Let $R$ be a non-unital ring and $M$ a $R$-module. If $x \in M$ has the property that $rx=0$ for all $r \in R$, does it follow that $x=0$?
(If $R$ is unital then the answer is trivially affirmative.)
Is an element in a module that is annihilated by all the scalars necessarily zero?
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$\begingroup$
abstract-algebra
ring-theory
modules
noncommutative-algebra
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3Let $R$ a non-unital ring, $M$ an abelian group, and define $rx = 0$ for $r\in R$ and $x\in M$. Does that make $M$ an $R$-module? – 2017-02-15
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0@DanielFischer: I am not entirely happy with that example, because it is trivial. It's like someone asking for an example of smooth manifold and receiving the empty set as an answer. Technically it's correct, but it's also clearly not a well chosen example. – 2017-02-15
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4Well, it is the quintessential example. Whenever you have a (non-unital) ring $R$ and an $R$-module $M$, let $M_0 = \{ x \in M : (\forall r\in R)(rx = 0)\}$. Then $M_0$ is an $R$-submodule of $M$, and you have $rx = 0$ for all $r\in R$ and $x\in M_0$. So every example contains the above. – 2017-02-15
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3But, let $R = 2\mathbb{Z}$, and let $M = \mathbb{Z}/(4\mathbb{Z})$, and make $M$ an $R$-module by restricting the multiplication of $\mathbb{Z}\times M$ to $R\times M$. Then $r\cdot (2 + 4\mathbb{Z}) = 0$ for all $r\in R$. – 2017-02-15
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0@DanielFischer: I understand your point, but it was not clear to me whether $M_0$ may be non-trivial. Your second example clarified this, thank you. – 2017-02-15
1 Answers
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$\newcommand{\Z}{\mathbb{Z}}$Consider the Prüfer group $M = Z(2^{\infty})$. It is an abelian group, thus a $\Z$-module.
It is also a module over $R = 2 \Z$. If $x \ne 0$ is the unique involution of $M$, then $R x = 0$.