# Algebraic integers in a quadratic number field

I want to show that an element in $$\alpha$$ in a quadratic number field, $$\mathbb{Q}(d)$$, is an algebraic integer if the Norm and the Trace of $$\alpha$$ are in $$\mathbb{Z}.$$

• What have you tried and where are you stuck? Dec 5 '18 at 18:05

Let $$\alpha\in\mathbb{Q}(\sqrt d)$$ be an algebraic integer; By definition, it is the root of a polynomial $$f(x)\in\mathbb{Z}[x]$$. One can show that in fact, the minimal polynomial of $$\alpha$$ over $$\mathbb{Q}$$, is with integral coefficients (please tell me if you want me to elaborate on this point). Note that if $$\alpha=a+b\sqrt d$$, then $$\alpha$$ is a root of $$x^2-2ax+a^2-db^2\in\mathbb{Q}[x]$$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $$\mathbb{Q}$$ has $$\alpha$$ as a root (otherwise, $$\alpha$$ would be a rational number, and this case is easy to handle with). But then it follows that $$2a,a^2-db^2\in\mathbb{Z}$$, and these are the trace and norm of $$\alpha$$, so we're finished.

Conversely, write $$\alpha=a+b\sqrt d$$. Then $$tr_{\mathbb{Q}(\sqrt d)/\mathbb{Q}}(\alpha)=2a$$ and $$Norm_{\mathbb{Q}(\sqrt d)/\mathbb{Q}}(\alpha)=a^2-db^2$$. If they're both in $$\mathbb{Z}$$, then $$f(x)=x^2-2ax+(a^2-db^2)\in\mathbb{Z}[x]$$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $$f(\alpha)=0$$ and so $$\alpha$$ is algebraic.

Conceptually, $$\alpha$$ is an algebraic integer, iff its minimal polynomial over $$\mathbb{Q}$$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $$\alpha$$.