# Arbitrary Ring and Field Contruction

Does anybody know if there has been any study in the construction of rings and/or fields out of groups, particularly with respect to certain operations? For example, the set of the reals and operation $a \cdot b := a + b - 1$ forms a commutative group, but how would one find a second operation that would create a field or ring? If it exist, could you construct it out of addition, multiplication and their inverses? How would you approach problems like these in general?

• In your specific example, $a*b:=(a-1)(b-1)$ works as multiplication to your addition "$\cdot$" :) – Hagen von Eitzen Jun 29 '18 at 6:41

In general when you have a structure $M$ (in some language L; in this case L is the language or ring and the L-structure is $\mathbb{R}$) and a set $S$ such that there exists a bijective map $f:M\to S$ then you can pass the structure of $M$ to the set $S$ and you get a new $L$-structure. In the case of the Rings you can define:

$+^\sim : S\times S\to S$ such that for every $(a,b)\in S\times S$ $a+^\sim b:=f(f^{-1}(a)+f^{-1}(b))$

$\cdot^\sim : S\times S\to S$ such that for every $(a,b)\in S\times S$ $a\cdot^\sim b:=f(f^{-1}(a)f^{-1}(b))$

So you can prove that $(S,+^\sim, \cdot^\sim, f(0), f(1))$ is Ring (with unity if $M$ is a ring with unity).

Now we can define a trivial bijective map to $f:\mathbb{R}\to \mathbb{R}$ , the function that maps every $a\in \mathbb{R}$ to $f(a):=a+1$

In this case $\mathbb{R}$ is a new Ring with unity with the new operations:

$+^\sim : \mathbb{R} \times \mathbb{R}\to \mathbb{R}$ such that for every $(a,b)\in \mathbb{R}\times \mathbb{R}$ $a+^\sim b:=f(f^{-1}(a)+f^{-1}(b))=$

$f((a-1)+(b-1))=f((a+b)-2)=a+b-1$

$\cdot^\sim : \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ such that for every $(a,b)\in \mathbb{R}\times \mathbb{R}$ $a\cdot^\sim b:=$

$f(f^{-1}(a)f^{-1}(b))=f((a-1)(b-1))=$

$f(ab-(a+b)+1)=ab-(a+b)+2$

You can observe that $f(0)=1$ so the neutral element is changed with the neutral element with the product (but $f(1)=2$ so 0 it is not the new neutral element with respect to the new product)

You can find a bijective map that map $f(0)=1$ and $f(1)=0$ in general for Rings.

What is this map?