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,

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

|N(\alpha)| = N(I)
—————————————————————–

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
———————————————————

Reply:

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.

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