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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My abstract book defines inverses (units) as solutions to the equation $ax=1$ then stipulates in the definition that $xa=1=ax$, even in non-commutative rings. But I'm having trouble understanding why this would be true in the generic case.

Can someone help me understand?

Book is "Abstract Algebra : An Introduction - Third Edition" By Thomas W. Hungerford. ISBN-13: 978-1-111-56962-4 (Chapter 3.2)


Ok, I think I got it. Left inverses are not necessarily also right inverses. However, if an element has a left inverse and a right inverse, then those inverses are equal:

$$ lx = 1 \\ xr = 1\\ lxr = r \\ lxr = l \\ r = l $$


Also, I assumed that 'unit' was synonymous with 'has a multiplicative inverse'. It is not; a 'unit' is an element that has both a left and right inverse, not just an inverse in general.

share|cite|improve this question

migrated from Oct 28 '13 at 7:46

This question came from our site for professional mathematicians.

This is false. You can find a counterexample in, say, the ring of linear operators on an infinite-dimensional vector space. – Qiaochu Yuan Oct 28 '13 at 7:32
(Every time you mention a book, give a complete reference)) – Mariano Suárez-Alvarez Oct 28 '13 at 7:41
I may be missing something but a ring (with identity) has a group structure with respect to multiplication and addition and the multiplicative and additive inverses are unique. – Mustafa Said Oct 28 '13 at 7:44
@MustafaSaid a ring does NOT have a group structure with respect to multiplication. consider $\mathbb{Z}/4\mathbb{Z}$, where $2x=2$ has more than one solution – TBrendle Oct 28 '13 at 7:58
Doesn't "stipulates" imply that Hungerford is including $ax = xa =1$ as part of his definition? – Robert Lewis Oct 28 '13 at 8:04

No, I have the book in front of me and he defines units as elements that have both a left and a right inverse, so $a$ is a unit if there exist elements $x$ and $y$ in $R$ such that $ax=ya=1$. Note $x$ need not equal $y$.

In his very next Remark he then proves that in this situation $x=y$, as a theorem, not as part of the definition.

Edit: Now you are editing your question and filling it with even more confusion. You might try a book with more examples, such as Dummit & Foote.

share|cite|improve this answer
Yes. I was tripped up by his definition; my mistake. – user84922 Oct 28 '13 at 8:06

Note: In a ring R with 1, if EVERY non-zero element x has a left inverse, then R is a division ring (so ax=1 implies xa=1). More generally, if EVERY non-zero element has either a left or a right inverse, then R is a division ring.

share|cite|improve this answer

Your Answer


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.