3.2 Ideals in rings of integers

Let me recall some definition you may have seen in other courses.

Definition 3.2.1

Let \(R\) be a commutative ring. Then \(R\) is called Noetherian any of the following equivalent conditions holds:

Every ideal is finitely generated.
Every increasing chain of ideals \(I_1 \subset I_2 \subset \dots \) is eventually constant.
Every non-empty set \(S\) of ideals contains a (not necessarily unique) maximal member.

Exercise 3.2.2

Check that these definitions are all equivalent.

Proposition 3.2.3

Let \(K\) be a number field and \(\mathcal{O}_K\) its ring of integers. Then \(\mathcal{O}_K\) is a Noetherian ring.

Proof ▼

We saw in Corollary 2.2.8 that \(\mathcal{O}_K\) is finitely generated as an additive group. Now, any ideal \(\mathfrak {a}\subset \mathcal{O}_K\) is an additive subgroup of the abelian group \(\mathcal{O}_K\) and therefore is also finitely generated. ¹ So \(\mathcal{O}_K\) is Noetherian.

Proposition 3.2.4

Let \(K\) be a number field and \(\mathfrak {a}\) a non-zero ideal in \(\mathcal{O}_K\). Then \(\mathfrak {a}\) contains some positive integer and moreover the quotient ring \(\mathcal{O}_K /\mathfrak {a}\) is finite.

Proof ▼

Let \(\alpha \in \mathfrak {a}\backslash 0\) and \(N=N_{K/\mathbb {Q}}(\alpha )\). Since \(\alpha \neq 0\), then \(N \neq 0\). Now, by Proposition 1.7.6 we see that \(N=\alpha \beta \) where \(\beta \) is a product of conjugates of \(\alpha \) so it is in \(\overline{\mathbb {Z}}\). Moreover, \(\beta \in \mathcal{O}_K\) since \(\beta =N/\alpha \in K\) (recall \(\mathcal{O}_K=K \cap \overline{\mathbb {Z}}\)). But since \(\mathfrak {a}\) is an ideal \(N=\alpha \beta \in \mathfrak {a}\).

Now, lets use this to prove the quotient is finite. First note that since \(N \in \mathfrak {a}\) then \((N) \subset \mathfrak {a}\) therefore \(|\mathcal{O}_K /\mathfrak {a}| \leq |\mathcal{O}_K / (N)|\). Secondly, since \(\mathcal{O}_K\) is finitely generated as an abelian group, by picking an integral basis we get an isomorphism (of additive abelian groups) \(\mathcal{O}_K \cong \mathbb {Z}^n\) where \(n=[K:\mathbb {Q}]\). Therefore \(\mathcal{O}_K/(N) \cong (\mathbb {Z}/N\mathbb {Z})^n\) which is finite. Therefore so is \(\mathcal{O}_K/\mathfrak {a}\).

Lemma 3.2.5

A finite integral domain is a field.

Proof ▼

Let \(R\) be a finite, non-trivial integral domain. Let \(r \in R\backslash 0\). We need to show that \(r\) has an inverse. Here is the trick: consider the sequence \(r,r^2,r^3,\dots .\) Since \(R\) is finite at some point we must have \(r^n=r^m\) for some \(m{\lt}n\). Then \(r^m(r^{n-m}-1)=0\), but this is where being an integral domain comes in, since this means either \(r^m=0\) or \(r^{n-m}-1=0\). Since \(r \neq 0\) and \(R\) is an integral domain \(r^m \neq 0\). Therefore \(r^{n-m}=1\). This means \(r^{-1}=r^{n-m-1}\) and therefore \(r\) has an inverse.

Corollary 3.2.6

Let \(K\) be a number field. Then every non-zero prime ideal in \(\mathcal{O}_K\) is maximal.

Proof ▼

By Proposition 1.1.22 we know that if \(\mathfrak {p}\subset \mathcal{O}_K\) is a prime ideal then \(\mathcal{O}_K/\mathfrak {p}\) is an integral domain. Now, by Proposition 3.2.4 \(\mathcal{O}_K /\mathfrak {p}\) is finite. But by Lemma 3.2.5 this is then a field. Now, using Proposition 1.1.22 again, \(\mathfrak {p}\) is maximal.

Definition 3.2.7

Let \(R,A\) be rings with \(A \subset R\) a subring and \(x \in R\). We say \(x\) is integral over \(A\) if there exist \(a_i \in A\) such that

\[ x^n+a_{n-1}x^{n-1}+\cdots +a_0=0 \]

for some \(n\).

Let \(A'\) be the set of all elements of \(R\) that are integral over \(A\). Then similarly to how we prove Corollary 2.1.16, \(A'\) is a ring which we call the integral closure of \(A\) in \(R\).

Moreover, if \(R\) is an integral domain and we let

\[ \mathrm{Frac}(R):=\{ a/b | a,b \in R, b \neq 0\} \]

be its fields of fractions (this can alternatively be defined as smallest field containing \(R\)), then we say \(R\) is integrally closed, if it is integrally closed in its field of fractions.