3
$\begingroup$

Does there exist an infinite commutative Artin ring (with identity) that has only finitely many units? If so, I would like to see an example, if not, I would like a hint for a proof of this.

The internet has been searched, and also Lang's Algebra, Jacobson's Basic Algebra and Atiyah-MacDonald's Commutative Algebra. No examples were found there.

$\endgroup$
2
$\begingroup$

Another way to see it:

If $R$ is local Artinian, then its maximal ideal $M$ consists of nilpotent elements and everything else in $R\setminus M$ is a unit.

It’s also well-known that $1+x$ is a unit when $x$ is a nilpotent. If there are only finitely many units, there are only finitely many things in $M$, and by assumption $R\setminus M$ is also finite, so we’ve accounted now for all elements of $R$ (finitely many of them.)

The problem reduces to products of local artinian rings just as explained elsewhere.

$\endgroup$
  • 1
    $\begingroup$ This is more elementary, thank you for posting this answer! $\endgroup$ – JSchoone Mar 2 '18 at 12:13
  • $\begingroup$ @JSchoone Glad to have helped! $\endgroup$ – rschwieb Mar 2 '18 at 14:16
2
$\begingroup$

A commutative artinian ring is a (finite) product of local rings. A unit in the product is a tuple of units. So you are asking whether there exists an infinite local artinian ring $A$ with finitely many units. Let $I$ be the maximal ideal; then $A/I$ is a finite field; moreover $I^n=0$ for some $n$. Consider the chain $$ 0=I^n\subseteq I^{n-1}\subseteq\dots\subseteq I^2\subseteq I\subseteq I^0=A $$ where each $I^{k-1}/I^k$ is a finite dimensional vector space over $A/I$.

$\endgroup$
  • 1
    $\begingroup$ I think I see what you're doing, but the assumptions and conclusion aren't completely clear to me. I think you are assuming $A$ is a local artinian ring with finitely many units and you are trying to prove that it is a finite ring, is that right? In which case, how do you conclude finiteness in the end? It seems to be by induction on $n$ plus the fact that $I^{k-1}/I^k$ is finite dimensional as an $A/I$-vector space (thus $I^k$ has finite index in $I^{k-1}$)? $\endgroup$ – Ben Blum-Smith Mar 1 '18 at 23:03
  • 1
    $\begingroup$ @BenBlum-Smith Each $I^{k-1}/I^k$ is finite. Now it's just standard (abelian) group theory: if a group has a normal series with finite factors, then it is finite. $\endgroup$ – egreg Mar 1 '18 at 23:18
  • $\begingroup$ Thank you both for your comments, and egreg especially for your answer! It is very insightful. $\endgroup$ – JSchoone Mar 1 '18 at 23:35

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.