$R$ is an arbitrary (non-unital) ring such that if $n\in \mathbb{Z}$ and $r\in R$, $nr=r+r+...+r$ ($n$ times) if $n\geq 0$ and $nr=(-r)+...+(-r)$ ($n$ times) if $n<0$, where $r+(-r)=0$. Now let $\widetilde{R}=R\times \mathbb{Z}$ with the operations $+$ and $\cdot$: $(a,n), (b,m)\in \widetilde{R}$, $(a,n)+(b,m)=(a+b,n+m)$ and $(a,n)\cdot(b,m)=(ab+nb+ma,nm)$. Show $1)$ that $\widetilde{R}$ is a unital ring and $2)$ that the elements of the form $(a,0)$ form a subring isomorphic to $R$.
Attempt of $(1)$:
To show $\widetilde{R}$ is unital, need to show $\exists 1_\widetilde{R}\in\widetilde{R}$. That is, I want to find a $(c,k)\in\widetilde{R}$ such that $(a,n)\cdot(c,k)=(a,n)$, i.e. I want $ac+nc+ka=a$ and $nk=n$. Since $n,k\in\mathbb{Z}$, $k=1$. Since $a\in R$ has additive inverse, I can solve this and get $c(a+n)=0$, so I get $1_\widetilde{R}=(c,k)=(0,1)$. So $\widetilde{R}$ is unital. $\Box$ (Is this right?)
Attempt of $(2)$:
Let $S\subset \widetilde{R}$ with $S=\{(a,0):a\in R\}$. I want to find $\phi:S\rightarrow R$ isomorphism. This is where I am confused. Can anyone help find this isomorphism?