# Is this module torsion?

Let $$p$$ be prime, $$R$$ be a ring and $$M_n$$ be a $$R$$-modules such that $$p^nM_n = 0, p^{n-1}M_n \neq 0$$. Set $$M' := \text{Hom}_R(\oplus_{n>0}M_n, \oplus_{n>0}M_n)$$.

Is $$M'$$ torsion? I.e. is $$M' \otimes \mathbb Q = 0$$?

I'm confused because of the following: take $$f = id \in M'$$. Then there is no $$t \in \mathbb N$$ such that $$p^tf = 0$$. Yet, for any $$t \in \mathbb N$$, $$f \otimes 1 = f \otimes \frac{p^t}{p^t} = p^tf \otimes \frac{1}{p^t} \in M' \otimes \mathbb Q$$. So if we want to evaluate $$f\otimes 1$$ at, say, $$m:=(m_1,\cdots,m_n,0,\cdots)$$ where $$m_i \in M_i$$, we get $$(f\otimes 1)(m) = (p^nf\otimes \frac{1}{p^n})(m) = f(p^n m) \otimes \frac{1}{p^n} = 0.$$

• $M'$ is "locally torsion", but not torsion, as you have observed. A specific counterexample is $R=\mathbb{Z}$ and $M_n = \mathbb{Z}/p^n$. Do you see it? Commented Jul 7, 2021 at 12:11
• @MartinBrandenburg I've never heard of "local torsion", do you maybe have a reference for it, what it means? The p-adic integers were essentially my motivation for the question. I still don't understand what's wrong with the second observation which appears to suggest that $f \otimes \mathbb Q = 0$. Commented Jul 7, 2021 at 12:24
• You seem to assume that $f \otimes 1$ is a function. Why? Commented Jul 7, 2021 at 12:29
• @MartinBrandenburg because $f \in M'$ is a homomorphism. Am I getting something very wrong here? Commented Jul 7, 2021 at 12:34

Let $$M = \bigoplus_n M_n$$. You only prove that the image of $$\hom(M,M) \otimes \mathbb{Q} \to \hom(M \otimes \mathbb{Q},M \otimes \mathbb{Q})$$ is zero (which is because we have $$M \otimes \mathbb{Q}=0$$), which is not enough to conclude that $$\hom(M,M) \otimes \mathbb{Q}$$ is zero.
In fact, if $$\hom(M,M) \otimes \mathbb{Q}$$ is zero, we must have $$z \cdot \mathrm{id}_M = 0$$ for some $$z \in \mathbb{Z} \setminus \{0\}$$, which means that $$z \cdot M_n = 0$$ for all $$n$$. By assumption, this means that $$p^n \mid z$$ for all $$n$$, which implies $$z=0$$, a contradiction.