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(so \(\delta^2 = d\) and \(\gamma_j\) is the same as \(\delta\) except that the \(j\)th column has been replaced by \(\sigma_i(\alpha)\).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

*blynn@cs.stanford.edu*💡