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 $d \neq 0$?


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 $\mathbb{Z}_2$!

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.