Number Fields

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 $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=\mathrm{disc}(R)$.

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 $KL$ is the smallest subfield of $ℂ$ containing both $K$ and $L$. In fact, $KL$ 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 $K,L,KL$. Clearly $T$ contains

$$RS=\{{\alpha }_1{\beta }_1+...+{\alpha }_r{\beta }_r:{\alpha }_i\in R,{\beta }_i\in S\}$$

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 $d=\gcd (\mathrm{disc}(R),\mathrm{disc}(S))$.

Theorem: Suppose that $[KL:\mathbb{Q}]=mn$. Then $T\subset \frac{1}{d}RS$.

Corollary: If $[KL:\mathbb{Q}]=mn$ and $d=1$, then $T=RS$.

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

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

Proof: $\sigma$ has $n$ extensions to embeddings of $\mathrm{KL}$ in $ℂ$ 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 $ℂ$.

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 $mn$ products ${\alpha }_i{\beta }_j$ form a basis for $RS$ over $\mathbb{Z}$ and for $KL$ over $\mathbb{Q}$. Any $\alpha \in T$ can be written in the form

$$\alpha =\sum _{i,j}\frac{m_{ij}}{r}{\alpha }_i{\beta }_j$$

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

By the lemma, every embedding of $\sigma$ of $K$ in $ℂ$ extends to an embedding of $KL$ in $ℂ$ that fixes $L$. Thus

$$\sigma (\alpha )=\sum _{i,j}\frac{m_{ij}}{r}\sigma ({\alpha }_i){\beta }_j$$

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

$$x_i=\sum _{j=1}^n\frac{m_{ij}}{r}{\beta }_j$$

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 ${\delta }^2=\mathrm{disc}(R)$. Setting $e=\mathrm{disc}(R)$ gives $e{x}_i=\delta {\gamma }_i\in 𝔸$, and thus

$$e{x}_i=\sum _{j=1}^n\frac{e{m}_{ij}}{r}{\beta }_j\in S$$

Since the ${\beta }_j$ form an integral basis for $S$, $e{m}_{ij}/r$ must all be integers, thus $r\mid e{m}_{ij}$. Since $r$ is coprime to $\gcd (m_{ij})$ we must have $r\mid e=\mathrm{disc}(R)$.

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

$${\omega }_1=e^{2\pi i/m_1},{\omega }_2=e^{2\pi i/m_2},K_1=\mathbb{Q}[{\omega }_1],K_2=\mathbb{Q}[{\omega }_2],R_1=𝔸\cap K,R_2=𝔸\cap K$$

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 $\varphi (m)=\varphi (m_1)\varphi (m_2)$ since $m_1,m_2$ are coprime. Also, recall we have shown that $\mathrm{disc}({\omega }_1)$ divides a power of $m_1$ and $\mathrm{disc}({\omega }_2)$ a power of $m_2$, thus their greatest common divisor is one. So we may apply the previous corollary to obtain

$$R=R_1{R}_2=\mathbb{Z}[{\omega }_1]\mathbb{Z}[{\omega }_2]=\mathbb{Z}[\omega ]$$