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(1) Is there a finite nilpotent ring $R$ such that the ratio $$c(R)=\frac{|\{(x,y)\in R\times R \; | \; xy=yx\}}{|R|}$$ is not integer?

Edit 1: The nilpotent condition is put later.

Edit/Answer: Answer is negative as Frieder Ladisch nicely observed below. That is $c(R)$ is always an integer. Actually $c(R)$ is the number of conjugacy classes of the group $1+R$.

(2) Is there a finite nilpotent Lie algebra $L$ such that the ratio $$c(L)=\frac{|\{(x,y)\in L\times L \; | \; [x,y]=0\}}{|L|}$$ is not integer?

Edit: The nilpotent condition on the Lie Algebra is put later.

The motivation is that the same question for finite groups has negative answer.

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    $\begingroup$ What have you tried? in a Lie algebra of dimension $n$ over a field of cardinal $q$ such that every commuting pair is collinear, the number of commuting pairs is $q^{n+1}+q^n-q$, so the ratio is $q+1-1/q$. There are such examples with $n=2$ or $n=3$ (so, the smallest possible non-abelian Lie algebra is a counterexample!). $\endgroup$
    – YCor
    Commented Jan 10, 2016 at 12:37
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    $\begingroup$ Counter examples for nilpotent Lie algebras will involve "small primes". If $p>n$, then $\exp$ is a well defined bijection from $N_n(\mathbb{F}_p)$ to $U_n(\mathbb{F}_p)$ (where $N_n$ and $U_n$ are upper triangular matrices with $0$'s and $1$'s on the diagonal respectively). The map $\exp$ takes sub-Lie-algebras to sub-groups and commuting pairs to commuting pairs. $\endgroup$
    – David E Speyer
    Commented Jan 10, 2016 at 12:48
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    $\begingroup$ The title is written is the positive while the question is in the negative, which makes "a negative answer" quite ambiguous. It's more natural to ask everything in the positive. $\endgroup$
    – YCor
    Commented Jan 10, 2016 at 14:05
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    $\begingroup$ @AlirezaAbdollahi I also got the nilpotency $\geq 3$ result. Proof for the curious: Let $L$ be a Lie algebra with nilpotency $2$. Let $Z$ be the center. Let $\ell=\dim L$ and $z=\dim Z$. We write $\bar{\ }$ for reduction from $L$ to $L/Z$. For $g \in L$, we have $[g,h]=0$ iff $\bar{h}$ is in $\mathrm{Ker}([\bar{g}, \ ] : L/Z \to Z)$. The dimension of the kernel is $\geq (\ell-z)-z=\ell-2z$. For each possible $\bar{g} \in L/Z$, this contributes $q^{\ell-2z} q^{2z}=|L|$ pairs $(g,h)$, since there are $q^z$ ways each to lift $\bar{g}$ and $\bar{h}$ to $L$. Sum over $\bar{g}$ to conclude. $\endgroup$
    – David E Speyer
    Commented Jan 10, 2016 at 17:19
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    $\begingroup$ Well, for (1) the answer is the same as for groups, simply because $ 1+R $ is a group, and $(1+x)(1+y) = (1+y)(1+x)$ iff $ xy=yx $. $\endgroup$
    – Frieder Ladisch
    Commented Jan 11, 2016 at 10:40

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