# Examples of $R$-modules $X$ such that $(X \setminus TX) \cup \{0\}$ isn't a submodule.

Work over an ambient commutative ring with unity. Given a module $X$, write $TX$ for its submodule of torsion elements. Suppose we want to find the "submodule" of torsion-free elements of $X$. So basically, we're interested in $(X \setminus TX) \cup \{0\}$. If I understand correctly, this basically doesn't work, so instead we deal with the quotient module $X/TX$. Actually, I don't quite see why it doesn't work:

Questions.

Q0. What are some examples of $\mathbb{Z}$-modules $X$ such that $(X \setminus TX) \cup \{0\}$ isn't a submodule?

Q1. I'm also interested in examples where $\mathbb{Z}$ is replaced by other commutative rings.

Q2. A truly complete answer would provide a characterization of those rings $R$ such that for all $R$-modules $X$, the set $(X \setminus TX) \cup \{0\}$ is an $R$-submodule. Obviously, every field has this property.

Consider $X = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Then both $(1,1)$ and $(-1,0)$ are in $(X \setminus TX) \cup \{0\}$, but their sum isn't. In fact let $R$ be any ring that has a module $M$ containing a nonzero torsion element $m$. Then in $R \times M$, both $(1, m)$ and $(-1,0)$ are torsion-free, but their sum isn't. In other words, there is no example other than rings for which no module has torsion, AKA fields (if $R$ is not a field let $I$ be a nontrivial ideal, then all of $R/I$ is torsion)
• Its weird, the first thing I tried was $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ but ended up concluding this wasn't an example. You're obviously right though. I think I'm visualizing it wrong. – goblin Jul 12 '16 at 12:53
• Thanks. I'll accept your answer later when I get home. The software won't let me do it right now. I think I was visualizing $(X \setminus TX)$ as the orthogonal complement of $TX$, which is why I was getting confused. – goblin Jul 12 '16 at 12:55
Let's ask when, given a module $M$ with a proper nonzero submodule $L$, the set $(M\setminus L)\cup\{0\}$ is a submodule.
Let $x\in M\setminus L$ and let $y\in L$, $y\ne0$. Then also $x+y\in M\setminus L$ and $-x\in M\setminus L$; on the other hand $$(x+y)+(-x)=y\notin (M\setminus L)\cup\{0\}$$ Therefore the set is not even an additive subgroup.
So your set is a submodule only in the trivial cases when $TX=0$ or $TX=X$.