Given a ring $A$, and an $A$-module $M$. Let $x_1, . . . , x_n ∈ M$. Say $\operatorname{rank}(M) = n$ if for all $m ∈ M$ there exist unique elements $a_i ∈ A$, $i = 1, . . . , n,$ such that $m=\sum_{i=1}^n a_i x_i$.

Give examples such that
1. $M$ is free of rank $1$ and has a submodule $M_0$ which cannot be generated by less than two elements, and
2. $M$ is finitely generated, but $M_0$ is not finitely generated.

I am stuck on this examples but I am unable to find one. I know that if $A$ is a field then its impossible since M is vector space. I tried to test with $A = \mathbb{Z}$ but it doesn't work for me. Can anyone give me some hints to work on? Thank you in advance!

  • $\begingroup$ For free modules see here. $\endgroup$ – Dietrich Burde Nov 5 '15 at 20:17
  • $\begingroup$ The title and the question seem to ask different things. $\endgroup$ – user26857 Feb 10 '16 at 17:08

Set $A=\Bbb Z[x,y]/(xy)$ with $M=A$ and $M_0=(x,y)$.

For an infinite example, instead use $A= \Bbb Z[x_1,x_2,\ldots]/I$ where $I$ is the ideal generated by all cross-products $x_ix_j, i\neq j$. Similarly use $M=A$ and $M_0=(x_1,x_2,\ldots)$.

  • $\begingroup$ Why the quotients? $A=\mathbb{Z}[x,y]$ works as does $A=\mathbb{Z}[x_1,x_2,\ldots]$. $\endgroup$ – David Hill Nov 5 '15 at 20:37
  • $\begingroup$ @David no, it doesn't, because the word "unique" in the problem. For instance, in the first example (but without quotient), if we have $m=xy\in M_0$, it could be represented both as $xy$ with $x\in A, y\in M_0$ and as $yx$ with $y\in A,x\in M_0$. $\endgroup$ – Arthur Nov 5 '15 at 20:44
  • $\begingroup$ I see. Thanks for pointing that out. $\endgroup$ – David Hill Nov 5 '15 at 20:52
  • $\begingroup$ @Arthur If I understand well, a module of rank $n$ is in fact a free module of rank $n$. Moreover, the OP seems to ask for a submodule $M_0$ of a free module of rank one such that $M_0$ is 2-generated, but not less. Then what has this to do with the uniqueness of the representation of $xy$ in your example, and why isn't ok the proposed example of David Hill? If instead (as the title seems to suggest) want $M_0$ having rank greater than one, then this is not possible. What I'm missing here? $\endgroup$ – user26857 Feb 10 '16 at 16:59
  • $\begingroup$ @user26857 I missed that $M_0$ didn't have to be free. As you said, that cannot be done, and I didn't realise that there is more than one way to decompose $0$ in my examples of $M_0$. So David Hill's example works just fine. $\endgroup$ – Arthur Feb 10 '16 at 18:10

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