# Can zero divisor in a non-associative ring be a unit?

Statement

Left or right zero divisors in ring can never be units.

is proved in Wikipedia this way:

If $a$ is invertible and $ab = 0$

$0 = a^{−1}0 = a^{−1}ab = b$

I'm confused by third transition.

I suppose that more detailed version of proof written this way

$0 = a^{−1}0 = a^{−1}(ab) = (a^{−1}a)b = b$

shows us that ring is required to be associative for us to prove the statement.

Is it true?

I also need to note that the whole question occured because of my math book, where ring by definition is not necessarily associative under multiplication.

-
It is. But associativity of multiplication is one of the axioms an algebraic structure has to fulfill to be a ring. – Damian Sobota Nov 28 '11 at 22:05
Sorry, I forgot to mention that in my algebra studybook rings arent necessarily associative under multiplication. – furikuretsu Nov 28 '11 at 22:13
@furikuretsu: Your studybook should call these structures algebras, not rings (and so should your question). While there is lack of agreement about whether rings should be defined to have a multiplicative identity, there is none (as far as I know) about their associativity. By the way "unit" supposes the existence of an identity; so your "rings" have an identity but are not associative? – Marc van Leeuwen Nov 29 '11 at 10:13
@Marc van Leeuwen: my book defines the ring without identity, associativity and commutativity under multiplication. Just abelian group under addition, having both distributive laws and multiplication defined on set. Sorry for the mess, anyway. I never thought the math definitions from different books can be so... different. – furikuretsu Nov 30 '11 at 1:45

## 1 Answer

The sedenions form a nonassociative ring with identity in which every nonzero element has a multiplicative inverse, but there are zero divisors.

-