Let given ring $R$ of finite rank. Is it true that for all primes $p$ large enough modules $Der_{\mathbb{Z}}(R/pR) = \{0\}$?

For every ring we define $Der_{\mathbb{Z}}(R)$ as set of linear operators $D : R\to R$, where $\forall a, b\in R, D(ab) = aD(b) + bD(a)$.

I can prove this in case $R/pR$ without nillpotent elements:

In this case for eny $r\in R\setminus\{0\}$, $D\in Der_{\mathbb{Z}}(R/pR)$ and any prime $p > n = rk(R)$, well known that $\exists f = a_0 +...+ a_qx^q\in \mathbb{Z}[x] : f(r) = 0$, $\deg f\leq n$, if $p|a_0,...,a_q$ then we can dactor $f$ by $p$, so we can think that $f\not= 0\in\mathbb{Z}_p[x]$. Let $\{P_i\}_{1\leq i\leq k}$ are different primes in $\mathbb{Z}_p[x]$, such that $f = P_1^{w_1}...P_k^{w_k}\in\mathbb{Z}_p[x]$. $R/pR$ without nillpotent elements so $g = P_1...P_k, g(r) = 0$. So $D(g(r)) = g'(r)D(r) = g(r)D(r) = 0$, $gcd(g, g') = 1$, so $D(r) = 0$. $\Box$

Let given ring $R$ of finite rank. Is it true that for all primes $p$ large enough modules $Der_{\mathbb{Z}}(R/pR) = \{0\}$?

For every ring we define $Der_{\mathbb{Z}}(R)$ as set of linear operators $D : R\to R$, where $\forall a, b\in R, D(ab) = aD(b) + bD(a)$.

I can prove this in case $R/pR$ without nilpotent elements:

In this case for any $r\in R\setminus\{0\}$, $D\in Der_{\mathbb{Z}}(R/pR)$ and any prime $p > n = rk(R)$, it is well known that $$\exists f = a_0 +...+ a_qx^q\in \mathbb{Z}[x] : f(r) = 0$$ with $$\deg f\leq n$$ Moreover if $p|a_0,...,a_q$ then $p|f$, so we can think that $f\not= 0\in\mathbb{Z}_p[x]$. 

Let $\{P_i\}_{1\leq i\leq k}$ be different primes in $\mathbb{Z}_p[x]$, such that $f = P_1^{w_1}...P_k^{w_k}\in\mathbb{Z}_p[x]$.

Let $R/pR$ be without nilpotents so that $g = P_1...P_k\implies g(r) = 0$. 

So $D(g(r)) = g'(r)D(r) = g(r)D(r) = 0$, $gcd(g, g') = 1$, so $D(r) = 0$. $\Box$

Let given ring $R$ of finite rank. Is it true that for all primes $p$ large enough modules $Der_{\mathbb{Z}}(R/pR) = \{0\}$?

For every ring we define $Der_{\mathbb{Z}}(R)$ as set of linear operators $D : R\to R$, where $\forall a, b\in R, D(ab) = aD(b) + bD(a)$.

I can prove this in case $R/pR$ without nillpotent elements:

In this case for eny $r\in R\setminus\{0\}$, $D\in Der_{\mathbb{Z}}(R/pR)$ and any prime $p > n = rk(n)$, well known that $\exists f = a_0 +...+ a_qx^q\in \mathbb{Z}[x] : f(r) = 0, \deg f\leq n$, if $p|a_0,...,a_q$ then we can dactor $f$ by $p$, so we can think that $f\not= 0\in\mathbb{Z}_p[x]$. Let $\{P_i\}_{1\leq i\leq k}$ are different primes in $\mathbb{Z}_p[x]$, such that $f = P_1^{w_1}...P_k^{w_k}\in\mathbb{Z}_p[x]$. $R/pR$ without nillpotent elements so $g = P_1...P_k, g(r) = 0$. So $D(g(r)) = g'(r)D(r) = g(r)D(r) = 0$, $gcd(g, g') = 1$, so $D(r) = 0$. $\Box$

Let given ring $R$ of finite rank. Is it true that for all primes $p$ large enough modules $Der_{\mathbb{Z}}(R/pR) = \{0\}$?

For every ring we define $Der_{\mathbb{Z}}(R)$ as set of linear operators $D : R\to R$, where $\forall a, b\in R, D(ab) = aD(b) + bD(a)$.

I can prove this in case $R/pR$ without nillpotent elements:

In this case for eny $r\in R\setminus\{0\}$, $D\in Der_{\mathbb{Z}}(R/pR)$ and any prime $p > n = rk(R)$, well known that $\exists f = a_0 +...+ a_qx^q\in \mathbb{Z}[x] : f(r) = 0$, $\deg f\leq n$, if $p|a_0,...,a_q$ then we can dactor $f$ by $p$, so we can think that $f\not= 0\in\mathbb{Z}_p[x]$. Let $\{P_i\}_{1\leq i\leq k}$ are different primes in $\mathbb{Z}_p[x]$, such that $f = P_1^{w_1}...P_k^{w_k}\in\mathbb{Z}_p[x]$. $R/pR$ without nillpotent elements so $g = P_1...P_k, g(r) = 0$. So $D(g(r)) = g'(r)D(r) = g(r)D(r) = 0$, $gcd(g, g') = 1$, so $D(r) = 0$. $\Box$