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.

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    $\begingroup$ This is false. You can find a counterexample in, say, the ring of linear operators on an infinite-dimensional vector space. $\endgroup$ – Qiaochu Yuan Oct 28 '13 at 7:32
  • $\begingroup$ (Every time you mention a book, give a complete reference)) $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '13 at 7:41
  • $\begingroup$ 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. $\endgroup$ – Mustafa Said Oct 28 '13 at 7:44
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    $\begingroup$ @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 $\endgroup$ – TBrendle Oct 28 '13 at 7:58
  • $\begingroup$ Doesn't "stipulates" imply that Hungerford is including $ax = xa =1$ as part of his definition? $\endgroup$ – 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.

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  • $\begingroup$ Yes. I was tripped up by his definition; my mistake. $\endgroup$ – 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.

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