4
$\begingroup$

I didn't see this problem before. I motivated by the questions

  1. Is every commutative group structure underlying at least one (unitary, commutative) ring structure

  2. A basic question about rings

Suppose A is an Abelian group such that it is possible to define an associative unitary ring structure on A. An element $x\in A$ is called potentially identity if there exists an associative ring $R$ such that $A$ is the additive group of $R$ and $x=1_R$, the identity element of $R$. Determine the set of all potentially identity elements of $A$.

$\endgroup$
  • $\begingroup$ Note that if a potential identity $x$ has finite order $n,$ then the group $A$ must have exponent $n.$ $\endgroup$ – Geoff Robinson Jan 20 '14 at 20:27
  • $\begingroup$ @GeoffRobinson: Very good point! $\endgroup$ – M. Shahryari Jan 20 '14 at 20:30
  • 1
    $\begingroup$ @GeoffRobinson: Actually, that gives a complete answer when $x$ has finite order $n$. If $A$ has exponent $n$, then it's a $\mathbb{Z}/n\mathbb{Z}$-module, and the subgroup $\langle x\rangle\cong\mathbb{Z}/n\mathbb{Z}$ generated by $x$ is injective as a $\mathbb{Z}/n\mathbb{Z}$-module, so $A=\mathbb{Z}/n\mathbb{Z}\oplus A'$ for some subgroup $A'$, which can be made into a ring with $A'^2=0$. $\endgroup$ – Jeremy Rickard Jan 21 '14 at 12:09
  • 1
    $\begingroup$ If $x$ has infinite order, there are some obvious restrictions on how divisible $x$ must be. The set of integers $n$ for which $ny=x$ has a solution must be closed under multiplication. $\endgroup$ – Jeremy Rickard Jan 21 '14 at 12:12
  • $\begingroup$ @Jeremy Rickard: The solution for the finite exponent case is now completed by your comment. The most complicated case is probably torsion free case. $\endgroup$ – M. Shahryari Jan 21 '14 at 13:05
1
$\begingroup$

Maybe it will be useful for you the notion of potentially invertible elements of a semigroup introduced by E. Shutov [E.S. Lyapin, "Semigroups" , Amer. Math. Soc.].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.