## The Additive Structure of a Number Ring

We shall need the fact that a subgroup $$H$$ of a free abelian group $$G$$ of rank $$n$$ is a free abelian group of rank $$\le n$$. Refer to the group theory notes, or use the following proof by induction. The case $$n = 1$$ is easy to show. Next, assume this fact is true for $$n-1$$, and let $$\Pi:G\rightarrow \mathbb{Z}$$ be the map that projects to the first coordinate. The kernel $$K$$ of this map has rank $$n-1$$ (it consists of all elements of $$G$$ that have the identity element as their first component). Thus by inductive assumption, $$H \cap K$$ is a free abelian group of rank $$\le n-1$$. Now $$\Pi(H) \subset \mathbb{Z}$$ is either $$\{0\}$$ or $$\mathbb{Z}$$. In the former case, we must have $$H = H \cap K$$. In the latter, pick $$h \in H$$ such that $$\Pi(h)$$ generates $$\Pi(H)$$. Then $$H = \mathbb{Z} h \oplus (H \cap K)$$.

Let $$K$$ be a number field of degree $$n$$ over $$\mathbb{Q}$$, and let $$R$$ be the ring of algebraic integers in $$K$$. Suppose $$\alpha \in K$$. Then there exists an integer $$m$$ such that $$m\alpha$$ is integral, because if $$a_n \alpha^n +...+ a_0 = 0$$ for $$a_0, ...,a_n \in \mathbb{Z}$$, then $$a_n \alpha$$ is a root of the monic polynomial

$x^n + a_{n-1} x^{n-1} + a_n a_{n-2} x^{n-2} ...+ a_n^{n-1} a_0$

Thus given any basis of $$K$$ over $$\mathbb{Q}$$, we can find a basis consisting of algebraic integers by multiplying by an appropriate constant. Then for such a basis $$\alpha_1,...,\alpha_n \in R$$ for $$K$$ over $$\mathbb{Q}$$, we have a free abelian group of rank $$n$$ inside $$R$$:

$\{ m_1 \alpha_1 +...+ m_n \alpha_n : m_i \in \mathbb{Z} \} = \mathbb{Z} \alpha_1 \oplus +...+ \oplus \mathbb{Z}\alpha_n$

Theorem: Let $$\{\alpha_1,...,\alpha_n\} \subset R$$ be a basis for $$K$$ over $$\mathbb{Q}$$, and let $$d = disc(\alpha_1,...,\alpha_n)$$. Then every $$\alpha \in R$$ can be written in the form

$\alpha = \frac{m_1\alpha_1 +...+m_n\alpha_n}{d}$

where $$m_j \in \mathbb{Z}$$ and $$d | m_j^2$$ for all $$j$$.

Proof: Write $$\alpha = x_1\alpha_1 +...+ x_n\alpha_n$$ where $$x_j \in \mathbb{Q}$$. Let $$\sigma_1,...,\sigma_n$$ be the embeddings of $$K$$ in $$\mathbb{C}$$. Then we have the system of equations

$\sigma_i(\alpha) = x_1\sigma_i(\alpha_1) +...+ x_n\sigma_i(\alpha_n)$

Using Cramer’s rule, we have $$x_j = \gamma_j / \delta$$ where $$\delta = |\sigma_i(\alpha_j)|$$ (so $$\delta^2 = d$$ and $$\gamma_j$$ is the same as $$\delta$$ except that the $$j$$th column has been replaced by $$\sigma_i(\alpha)$$. Thus $$d x_j = \delta \gamma_j$$ so $$d x_j$$ is an algebraic integer. As it is also rational, we must have $$m_j = d x_j \in \mathbb{Z}$$. Lastly, it can be seen that $$m_j^2 / d = \gamma_j^2$$ which is an algebraic integer, and since $$m_j^2 / d$$ is rational, it follows that $$m_j^2 / d$$ is an integer.

Corollary: $$R$$ is a free abelian group of rank $$n$$

Proof: We have already established that $$R$$ contains a free abelian group of rank $$n$$. The previous theorem shows that $$R$$ is contained within the free abelian group of rank $$n$$

$\mathbb{Z}\frac{\alpha_1}{d} \oplus ... \oplus \mathbb{Z}\frac{\alpha_n}{d}$

Thus $$R$$ has a basis over $$\mathbb{Z}$$, that is, there exist $$\beta_1,...,\beta_n \in R$$ such that every $$\alpha \in R$$ can be written as

$\alpha = m_1\beta_1 +...+ m_n\beta_n$

where $$m_i \in \mathbb{Z}$$. Such a basis is called an integral basis for $$R$$, or a basis for $$R$$ over $$\mathbb{Z}$$.

For example, in the quadratic field $$\mathbb{Q}[\sqrt{m}]$$ for squarefree $$m$$, an integral basis for its number ring $$R$$ is $$\{1,\sqrt{m}\}$$ when $$m = 2,3 (mod 4)$$ and $$\{1,(1+\sqrt{m})/2\}$$ when $$m = 1 (mod 4)$$.

Theorem: Let $$\omega = e^{2 \pi i /m}$$ where $$m = p^r$$ is a prime power. Then $$\mathbb{A} \cap \mathbb{Q}[\omega] = \mathbb{Z}[\omega]$$.

