[summary: A unit of a ring is an element with a multiplicative inverse.]

An element $~$x$~$ of a non-trivial ring%%note:That is, a ring in which $~$0 \not = 1$~$; equivalently, a ring with more than one element.%% is known as a unit if it has a multiplicative inverse: that is, if there is $~$y$~$ such that $~$xy = 1$~$. (We specified that the ring be non-trivial. If the ring is trivial then $~$0=1$~$ and so the requirement is the same as $~$xy = 0$~$; this means $~$0$~$ is actually invertible in this ring, since its inverse is $~$0$~$: we have $~$0 \times 0 = 0 = 1$~$.)

$~$0$~$ is never a unit, because $~$0 \times y = 0$~$ is never equal to $~$1$~$ for any $~$y$~$ (since we specified that the ring be non-trivial).

If every nonzero element of a ring is a unit, then we say the ring is a field.

Note that if $~$x$~$ is a unit, then it has a unique inverse; the proof is an exercise. %%hidden(Proof): If $~$xy = xz = 1$~$, then $~$zxy = z$~$ (by multiplying both sides of $~$xy=1$~$ by $~$z$~$) and so $~$y = z$~$ (by using $~$zx = 1$~$). %%

Examples

• In [48l $~$\mathbb{Z}$~$], $~$1$~$ and $~$-1$~$ are both units, since $~$1 \times 1 = 1$~$ and $~$-1 \times -1 = 1$~$. However, $~$2$~$ is not a unit, since there is no integer $~$x$~$ such that $~$2x=1$~$. In fact, the only units are $~$\pm 1$~$.
• [4zq $~$\mathbb{Q}$~$] is a field, so every rational except $~$0$~$ is a unit.