Show that $F(M)=M_{\mathrm{tor}}$ not exact I have a question in my Linear Algebra class that asks:

Give an example to show the functor $F(M) = M_{\mathrm{tor}}$ is not exact.

Here $M_{\mathrm{tor}}$ is referring to all torsion elements of the $R$-module $M$.
I am still trying to figure out exactly what something like this would look like, and I'm struggling with what $Hom_R(M,-)$ actually is (my teacher did not define it and I cant find a clear definition anywhere on here). So my questions are:


*

*Can you help me understand what $Hom_R(M,-)$ really means/ give me an intuition of how to use this in my example?

*Can you give me an idea of how to get started coming up with an example?


Thank you!
 A: The right context for this is homological algebra, and you should find all the details in any textbook on the topic.
For an $R$-module $M$, the submodule $M_{\mathrm{tor}}$ is formed by the torsion elements of $M$, i.e. elements $x\in M$ such that $r\cdot x = 0$ for some non-zero divisor $r\in R$ (in many situations $R$ is a domain, so the condition is just $r \ne 0$).
Left exactness of $(-)_{\mathrm{tor}}$ is a straightforward verification (see for instance Exercise 12 in chapter 3 of Atiyah-Macdonald). To see that it is not right exact, just consider the short exact sequence of $\mathbb{Z}$-modules
$$0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$$
$\operatorname{Hom}_R (M,-)$ is another functor, associating to every $R$-module $N$ the $R$-module of $R$-linear morphisms $M\to N$. It is also left exact (an easy verification or look up any homological algebra textbook), but not right exact. For a concrete example, consider the same short exact sequence as above and apply $\operatorname{Hom}_\mathbb{Z} (\mathbb{Z}/n\mathbb{Z}, -)$ to it.

It is hard to give useful explanations without knowing your background, but the above examples should be easy: for any pair of abelian groups $A$ and $B$, the homomorphisms $A\to B$ naturally form an abelian group $\operatorname{Hom} (A,B)$, and we claim that any short exact sequence of abelian groups
$$0\to B'\xrightarrow{i} B \xrightarrow{p} B''\to 0$$
induces an exact sequence
$$0\to \operatorname{Hom} (A,B')\xrightarrow{i_*} \operatorname{Hom} (A,B) \xrightarrow{p_*} \operatorname{Hom} (A,B'')$$
The example above shows that we can't always put "$\to 0$" on the right. The same with the torsion: for an abelian group $A$ you have its torsion subgroup $A_\mathrm{tor}$, and any short exact sequence
$$0 \to A \to B \to C \to 0$$
induces an exact sequence
$$0 \to A_\mathrm{tor} \to B_\mathrm{tor} \to C_\mathrm{tor}$$
without "$\to 0$" on the right in general.
This is a really long story, and it is homological algebra that systematically studies exactness (or rather failure to be exact).

