ClayMapsAndCouplingMaps

(1) Clay maps In [Clay 68], Clay uses the following general procedure to construct nearrings on a given group.

Let $(G,+)$ be a group and let $E=\Hom(G,G)$ be the set of all endomorphisms of $G$, which may not be a ring or even a group. If a mapping $f:G\to E$ is given, then one can define a binary operation $*_f$ on $G$ by $ a*_fb=f_a(b)$ where $f_a=f(a)\in E$. This binary operation $*_f$ is left distributive over $+$. Conversely, if $*$ is a binary operation on $G$ and is left distributive over $+$, then $*$ defines a mapping $f_*:G\to E$ via $(f_*(a))(b)=f_a(b)=a*b$. These correspondences between $E^G=\{f\mid f:G\to E\}$ and the binary operations which are left distributive over $+$ are inverse one-to-one correspondences.

So to make a nearring out of a group $(G,+)$, one takes a mapping $f:G\to E$ and hope that the binary operation $*_f$ is itself associative. But this is true if and only if $f$ also has the property that for every $x,y\in G$, $f(x*_fy)=f_x\circ f_y$ where $\circ$ is the composition of functions.

Two multiplications $*_1$ and $*_2$ on $G$ are <i>similar</i> if there is an automorphism $\alpha$ of $G$ such that $\alpha(x*_1y)=\alpha(x)*_2\alpha(y)$ for all $x,y\in G$. It is clear that $(G,+,*_1)$ and $(G,+,*_2)$ are isomorphic if and only if $*_1$ and $*_2$ are similar.

Theorem [Clay 68]. Let $*_1$ and $*_2$ be multiplications on a group $(G,+)$ ad let $f_1$ and $f_2$ be the corresponding mappings in $E^G$. The $*_1$ and $*_2$ are similar if and only if there exists an automorphism $\alpha$ of $G$ such that $f_2(\alpha(x))\circ\alpha=\alpha\circ f_1(x)$ for all $x\in G$.

The following constructing method for planar nearring is based on the above procedure:

Let $(F,+,\cdot)$ be a field and let $(V,+)$ be a vector space over $F$. Let $f:V\to F$ with $\vert f(V)\vert\geq 3$ and $f(f(u)v)=f(u)f(v)$ for all $u,v\in V$. Since we can identify $F$ with a subset of $\Hom(V,V)$, and if $*_f$ is the corresponding binary operation mentioned above, then $f(a*_fb)=f(f_a(b))=f(f(a)b)=f(a)f(b)$, so $(V,+,*_f)$ is a nearring. Actually, $\bm(V,+,*_f)$ is planar.

Remark. This generalizes the "field generated" method.

Examples. (a) Consider the $n$ dimensional vector $\mathbb R^n$ over $\mathbb R $, $n\geq 1$. For any real number $p > 0$, define $f_{n,p}: \mathbb R^n\to\mathbb R$ by $$f_{n, p}(\bm{x})={|x_1^p+ \cdots+x_n^p|}^{\frac1p}=\|\bm{x}\|_p,$$ for $\bm{x}=(x_1,x_2,\dots,x_n)\in\mathbb R^n$. Then for any $\bm x,\bm y\in\mathbb R^n$, we have $$f_{n,p}(f_{n,p}(\bm x)\bm y)=\|(\|\bm x\|_p\bm y)\|_p=\|\bm x\|_p\cdot\|\bm y\|_p=f_{n,p}(\bm x)f_{n,p}(\bm y).$$ So each $(\mathbb R^n,+,*_{f_{n,p}})$ is a planar nearring.

(b) Let $F$ be a field and $n\geq 0$. Denote $F_n[x]=\{\sum_{i=0}^n a_ix^i\}$, and $F_{-n}[ [x] ]=\{\sum_{i=-n}^\infty a_ix^i\mid a_i\in F\}$. We have $F_n[x]\subset F[x]\subset F_{-n}[ [x] ]$. Let $R$ be one of $F[x]$, $F_n[x]$, or $F_{-n}[ [x] ]$. For any $a(x)=\sum_{i\geq -n}a_ix^i\in R\setminus\{0\}$ define $m(a)=a_s$ where $s=\min\{i\geq -n\mid a_i\not=0\}$ and $m(0)=0$. Define $f:R\to F$ by $f(a(x))=m(a(x))$. Then $f(f(a(x))b(x))=f(a(x))f(b(x))$ for all $a(x),b(x)\in R$, and $(R,+,*_f)$ is a planar nearring. Note that the example on $\bR^2$ with $a*_1b=a_1b$ if $a_1\not=0$, and $a_2b$ if $a_1=0$ is a special case of this example.

[Clay 68] The near-rings on groups of low order, Math. Z. 104 (1968), 249--265. MR 37-258.

(2) Coupling maps