Are torsion submodules not unique? 
This was a statement as part a lemma in my book (Goodman) on page 388. I found this very ambiguous because

*

*What exactly is $A$ in relation to $M$? Do they mean $A = Tor(M) \cap S$ where $S$ is some submodule of $M$?


*What exactly is a torsion submodule? I thought there is only the torsion submodule $Tor(M).$


*This theorem 8.5.2 is the Structure Theorem (existence) where $M = R^k \oplus_{i = 1}^s R/(a_i)$ with $a_i |a_{i+1}$.
How do we know $\oplus_{i = 1}^s R/(a_i)$ is a torsion submodule?
 A: Let $R$ be a unital commutative ring, and $M$ an $R$-module. Recall:

*

*An element $m \in M$ is a torsion element if $rm = 0$ for some regular (non zero-divisor) $r \in R$.

*Let $\operatorname{Tor}(M) \subseteq M$ be the set of all torsion elements, then $\operatorname{Tor}(M)$ is a submodule.

*$M$ itself is called a torsion module if $M = \operatorname{Tor}(M)$. In other words, a torsion module is a module consisting only of torsion elements.

*$M$ is called torsion-free if $\operatorname{Tor}(M) = \{0\}$.

If $R$ is a PID, then we can replace the word "regular" with "nonzero". The theorem is trying to say that in any random direct sum decomposition $M = A \oplus B$ where $A$ happens to be torsion and $B$ happens to be free, then in fact $A$ is equal to the torsion submodule $\operatorname{Tor}(M)$.
As for how to see that $\bigoplus_{i} R / (a_i)$ is a torsion module, is it clear to see that each $R / (a_i)$ is a torsion module? Once you see that, consider the proof of the fact that $\operatorname{Tor}(M)$ is a submodule: there should be a similar looking trick.
A: They're being a bit vague about equality vs isomorphism (which is customary once one is more familiar with things) so here's a more precise way of saying what they mean:
First we define an $R$-module to be torsion if for all $m\in M$ there is some nonzero $r\in R$ with $rm=0$. The claim then is that if there is any isomorphism $\phi:M\to A\oplus B$ where $A$ is a torsion $R$-module and $B$ is free, then the image of $M_{\mathrm{tor}}$ under $\phi$ is exactly equal to $A\oplus\{0\}$.
So you see they are not requiring that $A$ literally be a submodule of $M$ but rather isomorphic to one.
A torsion submodule of $M$ is a submodule which happens to be a torsion module. The torsion submodule of $M$ is the submodule $M_{\mathrm{tor}}:=\{m\in M\mid\text{there exists $r\in R\smallsetminus\{0\}$ with $rm=0$}\}$. Alternatively it is the largest torsion submodule of $M$, and it is unique in this sense.
