]> Number Fields - The Additive Structure of a Number Ring

## The Additive Structure of a Number Ring

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 $\Pi :G\to ℤ$ 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 $\Pi \left(H\right)\subset ℤ$ is either $\left\{0\right\}$ or $ℤ$. In the former case, we must have $H=H\cap K$. In the latter, pick $h\in H$ such that $\Pi \left(h\right)$ generates $\Pi \left(H\right)$. Then $H=ℤh\oplus \left(H\cap K\right)$.

Let $K$ be a number field of degree $n$ over $ℚ$, 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 ${a}_{0},...,{a}_{n}\in ℤ$, 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 $ℚ$, 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 $ℚ$, we have a free abelian group of rank $n$ inside $R$:

$\left\{{m}_{1}{\alpha }_{1}+...+{m}_{n}{\alpha }_{n}:{m}_{i}\in ℤ\right\}=ℤ{\alpha }_{1}\oplus +...+\oplus ℤ{\alpha }_{n}$

Theorem: Let $\left\{{\alpha }_{1},...,{\alpha }_{n}\right\}\subset R$ be a basis for $K$ over $ℚ$, and let $d=\mathrm{disc}\left({\alpha }_{1},...,{\alpha }_{n}\right)$. Then every $\alpha \in R$ can be written in the form

$\alpha =\frac{{m}_{1}{\alpha }_{1}+...+{m}_{n}{\alpha }_{n}}{d}$

where ${m}_{j}\in ℤ$ and $d\mid {m}_{j}^{2}$ for all $j$.

Proof: Write $\alpha ={x}_{1}{\alpha }_{1}+...+{x}_{n}{\alpha }_{n}$ where ${x}_{j}\in ℚ$. Let ${\sigma }_{1},...,{\sigma }_{n}$ be the embeddings of $K$ in $ℂ$. Then we have the system of equations

${\sigma }_{i}\left(\alpha \right)={x}_{1}{\sigma }_{i}\left({\alpha }_{1}\right)+...+{x}_{n}{\sigma }_{i}\left({\alpha }_{n}\right)$

Using Cramer's rule, we have ${x}_{j}={\gamma }_{j}/\delta$ where $\delta =\mid {\sigma }_{i}\left({\alpha }_{j}\right)\mid$ (so ${\delta }^{2}=d$ and ${\gamma }_{j}$ is the same as $\delta$ except that the $j$th column has been replaced by ${\sigma }_{i}\left(\alpha \right)$. Thus $d{x}_{j}=\delta {\gamma }_{j}$ so $d{x}_{j}$ is an algebraic integer. As it is also rational, we must have ${m}_{j}=d{x}_{j}\in ℤ$. 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 $n$

$ℤ\frac{{\alpha }_{1}}{d}\oplus ...\oplus ℤ\frac{{\alpha }_{n}}{d}$

Thus $R$ has a basis over $ℤ$, 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 ${m}_{i}\in ℤ$. Such a basis is called an integral basis for $R$, or a basis for $R$ over $ℤ$.

For example, in the quadratic field $ℚ\left[\sqrt{m}\right]$ for squarefree $m$, an integral basis for its number ring $R$ is $\left\{1,\sqrt{m}\right\}$ when $m=2,3\left(\mathrm{mod}4\right)$ and $\left\{1,\left(1+\sqrt{m}\right)/2\right\}$ when $m=1\left(\mathrm{mod}4\right)$.

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

We shall need a couple of lemmas to prove this.

