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

1. Feb 12, 2017

### 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 ... ...

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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2. Feb 12, 2017

### Staff: Mentor

I'm not aware of an example and still try to find an argument, why this should be impossible. At least I've found some strange implications.

Let us assume $ab=1$ and $ba \neq 1$.

Then $a(ba-1)=0$ by associativity and you don't want to have a unit being a zero divisor.
In addition, $a(ba+b-1)=1=ab$ and $ba+b-1 \neq b$ and you don't want to have two different right inverses either.

So although I haven't found a direct contradiction (yet!?), the resulting ring has some really weird properties, at least $a \in R$ has.

3. Feb 12, 2017

### Math Amateur

Thanks for the post fresh_42 ...

But I need some help to see why a(ba-1) = 0 ...

Can you help ...

Peter

4. Feb 12, 2017

### Stephen Tashi

Yes.

The term "left inverse" is in use. Perhaps that's why we don't hear about "left units".

5. Feb 12, 2017

### lavinia

I think there are many rings with one sided units.

6. Feb 12, 2017

### pwsnafu

Consider the ring of all operators $C^\infty \to C^\infty$. What is the inverse of the derivative?

7. Feb 13, 2017

### Staff: Mentor

$a\cdot (ba-1)=a\cdot (b\cdot a)-a\cdot 1=(a\cdot b) \cdot a - a= 1 \cdot a - a = 0$ so $a$ is a unit ($ab=1$) and a zero divisor $a\cdot (ba-1)=0$.

$a(ba+b-1)=a\cdot (ba-1)+a\cdot b=0+1=1=ab$ by the above and the definition of $b$. So $(ba+b-1)$ and $b$ are both right inverse to $a$. But if they were equal, that is $ba+b-1=b$ then $ba-1=0$ or $ba=1$ which we excluded. Thus they are actually two different right inverses.

8. Feb 14, 2017

### lavinia

A classic example occurs in the ring of linear maps on the vector space of infinite tuples of numbers in some field.

The linear map $R$ that shifts the sequence once to the right and puts zero at the beginning of the sequence is inverted by the linear map $L$ that shifts the sequence once to the left and drops off the first number in the sequence. But $L$ is not invertible since it is not injective.

- I will look for some examples that do not involve infinite dimensional space such as this one or the example given above of the derivative operator. I think there are examples for some group rings over certain algebraic extensions of the rationals.

Last edited: Feb 16, 2017
9. Feb 15, 2017

### Math Amateur

Thanks to all who have posted ... it has been most helpful ... I am really grateful for the help ...

Peter