Number Fields Rings of Integers and Discriminants

If the minimal polynomial is monic over $\mathbb{Z}$, [F1] gives $\alpha \in {\mathcal{O}}_K$.

Conversely, a monic integral polynomial annihilating $\alpha$ is divisible by its monic minimal polynomial; Gauss's lemma makes the latter integral.

Write the monic minimal polynomial of $\alpha$ as $X^n+c_{n-1}{X}^{n-1}+\cdots +c_0\in \mathbb{Q}[X],$ and choose a positive common denominator $m$ of the $c_i$. Multiplying the equation for $\alpha$ by $m^n$ shows that $m\alpha$ satisfies $X^n+(m{c}_{n-1})X^{n-1}+(m^2{c}_{n-2})X^{n-2}+\cdots +m^n{c}_0,$ a monic polynomial in $\mathbb{Z}[X]$.

Thus $m\alpha \in {\mathcal{O}}_K$, and $\alpha =(m\alpha )\mathrm{/}m$ lies in its fraction field; the reverse inclusion is contained in $K$.

Let $f(X)=X^r+c_{r-1}{X}^{r-1}+\cdots +c_0$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Fact [F2] gives $c_j\in \mathbb{Z}$.

Put $h=[K:\mathbb{Q}(\alpha )]$. Regarding $K$ as a $\mathbb{Q}(\alpha )$-vector space, multiplication by $\alpha$ acts as the scalar $\alpha$ on each of $h$ basis directions. Hence its characteristic polynomial over $\mathbb{Q}$ is $f(X{)}^h$. By [F1], ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(\alpha )=-h{c}_{r-1},N_{K\mathrm{/}\mathbb{Q}}(\alpha )=(-1{)}^{rh}{c}_0^h,$ which are integers.

Fact [F1] makes ${\mathcal{O}}_K$ finitely generated; it is torsion-free because it is contained in the characteristic-zero field $K$.

The PID structure theorem, equivalently [F2] after embedding in a finite free module, makes it free. Fact [F3] gives ${\mathcal{O}}_K{\otimes }_{\mathbb{Z}}\mathbb{Q}\cong K$, so its rank is $[K:\mathbb{Q}]$.

The order has full rank by definition, so [F1] makes it a rank-$n$ submodule of a free rank-$n$ group.

Apply [F2]: the quotient is a finite direct sum of $\mathbb{Z}\mathrm{/}{a}_i\mathbb{Z}$, and the resulting basis of $\mathcal{O}$ is integral.

Bilinearity of trace gives $G_{\beta }=A^t{G}_{\alpha }A$ for the trace Gram matrices.

Taking determinants gives the asserted square factor.

With $A=({\sigma }_i({\alpha }_j))$, [F1] gives $G=A^{t}A$ for the trace Gram matrix.

If $A$ were singular, a nonzero linear combination of its rows would vanish on the basis and hence on every element of $K$, contradicting [F2]. Thus $\det A\mathrm{\ne }0$.

Taking determinants in step 1.1 gives $\det G=\det (A{)}^2$, and step 1.2 gives the asserted nonvanishing.

The integral change matrix is unimodular, so [F1] gives basis independence; integral traces make the value integral.

Distinct embeddings make the matrix in [F2] invertible, so its square is nonzero.

Fact [F1] makes the determinant the Vandermonde product of the conjugates of $\alpha$.

Squaring it and regrouping the root differences is the polynomial discriminant, equivalently the displayed derivative norm.

Choose the bases from [F1]; the absolute change determinant is the index.

Apply [F2] and square the determinant.

The power order is an order, and [F1] says its discriminant is the square of its index times $d_K$.

A square index dividing a squarefree integer is $1$, so the two orders coincide.

Write an integral element as $(a+b\sqrt{d})\mathrm{/}c$ in lowest terms. Fact [F1] forces $2a\mathrm{/}c,(a^2-d{b}^2)\mathrm{/}{c}^2\in \mathbb{Z}$.

These divisibilities force $c\mid 2$; the $c=2$ case occurs exactly when $d\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$, yielding the displayed bases.

In each basis of [F1], form the $2\times 2$ trace Gram matrix.

Its determinant is respectively $d$ and $4d$.

Apply [F1] with base ring $\mathbb{Z}$ and extension $K\mathrm{/}\mathbb{Q}$.

Its integral closure is precisely ${\mathcal{O}}_K$, so it is Dedekind.