Show that $F(M)=M_{\mathrm{tor}}$ not exact

Question

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:

1. Can you help me understand what $Hom_R(M,-)$ really means/ give me an intuition of how to use this in my example?
2. Can you give me an idea of how to get started coming up with an example?

Thank you!

Answer 1

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.

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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).

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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.

Answer 2

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,