# Distributivity of Exponents in Rings

Let $R$ be a ring with under $+$ and $*$

For $x \in R$, we define $x^m := \underbrace{x*\cdots*x}_{m \text{ terms}}$

Conjecture. $R$ is commutative if and only if:

For every positive integer m, and for all $x, y$ in $R$, $(x*y)^m = x^m * y^m$.

I was able to show that if $R$ is commutative, then the exponent is distributive, but not the backward implication. Is the conjecture true, and if not, what is a necessary and sufficient condition such that For all $x, y$ in $R$, $(x*y)^m = x^m * y^m$?

• @bof yes, it is bound by a universal quantifier. – rr01 Aug 8 '17 at 6:26
• I corrected my answer, it was false. – Idéophage Mar 5 '19 at 13:23

Yes, every ring of this form is commutative. For $$x,y ∈ R$$, let $$f(x,y) := x^2y^2-(xy)^2$$. Then we have the equality (true in any ring) $$xy-yx = f(x,y) - f(1+x,y) - f(x,1+y) + f(1+x,1+y) \text{.}$$
Note that if you require that your ring have no nonzero zero divisor, the reason is simpler: $$(xy)^2-x^2y^2 = x(yx-xy)y=0$$ and if $$x$$ and $$y$$ are nonzero, then $$yx-xy=0$$.
• This is not a counterexample. In your quotient, only the projections of $x$ and $y$ satisfy the $\left(uv\right)^n = u^n v^n$ identity, but the OP wants this identity to hold for every pair of elements $u$ and $v$. – darij grinberg Mar 5 '19 at 6:20