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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

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    $\begingroup$ Could you provide a bit more of context for this term -- such as a full sentence or paragraph where it appeared? $\endgroup$ Feb 24, 2016 at 11:41
  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Feb 24, 2016 at 11:43
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    $\begingroup$ 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. $\endgroup$ Feb 24, 2016 at 12:01
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    $\begingroup$ @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? $\endgroup$
    – DonAntonio
    Feb 24, 2016 at 12:10
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    $\begingroup$ 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. $\endgroup$
    – knsam
    Feb 24, 2016 at 13:24

1 Answer 1

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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 $\mathcal{O}_K$.

This is a fairly standard ordeal in number theory, look at any standard text in algebraic number theory.

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  • $\begingroup$ We haven't talked about modules or Z-free modules. $\endgroup$
    – Richard
    Feb 24, 2016 at 11:50
  • $\begingroup$ 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. $\endgroup$ Feb 24, 2016 at 12:07
  • $\begingroup$ But since O is a lattice, any Z-basis for O must also be a Q-basis for O_k. Then why is the index not simply 1? $\endgroup$
    – Richard
    Feb 24, 2016 at 22:52
  • $\begingroup$ @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. $\endgroup$ Feb 28, 2016 at 15:53
  • $\begingroup$ Oh right, makes sense. Another question I had was: the ring of integers of Q(root3) is Z[root3]. But the ring of integers of Q(root5) is Z[(1+root5)/2]. Why isn't it Z[root5]? I don't see why it is different since both 3 and 5 are prime. $\endgroup$
    – Richard
    Feb 29, 2016 at 21:59

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