Finite rank ring

Question

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 $D(r) = 0$. $\Box$

Answer 1

I found helpfull lemma

Lemma 1

Let given ring of finite rank $n$, let $p>(2n)!$ be a prime prove that next conditions are equivalent:

a) $Der_{\mathbb{Z}}(R/pR) = \{0\}$.

b) $Nil(R/pR)=\{0\}$, where $Nil(R)$ is nilradical of $R$.

Proof:

b) => a) is proved in previous post. So we need to prove a) => b).

Lemma 2

Let given ring $R$, number $n$ and prime number $p>(2n)!$, where $pR = \{0\}$ and for every element $a\in Nil(R), a^n = 0$. Prove that for some natural $k$, $(Nil(R))^k = \{0\}$.

Proof:

Induction by $n$. Let $I = <\{a^{n-1}|a\in Nil(R)\}>R$. For every $x, y\in Nil(R)$, $x+y\in Nil(R)$, so $0 = (x+y)^{2n-2} = \frac{(2n-2)!}{(n-1)!^2}x^{n-1}y^{n-1}$, so $x^{n-1}y^{n-1}=0$, $I^2 =\{0\}$. For every $a\in Nil(R), a^{n-1}\in I$, so $\forall r\in Nil(R/I) = Nil(R)/I$, $r^{n-1} = 0$. So by induction for some natural $k$, $(Nil(R/I))^k = (Nil(R)/I)^k = \{0\}$ and $Nil(R)^k\subseteq I$. So $Nil(R)^{2k} = \{0\}$. done

Lemma 3

Let given ring $R$ of finite rank $n$, let $p > (2n)!$ is prime. Prove that for some natural $k$, $(Nil(R/pR))^k = \{0\}$.

Proof:

Let $a = r + pR$ is any element from $Nil(R/pR)$, let $\lambda : a^{\lambda} =0\not= a^{\lambda - 1}$. Rank of $R$ is $n$, so exists $f = a_0+ ... +a_nx^n\in\mathbb{Z}[x] : f(r) = 0$. If $p|a_i, i=1,..., n$, then we can factor $f$ by $p$, so we can think that $f\not= 0\in \mathbb{Z}_p[x]$. Let $f = a_i'x^i + ... + a_n'x^n\in \mathbb{Z}_p[x]$, where $a_i'\not= 0\in \mathbb{Z}_p$. If $i < \lambda$, then $0=a^{\lambda - i - 1}f(a) = a_i'a^{\lambda - 1}\not= 0$, so $\lambda\leq i \leq n$. Now from lemma 2 we finish proof. done

Lemma 4

Let given ring $R$ and prime $p$, such that $pR =\{0\}$ and for some natural $k$, $(Nil(R))^k =\{0\}$. Prove that there exists ring $Q\subseteq R: Q\oplus Nil(R) = R$.

Now return to main lemma 1. From lemmas 3 and 4 we get that there exists ring $Q\subseteq R/pR: Q\oplus Nil(R/pR) = R/pR$. Consider linear projector $P : R/pR \to Nil(R/pR)$, where $\ker P = Q$, $Im P = Nil(R/pR)$. Let $Nil(R/pR)\not=\{0\}$ one can find element $a\in R/pR : (Nil(R/pR))^2\subseteq Ann(a), Nil(R/pR)\notin Ann(a)$, so we can define linear operator $D : R/pR\to aR/pR\subseteq R/pR$, $D(x) = aP(x)$, one can check that $D\in Der_{\mathbb{Z}}(R/pR)$, $Nil(R/pR)\notin Ann(a)$, so $D\not= 0$, but $Der_{\mathbb{Z}}(R/pR) = \{0\}$, so $Nil(R/pR)=\{0\}$. done