Let $A$ be a ring in which the product of any two nonzero elements is nonzero (we shall say that $A$ is an integral domain, even if $A$ is non commutative). It is well-known that if $A$ is commutative, then a non zero polynomial $P(X)\in A[X]$ has a finite number of zeros in $A$.

I deeply believe that the converse is also true. To state the conjecture, I need some natural definitions.

Define $A_L[X]$ the set of all left-polynomials, i.e expressions of the form $P(X)=a_nX^n+...+a_1X+a_0$ where $a_0, ..., a_n \in A$.

We say that $x\in A$ is a zero of $P$ if, and only if, $a_nx^n+...+a_1x+a_0=0$.

CONJECTURE : Let $A$ be a integral domain such that each $P(X)\in A_L[X]$ has a finite number of zeros in $A$. Then $A$ is commutative.

For instance, it is a direct generalization of Wedderburn's theorem : if $A$ is a finite field, then each left-polynomial over $A$ clearly have a finite number of zero in $A$.

I succeed to prove this result in the special case where $A$ is a field and the dimension of $A$ over its center $Z$ (seen as a $Z$-vector space) is finite. I used the article of Gordon and Motzkin.

Is this conjecture known ? True ? False ?

Many thanks for your help ! Have a great week-end :)