Let given ring $R$ without zero divizors, where adittive group of $R$ with zero torsion. Let given subring $R\_0\leq R$, and $p$ is prime number, such that $\forall r\in R, \exists i>0 : p^ir\in R\_0$. Is it true that if $Nil(R\_0/pR\_0)=\{0\}$, then $Nil(R/pR) =\{0\}$?


Second question:


Is previous problem true in case $rk(R\_0) <\infty$?



Subring of ring
No, this isn't true in general: 
$$R\_0:= \mathbb{Z}[X] \le \mathbb{Z}[X,Y]/(X^2-pY)=: R$$
is a counter-example because 
$R/pR = \mathbb{F}\_p[X,Y]/(X^2)$ has non-trivial radical. 


$R$ is a domain since $X^2-pY$ is irreducible in $\mathbb{Z}[X,Y]$. To see that $R\_0$ embedds into $R$ suppose that $f\in \mathbb{Z}[X]$ maps to zero. So there is $g \in \mathbb{Z}[X,Y]$ s.t. $f=g(X^2-pY)$. If $f\neq 0$, $f$ can be uniquely written as a product of irreducible polynomials from $\mathbb{Z}[X]$ (also irreducible in $\mathbb{Z}[X][Y]$). Thus, $f$ has two different factorizations into irreducible polynomials in $\mathbb{Z}[X,Y]$ which isn't possible. 


