Is the right annihilator of an element in a ring a subring? let $R$ be a ring and let $x$ not equal to $0$ be a fixed element in R. Then is $\{r \mid xr=0\}$ a subring of $R$?
The solution says yes, but I don't think so, because the multiplicative identity is not always included.(e.g. R = Z4, and x = 2) Any ideas?
 A: For $0 \ne x \in R$, setting
$S(x) = \{r \in R \mid xr = 0\}, \tag{1}$
then it is easy to see that i.) $0 \in S(x)$, since $x0 = 0$; ii.)  $r, s \in S(x) \Rightarrow r + s \in S(x)$, since $xr = xs = 0 \Rightarrow x(r + s) = xr + xs = 0$; iii.)  $r \in S(x) \Rightarrow -r \in S(x)$, since $x(-r) = -rx = 0$; iv.) $r, s \in S(x) \Rightarrow rs, sr \in S(x)$, since $xr = xs = 0 \Rightarrow x(rs) = (xr)s = 0s = 0$ and $x(sr) = (xs)r = 0r = 0$.  Thus $S(x)$ is closed under the operations of $R$ and contains $0 \in R$, the additive identity.  It thus follows that $S(x)$ satisfies all of the ring axioms with respect to the operations of $R$.  But if the one's accepted definition of "ring" includes necessarily the existence of a multiplicative unit, then $S(x)$ is not a subring of $R$, since $1_R x = x 1_R = x \ne 0$, so $1_R \notin S(x)$.  In my experience, most authors would not insist the definition of "ring" include a multiplicative identity, reserving a term such as "unital ring" for those that do have a $1_R \in R$.  So I would say that, under the usual, conventional definition of a ring, $S(x)$ is indeed a subring of $R$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: You can show that:


*

*Your set is always a right ideal of $R$. Moreover, you could consider it a rng contained in $R$ since it's closed under multiplication (and is an abelian subgroup of $(R,+)$.)

*There's no guarantee it will be a ring with identity. But it can happen: if $x$ commutes with elements of $R$ and $x^2=x$, then the set you gave is a ring with identity $1-x$.

*It cannot be a subring that shares the identity with $R$. That would imply that the identity $1$ of $R$ is in the set, so that $x=x1=0$, but by assumption $0\neq x$.

