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= |\sigma_i(\alpha_j)|$ (so $\delta^2 = d$ and $\gamma_j$ is the same as $\delta$ except that the $j$th column has been replaced by $\sigma_i(\alpha)$.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 $m$

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