Number Fields

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 $\Pi :G\to \mathbb{Z}$ 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

$$x^n+a_{n-1}{x}^{n-1}+a_n{a}_{n-2}{x}^{n-2}...+a_n^{n-1}{a}_0$$

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 $R$:

$$\{m_1{\alpha }_1+...+m_n{\alpha }_n:m_i\in \mathbb{Z}\}=\mathbb{Z}{\alpha }_1\oplus +...+\oplus \mathbb{Z}{\alpha }_n$$

Theorem: Let $\{{\alpha }_1,...,{\alpha }_n\}\subset R$ be a basis for $K$ over $\mathbb{Q}$, and let $d=\mathrm{disc}({\alpha }_1,...,{\alpha }_n)$. Then every $\alpha \in R$ can be written in the form

$$\alpha =\frac{m_1{\alpha }_1+...+m_n{\alpha }_n}{d}$$

where $m_j\in \mathbb{Z}$ and $d\mid m_j^2$ 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 $ℂ$. Then we have the system of equations

$${\sigma }_i(\alpha )=x_1{\sigma }_i({\alpha }_1)+...+x_n{\sigma }_i({\alpha }_n)$$

Using Cramer's rule, we have $x_j={\gamma }_j/\delta$ where $\delta =\mid {\sigma }_i({\alpha }_j)\mid$ (so ${\delta }^2=d$ and ${\gamma }_j$ is the same as $\delta$ except that the $j$th column has been replaced by ${\sigma }_i(\alpha )$.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 $n$

$$\mathbb{Z}\frac{{\alpha }_1}{d}\oplus ...\oplus \mathbb{Z}\frac{{\alpha }_n}{d}$$

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

$$\alpha =m_1{\beta }_1+...+m_n{\beta }_n$$

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 $m=\mathrm{2,3}(\mathrm{mod}4)$ and $\{1,(1+\sqrt{m})/2\}$ when $m=1(\mathrm{mod}4)$.

Theorem: Let $\omega =e^{2\pi i/m}$ where $m=p^r$ is a prime power. Then $𝔸\cap \mathbb{Q}[\omega ]=\mathbb{Z}[\omega ]$.

We shall need a couple of lemmas to prove this.

Lemma: For $m\ge 3$, $\mathbb{Z}[1-\omega ]=\mathbb{Z}[\omega ]$ and $\mathrm{disc}(1-\omega )=\mathrm{disc}(\omega )$.

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:

$$\begin{array}{ccc}\mathrm{disc}(\omega )& =& \prod _{1\le r<s\le n}({\alpha }_r-{\alpha }_s{)}^2\\ & =& \prod _{1\le r<s\le n}((1-{\alpha }_r)-(1-{\alpha }_s){)}^2\\ & =& \mathrm{disc}(1-\omega )\end{array}$$

Lemma: For $m=p^r$

$$\prod _{1\le k\le m,p\nmid k}(1-{\omega }^k)=p$$

Proof: Set

$$f(x)=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}=1+x^{p^{r-1}}+x^{2p^{r-1}}+...+x^{(p-1)p^{r-1}}$$

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

$$f(x)=\prod _k(x-{\omega }^k)$$

since there are exactly $\varphi (p^r)=(p-1)p^{r-1}$ values of $k$. Then set $x=1$.

Proof of Theorem: Recall every algebraic integer $\alpha$ can be expressed as

$$\alpha =\frac{m_1+m_2(1-\omega )+...+m_n(1-\omega {)}^{n-1}}{d}$$

where $n=\varphi (p^r)$, $m_i\in \mathbb{Z}$ and $d=\mathrm{disc}(\omega )$. We have previously shown that $\mathrm{disc}(\omega )\mid m^{\varphi (m)}$ 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

$$\beta =\frac{m_i(1-\omega {)}^{i-1}+m_{i+1}(1-\omega {)}^i+...+m_n(1-\omega {)}^{n-1}}{p}$$

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 $N(1-\omega )\mid N(m_i)$, 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 $R=𝔸\cap K$. Then $\mathrm{disc}({\beta }_1,...,{\beta }_n)=\mathrm{disc}({\gamma }_1,...,{\gamma }_n)$.

Proof: We may write

$$(\begin{array}{c}{\beta }_1\\ ⋮\\ {\beta }_n\end{array})=M(\begin{array}{c}{\gamma }_1\\ ⋮\\ {\gamma }_n\end{array})$$

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

$$\mathrm{disc}({\beta }_1,...,{\beta }_n)=\mid M{\mid }^2\mathrm{disc}({\gamma }_1,...,{\gamma }_n)$$

Now $\mid M\mid \in \mathbb{Z}$ thus $\mathrm{disc}({\beta }_1,...,{\beta }_n)$ divides $\mathrm{disc}({\gamma }_1,...,{\gamma }_n)$. A similar argument shows $\mathrm{disc}({\gamma }_1,...,{\gamma }_n)$ must also divide $\mathrm{disc}({\beta }_1,...,{\beta }_n)$ 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 $\mathrm{disc}({\beta }_1,...,{\beta }_n)=\mathrm{disc}({\gamma }_1,...,{\gamma }_n)$, and we may use this fact to define $\mathrm{disc}(G)$ 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 $\mathrm{disc}(R)$, and also $\mathrm{disc}(K)$ where $R$ is the number ring of $K$.

Example: We have for squarefree $m$

$$\mathrm{disc}(\mathbb{Q}[\sqrt{m}])=\{\begin{array}{c}\mathrm{disc}(\sqrt{m})=4m\text{if}m=\mathrm{2,3}(\mathrm{mod}4)\\ \mathrm{disc}(\frac{1+\sqrt{m}}{2})=m\text{if}m=1(\mathrm{mod}4)\end{array}$$