Subring of ring 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 $rk(R_0) <\infty$?
 A: 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. 
A: Solution to second question in particular case, when $R_0/pR_0$ without zero divisors:
Note that $R_0/pR_0$ is finite dimensional vector space, $\dim R_0/pR_0\leq rk(R_0)$. So easy to see that $R_0/pR_0$ is field. Let for some $x\in R\setminus pR, y\in R, x^2 = py$, let $i\geq0 : r:=p^ix\in R_0, p^{i-1}x\notin R$. From defenition of $r, r\in R_0\setminus pR_0$, $R_0/pR_0$ is field, so $\exists r'\in R_0 : rr'- 1\in pR_0$. So $xrr' = p^ix^2r' = p^{i+1}yr'\in pR$, $x = xrr' - (rr' - 1)x\in pR$, but $x\notin pR$. done 

Solution to second question in case $rk(R_0)\leq 2$:
Let $r\in R\setminus pR$, such that $r^2\in pR$. Let $i = min\{i: p^ir\in R_0\}$, $x=p^ir\in R_0\setminus pR_0$. $rk(R_0)\leq 2$, so for some $f = a_0 + a_1t + a_2t^2\in \mathbb{Z}[t], f(x) = 0$, if $p|a_0, a_1, a_2$, we can reduce $f$ by $p$. $a_0 = -a_1x -a_2x^2, a_0^2 = x^2(a_1^2 + 2a_1a_2x + a_2^2x^2)$, so $a_0^2\in p^{2i +1}R$, so $a_0\in p^{i+1}R$, $1/p\notin R$, so $p^{i+1}|a_0$. If $p|a_1$, then $p\nmid a_2$ and $x^2 =(1/a_2)(-a_0 - a_1x)\in pR_0$, $Nil(R_0/pR_0)\not=0$. So $p\nmid a_1$, $a_1x = -a_0 - a_2x^2\in p^{i+1}R$, so $x\in p^{i+1}R, r\in pR$. done
A: Solution of second question:
Lemma:
Let given ring $R$ of finite rank, where $R_+$ with zero torsion, let given that $Nil(R/pR) =\{0\}$ for some prime number $p$. Prove that for some finite fields $F_1, ..., F_k$ of characteristic $p$, $R = F_1\times ... \times F_k$.
Proof:
Easy to see that $R/pR$ is finite dimension vector space with base field $\mathbb{Z}_p$. If $R/pR$ without zero divizors, then it is field. So let $\exists a, b\in R/pR\setminus\{0\} : ab = 0$. $|R/pR|<\infty$, so for some $i>j$, $a^i = a^j$, $a(a^{i-j} - 1)\in Nil(R/pR)$, so $a(a^{i-j} - 1) = 0$. Let $a' = a^{i-j}$, $a'^2 = a'$, $Nil(R/pR) =\{0\}$, so $a(R/pR)\oplus Ann(a) = R/pR$, $b\in Ann(a)\not=\{0\}$. By some induction we get that for some fields $F_1,..., F_k$, $a(R/pR)=F_1×...×F_l$, $Ann(a) = F_{l+1}×...×F_k$, $R/pR = F_1×...×F_k$. done
Let $i = \min\{t|\exists r\in (p^tR\setminus p^{t+1}R)\cap (R_0\setminus pR_0), r^2\in p^{2t+1}R\}$, if $Nil(R/pR)\not= \{0\}$, then easy to see that such $i$ is correct defined. So for some $x\in (p^iR\setminus p^{i+1}R)\cap (R_0\setminus pR_0)$ we have $x^2\in p^{2i+1}R$. We now prove that $i = 0$. From lemma we know that for some finite fields of characteristic $p$, $F_1,..., F_k$, $R_0/pR_0 = F_1×...×F_k$. So for some $f_1\in F_1,..., f_k\in F_k$ we get that $x + pR_0 = (f_1,..., f_k)\in R_0/pR_0$. Well known that for every $m>0$, $f_m^{|F_m|} = f_m$, so exists $N>>1 : \forall m, f_m^N = f_m$, so $x^N + pR_0 = (f_1^N,..., f_k^N) = (f_1,..., f_k) = x + pR_0$, $x^N - x\in pR_0$. Easy to see that if $i>0$, then for $N$ large enough $x^{N - 1} - 1\equiv -1\mod p^{i+1}R$, so $x_1 := x^N - x\in p^iR\setminus p^{i+1}R$, $x_1\in pR_0$. Let $j>0 : x_1\in p^jR_0\setminus p^{j+1}R_0$, easy to see that $x_1\notin p^{i+1}R_0$, so $j\leq i$, if $x = p^iy, y\in R\setminus pR$, $y^2\in pR$, then $x_2:= p^{i-j}y(x^{N-1} - 1)\in R_0\setminus pR_0$, $x_2\in p^{i-j}R\setminus p^{i-j+1}R$, $x_2^2\in p^{2(i-j) + 1}R$, so from defenition of $i$, $i\leq i - j$, but $j>0$, so $i = 0$. $x\in (R\setminus pR)\cap (R_0\setminus pR_0)$, $x^2\in pR$, so $Nil(R_0/(pR\cap R_0))\not= \{0\}$. For some $q_1, ..., q_v, R_0/(pR\cap R_0)\cong F_{q_1}×...×F_{q_v}$, so $Nil(R_0/(pR\cap R_0))= \{0\}$. And we get that $Nil(R/pR) =\{0\}$. done
