## 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 $m_1, m_2 \gt 1$. 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

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