Proof. If $u$ and $v$ are two units, then there are the elements $r$ and $s$ of $R$ such that $ru = ur = 1$ and $sv = vs = 1$ . Then we get that$(sr)(uv) = s(r(uv)) = s((ru)v) = s(1v) = sv = 1$ , similarly $(uv)(sr) = 1$ . Thus also $uv$ is a unit, which means that $E$ is closed under multiplication. Because $1 \in E$ and along with $u$ also its inverse $r$ belongs to $E$ , the set $E$ is a group.

Corollary. In a commutative ring, a ring product is a unit iff all factors are units.

Examples

1. When $R = \mathbb{Z}$ , then $E = \{1,\,-1\}$ .
2. When $R = \mathbb{Z}[i]$ , the ring of Gaussian integers, then $E = \{1,\,i,\,-1,\,-i\}$ .
3. When $R = \mathbb{Z}[\sqrt{3}]$ , then $E = \{\pm(2\!+\!\sqrt{3})^n\,\vdots\,\,\, n\in\mathbb{Z}\}$ .
4. When $R = K[X]$ where $K$ is a field, then $E = K\!\smallsetminus\!\{0\}$ .
5. When $R = \{0\!+\!\mathbb{Z},\,1\!+\!\mathbb{Z},\,\ldots,\, m\!-\!1\!+\!\mathbb{Z}\}$ is the residue class ring modulo $m$ , then $E$ consists of the prime classes modulo $m$ , i.e. the residue classes $l\!+\!\mathbb{Z}$ satisfying $\gcd(l, m) = 1$ .