1
$\begingroup$

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 :)

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ What is an integer ring? In particular, are finite rings integer rings? What about the group algebra $\mathbb{Z}[G]$, where $G$ is a free group? $\endgroup$
    – David E Speyer
    Commented Dec 11, 2015 at 18:34
  • 1
    $\begingroup$ The contrapositive of your conjecture is that if $A$ is noncommutative, then there is some polynomial with an infinite number of zeroes. This is clearly false if $A$ is finite, so you'll have to tell us whether "integer ring" excludes that possibility. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 11, 2015 at 18:44
  • $\begingroup$ but English "integer" is not a translation of French "intègre" in any meaning! $\endgroup$
    – YCor
    Commented Dec 12, 2015 at 11:07
  • $\begingroup$ Of course... I meant not necessarily commutative Integral domains. $\endgroup$
    – Stabilo
    Commented Dec 12, 2015 at 11:17
  • $\begingroup$ Have you looked at the universal enveloping algebra of a nonabelian Lie algebra? $\endgroup$
    – Qiaochu Yuan
    Commented Dec 12, 2015 at 17:44

0

Reset to default

You must log in to answer this question.