Division ring: a ring with $1 \ne 0$ in which every nonzero element is a unit

A division ring (also skew field) is a ring $D$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) such that

• (V1) $1 \ne 0$ in $D$;
• (V2) every $x \in D$ with $x \ne 0$ is a unit, that is, has a two-sided multiplicative inverse (Left inverse, right inverse, and invertible element of a monoid, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

Equivalently, $D^{\times} = D \setminus \{0\}$: by (V2) every nonzero element is a unit, and by (V1) together with The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring the element $0$ is not a unit, since $0$ is a unit only when $1 = 0$. Consequently $(D \setminus \{0\}, \cdot, 1)$ is a group (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring); in particular $D \setminus \{0\}$ is closed under multiplication, so a division ring has no zero divisors.

A commutative division ring is a division ring whose multiplication is commutative. Those are exactly the fields (Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree, Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring).

• (V1) is not implied by (V2). In the one-element ring, where $1 = 0$, there is no nonzero element at all, so (V2) holds vacuously; (V1) is what excludes it, exactly as in Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors. Without (V1) the one-element ring would count as a division ring and the sentence "a division ring has no zero divisors" would still be true but useless.

• The definition is not vacuous beyond the commutative case. The quaternions $\mathbb{H}$ are a division ring that is not commutative, hence not a field ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1} = \bar q / N(q)$ for $q \ne 0$, while $ij = k$ and $ji = -k$); they are constructed on this page for that reason.

• "Has no zero divisors" is genuinely weaker. $\mathbb{Z}$ has no zero divisors and is not a division ring, since $2$ is not a unit; the companion page records this. So (V2) is a real strengthening of the domain condition, and no argument on this page derives one from the other.