Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addition, we cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray}\rm (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x + \color{#C00}{x\!+\!y} + y\\
\rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x + \color{#0A0}{y\!+\!x} + y\end{eqnarray}\bigg\rbrace\Rightarrow\, \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
In order to state this result concisely, recall that a SemiRing is
that generalization of a Ring whose additive structure is relaxed
from a commutative Group to merely a SemiGroup, i.e. here the only
hypothesis on addition is that it be associative (so in SemiRings,
unlike Rings, addition need not be commutative, nor need every
element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may
be stated as follows: a semiring with $\,1\,$ and cancellative addition
has commutative addition. Such semirings are simply subsemirings
of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative
semigroup embeds canonically into a commutative group, its group
of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$
i.e. the additive version of the fraction field construction).
Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals;
distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$;
$\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings
(e.g. formal power series); formal languages with union, concat; etc.
For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory.
pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference
held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
Zbl review,
AMS review
Best Answer
Hint $\rm\quad\rm x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \, $ in $\,\rm R, \, $ by evaluating at $\rm\ x = 0 $
Remark $\ $ This has a very instructive universal interpretation: if $\rm\, x\,$ is a unit in $\rm\, R[x]\,$ then so too is every $\rm\, R$-algebra element $\rm\, r,\,$ as follows by evaluating $\ \rm x \ f(x) = 1 \ $ at $\rm\ x = r\,.\,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $\rm R$-algebra. $ $ A natural choice is the nonunit $\,\rm 0\in R,\,$ which yields the above proof.