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Topic: Right inverse but no left inverse in a ring (Read 8228 times) |
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ecoist
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Right inverse but no left inverse in a ring
« on: Apr 3rd, 2006, 9:59am » |
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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.
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Pietro K.C.
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Re: Right inverse but no left inverse in a ring
« Reply #1 on: Apr 21st, 2006, 2:32am » |
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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:
Take a,b as in the problem statement, i.e. ab=1 and 'a' without a left inverse. Consider the expression
an-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 bn-1.
What's more, the above expression doesn't repeat values either. That is, if
f : N --> R is defined by
f(n) = an-1 - ban
then f is injective. To see this, suppose f(m) = f(n) for m>n to obtain the following equation:
am-1 - bam = an-1 - ban
Right-multiply everything by bn. The right side vanishes, giving us
am-n-1 - bam-n = 0
whence
am-n-1 = bam-n.
Right-multiply through by bm-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,
an-1 - ban
has the very convenient property of vanishing under left-multiplication by 'a'. Therefore, if ab=1, then
a(b + an-1 - ban) = 1
for all natural n. Since g(x) = b+x is also injective, the above is an infinite family of right inverses.
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Barukh
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Re: Right inverse but no left inverse in a ring
« Reply #2 on: Apr 25th, 2006, 10:53pm » |
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Very nice solution, Pietro! 
Please, tell your story.
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Michael Dagg
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Re: Right inverse but no left inverse in a ring
« Reply #3 on: Apr 30th, 2006, 2:37pm » |
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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.
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| « Last Edit: Apr 30th, 2006, 2:42pm by Michael Dagg » |
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Regards,
Michael Dagg
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Eigenray
wu::riddles Moderator
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Re: Right inverse but no left inverse in a ring
« Reply #4 on: Jan 7th, 2008, 12:31am » |
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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?
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