# Ring of Integers for $\mathbb{Z}[\sqrt{d}]$

I'm pretty sure this is a very basic thing, but my background is in physics and I have never previously done any number theory.

We have as a theorem that for an algebraic number field $$K$$, $$\alpha \in K$$ is an algebraic integer if and only if its minimal polynomial in $$\mathbb{Q}$$ has coefficients in $$\mathbb{Z}$$. The minimum polynomial for $$\alpha = a + b \sqrt{d}$$ in the field $$\mathbb{Q}[\sqrt{d}]$$ is then:

$$(X - (a + b \sqrt{d}))(X - (a - b\sqrt{d})) = X^{2} - 2aX + (a^{2} - b^{2}d)$$

Therefore $$\alpha$$ is an algebraic integer $$\iff 2a \in \mathbb{Z}, a^{2}-b^{2}d \in \mathbb{Z}$$.

Fine, but why is this claim true?

$$d = 2,3 \; \bmod \; 4 \implies \mathscr{O}_{\mathbb{Q}[\sqrt{d}]} = \mathbb{Z}[\sqrt{d}]$$ $$d= 1\; \bmod \; 4 \implies \mathscr{O}_{\mathbb{Q}[\sqrt{d}]} = \mathbb{Z} \bigg[\frac{1+\sqrt{d}}{2}\bigg]$$

Is this somehow related to the elementary theorem about when the sum of two squares is an integer?

• You mean $\alpha$ an algebraic integer $\iff \ldots$. May 23, 2019 at 0:03
• Yes, corrected thanks May 23, 2019 at 0:05
• Note $a^2\equiv0$ or $1\pmod4$ and same with $b^2$; cf. this question May 23, 2019 at 0:06

You got $$n=2a \in\Bbb Z$$ and $$a^2-db^2\in \Bbb Z.$$ These imply $$m=2b\in\mathbb Z$$ and $$n^2-dm^2\in\Bbb Z.$$ $$n^2$$ and $$m^2\equiv 0$$ or $$1\pmod4$$. If $$n^2\equiv dm^2\pmod4$$ and $$d\equiv2$$ or $$3\pmod4$$, the only solution is $$n^2\equiv m^2\equiv0\pmod4$$ so $$n\equiv m\equiv 0\pmod2$$ so $$a, b \in\Bbb Z$$. If $$d\equiv1\pmod4$$ there is also the solution $$n^2\equiv m^2\equiv1\pmod4$$, in which case $$a$$ and $$b$$ are half odd integers.

• Note: the fact that $0^2\equiv2^2\equiv0$ and $1^2\equiv3^2\equiv1\pmod4$ also means that if $p\equiv3\pmod4$ then $p$ cannot be expressed as a sum of two squares May 23, 2019 at 14:55

Yes, it is related to that result. If $$a$$ and $$b$$ are both rational numbers (not necessarily integers), and $$d$$ is an integer, then $$a^2 - db^2$$ is sure to be rational, but it may or may not be an integer.

If $$a$$ and $$b$$ are both halves of odd integers (call them $$\alpha$$ and $$\beta$$), and $$d \equiv 1 \pmod 4$$, it will happen that $$a^2 = \left(\frac{\alpha}{2}\right)^2 = \frac{\alpha^2}{4}$$ with $$\alpha^2 \equiv 1 \pmod 4$$, and likewise $$b^2 = \left(\frac{\beta}{2}\right)^2 = \frac{\beta^2}{4}$$ with $$\beta^2 \equiv 1 \pmod 4$$. Since $$d \equiv 1 \pmod 4$$ as well, we then have $$a^2 - db^2 = \frac{\alpha^2}{4} - \frac{d \beta^2}{4},$$ from which it obviously follows that $$\alpha^2 - d \beta^2$$ is a multiple of 4.

It often helps to work these things out with concrete examples. Since you mention a physics background, I'm going to assume you're well-versed in the arithmetic of complex numbers.

Try $$d = -3$$. Given the number $$\frac{-11}{2} + \frac{7 \sqrt{-3}}{2},$$ we readily see that the trace (the $$2a$$) is $$-11$$ and the norm (the $$a^2 - db^2$$) is 67. The relevant polynomial is then $$x^2 + 11x + 67$$.

Now try $$d = -5$$ and the number $$\frac{-11}{2} + \frac{7 \sqrt{-5}}{2}.$$ The trace is indeed $$-11$$ for this number as well, but the norm is... $$\frac{121}{4} + \frac{5 \times 49}{4} = \frac{183}{2},$$ and the polynomial is $$2x^2 + 22x + 183$$.