1
$\begingroup$

So, inspired by the question Finite integral domains are commutative?, i was wondering if the next generalization is also true:

  • Let $A$ be an integral domain (using the terminology of the linked question), such that every element $a \in A$ generates a finite subring. Then, is $A$ commutative? (WLOG, we can suppose that these subrigs are proper, because otherwise the answer is clear)

Now, here is what i tried (with little to no success):

  1. Strengthening the hypothesis by requiring that every proper subring of $A$ is finite, by following the answer to the linked question, $a,b \in A$ commute if the subring generated by these two elements is proper. Not an answer, but at least something.
  2. If we also ask that $A$ has a unit element $1$, then the subring $\widetilde{u}$ generated by $u \in A-\left \{ 0_A \right \}$ is finite and integral, so by https://proofwiki.org/wiki/Finite_Ring_with_No_Proper_Zero_Divisors_is_Field (this link was also mentioned in the question linked above) it is a field, so $u$ is invertible in $\widetilde{u}$, and so it is invertible in $A$. Then, $A$ would be a division ring. Still, not what i am looking for.

Any help would be appreciated.

$\endgroup$
1
$\begingroup$

If the ring generated by $a\ne0 \in A$ is finite then $a^k=a^{k+d}$ for some $k,d\ge 1$. Therefore since $a(a^{k-1}-a^{k-1+d})=0$,..., so $a=a^{d+1}$. Then for every $b\ne0 \in A$, $ab-a^{d+1}b=0$, hence $a^db=b$. Similarly, $ba^d=b$. So $a^d$ is the identity element 1. Thus for every $a\ne0\in A$ there exists $d=d(a)$ such that $a^d=1$. Thus $A$ is a periodic division ring. So $A$ is a field, hence commutative.

$\endgroup$
0
$\begingroup$

To state it in terms of the solution you linked to:

It's quite clear that a "domain" in which every element generates a finite subring is a division ring, because each such finite subring is a field and even share the same identity (by the linked solution.) Obviously then all nonzero elements are invertible.

So the trick is to prove commutativity. At the moment I can think of no better way than the one JCAA alludes to, which is appealing to Jacobson's theorem that periodic rings are commutative. (This ring is periodic since, by finiteness and cancellation, $a^{i_a}=a$ for some integer $i_a > 1$ depending on $a$.)

Maybe something more elementary can be worked out by assuming $aba^{-1}b^{-1}$ or $ab-ba$ is nonzero and then looking at the field generated by them...

$\endgroup$

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.