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.

Put $n=[K:\mathbb{Q}]$ and choose a $\mathbb{Q}$-basis $e_1,\dots ,e_n$ of $K$. By [F1], $K$ is an $n$-dimensional $\mathbb{Q}$-vector space.

We prove by induction on $m\ge 0$ that every additive subgroup $H\le {\mathbb{Z}}^m$ has a finite $\mathbb{Z}$-basis. For $m=0$, the zero subgroup has the empty basis. Now let $m>0$ and assume the assertion for $m-1$. Project $H$ to its first coordinate. By [F6], the image is either $\{0\}$ or $d\mathbb{Z}$ for some positive integer $d$.

Define $B(x,y)={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(xy)$. Multiplication in $K$ gives $m_{(ax+by)z}=a{m}_{xz}+b{m}_{yz}$ for $a,b\in \mathbb{Q}$, and matrix trace is linear; the same holds in the second variable since $K$ is commutative. Thus $B$ is a bilinear form.

For each $i$, [F3] gives a positive integer $m_i$ such that $u_i:=m_i{e}_i\in {\mathcal{O}}_K$. These elements remain a $\mathbb{Q}$-basis. Set $M={\sum }_{i=1}^n\mathbb{Z}{u}_i$. By [F5], $M\subseteq {\mathcal{O}}_K$.

If the image is zero, $H$ identifies with a subgroup of ${\mathbb{Z}}^{m-1}$ and has a finite basis by induction. If the image is $d\mathbb{Z}$, its kernel $H_0$ identifies with a subgroup of ${\mathbb{Z}}^{m-1}$ and has a finite basis $k_1,\dots ,k_r$ by induction. Choose one $h\in H$ whose first coordinate is $d$.

If $x\mathrm{\ne }0$, take $y=x^{-1}$. Then $B(x,y)={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(1)=n$, because multiplication by $1$ is the identity on the $n$-dimensional space $K$. As $n>0$ in $\mathbb{Q}$, this proves that the radical of $B$ is zero.

In the nonzero-image case of step 2.3, every $g\in H$ has first coordinate $qd$ for some $q\in \mathbb{Z}$, so $g-qh\in H_0$; hence $h,k_1,\dots ,k_r$ span $H$. If $ah+{\sum }_i{b}_i{k}_i=0$, its first coordinate gives $ad=0$, hence $a=0$, and the independence of the basis of $H_0$ then gives every $b_i=0$. Thus this spanning list is a basis. The induction uses only one lift for each fixed subgroup at this finite stage, not a choice of lifts for a family.

Let $A=({\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_i{u}_j){)}_{i,j}$. If $Az=0$ for $z\in {\mathbb{Q}}^n$, put $x={\sum }_j{z}_j{u}_j$. Then ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_{i}x)=0$ for every $i$, and bilinearity makes ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(yx)=0$ for every $y\in K$, since the $u_i$ form a $\mathbb{Q}$-basis. By step 3.1, $x=0$, so $z=0$. Thus $A$ has zero kernel and [F7] makes it invertible.

For each $j$, solve $A{v}_j={\epsilon }_j$ over $\mathbb{Q}$, where ${\epsilon }_j$ is the $j$th standard coordinate vector, and put $u_j^{\ast }={\sum }_k(v_j{)}_k{u}_k$. By [F7], these solutions exist uniquely and satisfy ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_i{u}_j^{\ast })={\delta }_{ij}$ for every $i,j$.

The coordinates of the finitely many $u_j^{\ast }$ in the basis $u_1,\dots ,u_n$ are rational. Choose a positive integer $c$ clearing all their denominators. Then $c{u}_j^{\ast }\in M\subseteq {\mathcal{O}}_K$ for every $j$.

If $x={\sum }_i{a}_i{u}_i\in {\mathcal{O}}_K$, the trace-dual equations give $c{a}_j={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(xc{u}_j^{\ast })$. The product is integral by [F5], so [F4] makes this trace an integer. Thus every $a_j\in c^{-1}\mathbb{Z}$, proving $M\subseteq {\mathcal{O}}_K\subseteq c^{-1}M$.

Since $u_1\mathrm{/}c,\dots ,u_n\mathrm{/}c$ is a $\mathbb{Z}$-basis of $c^{-1}M\cong {\mathbb{Z}}^n$, the inclusion in step 7.1 identifies ${\mathcal{O}}_K$ with an additive subgroup of ${\mathbb{Z}}^n$. By the induction, it has a finite $\mathbb{Z}$-basis $v_1,\dots ,v_r$.

The $v_i$ span $K$ over $\mathbb{Q}$: their integral span contains $M$, which spans $K$. Their $\mathbb{Z}$-independence implies $\mathbb{Q}$-independence after clearing denominators. Thus they are a $\mathbb{Q}$-basis and $r=n=[K:\mathbb{Q}]$. This proves the claim.

