Question

Let S be a set, R a ring and f be a 1-1 mapping of S onto R. Define + and · on S by: )) x y f (f(x)) f(y 1 + = + − )) x y f (f(x) f(y 1 ⋅ = ⋅ − . ∀ x, y∈S Show that ) (S, +, ⋅ is a ring isomorphic to R. 

Answer 1

Let $S$ be a set, $R$ a ring and $f$ be a 1-1 mapping of $S$ onto $R$. Define $+$ and $·$ on $S$ by:

$x+ y=f^{-1}(f(x)+_Rf(y))$, $x\cdot y=f^{-1}(f(x)\cdot_Rf(y))$ for all $x, y\in S$. Let us show that $(S, +, ⋅)$ is a ring isomorphic to $R$. Taking into account that

$f(x+ y)=f(f^{-1}(f(x)+_Rf(y)))=f\circ f^{-1}(f(x)+_Rf(y))=f(x)+_Rf(y)$ and

$f(x\cdot y)=f(f^{-1}(f(x)\cdot _Rf(y)))=f\circ f^{-1}(f(x)\cdot _Rf(y))=f(x)\cdot _Rf(y)$, we conclude that $f:S\to R$ is a ring homomorphism. Since $f$ is bijection, $f$ is also a ring isomorphism. It follows that $(S, +, ⋅)$ is isomorphic to the ring $R$, and hence $(S, +, ⋅)$ is also a ring.