## The Discriminant: Applications

We can identify integral bases using the discriminant. Let \(\alpha_1,...,\alpha_n \in R\). Then they form an integral basis for \(R\) if and only if \(disc(\alpha_1,...,\alpha_n) = disc(R)\).

We may also prove that \(\mathbb{Z}[\omega]\) is the number ring of \(\mathbb{Q}[\omega]\) for \(\omega = e ^{2\pi i /m}\) for any \(m\) (previously it was only for prime powers).

Let \(K, L\) be two number fields. Their composite field \(K L\) is the smallest subfield of \(\mathbb{C}\) containing both \(K\) and \(L\). In fact, \(K L\) consists of elements of the form

for all \(\alpha_i \in K, \beta_i \in L\).

Let \(R, S, T\) be the number rings of \(K, L, K L\). Clearly \(T\) contains

However in general, equality does not hold, though under certain conditions it does. Let \(m, n\) be the degrees of \(K, L\) over \(\mathbb{Q}\). Let \(d = gcd(disc(R), disc(S))\).

**Theorem:** Suppose that \([K L:\mathbb{Q}] = m n\).
Then \(T \subset \frac{1}{d} R S\).

**Corollary:** If \([K L:\mathbb{Q}] = m n\) and \(d = 1\), then \(T = R S\).

To prove the theorem, first we need the following lemma.

**Lemma:** Suppose \([K L:\mathbb{Q}] = m n\). Let \(\sigma\) be an embedding
of \(K\) in \(\mathbb{C}\) and let \(\tau\) be an embedding of \(L\) in \(\mathbb{C}\).
Then there is an embedding of \(KL\) in \(\mathbb{C}\) that restricts to \(\sigma\)
on \(K\) and to \(\tau\) on \(L\).

**Proof:** \(\sigma\) has \(n\) extensions to embeddings of \(KL\) in \(\mathbb{C}\)
and no two of them can agree on \(L\) hence they have distinct restrictions to
\(L\). One of these must be \(\tau\) because \(L\) has exactly \(n\) embeddings in
\(\mathbb{C}\).

**Proof of Theorem:** Let \(\{\alpha_1,...,\alpha_m\}\) be a basis for \(R\)
over \(\mathbb{Z}\) and let \(\{\beta_1,...,\beta_n\}\) be a basis for \(S\) over
\(\mathbb{Z}\). Then the \(m n\) products \(\alpha_i \beta_j\) form a basis for
\(R S\) over \(\mathbb{Z}\) and for \(K L\) over \(\mathbb{Q}\).
Any \(\alpha \in T\) can be written in the form

where \(r, m_{i j} \in \mathbb{Z}\), and \(gcd(r, gcd(m_{i j})) = 1\). We want to show that for all \(\alpha\), \(r|d\). It suffices to show \(r | disc(R)\) because then by symmetry, \(r | disc(S)\) and then we are done.

By the lemma, every embedding of \(\sigma\) of \(K\) in \(\mathbb{C}\) extends to an embedding of \(K L\) in \(\mathbb{C}\) that fixes \(L\). Thus

For \(i = 1,...,m\) set

Thus

for each \(\sigma\). By Cramer’s rule, \(x_i = \gamma_i / \delta\) where \(\delta\) is the determinant formed by the coefficients \(\sigma(\alpha_i)\), and \(\gamma_i\) is the same except that the \(i\)th column has been replaced by \(\sigma(\alpha)\). Now \(\delta\) and the \(\gamma_i\) are algebraic integers, and \(\delta^2 = disc(R)\). Setting \(e = disc(R)\) gives \(e x_i = \delta \gamma_i \in \mathbb{A}\), and thus

Since the \(\beta_j\) form an integral basis for \(S\), \(e m_{i j}/r\) must all be integers, thus \(r | e m_{i j}\). Since \(r\) is coprime to \(gcd(m_{i j})\) we must have \(r | e = disc(R)\).

**Corollary:** Let \(K=\mathbb{Q}[\omega], \omega=e^{2 \pi i /m}, R = \mathbb{A}\cap K\). Then \(R = \mathbb{Z}[\omega]\).

**Proof:** We know
this is true if \(m\) is a prime power. If \(m\) is not a prime power, write
\(m = m_1 m_2\) for coprime \(m_1, m_2 > 1\). Suppose the result is true for
\(m_1, m_2\). Then let

By inductive hypothesis \(R_1 = \mathbb{Z}[\omega_1], R_2 = \mathbb{Z}[\omega_2]\). Since \(\omega^{m_1} = \omega_2, \omega^{m_2} = \omega_1\) we have \(\omega = \omega_1^r \omega_2^s\) for some \(r,s \in \mathbb{Z}\), and hence \(K = K_1 K_2\), and also \(\mathbb{Z}[\omega] = \mathbb{Z}[\omega_1] \mathbb{Z}[\omega_2]\). We also have \(\phi(m) = \phi(m_1)\phi(m_2)\) since \(m_1, m_2\) are coprime. Also, recall we have shown that \(disc(\omega_1)\) divides a power of \(m_1\) and \(disc(\omega_2)\) a power of \(m_2\), thus their greatest common divisor is one. So we may apply the previous corollary to obtain

*blynn@cs.stanford.edu*💡