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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 $R$, $(x*y)^m = x^m * y^m$?

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  • $\begingroup$ @bof yes, it is bound by a universal quantifier. $\endgroup$ – rr01 Aug 8 '17 at 6:26
  • $\begingroup$ I corrected my answer, it was false. $\endgroup$ – Idéophage Mar 5 '19 at 13:23
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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) $$xy-yx = f(x,y) - f(1+x,y) - f(x,1+y) + f(1+x,1+y) \text{.}$$

I have no explanation for this formula, I just found it by chance.

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 $yx-xy=0$.

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  • $\begingroup$ 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 $v$. $\endgroup$ – darij grinberg Mar 5 '19 at 6:20
  • $\begingroup$ @darijgrinberg Thank you, it is corrected. $\endgroup$ – Idéophage Mar 5 '19 at 13:22

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