# does there always exist a nonzero idempotent in a ring $R$

I was studying about integral domain and wondered that why it should be a commutative ring with unity; that is, I was wondering if there is a noncommutative ring with no zero divisors (1) with/ or (2) without unity, or a (3) commutative ring with no zero divisors with unity.

and I guess this page provided a partial answer; it proved that if a ring $R$ has no zero divisors, a nonzero idempotent($a$ is an idempotent if $a^2=a$) of $R$ must be the unity of $R$. so if I can prove that a ring $R$ with no zero divisors has a nonzero idempotent, then every ring with no zero divisors must have a unity.

every ring has an idempotent since it contains $0$. but (4) is there a nonzero one if the ring has no zero divisors?

I guess the commutativity is less connected with the condition having no zero divisors since the division ring can be noncommutative and has no zero divisors. so the case (1) is solved and (2), (3) are relying on (4).

hence, what I want to know is just (4) as in the title.

• 4 is false if you allow non-unital rings (Fondly called rngs). The set of even integers is one such object and it has no non-zero idempotents. – RKD May 1 '17 at 2:57

No, there does not always exist a nontrivial idempotent in a ring, for example the ring of integers has $0$ and $1$ as its only idempotents.
If the ring has no zero divisors, there is no nontrivial idempotent since $a^2 = a$ rearranges as $a(a-1) = 0$, and since there are no zero divisors we can apply cancellation to get $a = 0$ or $a = 1$.
The place where nontrivial idempotents show up is in products of rings: examples of this are $(0, 1)$ and $(1, 0)$ in any product of unitial rings $R \times S$. I know that for unital commutative rings, the presence of a nontrivial idempotent means the ring factorises into a product, and this should be true in more general contexts as well.