What do you call elements of a commutative ring whose multiplication is injective?

Question

What do you call an element $r$ of a commutative ring $R$ such that multiplication by $r$ is injective, regardless of whether $r$ is a unit or not?

For example, in $\mathbb{Z}$, multiplication by a fixed nonzero integer $a$ is injective.

However, in $\mathbb{Z}/10\mathbb{Z}$, only the following elements have injective multiplication: $\{1, 3, 7, 9\}$ and they are all units (because the ring is finite and they are not zero divisors).

"Injectivity" of elements seems like a useful notion.

In any integral domain, all nonzero elements are "injective".

Additionally, if all nonzero elements are "injective", then $R$ is an integral domain. For example, let $x$ be an arbitrary element. If there exists a nonzero $y$ (possibly equal to $x$) such that $xy = 0$, then $x$ is not injective since $x0 = xy = 0$. Therefore there are no zero divisors in $R$ and $R$ is an integral domain.

Answer 1

Multiplication by $r$ is not injective if and only if $r$ is a zero-divisor.

If $r$ is a zero-divisor then there is some $s\ne0\in R$ for which $rs=0$, so $0$ and $s$ are both sent to $0$ by multiplcation by $r$.

On the other hand, if multiplication by $r$ is not injective there are $s_1\ne s_2$ such that $rs_1=rs_2$. But then $r(s_1-s_2)=0$ where $s_1-s_2\ne0$, so $r$ is a zero-divisor.