I'm prepping for prelim exams and realized I have a fundamental gap in understanding of free modules. I'm trying to answer the following question:

Let $M = \mathbb{Z}^n$ and denote by $pM = p\mathbb{Z}^n$. Suppose that $L$ is a submodule of $M$ with $pM ⊂ L ⊆ M.$
(a) Show that $L$ is a free $\mathbb{Z}$-module of rank $n$.
(b) Show that index $[M : L]$ is finite, and in terms of the index describe the invariant factors (or elementary divisors) of L in M as a $\mathbb{Z}$-module.

I realized that I don't understand what a free module of rank n that doesn't equal $\mathbb{Z}^n$ looks like.

I know that if $N$ is a submodule of a free $\mathbb{Z}^n$ module, then $N\cong \mathbb{Z}^m$ for some $m\le n$ (this is a theorem from our class).

But if $L$ is a free $\mathbb{Z}$ module of rank $n$, then applying this theorem, wouldn't $L=M$ in the problem? What are free $\mathbb{Z}$ modules of rank $n$ that are not $\mathbb{Z}^n$?

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    $\begingroup$ Is $pM=M$?. But isn't $pM\cong M$? $\endgroup$ – Mohan Apr 6 '16 at 0:55
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    $\begingroup$ $pM\neq M$ but $pM\cong M$ as groups. $\endgroup$ – Laura Apr 6 '16 at 12:25

Any two free abelian groups $A$ and $B$ of rank $n$, with $A$ a subgroup of $B$, are isomorphic as abelian groups: both are isomorphic to $\mathbb{Z}^n$. But you can still have $A \subsetneq B$; in other words, $A$ and $B$ can be abstractly isomorphic (that is, there is some function between them which is a group isomorphism), but the inclusion homomorphism of $A$ into $B$ doesn't have to be an isomorphism.

For example, $A = 2\mathbb{Z} \times 3\mathbb{Z}$ is properly contained in $B = \mathbb{Z} \times \mathbb{Z}$, and $A$ is free of rank two: $(2,0)$ and $(0,3)$ form a basis for $A$. This is no longer a basis for $B$.

  • $\begingroup$ Thanks! How would I go about describing the invariant factors in terms of the index? $\endgroup$ – Laura Apr 6 '16 at 12:27
  • $\begingroup$ Ummm...I forgot $\endgroup$ – D_S Apr 6 '16 at 15:14

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