4

Suppose $M$ is an $R$-module where $R$ is an integral domain.Define $Tor(M)$ be the set containing torsion elements of $M$. Prove that $M/Tor(M) $ is torsion-free.

I have manage to prove that $Tor(M)$ is a submodule of $M$. Then my aim is to prove $Tor(M/Tor(M)) \cong \lbrace Tor(M) \rbrace$

My attempt: Let $m + Tor(M) \in Tor(M/Tor(M))$. Then there exists an $r \in R$, $r \neq 0$ such that $r(m+Tor(M))=rm+Tor(M)=Tor(M) \Rightarrow rm \in Tor(M)$. Then there exists an $s \in R,s \neq 0$ such that $srm=0$. Hence, we have $sr \neq 0 \Rightarrow m \in Tor(M)$. This tells us that all torsion elements of $M/Tor(M)$ is of the form $Tor(M)$, which means $Tor(M/Tor(M)) \subset \lbrace Tor(M) \rbrace$.

I don't know how to prove another direction. Can anyone help me?

Idonknow
  • 15,643
  • 1
    You're already done. You can't go any smaller than the trivial subgroup! – anon Sep 13 '13 at 14:53
  • If you wanted to be crazy and actually "prove" the other direction: a coset $a + Tor(M)$ is in $Tor(M/Tor(M))$ iff there exists a non-zero $r \in R$ such that $ra + Tor(M) = Tor(M)$, i.e. iff there exists a non-zero $r \in R$ such that $ra \in Tor(M)$. Taking $a = 0$ (and any non-zero $r$) you get the "other direction" you are looking for. – tkr Sep 13 '13 at 15:29
  • I am not very sure about the way i do... but i would see that as we have making all torsion elements to be identity.. we will be left with torsion free elements in $M/Tor(M)$... Is my way considerable??? –  Sep 13 '13 at 15:49
  • Note that $\text{Tor}(M)$ is the zero-element of the quotient module $M / \text{Tor}(M)$. So the other direction is just the observation that (in any module) 0 is a torsion element. – Magdiragdag Sep 13 '13 at 18:10
  • @anon Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 14 '13 at 14:39

1 Answers1

1

The identity element of a group and the zero element of a module are both torsion elements. In the second case one may multiply the zero element by any nonzero scalar of the ring (say $1$) and get zero, making zero a torsion element. The first case uses similar reasoning. This tells us that the zero element is always contained in the torsion subset of any module. In particular this allows us to go further and prove that the torsion subset is actually a submodule (for, how could we accept it is a submodule without believing it has the zero element?).

When we take any group, module or ring quotient $A/I$, the trivial coset $I$ itself is the identity or zero element of the quotient structure. In particular ${\rm Tor}(M)$ is the zero of $M/{\rm Tor}(M)$, so it is automatically contained in ${\rm Tor}(M/{\rm Tor}(M))$. Usually the facts that zero is torsion and that the torsion subset is actually a submodule are taken for granted and go without saying, so in a proof you would only need to show one direction: that assuming $x\in{\rm Tor}(M/{\rm Tor}(M))$ leads to the conclusion $x={\rm Tor}(M)$ (not "of the form," actually equal), which is what you've done already.

anon
  • 151,657