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 $4d$.

By [F1], $\mathbb{Z}$ is a domain. A fraction $a\mathrm{/}b$ equals $c\mathrm{/}d$ in $\mathrm{Frac}(\mathbb{Z})$ exactly when $u(ad-cb)=0$ for some nonzero $u\in \mathbb{Z}$; the domain property makes this equivalent to $ad=cb$, the relation defining $\mathbb{Q}$. The operations agree under $(a,b)\mapsto j(a)\mathrm{/}j(b)$, so this gives the canonical identification $\mathrm{Frac}(\mathbb{Z})=\mathbb{Q}$.

Let $I$ be an ideal of $\mathbb{Z}$. It is an additive subgroup, so [F2] gives $I=n\mathbb{Z}$ for some $n\in \mathbb{N}$; as a $\mathbb{Z}$-module, $I$ is generated by that one element, including when $n=0$. Every submodule of the left regular module $\mathbb{Z}$ is such an ideal. Therefore this module is Noetherian, and $\mathbb{Z}$ is a Noetherian ring by [F11].

The field $\mathbb{Q}$ has characteristic zero: for every positive integer $m$, $m\cdot 1_{\mathbb{Q}}=j(m)\mathrm{\ne }0$ by the injectivity in [F10]. By [F5], $\mathbb{Q}$ is perfect.

Let $P$ be a nonzero prime ideal of $\mathbb{Z}$. By [F2], $P=p\mathbb{Z}$ for a positive integer $p$; since $P$ is proper, $p\mathrm{\ne }1$, so $p>1$. If $p\mid ab$, then $ab\in P$, and [F13] gives $a\in P$ or $b\in P$, so $p\mid a$ or $p\mid b$. The Euclid-property characterization therefore makes $p$ a prime integer.

If $x\in \mathrm{Frac}(\mathbb{Z})$ is integral over $\mathbb{Z}$, step 1.1 identifies it with a rational number, and [F3] gives $x\in \mathbb{Z}$. Hence $\mathbb{Z}$ is integrally closed by [F12].

Put $n=[K:\mathbb{Q}]$, which is finite and at least $1$ by [F4] and because $K$ is a field extension of $\mathbb{Q}$. For each $\alpha \in K$, the $n+1$ vectors $1,\alpha ,\dots ,{\alpha }^n$ in the $n$-dimensional $\mathbb{Q}$-vector space $K$ are linearly dependent, so a nonzero polynomial over $\mathbb{Q}$ vanishes at $\alpha$. Thus $K\mathrm{/}\mathbb{Q}$ is algebraic; [F6] and step 1.3 make it separable.

For the prime $p$ from step 1.4, [F14] makes $\mathbb{Z}\mathrm{/}p$ a field and identifies it with the quotient ring $\mathbb{Z}\mathrm{/}p\mathbb{Z}=\mathbb{Z}\mathrm{/}P$. The quotient criterion in [F14] then makes every nonzero prime ideal of $\mathbb{Z}$ maximal.

First, [F16] gives $0<1<2$, so $2>1$. If $d$ is any positive divisor of $2$, [F15] gives $1\le d\le \mathrm{\mid }2\mathrm{\mid }=2$. If $d\mathrm{\ne }1$, then $1<d$; translation compatibility of the order gives $0<d-1$, and integer discreteness in [F15] gives $1\le d-1$. Hence $2\le d$, so antisymmetry and $d\le 2$ give $d=2$. Thus the only positive divisors of $2$ are $1$ and $2$, and [F17] proves that $2$ is prime. Now [F14] makes $2\mathbb{Z}$ maximal, and [F18] makes it prime. Since $\mathbb{Z}$ is a domain, $(0)$ is prime; and $2\mathrm{\ne }0$ by [F16], so $(0)⊊2\mathbb{Z}$. Every prime chain with positive length starts at $(0)$ and has length at most $1$, because each nonzero prime is maximal by step 2.3. Thus [F19] gives $\dim \mathbb{Z}=1$.

Steps 1.1, 1.2, 2.1, and 3.1 show that $\mathbb{Z}$ is a Noetherian integrally closed domain of dimension $1$. Hence [F7] makes $\mathbb{Z}$ a Dedekind domain.

By [F9], ${\mathcal{O}}_K$ is the integral closure of $\mathbb{Z}$ in $K$. Apply [F8] with $R=\mathbb{Z}$, fraction field $\mathbb{Q}=\mathrm{Frac}(\mathbb{Z})$ from step 1.1, and finite separable extension $K\mathrm{/}\mathbb{Q}$ from step 2.2. Under the given Axiom of Choice, this proves that ${\mathcal{O}}_K$ is Dedekind.