infinitude of inverses

infinitude of inverses

Proposition 1

.

Let

R

be a ring with 1.

1.

If

a

∈

R

has a

right inverse

but no

left inverses

, then

a

has infinitely many right inverses.

2.

If

a

∈

R

has more than one right inverse, then

a

has infinitely many right inverses.

Proof.

1.

Let

a

⁢

b

=

1

. Define

b

0

=

b

,

b

1

=

1

-

b

0

⁢

a

+

b

0

,

…

,

b

i

+

1

=

1

-

b

i

⁢

a

+

b

i

,

…

Then, by

induction

, we see that

a

⁢

b

i

=

a

-

a

⁢

b

i

-

1

⁢

a

+

a

⁢

b

i

-

1

=

a

-

a

+

1

=

1

. Next we want to show that

b

i

≠

b

j

if

i

≠

j

. Suppose

i

>

j

and

b

i

=

b

j

. Again by induction, we have

b

j

=

b

i

=

1

+

(

1

-

a

)

+

⋯

+

(

1

-

a

)

i

-

j

-

1

+

b

j

⁢

(

1

-

a

)

i

-

j

(1)

If we let

c

=

1

+

(

1

-

a

)

+

⋯

+

(

1

-

a

)

i

-

j

-

1

then

(

1

-

a

)

⁢

c

=

c

⁢

(

1

-

a

)

=

(

1

-

a

)

+

(

1

-

a

)

2

+

⋯

+

(

1

-

a

)

i

-

j

=

c

-

1

+

(

1

-

a

)

i

-

j

.
So

Equation

3 can be rewritten as

c

=

b

j

-

b

j

⁢

(

1

-

a

)

i

-

j

=

b

j

⁢

(

1

-

(

1

-

a

)

i

-

j

)

=

b

j

⁢

c

⁢

a

. Then

c

⁢

b

j

=

b

j

⁢

c

⁢

a

⁢

b

j

=

b

j

⁢

c

.
Now, note that for

m

≤

n

,

(

1

-

a

)

n

⁢

b

j

m

=

(

1

-

a

)

n

-

m

⁢

(

b

j

-

1

)

m

. This implies that

c

⁢

b

j

i

-

j

-

1

=

b

j

i

-

j

-

1

+

(

b

j

-

1

)

⁢

b

j

i

-

j

-

2

+

⋯

+

(

b

j

-

1

)

i

-

j

-

1

=

g

⁢

(

b

j

)

+

(

b

j

-

1

)

i

-

j

-

1

.

On the other hand, we also have

c

⁢

b

j

i

-

j

-

1

=

b

j

⁢

c

⁢

b

j

i

-

j

-

2

=

b

j

⁢

(

b

j

i

-

j

-

2

+

(

b

j

-

1

)

i

-

j

-

3

+

⋯

+

(

1

-

a

)

⁢

(

b

j

-

1

)

i

-

j

-

2

)

=

g

⁢

(

b

j

)

+

b

j

⁢

(

1

-

a

)

⁢

(

b

j

-

1

)

i

-

j

-

2

.

So combining the above two equations, we get

(

b

j

-

1

)

i

-

j

-

1

=

b

j

⁢

(

1

-

a

)

⁢

(

b

j

-

1

)

i

-

j

-

2

. Let

d

=

(

b

j

-

1

)

i

-

j

-

2

, then

(

b

j

-

1

)

⁢

d

=

b

j

⁢

(

1

-

a

)

⁢

d

=

b

j

⁢

d

-

b

j

⁢

a

⁢

d

. Simplify, we have

d

=

b

j

⁢

a

⁢

d

. Expanding

d

, then

b

j

i

-

j

-

2

+

⋯

+

(

-

1

)

i

-

j

-

2

=

(

b

j

⁢

a

)

⁢

(

b

j

i

-

j

-

2

+

⋯

+

(

-

1

)

i

-

j

-

2

)

=

b

j

⁢

a

⁢

b

j

i

-

j

-

2

+

⋯

+

b

j

⁢

a

⁢

(

-

1

)

i

-

j

-

2

=

b

j

i

-

j

-

2

+

⋯

+

(

-

1

)

i

-

j

-

2

⁢

b

j

⁢

a

.

Then

1

=

b

j

⁢

a

and we have reached a

contradiction

.

2.

For the next part, notice that if

b

and

c

are two distinct right inverses of

a

, then neither one of them
can be a left inverse of

a

, for if, say,

b

⁢

a

=

1

, then

c

=

(

b

⁢

a

)

⁢

c

=

b

⁢

(

c

⁢

a

)

=

b

. So we can apply the same technique
used in the previous portion of the problem. Note that if

b

j

⁢

a

=

1

, then

1

=

b

j

⁢

a

=

(

1

-

b

j

-

1

⁢

a

+

b

j

-

1

)

⁢

a

=

a

-

b

j

-

1

⁢

a

2

+

b

j

-

1

⁢

a

.

Multiply

b

j

-

1

from the right, we have

b

j

-

1

=

a

⁢

b

j

-

1

-

b

j

-

1

⁢

a

2

⁢

b

j

-

1

+

b

j

-

1

⁢

a

⁢

b

j

-

1

=

1

-

b

j

-

1

⁢

a

+

b

j

-

1

Thus

b

j

-

1

⁢

a

=

1

. Keep going until we reach

b

⁢

a

=

1

, again a contradiction.

∎

Remark

. The first part of the above

proposition

implies that a

finite ring

is

Dedekind-finite

.

Title

infinitude of

inverses

Canonical

name

InfinitudeOfInverses

Date of creation

2013-03-22 18:17:30

Last modified on

2013-03-22 18:17:30

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 16U99

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