Particular example of a quadratic extension of a nonunital ring

Question

I want to construct a concrete non-unital ring $R$ with the following properties:

1. $R$ is a noncommutative non-unital ring with a right unite $r$ i.e $t.r=t$ for any $t\in R$.
2. $S\subset R$ is a commutative subring (with unit $r$) and the inclusion $S\subset R$ has a section $f: R\rightarrow S$ in the category of nonunital rings sending right unit $r$ to the unit $r$.
3. The kernel of $f$, the ideal $I$ has the property : the multiplication of any two elements of $I$ is equal to zero. And the inclusion $I\subset R$ has a section in the category of nonunital rings
4. For any element $s\in S$ and any element $i\in I$ we have $s.i=0$

Is there such triple $(R,S,I)$ verifying all condition 1-4 such that $R$ is non unital.

Answer 1

Apparently by "section" you mean "left inverse". Usually this word means "right inverse".

No. Let $g:R\to I$ be that left inverse to the inclusion. Since it is a (non-unital) ring map, and since products in $I$ are zero, and since $I$ is an ideal in $R$, we have $0=g(i)g(s)=g(is)=is$ for every $s\in S$ and $i\in I$. Therefore the ring $R$ is simply the ring $S\oplus I$ that we get from the commutative ring $S$ and the abelian group $I$ by writing $R=S\oplus I$ and letting $si$, $is$, and $i_1i_2$ be $0$. It is commutative and unital.