# derivation in near-ring

Let $N$ be any near-ring. An additive mapping $d:N\rightarrow N$ is called derivation if $d(xy)=d(x)y+xd(y)$ for any $x,y\in N$. Recently, i read some paper about derivation and there are many paper that discuss "A near-ring $N$ is commutative ring if it is satisfy one of derivation" and the famous one and oftenly cited is on the paper of Bell, et al (1997). Instead of direct proving a near-ring is commutative ring by its abelian, multiplication is commutative and both distributive holds, why using derivation? May be, they want to show that proving a near-ring is commutative ring, is not always using the same way, that there is another way to reach that, in this case using derivation. But, is there any reason why using derivation? And my other question is, what is the other useness of derivation in a ring/near-ring beside for commutativity? Thank you!

## 1 Answer

I am not sure whether I completely understood your question, but you asked why people use derivations to show commutativity. Well, I guess that some people working in near-rings tried to follow the results obtained in ring theory and there is a bunch of classical results which says that if a ring has a derivation with certain properties, then it is commutative. For example Herstein proved in 1978 in "A Note on derivations" that a prime ring R with derivation $d$ is commutative if $d(x)d(y)=d(y)d(x)$ for all $x,y \in R$. More on commutativity condition for rings can be found in the article "Commutativity conditions for rings: 1950–2005" by JamesPinter-Lucke.

• yes, I read it, showing the commutativity by involving their derivation, not only on ring, but also on near-ring. The problem is, from all paper I read, I can't even find the example that illustrate the theorem. Take paper by Boua, et al. Given N is prime and zero symmetry near-ring, if N admits non-zero derivation such that d([x,y])=[x,y] for all x,y in N, then N is commutative ring. Is there any example that satisfies the theorem? I know, when I can't find an example, it doesn't mean it does'nt exist, it only hasn't be discovered. But, if finding the example is hard, why such theorem exist? – Reza Habibi Jun 12 '18 at 11:46
• I don't understand what you mean by "I can't even find the example that illustrate the theorem." Do you want a ring and a derivation such that the hypothesis is not fulfilled? Or do you want a ring and a derivation that do satisfy the hypothesis, in which case your ring must be commutative and then the commutators are zero. For example in Herstein's result that I mention you have an equivalence: A prime ring with (non-zero) derivation d is commutative if and only if $d(x)d(y)=d(y)d(x)$, for all $x,y$. Hence any commutative prime ring with non-zero derivation d yields an example for you. – Christian Lomp Jun 12 '18 at 17:17
• "I can't even find the example that illustrate the theorem." means I get trouble to find the example that satisfy the theorem, moreover in most paper I read, there is only counter example that show the prime hypothesis is not superfluous, but there is no example that satisfy the theorem. I do want a ring and a derivation that satisfy the hypothesis. Moreover, I do want a prime and zero symmetry near-ring with (non-zero) derivation that satisfy d[x,y] = [x,y] for all x,y in N – Reza Habibi Jun 12 '18 at 20:27
• So, take the polynomial ring $R=\mathbb{Q}[x]$ over the rationals and take any non-zero derivation, for instance the canonical one $d=\frac{\partial}{\partial x}$. This satisfies obviously all what you want. – Christian Lomp Jun 13 '18 at 8:39