---
tag: 486GPTW3M8
kind: definition
name: Near-field
msc: [12K05, 16Y30]
defines: DSE72WQ5A2
uses: [082GHNPJ53, G5S1M9, 0EA5BTA0CX, 0ZXZ82FTX2, 0T3WVTD675]
sources:
  - url: https://en.wikipedia.org/wiki/Near-field_(mathematics)
---

## Definition

A *near-field* is a set $N$ with two binary operations $+$ and $\cdot$ such that:

1. $(N, +)$ is an [[abelian group|082GHNPJ53]], with identity element $0$.
2. $N \setminus \set{0}$ is closed under $\cdot$ and is a [[group|G5S1M9]] under $\cdot$.
3. $a \cdot 0 = 0 \cdot a = 0$ for all $a \in N$.
4. $(a + b) \cdot c = a \cdot c + b \cdot c$ for all $a, b, c \in N$.

## Remarks

Clauses 2 and 3 make multiplication associative on all of $N$, with an identity $1$.
Some authors use the left distributive law $a(b + c) = ab + ac$ in clause 4 instead; the two versions are mirror images of each other.
A near-field that also satisfies the other distributive law is a [[division ring|0EA5BTA0CX]].
The commutativity of $+$ in clause 1 is often omitted, because it follows from the other clauses.

Example: let $F = \F_9$ be the [[finite field|0ZXZ82FTX2]] with $9$ elements, and let $\phi(a) = a^3$ be its [[Frobenius automorphism|0T3WVTD675]].
Define $a \circ b = a b$ if $b$ is a square in $F$ (including $b = 0$) and $a \circ b = \phi(a) b$ otherwise.
Clause 4 holds because $\phi$ is additive.
Clause 2 holds because $\phi$ preserves squares.
So $b \circ c$ is a nonzero square exactly when $b$ and $c$ are both squares or both non-squares, and this gives associativity.
The element $1$ is the identity.
A nonzero $b$ has the inverse $c$ with $\phi^{k}(c) = b^{-1}$, where $k = 0$ if $b$ is a square and $k = 1$ otherwise.
Write $\F_9 = \F_3(i)$ with $i^2 = -1$; then $i$ is a square and $1 + i$ is not, and $i \circ (1 + i) = 1 - i$ while $(1 + i) \circ i = i - 1$.
So $(F, +, \circ)$ is a near-field with non-commutative multiplication; since every finite division ring is commutative, it is not a division ring.

Non-example: $\Z$ with its usual operations satisfies clauses 1, 3 and 4 but fails clause 2, because $2$ has no multiplicative inverse in $\Z$.