Algebraic Number Theory – II. Rings of Integers are Finitely Generated Abelian Groups

Geometrically, one pictures ${\mathbb{Z}}$ as a discretization of ${\mathbb{Q}}$, with evenly spaced integers on a number line punctuating the dense rationals. More generally, ${\mathbb{Z}^n}$ forms a lattice inside ${\mathbb{Q}^n}$, obtained by taking ${\mathbb{Z}}$-linear combinations of the standard basis, rather than ${\mathbb{Q}}$-linear combinations. We should like to view ${\mathcal{O}_K}$ similarly in the context of ${K}$. The following theorem validates this intuition.

Theorem 2.1: ${\mathcal{O}_{K}}$ is a finitely generated abelian group under addition (i.e., a ${\mathbb{Z}}$-module), with rank ${[K:\mathbb{Q}]}$.

The approach one takes to prove this result is to embed ${\mathcal{O}_K}$ in some larger object, and then to exploit the properties of this larger object. This is done abstractly, by embedding ${\mathcal{O}_K}$ in a larger ${\mathbb{Z}}$-module and invoking the structure theorem for finitely generated abelian groups.

1. Field norm and trace

We will need to develop some properties of the field norm and field trace, which are useful linear algebraic invariants that one may associate with any field extension ${L/F}$. For any element ${\alpha\in L}$, there is a natural linear transformation ${L\rightarrow L}$ given by ${x\mapsto \alpha x}$. With respect to the ${F}$-basis of ${L}$, this transformation has a matrix ${M_{\alpha}}$ whose entries are all in ${F}$, yielding the maps

$\displaystyle N_{L/F} : L\rightarrow F \text{ given by } \alpha \mapsto \det(M_{\alpha})$

$\displaystyle \text{Tr}_{L/F} : L \rightarrow F \text{ given by } \alpha \mapsto \text{Tr}(M_{\alpha})$

By definition, these maps inherit the multiplicative and additive properties of determinant and trace, respectively.

As it turns out, given ${\alpha\in L}$, the elements ${N_{L/F} (\alpha)}$, ${\text{Tr}_{L/F} (\alpha) \in K}$ may be expressed in terms of the roots of the minimal polynomial of ${\alpha}$ over ${F}$. Recall this minimal polynomial exists since the extension ${L/F}$ is finite, hence algebraic. We denote the minimal polynomial of ${\alpha}$ as ${\text{Irr}_{\alpha, F} (x) \in F[x]}$.

To proceed, define the characteristic polynomial of ${\alpha}$ as ${\text{Ch}_{\alpha, L/F} (x) = \det(x \cdot I_n - M_{\alpha}) \in F[x]}$. A simple manipulation shows that for any ${f\in F[x]}$, one has ${f(M_{\alpha}) = M_{f(\alpha)}}$. Since the map ${\alpha \mapsto M_{\alpha}}$ is injective, the Cayley-Hamilton theorem then implies ${\alpha}$ is a root of its characteristic polynomial.

Hence, ${\text{Irr}_{\alpha, F}}$ divides ${\text{Ch}_{\alpha, L/F}}$. In general, they need not be equal, as the degree of ${\text{Irr}_{\alpha, F}}$ is ${[F(\alpha):F]}$ whereas the degree of ${\text{Ch}_{\alpha, L/F}}$ is ${[L:F]}$, which do not have to be same. However, if ${L = F(\alpha)}$, then it follows that ${\text{Irr}_{\alpha, F} = \text{Ch}_{\alpha, L/F}}$. This fact may be generalized.

Proposition 2.2: Let ${d = [F(\alpha): F]}$. Then ${\text{Ch}_{\alpha, L/F} = (\text{Irr}_{\alpha, F})^{n/d}}$.

Proof: A basis for ${F(\alpha)/F}$ is ${\{1, \alpha, \alpha^2, \cdots, \alpha^{d-1}\}}$. Let ${\{\beta_1, \cdots, \beta_m\}}$ be a basis for ${L/F(\alpha)}$, where ${m = n/d}$. Then an ordered basis for ${L/F}$ is ${\{\alpha^{i} \beta_{j} \}_{0\le i \le d-1, 1\le j \le m}}$.