We shall need a couple of lemmas to prove this.

Lemma: For $$m \ge 3$$, $$\mathbb{Z}[1 - \omega] = \mathbb{Z}[\omega]$$ and $$disc(1-\omega) = disc(\omega)$$.

Proof: The first equation follows directly from $$\omega = 1 - (1-\omega)$$. We shall see this implies the second equation (using a generalization of the next theorem), but it is easy to show it directly:

$\array { disc(\omega) & = & {\prod_{1\le r \lt s \le n}(\alpha_r-\alpha_s)^2} \\ & = & {\prod_{1\le r \lt s \le n}((1-\alpha_r)-(1-\alpha_s))^2} \\ & = & disc(1-\omega) }$

Lemma: For $$m = p^r$$

${\prod_{1\le k \le m, p \nmid k}(1-\omega^k)} = p$

Proof: Set

$f(x) = \frac{x^{p^r} - 1}{x^{p^{r-1}}-1} =1 +x^{p^{r-1}} + x^{2p^{r-1}} + ... + x^{(p-1)p^{r-1}}$

Note all $$\omega^k$$ (for $$p \nmid k$$) are roots of $$f$$ since they are roots of $$x^{p^r}-1$$ but not of $$x^{p^{r-1}}-1$$. So we must have

$f(x) = {\prod_k(x-\omega^k)}$

since there are exactly $$\phi(p^r) = (p-1)p^{r-1}$$ values of $$k$$. Then set $$x = 1$$.

Proof of Theorem: Recall every algebraic integer $$\alpha$$ can be expressed as

$\alpha = \frac{m_1 + m_2(1-\omega) +...+m_n(1-\omega)^{n-1}}{d}$

where $$n = \phi(p^r)$$, $$m_i \in \mathbb{Z}$$ and $$d = disc(\omega)$$. We have previously shown that $$disc(\omega) | m^{\phi(m)}$$ so $$d$$ must be a power of $$p$$. We wish to show $$R = \mathbb{Z}[\omega] = \mathbb{Z}[1-\omega]$$. If this were not true, then $$R$$ would contain some element

$\beta = \frac{m_i(1-\omega)^{i-1}+m_{i+1}(1-\omega)^i+...+ m_n(1-\omega)^{n-1}}{p}$

for some $$i \le n$$, and $$m_j \in \mathbb{Z}$$ with $$p \nmid m_i$$.

By the second of the above lemmas, $$p /(1-\omega^n) \in \mathbb{Z}[\omega]$$ since each $$1-\omega^k$$ is divisible by $$1-\omega$$. Thus $$\beta p / (1-\omega^i) \in R$$. This leads to $$m_i/(1-\omega)\in R$$, from which it follows $$N(1-\omega)|N(m_i)$$, which is impossible since $$N(m_i) = m_i^n$$ while the lemma shows $$N(1-\omega)=p$$.

Theorem: Let $$\{\beta_1,...,\beta_n\}$$ and $$\{\gamma_1,...,\gamma_n\}$$ be two integral bases for $$R = \mathbb{A}\cap K$$. Then $$disc(\beta_1,...,\beta_n)=disc(\gamma_1,...,\gamma_n)$$.

Proof: We may write

${ ( \array { \beta_1 \\ \vdots \\ \beta_n } ) } = M { ( \array { \gamma_1 \\ \vdots \\ \gamma_n } ) }$

where $$M$$ is an $$n \times n$$ matrix over $$\mathbb{Z}$$. From here we can derive the matrix equation $$[\sigma_j(\beta_i)] = M[\sigma_j(\gamma_i)]$$, and taking determinants and squaring gives

$disc(\beta_1,...,\beta_n) =|M|^2 disc(\gamma_1,...,\gamma_n)$

Now $$|M| \in \mathbb{Z}$$ thus $$disc(\beta_1,...,\beta_n)$$ divides $$disc(\gamma_1,...,\gamma_n)$$. A similar argument shows $$disc(\gamma_1,...,\gamma_n)$$ must also divide $$disc(\beta_1,...,\beta_n)$$ thus they must be equal.

Note we may use the same argument to show that if $$\{\beta_1,...,\beta_n\}$$ and $$\{\gamma_1,...,\gamma_n\}$$ are elements of $$K$$ that generate the same additive subgroup of $$K$$ then $$disc(\beta_1,...,\beta_n) = disc(\gamma_1,...,\gamma_n)$$, and we may use this fact to define $$disc(G)$$ for any additive subgroup $$G$$ of $$K$$ generated by $$n$$ elements.

By the theorem, the discriminant of an integral basis is an invariant of $$R$$, and we denote it by $$disc(R)$$, and also $$disc(K)$$ where $$R$$ is the number ring of $$K$$.

Example: We have for squarefree $$m$$

$disc(\mathbb{Q}[\sqrt{m}]) = { \{ \array { disc(\sqrt{m}) = 4m \text { if } m=2,3 (mod 4) \\ disc(\frac{1+\sqrt{m}}{2}) = m \text { if } m=1 (mod 4) } }$

Ben Lynn blynn@cs.stanford.edu 💡