Order of Algebraic Number Field

Question

I'm having troubles understanding the definition of an order of an algebraic number field. The formal definition should be:

Let $K$ be a number field. An order $O$ of $K$ is a subring of $K$ that is finitely generated as a $\mathbb{Z}-$module and satisfies $O\otimes \mathbb{Q}=K.$

I understand what it means that $O$ is finitely generated as a $\mathbb{Z}-$module, but the second condition is not very intuitive. Can someone help me with that?

Answer 1

The point is that an order in a number field $K$ is supposed to be a "big" subring of $\mathcal O_K$ (the integers of $K$), and being "big" can be described in several equivalent ways (the equivalence is not always obvious):

1. having the same rank $[K:\mathbf Q]$ as a $\mathbf Z$-module that $\mathcal O_K$ has,

2. having finite index in $\mathcal O_K$,

3. containing a basis of $K/\mathbf Q$.

Maybe it's worth seeing some rings that are not orders. In $\mathbf Q(i)$, $\mathbf Z[i]$ is the full ring of integers (the maximal order in $\mathbf Q(i)$) and its subring $\mathbf Z[2i] = \mathbf Z + \mathbf Z(2i)$ has rank $2$ and is an order, while $\mathbf Z$ is not an order in $\mathbf Q(i)$: it's too small because its fraction field is $\mathbf Q$ rather than $\mathbf Q(i)$.

In $\mathbf Q(\sqrt{2},\sqrt{3})$, which has degree $4$ over $\mathbf Q$, the ring $\mathbf Z[\sqrt{2}]$ is not an order since its $\mathbf Z$-rank is $2$ rather than $4$ (and its fraction field $\mathbf Q(\sqrt{2})$ is too small).

Answer 2

It simply means that its field of fractions is $K$.

You have to remember for any integral domains $A\subset B$, if $B$ is an $A$-algebra which is finite as an $A$-module, $B$ is a field if and only if $A$ is a field.

Now $O\otimes \mathbf Q$ is the localisation of $O$ w.r.t. the multiplicative system $S=\mathbf Z^*$, and as $O$ is a finitely generated $A$-module, $O\otimes \mathbf Q\subset K$ is an integral domain and a finite-dimensional $\mathbf Q$-vector space. Thus, it is a field containing $O$ and contained in the field of fractions of $O$. By definition, it is therefore this field of fractions.

Answer 3

An equivalent definition is that $\mathcal{O}$ is a ring, which is a free abelian group of finite rank, generated by a $\mathbb{Q}$-basis of $K$.

Answer 4

First of all, the condition should say more precisely: "The natural map $O \otimes_{\mathbb Z} \mathbb Q \to K$ is an isomorphism."

Here, the "natural map" is the map given by the universal property of the tensor product, associated to the inclusions $O \to K$ and $\mathbb Q \to K$.

That map sends $a \otimes q$ to $aq \in K$. Because every element of $O \otimes_{\mathbb Z} \mathbb Q$ is a finite sum of elements of the form $a \otimes q$, the surjectivity of the map $O \otimes_{\mathbb Z} \mathbb Q \to K$ is equivalent to saying that $O$ is a generating set of $K$ as a $\mathbb Q$-vector space. (One can show that injectivity is automatic when $O$ is a finitely generated $\mathbb Z$-module.)

So, every subring $O \subset K$ that is a finitely generated $\mathbb Z$-module, is an order inside its fraction field $Frac(O) \subset K$. But to be an order of $K$, it needs to generate the entire field $K$ over $\mathbb Q$.