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  "body": "\n## Definition\n\nA *near-field* is a set $N$ with two binary operations $+$ and $\\cdot$ such that:\n\n1. $(N, +)$ is an [[abelian group|082GHNPJ53]], with identity element $0$.\n2. $N \\setminus \\set{0}$ is closed under $\\cdot$ and is a [[group|G5S1M9]] under $\\cdot$.\n3. $a \\cdot 0 = 0 \\cdot a = 0$ for all $a \\in N$.\n4. $(a + b) \\cdot c = a \\cdot c + b \\cdot c$ for all $a, b, c \\in N$.\n\n## Remarks\n\nClauses 2 and 3 make multiplication associative on all of $N$, with an identity $1$.\nSome authors use the left distributive law $a(b + c) = ab + ac$ in clause 4 instead; the two versions are mirror images of each other.\nA near-field that also satisfies the other distributive law is a [[division ring|0EA5BTA0CX]].\nThe commutativity of $+$ in clause 1 is often omitted, because it follows from the other clauses.\n\nExample: let $F = \\F_9$ be the [[finite field|0ZXZ82FTX2]] with $9$ elements, and let $\\phi(a) = a^3$ be its [[Frobenius automorphism|0T3WVTD675]].\nDefine $a \\circ b = a b$ if $b$ is a square in $F$ (including $b = 0$) and $a \\circ b = \\phi(a) b$ otherwise.\nClause 4 holds because $\\phi$ is additive.\nClause 2 holds because $\\phi$ preserves squares.\nSo $b \\circ c$ is a nonzero square exactly when $b$ and $c$ are both squares or both non-squares, and this gives associativity.\nThe element $1$ is the identity.\nA nonzero $b$ has the inverse $c$ with $\\phi^{k}(c) = b^{-1}$, where $k = 0$ if $b$ is a square and $k = 1$ otherwise.\nWrite $\\F_9 = \\F_3(i)$ with $i^2 = -1$; then $i$ is a square and $1 + i$ is not, and $i \\circ (1 + i) = 1 - i$ while $(1 + i) \\circ i = i - 1$.\nSo $(F, +, \\circ)$ is a near-field with non-commutative multiplication; since every finite division ring is commutative, it is not a division ring.\n\nNon-example: $\\Z$ with its usual operations satisfies clauses 1, 3 and 4 but fails clause 2, because $2$ has no multiplicative inverse in $\\Z$.\n",
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