1
$\begingroup$

Let $L$ be a number field and $\mathcal{O}_L$ its ring of integers. If $M\subset M'$ are non-zero finitely generated $\mathcal{O}_L$-submodules of $L$, then:

i) Prove that $(M':M)<\infty$;

ii) Assuming $M$ and $M'$ are free $\mathbb{Z}$-modules with same rank and basis $B=(b_1, ..., b_n)$ and $B'=(b'_1, ..., b'_n)$ respectively, prove that $(M':M)=|\det (T)|$, where $T$ is the transition matrix from $B$ to $B'$.

I already know a theorem which states that if $M$ is a non trivial $\mathcal{O}_L$-submodule of $L$, then $M\simeq \mathbb{Z}^n$, (meaning that M is isomorphic to a free $\mathbb{Z}$-module of rank $n$), where $n=[L:\mathbb{Q}]$. I think that makes part i) easy and explains the assumption on part ii). However, I have no idea how to relate $\det(T)$ to the index $(M':M)$. Any tips? Thanks!

  • 1
    Do you know that any square matrix $T$ over $\mathbb{Z}$ can be diagonalized by multiplying on right and left with elementary matrices? – 2017-01-15
  • 0
    @Mohan, if I'm not mistaken, I think what you're saying means that we may consider $M=$, for some $a_1,...,a_n\in \mathbb {Z}$, right? How do I get the índex from that? – 2017-01-16
  • 0
    Correction: $M=$ (considering $M'=$) – 2017-01-16
  • 0
    Is your definition of index the cardinality of $M'/M$? If so, you should be able to check this. – 2017-01-16

0 Answers 0