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        riddles >> putnam exam (pure math) >> Right inverse but no left inverse in a ring
        
(Message started by: ecoist on Apr 3^rd, 2006, 9:59am)

Title: Right inverse but no left inverse in a ring
Post by ecoist on Apr 3^rd, 2006, 9:59am

Let R be a ring with 1 and let a be an element of R with right inverse b (ab=1) but no left inverse in R. Show that a has infinitely many right inverses in R.

Title: Re: Right inverse but no left inverse in a ring
Post by Pietro K.C. on Apr 21^st, 2006, 2:32am

Jolly good problem! After I get some sleep, and if there seems to be interest, I'll post my tortuous way towards the solution I found. For now, let me just state the results:
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Take a,b as in the problem statement, i.e. ab=1 and 'a' without a left inverse. Consider the expression

a^n-1 - baⁿ

It is never zero, as that would imply ba=1, contrary to our hypothesis. To see this, just suppose it WAS equal to zero and right-multiply everything by b^n-1.

What's more, the above expression doesn't repeat values either. That is, if

f : N --> R is defined by

f(n) = a^n-1 - baⁿ

then f is injective. To see this, suppose f(m) = f(n) for m>n to obtain the following equation:

a^m-1 - ba^m = a^n-1 - baⁿ

Right-multiply everything by bⁿ. The right side vanishes, giving us

a^m-n-1 - ba^m-n = 0

whence

a^m-n-1 = ba^m-n.

Right-multiply through by b^m-n-1 to obtain ba=1, again contrary to initial supposition. So f is injective. Why is all this relevant? To allow us to construct an infinite family of right inverses to 'a'. It should be obvious that the expression under consideration,

a^n-1 - baⁿ

has the very convenient property of vanishing under left-multiplication by 'a'. Therefore, if ab=1, then

a(b + a^n-1 - baⁿ) = 1

for all natural n. Since g(x) = b+x is also injective, the above is an infinite family of right inverses.

[/hide]

Title: Re: Right inverse but no left inverse in a ring
Post by Barukh on Apr 25^th, 2006, 10:53pm

Very nice solution, Pietro! :D

Please, tell your story.

Title: Re: Right inverse but no left inverse in a ring
Post by Michael_Dagg on Apr 30^th, 2006, 2:37pm

Pietro K.C. result is on track with the classical result.

It also follows that right ideal is infinite as well.

The elements b + ( 1 - b a ) a^i are all right inverses of [a], and (assuming [ba] is not equal to 1) are distinct.

If the number of right inverses of [a] is finite, it follows that b + ( 1 - b a ) a^i = b + ( 1 - b a ) a^j for some i < j.
Subtract [b], and then multiply on the right by b^j; from ab=1 (and thus (1-ba)b = 0) we conclude 1 - ba = 0.

So if there are only finitely many right inverses, it's because there is a 2-sided inverse.

This problem is well known to the ring cognoscenti and it first appeared in N. Jacobson's algebra text and is attributed to Ivan Kaplansky.

Title: Re: Right inverse but no left inverse in a ring
Post by Eigenray on Jan 7^th, 2008, 12:31am

This actually came up for an algebra course I was grading. A student had assumed that:

Any non-unit of R is contained in a maximal left-ideal.

While false in general, before deciding how to grade it I wanted to know whether it was even true in the case under discussion (and how obviously true/false it was):

G is a finite group, F is a field, and R = F[G] is the group ring.

Of course if F were finite it would follow from the proof in this thread, but there was no such assumption.

So, is it true in this case? And as a followup: can you think of more classes of rings in which it is true that left-unit implies unit?