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Let $M$ be a finitely generated R-Module; $m\in M$. $b_{1}, b_{2}\ldots b_{n}$ is the minimal generating set of $M$. I read somewhere that if $m=s_{1}b_{1}+s_{2}b_{2}\ldots s_{q}b_{q}=s_{1}'b_{1}+s_{2}'b_{2}\ldots s_{q}'b_{q}$, then it is not necessary that $s_{1}=s_{1}'\ldots s_{n}=s_{n}'$, but it is necessary that $s_{1}b_{1}=s_{1}'b_{1}\ldots s_{n}b_{n}=s_{n}'b_{n}$.

Since $s_{i}$ belongs to the ring $R$ and $b_{i}$ belongs to the module $M$ which is not a subset of the ring $R$, is it always true that if $s_{i}.b_{i}=0$, then we can say nothing about whether either of them is $0$ or not?

My logic: $s_{1}b_{1}+s_{2}b_{2}\ldots s_{q}b_{q}=s_{1}'b_{1}+s_{2}'b_{2}\ldots s_{q}'b_{q}$

Therefore, $(s_{1}-s_{1}')b_{1}+(s_{2}-s_{2}')b_{2}\ldots (s_{q}-s_{q}')b_{q}=0$. None of $b_{1}\ldots b_{n}$ can be $0$, as they generate the non-zero $M$. I'm stuck at this point.

  • You may care to investigate faithful modules. – Lord_Farin Apr 25 '13 at 11:00
  • You need to be more careful with your terminology. Is $R$ an arbitrary commutative ring? Is $b_1,\ldots,b_q$ a minimal set of generators? And the paragraph which includes the word 'group' does not make any sense. – Rhys Apr 25 '13 at 11:03
  • I suppose you wished to write $s_1b_1=s_1'b_1$. Anyway, please note that, in general, you cannot "multiply" together elements of different groups. Here you have a "product" by elements of $R$, which is actually an action of $R$ on $M$. – A.P. Apr 25 '13 at 11:03
  • @Rhys yes $R$ is an arbitrary commutative ring. $b_{1},....b_{q}$ is also the minimal set of generators of $M$. The paragraph has also been modified. Apologies for the unclear language. –  Apr 25 '13 at 11:11

1 Answers1

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It is certainly possible to have $s b = 0$, with neither $s\in R$ nor $b \in M$ being zero. For example, let $R = \mathbb{Z}$, and let $M$ be a module with one generator $b$, satisfying $3b = 0$ (check that this really is a $\mathbb{Z}$-module!).

The statement in your first paragraph is actually wrong, at least without extra assumptions. Again, it's nice to just do a concrete example: Let $R = \mathbb{Z}$, and let $M$ be generated by $b_1, b_2$, satisfying $2b_1 = 3b_2$. Then we have $$ 3b_1 = b_1 + 3b_2~, $$ but $3b_1 \neq b_1$, and $0 \neq 3b_2$.

Rhys
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  • Does the assumption that $2b_{1}+b_{1}=3b_{1}$ always hold? I've seen this sort of thing in multiple places and it always confuses me. This does not follow from the axioms. –  Apr 25 '13 at 12:57
  • Yes; for any $r_1, r_2 \in R$, and $m \in M$, we have $(r_1 + r_2)m = r_1 m + r_2 m$. This is indeed one of the axioms for a module. – Rhys Apr 25 '13 at 13:00
  • True true. You have assumed $Z$ to be the ring here. I was making the more general statement that if $2$ and $3$ are part of another ring, the $2+1\neq 3$ necessarily. But this is irrelevant here. Edit: I'm assuming a unital ring of course. –  Apr 25 '13 at 13:10
  • The textbook says that 'under proper conditioning', $M$ will behave this way. Under what conditions will $M$ necessitate $s_{1}b_{1}=s_{1}'b_{1}\ldots s_{n}b_{n}=s_{n}'b_{n}$ if $s_{1}b_{1}+s_{2}b_{2}\ldots s_{q}b_{q}=s_{1}'b_{1}+s_{2}'b_{2}\ldots s_{q}'b_{q}$?

    Terribly sorry for the incessant questioning.

     –  Apr 25 '13 at 13:40
  • Without thinking too hard, it seems like that will only be true if $M$ is a direct sum of submodules, each generated by a single $b_i$. – Rhys Apr 25 '13 at 14:34
  • If that is so, why is it impossible that $2b_{1}=3b_{2}$? $b_{1}$ and $b_{2}$ are still part of the module $M$! –  Apr 25 '13 at 14:42
  • I realise I might be asking a lot of stupid questions, but I've been stuck at this part for quite some time, and it is a little irritating. I would completely understand if you chose not to answer these queries. –  Apr 25 '13 at 15:02
  • It's okay. :-)

    Notice that I said 'direct sum'. In the example I gave, $b_1$ and $b_2$ each generate a submodule, but together they do not generate a direct sum, precisely because of the given relation.

     – Rhys Apr 25 '13 at 15:24
  • I don't think that the book is saying that for a "properly conditioned" $M$ it will always be true that $s_1b_1+\ldots+s_qb_q=s'_1b_1+\ldots+s'_qb_q$ implies $s_jb_j=s'_jb_j$ for all $j$. What it's saying is that for properly conditioned $M$ a generating set ${b_1,\ldots,b_q}$ will exist for which that is true. "Properly conditioned" means that $R$ is a principal ideal domain, although in the book you are using, the author only proves this for the special case where the PID is a Euclidean domain. – Will Orrick Apr 26 '13 at 14:56