9

Let $R \neq 0$ be a ring which may not be commutative and may not have an identity. Suppose $R$ satisfies the following conditions.

1) $a^2 = a$ for every element $a$ of $R$.

2) $ab \neq 0$ whenever $a \neq 0$ and $b\neq 0$.

Is $R$ isomorphic to the field $\mathbb{Z}/2\mathbb{Z}$?

Prism
  • 11,782
Makoto Kato
  • 44,216
  • 3
    Every Boolean ring is commutative. See here. So the part "may not be commutative" is immaterial. – Prism Jul 21 '13 at 07:58
  • I noticed that someone serially upvoted for my questions and answers including this one. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. – Makoto Kato Nov 27 '13 at 07:15

2 Answers2

14

Assume that $a,b\neq0$. We have $$ 0=abab-ab=(aba-a)b, $$ so $a=aba$. Therefore $$ 0=a-aba=a^2-aba=a(a-ba). $$ So we also see that $a=ba$. Repeating the dose once more $$ 0=ba-a=ba-a^2=(b-a)a\implies b=a. $$ Thus your rng has only one non-zero element. As there are no zero-divisors that non-zero element is an identity, and it follows that $R\cong\mathbb{Z}/2\mathbb{Z}$.

Jyrki Lahtonen
  • 140,891
  • Beautiful proof! – Prism Jul 21 '13 at 08:20
  • 3
    I think I am going to dub proofs of this sort "washing machine argument" because you wash, rinse, spin, repeat. After two-three cycles the proof comes out clean from all the dirt it had at first! – Asaf Karagila Jul 21 '13 at 12:19
2

Let $a \neq 0$ be an element of $R$. It suffices to show that $a$ is an identity. So we would like to show first that $ab = b$ for any element $b$ of $R$. It suffices to show that $a(ab) = ab$ thanks to the condition 2). But this is obvious thanks to the condition 1). Similiarly we can show that $ba = b$.

Makoto Kato
  • 44,216