Infinite Artin ring with only finitely many units

Question

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.

Answer 1

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.

Answer 2

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 $A/I$.