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Hey everyone!

Lately I remembered an exercise from an algebra class from Jacobson's book: Prove that if an element has more than one right inverse then it has infinitely many, Jacobson attributes this excercise to Kaplansky. Regardless of the solution I began to wonder:

Does anybody know any explicit examples of rings that have this property of having elements with infinitely many (or, thanks to Kaplansky, multiple) right inverses? Is the same true for left inverses? I came across an article from the AMS Bulletin that studied this topic but skimming through it I could not find an explicit example, sorry I cant remember the author. Anyways, thanks and good luck!

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3 Answers 3

up vote 5 down vote accepted

Let $M$ be a module (over some ring) such that $M$ is isomorphic to $M\oplus M$, for example an infinite-dimensional vector space over a field. Let $R$ be the ring of endomorphisms of $M$. Let $f\in R$ be projection of $M\oplus M$ on the first factor composed with an isomorphism $M\to M\oplus M$. Then $f$ has as many right inverses as there are homomorphisms $M\to M$.

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If I use an isomorphism between $M$ and $M\oplus N$ instead, then my example becomes a bit simpler conceptually and also more general. It now includes examples mentioned in the other answers. –  Tom Goodwillie Oct 15 '10 at 10:30
    
True, that is why I chose your answer, it's more general thanks! –  Juan OS Oct 17 '10 at 2:18

Consider the space $\mathbb{Z}^\mathbb{N}$ of integer sequences $(n_0,n_1,\ldots)$, and take $R$ to be its ring of endomorphisms. Then the ``left shift'' operator $$(n_0,n_1,\ldots) \mapsto (n_1,n_2,\ldots)$$ has plenty of right inverses: a right shift, with anything you want dropped in as the first co-ordinate, gives a right inverse.

I recall finding this example quite helpful with the exercise ``two right inverses implies infinitely many'' — taking a couple of the most obvious right inverses in this case, and seeing how one can generate others from them.

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@Peter: Ironically, I think your example is essentially the same as mine but with the other convention on the order of the product x*y: for me, since these are functions, I read them as first do y, then do x, but your convention makes just as much sense. Thus we are working in opposite rings, as in my answer above. [To be precise, your ring is not literally the same as mine, but they are similar, and it is well known that the derivative is a rescaled shift operator.] –  Pete L. Clark Oct 15 '10 at 13:54
    
@Pete: ah, of course; I guess the precise differences are just rescaling and a change of scalars from $\mathbb{Z}$ to $\mathbb{R}$. Though I'm confused about what you say regarding the order of the product: I also read $x \cdot y$ as “first $y$ then $x$”; maybe we’re using left/right inverse opposite ways round? As I understand the convention, if $l\cdot r = 1$, then $l$ is a left inverse for $r$, and $r$ a right inverse for $l$. So a left inverse is epimorphic, like the left shift or the derivative? –  Peter LeFanu Lumsdaine Oct 15 '10 at 16:29
    
@Peter: yes, it looks we are using left/right inverse in different senses when the ring operation is function composition. As I say though, no matter. (I'm willing to believe that your convention is the right one. I am mildly dyslexic on this kind of mathematical issue: e.g. I used to have a hard time remembering which were left and which were right cosets.) –  Pete L. Clark Oct 17 '10 at 2:53
    
@Pete: what I always have the most trouble with is remembering which way round the subscripts for matrix entries go :-) But I guess I've been doing category theory long enough now that function-composition conventions are burned into my brain… –  Peter LeFanu Lumsdaine Oct 19 '10 at 2:45

For a memorable explicit example, let $V = \mathbb{R}[x]$ be the real vector space of polynomial functions, and let $R = \operatorname{End}(V)$ be the ring of $\mathbb{R}$-linear endomorphisms (aka linear operators) of $V$. Then the operator $D$ which sends a polynomial to its derivative has infinitely many left inverses. I would like for you to convince yourself of this, so I won't give the proof, but a hint is that this is connected to the additive constant attached to an indefinite integral.

Now, you originally asked about right inverses and then later asked about left inverses. This brings me to the second point in my answer. Definitely the theorem for right inverses implies that for left inverses (and conversely!): one needs only to consider the opposite ring $R^{\operatorname{op}}$ of $R$, which has the same underlying set and the same addition operation, but with mirror-image multiplication: for $x,y \in R^{\operatorname{op}}$, $x \bullet y := yx$.

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Thanx Pete! Your example is very concrete. –  Juan OS Oct 17 '10 at 2:19

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