In general, whenever you see some property of $R$-modules, it is useful to understand first what does it mean for $R = k$ a field (for vector spaces) and then for $R = \mathbb{Z}$ (that is, for abelian groups). Usually for vector spaces everything is trivial, but for abelian groups you can already see many interesting phenomena.
A: Perhaps a few interesting facts about this functor are worth mentioning. 
Given an $R$-module $M$, $M_{\mathrm{tor}}$ is the subset defined as
$$
M_{\mathrm{tor}} \overset{def}= \{ m \in M \, | \, \exists r \in R \setminus \mathrm{ZD}(R) \quad \mathrm{s.t.} \quad r \cdot m = 0\} \subseteq M. 
$$
I assumed you took your rings commutative ; in this case $\mathrm{ZD}(R)$ is the set of zero divisors of $R$, defined as
$$
\mathrm{ZD}(R) \overset{def}= \{ r \in R \, | \, \exists s \in R \setminus \{0\} \quad \mathrm{s.t.} \quad rs = 0 \}. 
$$
(In the non-commutative case, you would have the left and right zerodivisors, and then you have to worry about many trivial but numerous details.) Its complement is the set of non-zero divisors. 
Note that $(-)_{\mathrm{tor}}$ is one of these nice functors which comes with a natural transformation to the identity functor $\tau : (-)_{\mathrm{tor}} \to \mathrm{id}$. In other words, given a morphism of $R$-modules $f : M \to N$, we have a commutative square 
$$
\require{AMScd}
\begin{CD}
M_{\mathrm{tor}} @>{f_{\mathrm{tor}}}>> N_{\mathrm{tor}} \\
@VVV @VVV \\
M @>{f}>> N
\end{CD}
$$
The commutativity of this square essentially says that torsion elements map to torsion elements. This is because we define $f_{\mathrm{tor}} \overset{def}= f|_{M_{\mathrm{tor}}}$ since torsion elements map to torsion elements (if $r \cdot m = 0$, then $r \cdot f(m) = f(r \cdot m) = f(0) = 0$). Therefore, given an exact sequence, we can automatically produce a commutative diagram with exact rows : 
$$
\require{AMScd}
\begin{CD}
0 @>>> M_{\mathrm{tor}} @>{f_{\mathrm{tor}}}>> N_{\mathrm{tor}} @>{g_{\mathrm{tor}}}>> P_{\mathrm{tor}} \\
{} @VVV @VVV @VVV \\
0 @>>> M @>{f}>> N @>{g}>> P @>>> 0
\end{CD}
$$
The morphism $g_{\mathrm{tor}}$ will, however, not always be exact. However, when one is interested in studying torsion over an integral domain $R$, the notion of tensor product is usually more useful since if we denote the quotient field of $R$ by $Q$, $M_{\mathrm{tor}} = M$ is equivalent to $Q \otimes_R M = 0$. The functor $Q \otimes_R(-)$ is right-exact and has better understood left-derived functors called $\mathrm{Tor}$ functors, which produces a diagram of the following form (with exact rows) :
$$
\require{AMScd}
\begin{CD}
{\phantom{x}} @>>> {\phantom{x}} 0 @>>> M @>{f}>> N @>{g}>> P @>>> 0 \\
{} {} @VVV @VVV @VVV @VVV \\
\cdots @>>>  \mathrm{Tor}^R_1(P,Q) @>>> Q \otimes_R M @>{f}>> Q \otimes_R N @>{g}>> Q \otimes_R P @>>> 0
\end{CD}
$$
(forget the two extra arrows on the zero on the top left, I just couldn't use that package correctly... if someone knows how to remove them, go ahead and edit my answer!) The module $Q \otimes_R M$ is called the extension of scalars of $M$ by $Q$, otherwise known as the tensor product of the $R$-modules $Q$ and $M$. The morphism $M \to Q \otimes_R M$ is given by $m \mapsto 1 \otimes m$. (I'm essentially shooting names to attract your curiosity.)
Understanding this construction is perhaps a bit too advanced at this point, but it is nice to know that it is there. I recommend P.J. Hilton and U. Stammbach's book "A Course in Homological Algebra" when you will be interested in such functors in the future. 
A similar story happens with $\mathrm{Hom}$. To each pair of $R$-modules $(M,N)$, one associates the set 
$$
\mathrm{Hom}_R(M,N) \overset{def}= \{ f : M \to N \, | \, f \text{ is a morphism of } R\text{-modules} \}. 
$$
Two morphisms $f,g \in \mathrm{Hom}_R(M,N)$ are added pointwise and multiplied by elements of $r$ via $(r \cdot f)(m) \overset{def}= f(r \cdot m) = r \cdot f(m)$. This turns $\mathrm{Hom}_R(M,N)$ into an $R$-module. Given three $R$-modules $M,N,P$ and a morphism $g : N \to P$, we obtain a morphism $\mathrm{Hom}_R(M,g) : \mathrm{Hom}_R(M,N) \to \mathrm{Hom}_R(M,P)$ given by $(M \overset{f}{\to} N) \mapsto (M \overset{g \circ f}{\to} P)$. This definition is functorial in the sense that if we have morphisms 
$$
N_1 \overset{g_1}{\to} N_2 \overset{g_2}{\to} N_3, 
$$
then $\mathrm{Hom}_R(M,g_2 \circ g_1) = \mathrm{Hom}_R(M,g_2) \circ \mathrm{Hom}_R(M,g_1)$. You are welcome to verify this statement (it's all about understanding the notation, nothing happens here). 
Again, given an exact sequence 
$$
\require{AMScd}
\begin{CD}
0 @>>> N_1 @>{f}>> N_2 @>{g}>> N_3 @>>> 0
\end{CD}
$$
one obtains the following left-exact sequence 
$$
\require{AMScd}
\begin{CD}
0 @>>> \mathrm{Hom}_R(M,N_1) @>{\mathrm{Hom}(M,f)}>> \mathrm{Hom}_R(M,N_2) @>{\mathrm{Hom}(M,g)}>> \mathrm{Hom}_R(M,N_3)
\end{CD}
$$
but it is not right-exact in general. A short introduction to the case where this sequence can also be right-exact that I liked is given in Dummit & Foote's "Abstract Algebra". The book "A Course in Homological Algebra" I quoted earlier also explains how the $\mathrm{Hom}$ functor left-exact sequence extends to a long exact sequence on the right, giving rise to the $\mathrm{Ext}$ functors. Then again, a little bit of experience is recommended before diving into those things. 
Hope that helps,
