! :-) – DurhamManiac Apr 23 '17 at 18:12