Originally Posted by

**ymar** Hi, NonCommAlg.

Thank you for your answer. I know about the converse and it's easy to prove.

I've been thinking about your hint the whole day today and I'm afraid I'm still very far from seeing it. I noticed instead that free modules are very different from vector spaces. I was shocked to notice that $\displaystyle \mathbb{Z}$ doesn't have any linearly independent subset when viewed as a submodule of the (free) module $\displaystyle \left(\mathbb{Z}^2\right)_{\mathbb{Z}^2}.$ This is awful.

The first thing I would like to ask is if it's obvious for some reason that taking the free module as $\displaystyle B$ has to work given that we know that the theorem is true. In other words, why should I have thought of the free module when I was trying to prove it?

The second thing is, is it enough to prove this for the elements of a basis of $\displaystyle B$? I don't even see this.

OK, so I know that

$\displaystyle \left(\forall m'\in B\right) \left(\forall r\in R\setminus\{0\}\right) \left(\exists s\in R\right) \; m'rs\in A.$

What I want to know is that

$\displaystyle \left(\forall m'\in B\setminus\{0\}\right) \left(\exists s\in R\right) \; m's\in A \mbox{ and } m's\neq 0.$

You told me to assume that $\displaystyle B$ has a basis. To me it just means, "Write $\displaystyle m'$ in the basis." OK. Let $\displaystyle X=\left\{x_i\right\}_{i\in I}$ be a basis of $\displaystyle B.$ Let

$\displaystyle m'=\sum_{k=1}^t x_{i_k}a_k,$ where $\displaystyle a_k\in R.$

If I knew that can "push" the summands into $\displaystyle A\setminus\{0\}$ would I know that I can "push" the sum too? If I had $\displaystyle s_1,...,s_t\in R$ such that

$\displaystyle x_{i_k}a_k s_k\in A\setminus\{0\}$ for $\displaystyle k=1,..,t,$

would I be able to find $\displaystyle S\in R$ such that

$\displaystyle m's\in A\setminus\{0\}?$

If the ring were an integral domain, I could use the product $\displaystyle \prod_{k=1}^{t}s_k,$ but it's not...

almost! so, as i said, there is an onto homomorphism from some free module $\displaystyle F$ to $\displaystyle M$ and so if $\displaystyle K$ is the kernel, then $\displaystyle F/K \cong M.$ so $\displaystyle F/K$ is singular and we only need to prove that $\displaystyle F$ is an essential extension of $\displaystyle K.$ first note that $\displaystyle K \neq \{0\}$ because otherwise $\displaystyle M$ would be a free singular module and that's not possible (why?). now let $\displaystyle \{x_i: \ i \in I \}$ be an $\displaystyle R$-basis for $\displaystyle F$ and let $\displaystyle 0 \neq x \in F.$ we need to show that there exists $\displaystyle s \in R$ such that $\displaystyle 0 \neq xs \in K.$ we can write, after renaming the indices if necessarily,

$\displaystyle x = \sum_{i=1}^n x_ir_i, \ \ \ \ \ \ \ \ \ \ (1)$

where $\displaystyle r_1 \neq 0.$ for any $\displaystyle y \in F,$ let $\displaystyle \overline{y}=y+K \in F/K.$ now, since $\displaystyle F/K$ is singular, $\displaystyle \text{ann}(\overline{x_1}) \subseteq_e R$ and so $\displaystyle \text{ann}(\overline{x_1}) \cap r_1R \neq \{0\}.$ so there exists $\displaystyle s_1 \in R$ such that $\displaystyle r_1s_1 \neq 0$ and $\displaystyle x_1r_1s_1 \in K.$ hence $\displaystyle (1)$ gives us

$\displaystyle xs_1= x_1r_1s_1 +\sum_{i=2}^n x_ir_is_1. \ \ \ \ \ \ \ \ \ (2)$

note that $\displaystyle xs_1 \neq 0$ because $\displaystyle r_1s_1 \neq 0.$ now, if $\displaystyle r_is_1 = 0$ for all $\displaystyle 2 \leq i \leq n,$ then $\displaystyle xs_1 =x_1r_1s_1 \in K$ and we are done. otherwise, after renaming the indices in the sum on the right hand side of $\displaystyle (2)$ if necessary, we may assume that $\displaystyle r_2s_1 \neq 0.$ now repeat the above process to get $\displaystyle s_2 \in R$ such that $\displaystyle r_2s_1s_2 \neq 0$ and $\displaystyle x_2r_2s_1s_2 \in K.$ then $\displaystyle (2)$ will give us

$\displaystyle xs_1s_2 = x_1r_1s_1s_2 + x_2r_2s_1s_2 + \sum_{i=3}^n x_ir_is_1s_2. \ \ \ \ \ \ \ (3)$

the first two terms on the right hand side of $\displaystyle (3)$ are now in $\displaystyle K$ and $\displaystyle xs_1s_2 \neq0$ because $\displaystyle r_2s_1s_2 \neq 0.$ if we continue this process, we will eventually have a positive integer $\displaystyle 1 \leq m \leq n$ and $\displaystyle s = s_1s_2 \cdots s_m \in R$ such that $\displaystyle 0 \neq xs \in K,$ as desired. note that i wrote $\displaystyle m$ instead of $\displaystyle n$ because some terms might vanish in the process.