5
$\begingroup$

I'm currently reading Marcus' "Number Fields," and I'm having difficulty proving the following result:

Corollary 2.2: Let $m$ be a squarefree integer. The set of algebraic integers in the quadratic field $\mathbb{Q}(\sqrt{m})$ is $$ \left\{ a + b \sqrt{m} : a,b \in \mathbb{Z} \right\} $$ if $m \equiv 2,3 \pmod{4}$; and$$ \left\{ \frac{a+b\sqrt{m}}{2} : a,b \in \mathbb{Z} ~,~ a \equiv b \pmod{2} \right\}$$ when $m \equiv 1 \pmod{4}$. $\square$

It is easy enough to see that the claimed sets do, indeed, consist of algebraic integers. The other half of the proof proceeds something like this: suppose that $\alpha = r + s \sqrt{m}$ is an element of $\mathbb{Q}(\sqrt{m})$. Then $\alpha$ is a root of the polynomial $$ x^2 - 2rx + r^2 - s^2m $$ In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers. At this point, the proof stops, and says that this implies the result. And therein lies the problem: I don't see how to make the connection. I do know, for example, that if one of $r,s$ is an integer, then the other one must be, too, i.e., either both are integers, or both are not. However, I cannot get any further than that. Can anyone give me a gentle nudge in the right direction? Thank you very much!

$\endgroup$
2
$\begingroup$

In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers.

This requires more work than what you've done so far. You need to further know that $x^2 - 2rx + (r^2 - s^2 m)$ is the minimal polynomial of $\alpha$. Fortunately this is clear as long as $s \neq 0$, which is the interesting case anyway.

I don't see how to make the connection.

$2r$ is an integer; let's call it $n$. Then

$$r^2 - s^2 m = \frac{n^2}{4} - s^2 m$$

must also be an integer, or equivalently,

$$4r^2 - 4s^2 m = n^2 - (2s)^2 m \equiv 0 \bmod 4$$

must be an integer divisible by $4$. If $n$ is even, this means that $(2s)^2 m$ must be an integer divisible by $4$, or equivalently $s^2 m$ must be an integer, and since $m$ is squarefree this implies that $s$ must be an integer. (Otherwise, if the denominator of $s$ in lowest terms was $d$, then $m$ would have to be divisible by $d^2$.)

So the interesting case is when $n$ is odd, in which case $n^2 \equiv 1 \bmod 4$, and hence

$$(2s)^2 m \equiv 1 \bmod 4.$$

Since $m$ is squarefree this implies that $2s$ must be an integer, by the same argument as above, and moreover it must be an odd integer (so $s$ is half an odd integer, the same as $r$). It follows that $(2s)^2 \equiv 1 \bmod 4$, hence

$$m \equiv 1 \bmod 4.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.