[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$~$).
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# 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.