## A subtle question about principality

Dear Professor Kim,

I just have a few small questions. In lectures, we calculated the class group of $Q(\sqrt{10})$, which has ring of algebraic integers $Z[\sqrt{10}]$. We then found that that (maximal ideal) $P_2 = (2,\sqrt{10})$, $N(P_2) = 2.$

After a bunch of calculations we had to see whether $P_2$ was principal or not. Using the result,

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$I$ (non-zero) is principal iff there exists $\alpha \in I$ s.t

$|N(\alpha)| = N(I)$
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We had to consider if there was $n,m \in Z$ s.t

$|N(n+m\sqrt{10})| = 2$

Since the general element of the ideal $P_2$ is $2n+m\sqrt{10}$, is it ‘more correct’ to consider if there was $n,m$ s.t

$|N(2n+m\sqrt{10})| = 2$

I know this doesn’t make a whole lot of difference, it’s just one of those things.

So

$n^2 - 10m^2 = \pm2$

which was equivalent to the statement

$n^2 = 2 \mod 5$ or $n^2 = 3 \mod 5$

In general, do we consider any ‘modulo n’ so that the statement is simplified?

Many Thanks
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You are absolutely right about the principality question. That is, when we have an ideal $I$ in the ring of integers $O_K$ of an algebraic number field $K$ such that $N(I)=n$, then we are led to consider solutions to the equation

$N(z)=\pm n$

for various $z$. As we’ve seen in many examples, once $z$ is expressed in terms of an integral basis, this becomes an equation in $d=[K:Q]$ variables with integer coefficients to which we can consider solutions. Now, if there are no solutions with $z\in O_K$, then, a fortiori, there are no solutions $z\in I$ and we can conclude that $I$ is not principal. However, although I’m too lazy now to cook up an example, there are situations where there *is* a solution $z\in O_K$, but no solution $z\in I$. In the example you mention, the equation corresponding to $z\in I$ is easily found to be

$4n^2-10m^2=\pm 1$

which is obviously more resrictive than

$n^2-10m^2=\pm 1.$

In our case, the latter already has no solution, so we don’t need to consider the more refined equation.

Regarding your second question, firstly, the equation is not *equivalent* to the congruence equation but just implies it. So if the congruence equation has no solution, neither does the original, which is how we used it. Now, I don’t quite understand your final question, but perhaps I should remark that considering congruences is a standard way of investigating solutions to quadratic equations. In fact, it is useful for *any* Diophantine equation. However, a rather deep theorem says that for quadratic equations, sufficiently many congruence equations completely determine whether or not the original equation has rational solutions.