## 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

$\alpha_1\beta_1 + ...+\alpha_r\beta_r$

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

$R S =\{\alpha_1\beta_1+...+\alpha_r\beta_r: \alpha_i\in R, \beta_i\in S\}$

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

$\alpha = \sum_{i,j}\frac{m_{i j}}{r}\alpha_i\beta_j$

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

$\sigma(\alpha) = \sum_{i,j}\frac{m_{i j}}{r}\sigma(\alpha_i)\beta_j$

For $$i = 1,...,m$$ set

$x_i = \sum_{j=1}^n \frac{m_{i j}}{r}\beta_j$

Thus

$\sum_{i=1}^m \sigma(\alpha_i)x_i = \sigma(\alpha)$

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

$e x_i = \sum_{j=1}^n \frac{e m_{i j}}{r}\beta_j \in S$

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

$\omega_1 = e^{2\pi i/m_1}, \omega_2 = e^{2\pi i/m_2}, K_1 =\mathbb{Q}[\omega_1], K_2=\mathbb{Q}[\omega_2], R_1 = \mathbb{A}\cap K, R_2 = \mathbb{A}\cap K$

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

$R = R_1 R_2 = \mathbb{Z}[\omega_1]\mathbb{Z}[\omega_2] = \mathbb{Z}[\omega]$

Ben Lynn blynn@cs.stanford.edu 💡