Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer 1

up vote 5 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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