A ring which has many distinct left inverses

Question

I'm trying to find a ring which has an element that has many(more than two) distinct left inverses.

I first tried the (square) matrices but then realized that if a matrix is a left unit, then it has determinant greater than 0, thus it is also a right unit and has unique inverse.

And then next I tried to do it with functions with usual addition and composition, and I found this example(in the first answer, by user23211): Kaplansky's theorem of infinitely many right inverses in monoids?

But I realized that the space of functions does not form a ring since the distribution law fails.

To make distribution law hold, we can only have homomorphisms; while the elements in the above example are not.

I also have seen that we can construct a similar examples with infinite vector spaces, but they are not rings.

Is there any ring like this?

Answer 1

Consider a countably infinite dimensional vector space $V$ and let $R$ be the endomorphism ring of $V$.

Suppose the vector space has a basis $\{v_k:k\in\mathbb{N}\}$; define the endomorphism $f$ by $$ f(v_k)=v_{k+1} $$ and, for every $v\in V$, the endomorphism $g_v$ by $$ g_v(v_0)=v,\qquad g_v(v_{k+1})=v_k $$ Then, for every $k\in\mathbb{N}$, $$ g_v(f(v_k))=v_k $$ so $g_v$ is a left inverse of $f$ in $R$.

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It is an interesting exercise proving that, if the element $a\in R$ ($R$ any ring) has two distinct left inverses, then it has infinitely many.