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

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SUMMARY

The discussion centers on the definition of a unit in a ring, specifically in the context of noncommutative rings. According to Marlow Anderson and Todd Feil in "A First Course in Abstract Algebra: Rings, Groups and Fields (Second Edition)", an element \(a\) of a ring \(R\) with \(1\) is a unit if there exists an element \(b\) such that \(ab = ba = 1\). The conversation clarifies that in noncommutative rings, if \(ab = 1\) and \(ba \neq 1\), \(a\) is not a unit. The terms 'left inverse' and 'right inverse' are introduced as useful concepts, illustrated by the example of shift operators in the ring of linear transformations on \(\mathcal{l}^1(\mathbb{C})\).

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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 $$a$$ of a ring $$R$$ with $$1$$ is a unit if there is an element $$b$$ of $$R$$ such that
$$ab = ba = 1$$ ... ... So ... if, in the case where $$R$$ was noncommutative, $$ab = 1$$ and $$ba \neq 1$$ then $$a$$ would not be a unit ... is that right?Presumably it is not 'useful' to describe $$a$$ 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
 
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Hi Peter,
Peter said:
So ... if, in the case where $$R$$ was noncommutative, $$ab = 1$$ and $$ba \neq 1$$ then $$a$$ would not be a unit ... is that right?

Yes, that's correct.
Peter said:
Presumably it is not 'useful' to describe $$a$$ 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 $R$.
 
Euge said:
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
 

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