Since ${\text{Irr}_{\alpha, F}}$ has degree ${d}$, the characteristic polynomial ${\text{Ch}_{\alpha, F(\alpha)/F} = \text{Irr}_{\alpha, F}}$. With respect to the chosen basis for ${L/F}$, the matrix ${M_{\alpha, L/F}}$ is a diagonal block matrix comprised of ${n/d}$ copies of the matrix ${M_{\alpha, F(\alpha)/F}}$. Thus,

$\displaystyle \text{Ch}_{\alpha, L/F} (X)= \text{det}(X \cdot I_{d} - M_{\alpha, F(\alpha)/F})^m = (\text{Ch}_{\alpha, F(\alpha)/F})^m = (\text{Irr}_{\alpha, F})^{n/d}$

$\Box$

Let ${\alpha = \alpha_1}$, ${\alpha_2}$, ${\cdots}$, ${\alpha_d}$ be the roots of ${\text{Irr}_{\alpha, F}}$. These are called the algebraic conjugates of ${\alpha}$, and the proposition says that the eigenvalues of ${M_{\alpha}}$ are exactly the algebraic conjugates of ${\alpha}$, each counted with multiplicity ${[L:F]/[F(\alpha):F] = n/d}$. Therefore,

$\displaystyle \text{Tr}_{L/F} (\alpha) = \frac{n}{d} (\alpha_1 + \alpha_2 + \cdots + \alpha_d)$

$\displaystyle N_{L/F} (\alpha) = (\alpha_1 \alpha_2 \cdots \alpha_n)^{n/d}$

Returning to the case of ${K/\mathbb{Q}}$ a number field, suppose ${\alpha \in \mathcal{O}_K}$. Then ${\text{Irr}_{\alpha, K}}$ is monic with integer coefficients, so the norm and trace of ${\alpha}$ are integers.

Corollary 2.3: The restrictions ${N_{K/\mathbb{Q}} \vert_{\mathcal{O}_K}}$ and ${\text{Tr}_{K/\mathbb{Q}} \vert_{\mathcal{O}_K}}$ have range ${\mathbb{Z}}$.

2. A nondegenerate symmetric bilinear form

To embed ${\mathcal{O}_K}$ in a larger ${\mathbb{Z}}$-module is to embed ${\mathcal{O}_K}$ in ${\mathbb{Z}^n}$ for some ${n\in\mathbb{N}}$, since ${\mathcal{O}_K}$ has no nontrivial elements of finite order. The goal, then, is to produce an injective homomorphism ${\mathcal{O}_K \rightarrow \mathbb{Z}^n}$. The trace function restricted to ${\mathcal{O}_K}$ is an additive homomorphism into ${\mathbb{Z}}$, so one might naturally think of exploiting its properties. Traditionally, this is done via the trace pairing.

We must review a little linear algebra before describing the trace pairing. Recall that a bilinear form on a vector space ${V}$ over field ${K}$ is a map ${B: V\times V \rightarrow K}$ such that ${B(\cdot, v) : V \rightarrow K}$ and ${B(v, \cdot) : V \rightarrow K}$ are both linear, for any choice of ${v\in V}$. The bilinear form ${B}$ is symmetric if ${B(v,w) = B(w,v)}$.

If one fixes bases ${\mathcal{B}_1 = \{e_i\}}$ and ${\mathcal{B}_2 = \{f_j\}}$ of ${V}$, each bilinear form ${B}$ has an associated matrix ${M_{B, \mathcal{B}_1, \mathcal{B}_2} = M_{B} = \{M_{ij}\}}$. The entries of this matrix are ${M_{ij} = B(e_i, f_j)}$, and if ${B}$ is symmetric, then ${M_B}$ is a symmetric matrix. Assuming ${V}$ is finite dimensional, of dimension ${m}$, then one may identify ${V \cong K^n}$ via the isomorphisms ${\varphi_1, \varphi_2 : V \rightarrow K^m}$, which give the coordinates of a vector in bases ${\mathcal{B}_1}$ and ${\mathcal{B}_2}$ respectively. Then one can concretely write

$\displaystyle B(v,w) = \varphi_1 (v)^{T} \,\cdot \, M_B \, \cdot \, \varphi_2 (w)$

The associated matrix ${M_{B}}$ therefore uniquely determines ${B}$.

