Question about Thm 1.4.4 in ANT Alaca/Williams

Question

I am studying Introductory ANT by Alaca/Williams, p12, theorem 1.4.4:

"Let $m$ be a nonsquare integer such that $\mathbb {Z}+\mathbb{Z}\sqrt {m}$ is a PID. Let $p $ be an odd prime for which the Legendre symbol $\left (\frac {m}{p}\right)=1. $ Then there exist integers such that $ p=u^2-mv^2$ if $m <0$, or if ( ... )"

Considering that $\mathbb {Z}+\mathbb{Z}\sqrt {-11} \ $ is a PID and that $\left (\frac {-11}{23}\right)=1 \ $ I would have expected to be able to find integers $ u, v $ such that $23=u^2+11v^2$ but I can't. Please explain what I do wrong or misunderstand.

Answer 1

You overlooked the fact that $-11 = 1 \mod 4$. (This is somewhere in that first chapter, but I don't have the book with me at the moment).

That means you also have to look at the so-called "half-integers," and you're looking to solve $23 = \frac{u^2}{4} + \frac{11v^2}{4}$. The answer is $u = \pm9, v = \pm1$. Indeed, $$\left(\frac{9}{2} - \frac{\sqrt{-11}}{2}\right)\left(\frac{9}{2} + \frac{\sqrt{-11}}{2}\right) = 23.$$

Answer 2

That book really needs a Chapter 0, in my opinion. If you purport to write an introduction to algebraic number theory, you can't assume your readers know anything other than elementary number theory, basic algebra and basic set theory. And by "basic algebra," I mean really basic, like solving an equation in one variable, multiplying binomials and using the quadratic formula to solve $ax^2 + bx + c$. My proposed Chapter 0 would cover things like the difference between algebraic numbers and algebraic integers; what is a homomorphism, isomorphism, and all those whatever-morphisms; and aspects of notation, like why you bother to write $\mathbb{Z} + \mathbb{Z}\sqrt{m}$ rather than $\mathbb{Z}[\sqrt{m}]$, and why would anyone would want to go through the trouble of defining $K = \mathbb{Q}(\sqrt{m})$ and then $\mathcal{O}_K$.

But anyway, it would do you a lot of good to carefully reread Example 1.1.5. Still, I would add to Theorem 1.4.4 the stricture that $m \not\equiv 1 \pmod 4$ (understanding of course that $\mu(m) \neq 0$ [the Möbius function]). If $\mathbb{Z}[\sqrt{m}]$ is not integrally closed, then it can't be a PID. And if $m \equiv 1 \pmod 4$, then we know for a fact that $\mathbb{Z}[\sqrt{m}]$ is not integrally closed and that it is a subdomain of $\mathbb{Z}[\frac{1 + \sqrt{m}}{2}]$, or $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$. So Theorem 1.4.4 does not apply in this case, and that is why it is immediately followed by Theorem 1.4.5.

The reason you can't find a solution to $u^2 + 11v^2 = 23$ in integers is because there is no such solution to be found. Try the following query on Wolfram Alpha: solve u^2 + 11v^2 = 23 in integers. Now try this one: solve u^2/4 + (11v^2)/4 = 23 in integers. That works because the norm of an algebraic integer in $\mathbb{Z}[\frac{1 + \sqrt{m}}{2}]$ is $\frac{u^2 + mv^2}{4}$ and $\mathcal{O}_{\mathbb{Q}(\sqrt{-11})}$ is a principal ideal domain, and that domain includes $\mathbb{Z}[\sqrt{-11}]$ as a subdomain.

Now, I can't guarantee that you will never find an example of a non-PID where nevertheless one particular case of $(\frac{m}{p}) = 1$ does have a solution for $u^2 - mv^2 = p$. But if there is such a case, I do guarantee that in that same non-PID you will also find a case where that does not hold.