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

Proof: The first equation follows directly from $\omega =1-\left(1-\omega \right)$. 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}\left(\omega \right)& =& \prod _{1\le r

Lemma: For $m={p}^{r}$

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

Proof: Set

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

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

$f\left(x\right)=\prod _{k}\left(x-{\omega }^{k}\right)$

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

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

$\alpha =\frac{{m}_{1}+{m}_{2}\left(1-\omega \right)+...+{m}_{n}\left(1-\omega {\right)}^{n-1}}{d}$

where $n=\varphi \left({p}^{r}\right)$, ${m}_{i}\in ℤ$ and $d=\mathrm{disc}\left(\omega \right)$. We have previously shown that $\mathrm{disc}\left(\omega \right)\mid {m}^{\varphi \left(m\right)}$ so $d$ must be a power of $p$. We wish to show $R=ℤ\left[\omega \right]=ℤ\left[1-\omega \right]$. If this were not true, then $R$ would contain some element

$\beta =\frac{{m}_{i}\left(1-\omega {\right)}^{i-1}+{m}_{i+1}\left(1-\omega {\right)}^{i}+...+{m}_{n}\left(1-\omega {\right)}^{n-1}}{p}$

for some $i\le n$, and ${m}_{j}\in ℤ$ with $p\nmid {m}_{i}$.

By the second of the above lemmas, $p/\left(1-{\omega }^{n}\right)\in ℤ\left[\omega \right]$ since each $1-{\omega }^{k}$ is divisible by $1-\omega$. Thus $\beta p/\left(1-{\omega }^{i}\right)\in R$. This leads to ${m}_{i}/\left(1-\omega \right)\in R$, from which it follows $N\left(1-\omega \right)\mid N\left({m}_{i}\right)$, which is impossible since $N\left({m}_{i}\right)={m}_{i}^{n}$ while the lemma shows $N\left(1-\omega \right)=p$.

Theorem: Let $\left\{{\beta }_{1},...,{\beta }_{n}\right\}$ and $\left\{{\gamma }_{1},...,{\gamma }_{n}\right\}$ be two integral bases for $R=𝔸\cap K$. Then $\mathrm{disc}\left({\beta }_{1},...,{\beta }_{n}\right)=\mathrm{disc}\left({\gamma }_{1},...,{\gamma }_{n}\right)$.

Proof: We may write

$\left(\begin{array}{c}{\beta }_{1}\\ ⋮\\ {\beta }_{n}\end{array}\right)=M\left(\begin{array}{c}{\gamma }_{1}\\ ⋮\\ {\gamma }_{n}\end{array}\right)$

where $M$ is an $n×n$ matrix over $ℤ$. From here we can derive the matrix equation $\left[{\sigma }_{j}\left({\beta }_{i}\right)\right]=M\left[{\sigma }_{j}\left({\gamma }_{i}\right)\right]$, and taking determinants and squaring gives

$\mathrm{disc}\left({\beta }_{1},...,{\beta }_{n}\right)=\mid M{\mid }^{2}\mathrm{disc}\left({\gamma }_{1},...,{\gamma }_{n}\right)$

Now $\mid M\mid \in ℤ$ thus $\mathrm{disc}\left({\beta }_{1},...,{\beta }_{n}\right)$ divides $\mathrm{disc}\left({\gamma }_{1},...,{\gamma }_{n}\right)$. A similar argument shows $\mathrm{disc}\left({\gamma }_{1},...,{\gamma }_{n}\right)$ must also divide $\mathrm{disc}\left({\beta }_{1},...,{\beta }_{n}\right)$ thus they must be equal.

Note we may use the same argument to show that if $\left\{{\beta }_{1},...,{\beta }_{n}\right\}$ and $\left\{{\gamma }_{1},...,{\gamma }_{n}\right\}$ are elements of $K$ that generate the same additive subgroup of $K$ then $\mathrm{disc}\left({\beta }_{1},...,{\beta }_{n}\right)=\mathrm{disc}\left({\gamma }_{1},...,{\gamma }_{n}\right)$, and we may use this fact to define $\mathrm{disc}\left(G\right)$ 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 $\mathrm{disc}\left(R\right)$, and also $\mathrm{disc}\left(K\right)$ where $R$ is the number ring of $K$.

Example: We have for squarefree $m$

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