# Infinite Artin ring with only finitely many units

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.

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.

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

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 $A/I$.

• 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 $I^{k-1}$)? – Ben Blum-Smith Mar 1 '18 at 23:03
• @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. – egreg Mar 1 '18 at 23:18
• Thank you both for your comments, and egreg especially for your answer! It is very insightful. – JSchoone Mar 1 '18 at 23:35