# Cancellation law in a ring without unity

When discussing rings, integral domains, fields etc, I'm told that the cancellation law holds in any ring that has no zero divisors. By cancellation law, I mean that if we have no zero divisors, we can look at the equation $ab = ac$ and "cancel" the a on the left-hand side, and thus know that $b=c$. (I believe that the proof of this comes out of saying that $(ab-ac) = a(b-c) = 0$ thus implying that $(b-c) = 0$ if $a \neq 0$, and so $b=c$).

What I'm confused about, is that the requirement for the cancliation law is simply that our ring has no zero divsors, there's no mention of our ring containg unity. However, if our ring doesn't contain unity, but the cancellation law holds, then what can we make of the equation $a^2 = a$? Every time I try to simplify this, I find the need to use unity, which I don't believe I am guaranteed to have in my ring. Does saying that our ring has no zero divisors, in fact, imply that our ring contains unity?

However, if our ring doesn't contain unity, but the cancellation law holds, then what can we make of the equation $a^2=a$?

In a ring which satisfies cancellation on both sides, $a$ can only be one of two things: $0$, or else the identity for the ring.

Obviously $0$ satisfies $0^2=0$, and if $a$ is nonzero and satisfies $a^2=a$, then $a(ar-r)=(ra-r)a=0$ for arbitrary $r$, whereupon you must conclude that $ar=r=ra$ for all $r$, and that $a$ is the identity.

Of course, depending on your ring, $0$ may be the only element that satisfies $a^2=a$ (like the rng $(X)$ inside the ring $F[X]$.)

• Ahh I see now! Thank you so much! – P. Reinecke Oct 31 '17 at 20:26
• @G.Richardson 👍 ${}{}{}{}{}{}$ – rschwieb Oct 31 '17 at 20:39

What the "paradox" suggests is that in a ring with cancellation and no identity element the only solution to the equation $x^2 = x$ is $x=0$.

If there is a nonzero solution $a$ to that equation then for any $b$ in the ring $$0 = (a^2 - a)b = a^2 b - ab$$ so $$a^2 b = ab .$$ Then cancelling $a$ implies $$ab = b .$$ The same argument works for right multiplication so $a$ is a (hence the) multiplicative identity.

The usual definition of an integral domain requires an identity element. https://en.wikipedia.org/wiki/Integral_domain.

It's possible for a ring without unity to have no nonzero zero-divisors.

For example, let $R$ be the ideal $(2)$ of $\mathbb{Z}$. Then $R$ is a ring with no multiplicative identity, and no nonzero zero-divisors.

For a finite example, let $R$ be the ideal $(2)$ of $\mathbb{Z_6}$. Then once again, $R$ is a ring with no multiplicative identity, and no nonzero zero-divisors.

For the above examples, given the lack of nonzero zero-divisors, we get a cancellation law:

$\;\;{\small{\bullet}}\;\,$If $ab=ac$ and $a \ne 0$, then $b=c$.

However, as the answers by rschwieb and Ethan Bolker make clear, if a ring $R$ has a nonzero idempotent element $a$ (i.e., an element $a\;$such that $a^2=a$, and $a\ne 0$), then the element $a$ must be a multiplicative identity for $R$.