## 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

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 $R$:

**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

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

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= |\sigma_i(\alpha_j)|$ (so $\delta^2 = d$ and $\gamma_j$ is the same as $\delta$ except that the $j$th column has been replaced by $\sigma_i(\alpha)$.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

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

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:

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

**Proof:** Set

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

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

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

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

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

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 $m$