Can we find ( characterize ) all non-zero commutative Artinian rings $R$ for which $-1 , 1$ are the only units of $R$ ?

Question

Can we find ( characterize ) all non-zero commutative Artinian rings $R$ for which $-1 , 1$ are the only units of $R$ ?

Answer 1

A commutative artinian ring is a finite product of local artinian rings. If the only units in $R$ are $1$ and $-1$, then the same is true of any direct factor of $R$. If $S$ is a local artinian ring with maximal ideal $m$, then every element of $S\setminus m$ is a unit. If the only units are $1$ and $-1$, this means that either $m$ has $1$ element and $S$ has at most $3$ elements or $m$ has $2$ elements and $S$ has $4$ elements (note that $S\setminus m$ is a union of cosets of $m$, so it has at least as many elements as $m$). In the first case, $S\cong\mathbb{Z}/2$ or $S\cong\mathbb{Z}/3$. In the second case, $S/m\cong\mathbb{Z}/2$ but $1\not=-1$ in $S$, which implies $S\cong\mathbb{Z}/4$.

So $R$ must be a product of finitely many copies of $\mathbb{Z}/2$, $\mathbb{Z}/3$, and $\mathbb{Z}/4$. But for the only units in $R$ to be $1$ and $-1$, there can only be at most one factor that is $\mathbb{Z}/3$ or $\mathbb{Z}/4$, since otherwise you could take a unit that is $1$ on one factor and $-1$ on the other (and $-1\neq 1$ in both the factors). So the rings of this form are $(\mathbb{Z}/2)^n$, $(\mathbb{Z}/2)^n\times\mathbb{Z}/3$, and $(\mathbb{Z}/2)^n\times\mathbb{Z}/4$ (for any $n\in\mathbb{N}$).