I am wondering that as we define $Z(R)$, center of a ring is a subring of $R$. Can we define a subset which is collection of all those elements which semicommute with other elements. Will this set form a subring of $R$.
Note : $R$ is said to be semicommutative if $ab=0\implies aRb=(0)$.
So if we can define a set like $S(R)=\lbrace a \in R | ab=0 \implies aRb=(0)\rbrace $. Will this $S(R)$ forms a subring of $R$. Let us try to prove it, take $x,y\in R$, we have to show that $xy\in S(R)$, so suppose $xyb=0$ for some $b\in R$, then $xyrb=$, how can I proceed further?
No this is not a subring in general. Take $R=M_2(k)$ the ring of $2\times 2$ matricies over a field $k$. Then clearly the zero matrix is in $S(R)$, as well as every invertible matrix. Let us show that no matrix of rank $1$ is in $S(R)$: let $a\in R$ have rank $1$. Then there exist invertible $P,Q\in R$ such that $PaQ=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$. Then $ab=0$ if and only if $\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}Q^{-1}b=0$, which is the case precisely when $b$ is of the form $b=Q\begin{pmatrix}0 & 0 \\ \beta & \beta'\end{pmatrix}$ for some $\beta,\beta'\in k$. For example, if we choose $b=Q\begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}$ then $ab=0$. But then $Rb=RQ\begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}=R\begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}=\{\begin{pmatrix}\alpha & \alpha \\ \alpha' & \alpha'\end{pmatrix}\mid \alpha,\alpha'\in k\}$. In particular, as there exist $\alpha,\alpha'\in k$ such that $a\begin{pmatrix}\alpha\\ \alpha'\end{pmatrix}\neq 0$ (as $a$ has rank $1$), we obtain that $aRb\neq 0$.
Therefore, we have $S(R)=\{0\}\cup\operatorname{GL}_2(k)$, which is not a subring, as it isn't closed under addition.