The form ${B}$ is called nondegenerate if ${M_B}$ is invertible. As with most nice things in linear algebra, nondegeneracy does not depend on the choice of bases ${\mathcal{B}_1}$ and ${\mathcal{B}_2}$ (exercise!). A coordinate-free formulation of the condition is that there is no nonzero ${w\in V}$ such that ${B(\cdot, w) : V \rightarrow K}$ is the zero map. In particular, ${B}$ is a nondegenerate form iff

$\displaystyle (B(e_1, \cdot), B(e_2, \cdot), \cdots, B(e_m, \cdot)) : V \rightarrow K^m$

is injective.

Now, we can look at the trace pairing, which is the symmetric bilinear form ${\text{TrP}_{K/\mathbb{Q}}: K \times K \rightarrow \mathbb{Q}}$ given by

$\displaystyle \text{TrP}_{K/\mathbb{Q}} (\alpha, \beta) = \text{Tr}_{K/\mathbb{Q}} (\alpha\beta)$

Observe that if ${x\in K}$ is nonzero, then

$\displaystyle \text{TrP}_{K/\mathbb{Q}}(x, x^{-1}) = \text{Tr}_{K/\mathbb{Q}}(x\cdot x^{-1}) = \text{Tr}_{K/\mathbb{Q}} (1) = [K:\mathbb{Q}] > 0$

Hence, the trace pairing is nondegenerate.

We can now give a proof that ${\mathcal{O}_K}$ is a finitely generated ${\mathbb{Z}}$-module.

Proof 1 (Injecting into a module): In the previous blog post, we found that for every ${x\in K}$, there is an integer ${a}$ such that ${ax \in \mathcal{O}_{K}}$. Accordingly, if ${\{e_i\}_{1\le i\le n}}$ is a basis for ${K}$ as a ${\mathbb{Q}}$-vector space, there are integers ${a_i}$ such that ${\{a_i e_i\}_{1\le i \le n}}$ is also a basis for ${K}$, but ${\{a_i e_i\} \subset \mathcal{O}_K}$. Setting ${b_i = a_i e_i}$, one may define the homomorphism

$\displaystyle (\text{TrP}(b_1, \cdot), \text{TrP}(b_2, \cdot), \cdots, \text{TrP}(b_n, \cdot)) : K \rightarrow \mathbb{Q}^n$

which is injective, thanks to nondegeneracy of the trace pairing. Restricting this homomorphism to ${\mathcal{O}_K}$ will inject into ${\mathbb{Z}^n }$.

By the structure theorem on finitely generated abelian groups, ${\mathcal{O}_{K}}$ is a free abelian group of rank at most ${\text{rank}(\mathbb{Z}^n) = n}$. But ${\mathcal{O}_K}$ contains ${\{b_i\}}$, a linearly independent set of size ${n}$, so the rank of ${\mathcal{O}_K}$ is at least ${\text{rank}(\langle b_1, \cdots, b_n \rangle) = n}$. Hence, we conclude ${\mathcal{O}_K}$ is a free abelian group of rank ${n}$$\Box$

This proves the existence of some linearly independent set ${\{\alpha_1, \cdots, \alpha_n\} \subset \mathcal{O}_K}$ whose span in ${\mathbb{Z}}$-linear combinations is ${\mathcal{O}_K}$. One calls such a set an integral basis for ${\mathcal{O}_K}$.

3. Discriminant of a number field

Let us define the discriminant on ${K/\mathbb{Q}}$. This is the map ${\Delta : K^n \rightarrow \mathbb{Q}}$ given by

$\displaystyle \Delta(\alpha_1, \alpha_2, \cdots, \alpha_n) = \det(\{\text{Tr}_{K/\mathbb{Q}} (\alpha_i \alpha_j)\})$

In particular, if ${\{\alpha_i\} = \mathcal{B}}$ is a basis for ${K/\mathbb{Q}}$, the discriminant of ${\{\alpha_i\}}$ is the determinant of the trace pairing matrix ${M_{\text{TrP}, \mathcal{B}, \mathcal{B}}}$. This determinant is nonzero since the trace pairing form is nondegenerate. On the other hand, one can show that if ${\{\alpha_i\}}$ is linearly dependent, then ${\Delta(\alpha_1, \cdots, \alpha_n) = 0}$ (exercise!).

