Proof: If $ϕ: R \to R'$ is a ring isomorphism, then its inverse $ϕ^{-1} : R' \to R$ is a ring homomorphism.

Question

Proposition. If $ϕ: R \to R'$ is a ring isomorphism, then its inverse $ϕ^{-1} : R' \to R$ is a ring homomorphism.

How would you start a proof for this proposition?

Answer 1

Since $\phi$ is a bijective function, proving that $a=\phi^{-1}(b)$ is the same thing as proving that $\phi(a)=b$. Therefore if you want to prove, for instance, that $\phi^{-1}(a-b)=\phi^{-1}(a)-\phi^{-1}(b)$ for all $a$, $b$, then you only need to prove that $\phi(\phi^{-1}(a-b))=\phi(\phi^{-1}(a)-\phi^{-1}(b))$ for all $a$, $b$.

Answer 2

$x,y \in R', \phi^{-1}(x)\phi^{-1}(y)=\phi^{-1}(\phi(u))\phi^{-1}(\phi(v))=uv$ for some $u,v\in R$

$\phi^{-1}(x)+\phi^{-1}(y)=\phi^{-1}(\phi(u))+\phi^{-1}(\phi(v))=u+v$ for some $u,v\in R$

$R\cong R', \phi(1)=1, \phi^{-1}(1)=1$

Hence proven.