Rings with right inverses - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:30:57Z http://mathoverflow.net/feeds/question/42235 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42235/rings-with-right-inverses Rings with right inverses Juan OS 2010-10-15T00:15:53Z 2010-10-15T05:17:06Z <p>Hey everyone!</p> <p>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:</p> <p>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!</p> http://mathoverflow.net/questions/42235/rings-with-right-inverses/42237#42237 Answer by Tom Goodwillie for Rings with right inverses Tom Goodwillie 2010-10-15T00:28:35Z 2010-10-15T00:28:35Z <p>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 $M\to M$. </p> http://mathoverflow.net/questions/42235/rings-with-right-inverses/42246#42246 Answer by Pete L. Clark for Rings with right inverses Pete L. Clark 2010-10-15T02:04:19Z 2010-10-15T02:04:19Z <p>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 <em>left</em> 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.</p> <p>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 <strong>opposite ring</strong> $R^{\operatorname{op}}$ of $R$, which has the same underlying set and the same addition operation, but with mirror-image multiplication: for $x,y \in R^{\operatorname{op}}$, $x \bullet y := yx$. </p> http://mathoverflow.net/questions/42235/rings-with-right-inverses/42252#42252 Answer by Peter LeFanu Lumsdaine for Rings with right inverses Peter LeFanu Lumsdaine 2010-10-15T05:17:06Z 2010-10-15T05:17:06Z <p>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.</p> <p>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.</p>