Number Fields

A Dedekind domain is an integral domain $R$ such that

1. Every ideal is finitely generated.

2. Every nonzero prime ideal is a maximal ideal.

3. $R$ is integrally closed in its field of fractions.

The last condition means that if $\alpha /\beta \in K$ is a root of a monic polynomial over $R$, then $\alpha /\beta \in R$, that is, $\beta \mid \alpha$ in $R$.

The first condition is equivalent to both of the following:

• Every increasing sequence of ideals is eventually constant, that is, given $I_1\subset I_2\subset I_3...$, there exists some $k$ for which $I_k=I_n$ for all $n\ge k$.

• Every nonempty set $S$ of ideals has a maximal member, that is, there exists $M\in S$ such that $M\subset I\in S\Rightarrow M=I$.

A ring satisfying these conditions is called a Noetherian ring.

It is not difficult to show that these three conditions are equivalent. First suppose we are given some sequence $I_1\subset I_2\subset I_3...$. Then consider the ideal generated by all the ideals in the sequence. If it is finitely generated, then for some $k$, $I_k$ contains all the generators, and thus $I_k=I_n$ for all $n\ge k$.

Now suppose we are given a nonempty set $S$ of ideals. Then take any ideal $I_1\in S$, and look for an ideal $I_2$ that strictly contains $I_1$. Iterating this procedure produces a sequence $I_1\subset I_2\subset ...$. If there exists some $k$ for which $n\ge k$ implies $I_k=I_n$, then $I_k$ is a maximal ideal in $S$.

Next suppose we are given some ideal $I$. Consider the set $S$ of subideals of $I$ that are finitely generated. If it has a maximal element $I\prime$, then we must have $I=I\prime$.

Theorem: Every number ring is a Dedekind domain.

Proof: Since a number ring is a free abelian group of finite rank, any ideal must also be a free abelian group of finite rank (because it is a additive subgroup) thus every ideal is finitely generated.

For the second condition, it suffices to show that for every nonzero prime ideal $P$, the integral domain $R/P$ is a field. We do so by showing $R/P$ is finite, which implies it is a field because every finite integral domain is a field (because for any element $a$, we must have $a^n=1$ for some $n$, thus $a^{n-1}$ is the inverse of $a$).

We shall show that $R/I$ is finite for any nonzero ideal $I$. Take any nonzero $\alpha \in I$, and let $m=N^K(\alpha )$ where $K$ is the number field of $R$. So $m=\alpha \beta$ where $\beta$ is the product of the conjugates of $\alpha$. Hence $\beta$ is also be an algebraic integer. Now $m$ is a nonzero integer, so $\beta =m/\alpha \in K$, so we deduce $\beta \in R$ which means $m\in I$. Now $R/(m)$ is finite (it contains $m^n$ elements), hence $R/I$ is finite (its order divides $m^n$).

Lastly, suppose $\alpha$ is a root of a monic polynomial $f(x)=a_0+...+a_{n-1}{x}^{n-1}+x^n$ over $R$. Now $R=\mathbb{Z}[a_0,...,a_{n-1}]$ is clearly finitely generated, and since $\mathbb{Z}[a_0,...,a_{n-1},\alpha ]$ is also finitely generated since any expression containing ${\alpha }^m$ for $m\ge n$ can be rewritten using smaller powers of $\alpha$. Hence $R$ is integrally closed.