Suppose that ${\{\alpha_i\} = \mathcal{B}_1}$ and ${\{\beta_i\} = \mathcal{B}_2}$ are bases for ${K/\mathbb{Q}}$, with change of basis matrix ${C}$ taking coordinates in ${\mathcal{B}_1}$ to coordinates in ${\mathcal{B}_2}$. Then, by considering the bilinear form ${\text{TrP}_{K/\mathbb{Q}}}$, we have the matrix equality

$\displaystyle M_{\text{TrP}, \mathcal{B}_1, \mathcal{B}_1} = C^T \cdot M_{\text{TrP}, \mathcal{B}_2, \mathcal{B}_2} \cdot C$

which, upon taking determinants, gives the identity

$\displaystyle \Delta(\alpha_1, \alpha_2, \cdots, \alpha_n) = \det(C)^2 \Delta(\beta_1, \beta_2, \cdots, \beta_n)$

If ${\mathcal{B}_1}$ and ${\mathcal{B}_2}$ are both actually integral bases of ${\mathcal{O}_K}$, then the change of basis matrix ${C}$ is an invertible integer matrix, hence it has determinant with absolute value ${1}$. In this case, ${\Delta(\mathcal{B}_1) = \Delta(\mathcal{B}_2)}$. The common value is called the discriminant of ${K}$.

Using properties of the discriminant, we give another proof that ${\mathcal{O}_K}$ has an integral basis.

Proof 2 (Minimal discriminant): As in the previous proof, let us begin with a basis ${\{b_i\} \subset \mathcal{O}_K}$ that spans ${K/\mathbb{Q}}$. Since the restriction to ${\mathcal{O}_K}$ of trace has range ${\mathbb{Z}}$, the discriminant ${\Delta(b_1, \cdots, b_n)}$ is an integer. Hence, we may choose a basis ${\{c_i\} \subset \mathcal{O}_K}$ of ${K/\mathbb{Q}}$ such that ${|\Delta(c_1, \cdots, c_n)|}$ is minimal.

For some ${x\in \mathcal{O}_K}$, write ${x = \gamma_1 c_1 + \gamma_2 c_2 + \cdots + \gamma_n c_n}$. If ${\gamma_1 \not\in\mathbb{Z}}$, then it can be decomposed ${\gamma_1 = \lfloor \gamma_1 \rfloor + \theta}$ where ${\lfloor \gamma_1 \rfloor}$ is the greatest integer less than ${\gamma_1}$, and ${0 <\theta < 1}$ is the fractional part.

Then, setting ${d_1 = x - \lfloor \gamma_1 \rfloor c_1}$, and ${d_i = c_i}$ for ${2\le i \le n}$, we have another basis ${\{d_i\}}$ for ${K/\mathbb{Q}}$. One easily computes the determinant of the change of basis matrix to be ${\theta}$, so that the above identity on discriminants gives that

$\displaystyle |\Delta(d_1, \cdots, d_n)| = \theta^2 |\Delta(c_1, \cdots, c_n)| < |\Delta(c_1, \cdots, c_n)|$

contradicting the choice of ${\{c_i\}}$. Thus, each ${\gamma_i \in \mathbb{Z}}$, and the set ${\{c_i\}}$ forms an integral basis for ${\mathcal{O}_K}$. $\Box$

This argument yields a useful corollary. To simplify its statement, let us introduce the following definition: if ${\mathcal{B} \subset \mathcal{O}_K}$ is a basis for ${K}$ as a ${\mathbb{Q}}$-vector space, we call ${\mathcal{B}}$ an almost-integral basis.

Corollary 2.4: Let ${\mathcal{B}}$ be an almost-integral basis. Then, the following are equivalent:

• ${|\Delta(\mathcal{B})| \le |\Delta(\mathcal{B}')|}$ for all almost-integral bases ${\mathcal{B}'}$
• ${|\Delta(\mathcal{B})| = |\delta_K|}$
• ${\mathcal{B}}$ is an integral basis.