Question

If R is itself a field, show that$R=F$ .

Short answer

It is proved that$R=F$.

Step-by-step solution

Definition of Isomorphism

Ring Isomorphism

A ring isomorphism f from a ring R to a ring S is a mapping from R to S such that

1.) f is a homomorphism from R into S

2.) f is injective

Prove that R=F

Define a map $\varphi :R\to F$with $\varphi \left(r\right)=\left[r,1\right]$.

Now, it is a ring homomorphism.

Its kernel is composed of the $r\in R$such that $\left[r,1\right]=\left[0,1\right],$i.e., such that, $r=0$.

This means that, $\varphi$is injective.

Now, supposing R is a field then, for all $\left[a,b\right]\in F$we have that $\left[a,b\right]=\left[b^{-1}a,1\right]$.

Since $a=b{b}^{-1}a$, so in this case $\varphi$is surjective and isomorphism.

Therefore, $R=F$.

Hence proved.