Thinking about the Definition of a Unit of a ring R .... ....

[Math Amateur]

I have been thinking around the definition of a unit in a ring and trying to fully understand why the definition is the way it is ... ...

Marlow Anderson and Todd Feil, in their book "A First Course in Abstract Algebra: Rings, Groups and Fields (Second Edition), introduce units in a ring with 1 in the following way ... ...
https://www.physicsforums.com/attachments/6402
So ... an element \(\displaystyle a\) of a ring \(\displaystyle R\) with \(\displaystyle 1\) is a unit if there is an element \(\displaystyle b\) of \(\displaystyle R\) such that
\(\displaystyle ab = ba = 1\) ... ... So ... if, in the case where \(\displaystyle R\) was noncommutative, \(\displaystyle ab = 1\) and \(\displaystyle ba \neq 1\) then \(\displaystyle a\) would not be a unit ... is that right?Presumably it is not 'useful' to describe \(\displaystyle a\) as a 'left unit' in such a case ... that is, presumably, one-sided units are not worth defining ... is that right?
Could someone please comment on and perhaps clarify/correct the above ...Hope someone can help ...

Peter

 

[Euge]

Hi Peter,

So ... if, in the case where \(\displaystyle R\) was noncommutative, \(\displaystyle ab = 1\) and \(\displaystyle ba \neq 1\) then \(\displaystyle a\) would not be a unit ... is that right?

Yes, that's correct.

Presumably it is not 'useful' to describe \(\displaystyle a\) as a 'left unit' in such a case ... that is, presumably, one-sided units are not worth defining ... is that right?

No. The common terms are 'left inverse' and 'right inverse'. These concepts are useful, and you have already encountered them. As an example, let $R$ be the ring of linear transformations on $\mathcal{l}^1(\Bbb C)$, the space of summable complex sequences. The shift operators $S : (a_1,a_2,a_3,\ldots) \mapsto (a_2,a_3,a_4,\ldots)$ and $T : (a_1,a_2,a_3,\ldots) \mapsto (0,a_1,a_2,\ldots)$ are elements of $R$ that satisfy $ST = \operatorname{id}$ (so then $T$ is a right-inverse for $S$ and $S$ is a left-inverse for $T$), but $S$ and $T$ are not units in $R$.

 

[Math Amateur]

Hi Peter,

Yes, that's correct.

No. The common terms are 'left inverse' and 'right inverse'. These concepts are useful, and you have already encountered them. As an example, let $R$ be the ring of linear transformations on $\mathcal{l}^1(\Bbb C)$, the space of summable complex sequences. The shift operators $S : (a_1,a_2,a_3,\ldots) \mapsto (a_2,a_3,a_4,\ldots)$ and $T : (a_1,a_2,a_3,\ldots) \mapsto (0,a_1,a_2,\ldots)$ are elements of $R$ that satisfy $ST = \operatorname{id}$ (so then $T$ is a right-inverse for $S$ and $S$ is a left-inverse for $T$), but $S$ and $T$ are not units in $R$.

Thanks for the help, Euge ...

Appreciate your assistance ...

Peter