Let $R$ be a ring. An endomorphism $\alpha:R\to R$ is said to be right central reflexive if for all $a,b\in R,$ $aRb=(0)\implies bR\alpha (a)\subset Z(R)$, where $Z(R)$ denotes the centre of the ring. 


The ring $R$ is called right $\alpha$-central reflexive if $\alpha$ is a right central endomorphism for $R$. 


Let us consider a statement as follows:


For all $a,b\in R$, $aR\alpha (b)=(0)\implies bRa\subset Z(R)$.....(\*)


It can be proved that if $\alpha:R\to R$ is onto, $R$ is right central $\alpha$-reflexive implies (\*) is true. I want to show show that surjectivity of $\alpha$ is not superfluous. But I could not find an into endomorphism $\alpha$ such that $(\*)$ does not hold but $R$ is a right central $\alpha$-reflexive ring.


Please help me in this regard.



Regarding central reflexive rings
Here's a simple example.


Fix a domain $k$. Consider the (associative unital) ring $R=k\langle x,y\rangle/\langle\!xy=0\rangle\!\rangle$. Let $t$ be its endomorphism $x\mapsto 0$, $y\mapsto y$.


0) A basis of $R$ is given by the $y^nx^m$, $n,m\ge 0$.


1) I claim that $R$ right central reflexive, and even satisfies $aRb=0$ implies $bRt(a)=0$.


Indeed, suppose $aRb=0$ with $a,b$ nonzero. 
Write $a=P(y)+Q(y)x$ and $b=\sum\_{i\ge 0}R\_i(y)x^i$. Since $ayb=0$, we have $0=(P+Qx)y(\sum\_{i\ge 0}R\_ix^i)=\sum\_iP(y)yR\_i(y)$. Projecting (with respect to the basis) to $k[y]x^i$, we deduce that $P(y)yR\_i(y)=0$ for all $i$. Since $b\neq 0$, there exists $i$ such that $R\_i\neq 0$ and we deduce $P(y)=0$, so $a=Q(y)x$. Thus $t(a)=0$, so $bRt(a)=0$.


2) $(\*)$ fails with $(a,b)=(1,x)$: indeed $1Rt(x)=0$, but $xR1=xR$ contains $x$ and is not central as $0=xy\neq yx$.


