# Index of an order in a number field

I'm taking a course in algebraic number theory and the lecturer mentioned the 'index' of an order in a number field without defining what an 'index' is. Can somebody please clarify this for me? Thanks

• Could you provide a bit more of context for this term -- such as a full sentence or paragraph where it appeared? Feb 24, 2016 at 11:41
• It would have a meaning in the context of a subgroup, for example, just as "order" would have a meaning. It is only possible to guess without more context what the lecturer was discussing. Feb 24, 2016 at 11:43
• knsam's answer hits this nail on the head. As a concrete example you can consider the order $O=\Bbb{Z}[\sqrt5]$ of the number field $K=\Bbb{Q}(\sqrt5)$. $O$ is not the full ring of integers of $K$, but it is a ring all right, and it contains a $\Bbb{Q}$-basis of $K$, so it is an order of $K$. Here the full ring of integers is $$\mathcal{O}_K=\Bbb{Z}[\frac{1+\sqrt5}2].$$ We see that the index of $O$ as a subgroup of $\mathcal{O}_K$ is two. Therefore $f=2$ in this case. Feb 24, 2016 at 12:01
• @Richard That already is another question, belonging more to ring theory (or even to group theory). Do you know how to calculate the index of a subgroup of a finitely generated free-abelian group? With matrices and all that? Feb 24, 2016 at 12:10
• As a complement to Jyrki's comment, in general, the computation above would be done by writing the matrix of the inclusion (which is $\mathbf{Z}$-linear!) and finding its determinant. Feb 24, 2016 at 13:24

An order $\mathcal{O}$ in a number field $K$ is a free $\mathbf{Z}$-submodule of $\mathcal{O}_K$ of rank $[K:\mathbf{Q}]$. Since $\mathcal{O}_K$ is also a free $\mathbf{Z}$-module of rank $[K:\mathbf{Q}]$, it follows from the structure theorem for $\mathbf{Z}$-modules that the quotient $\mathcal{O}_K/\mathcal{O}$ is a finite abelian group. The order of this quotient is called the index of the order $\mathcal{O}$ in $\mathcal{O}_K$.
• Richard, the structure theory of $\Bbb{Z}$-modules = the structure theory of finitely generated abelian groups is something that is more often than not covered in a course that is a prerequisite for a course in algebraic number theory. Of course, students/teachers often have to deal with the problem of not having all the desirable background. You really should ask your teacher how much material they assume you to know from earlier courses in abstract algebra. Only your teacher is familiar with what has been offered locally, and can advice you about what remedial action (if any) is required. Feb 24, 2016 at 12:07
• @Richard: No. $\mathcal{O}_K$ is itself an order, and not a vector space over $\Bbb{Q}$. It is a free module over $\Bbb{Z}$ though (all orders are). So it has basis over $\Bbb{Z}$, but a $\Bbb{Z}$-basis of $O$ need not generate all of $\mathcal{O}_K$. See my example under the main question. Feb 28, 2016 at 15:53