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A common mistake by students in introductory algebra is (essentially) the following: $$\frac{x + y}{d} = \frac{x}{d} + y.$$ (See: "Cancelling Everything in Sight.") I want to justify this mistake.

The students are working in a field. Generalizing the situation to rings, we might write their mistake as $(x + y)d = xd + y$. This is equivalent to $yd = y$, hence the question: In what rings does $yd = y$ for all elements $y$ and $d$ with $d \neq 0$?

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I think that the only nontrivial ring in which this works is $\mathbb{Z}_2$.

We have $x^2 = x$ for every $x \in R$. ($x \neq 0$ follows from $x^2 = xx = x$, and $x = 0$ is trivial.) It is not hard to show that $R$ is commutative (consider $(a + b)^2$) so $x = xy = yx = y$ for all nonzero $x, y \in R$. This gives $R = \{0, z\}$ for some $z \neq 0$. There are two rings of order $2$, but $z^2 = z \neq 0$ implies that $R \simeq \mathbb{Z}_2$.

It might be worth noting that this is also a field, so any confused students would be justified in writing $\frac{x + y}{d} = \frac{x}{d} + y$ so long as they mentioned that they were working in $\mathbb{Z}_2$!

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    $\begingroup$ Or, said a slightly different way, "$d$ is an identity, so we have shown that all nonzero elements, if there are any, are equal to the identity." $\endgroup$ – rschwieb May 3 '18 at 13:59

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