Question

Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be an isomorphism of rings and let$\mathit{g}\mathbf{:}\mathit{S}\mathbf{\to }\mathit{R}$be the inverse function off(as defined in Appendix B). Show thatis also an isomorphism.

[Hint: To show$\mathit{g}\left(a+b\right)\mathbf{=}\mathit{g}\left(a\right)\mathbf{+}\mathit{g}\left(b\right)$, consider the images of the left-and right-hand side underfand use the facts that fis a homomorphism and is the identity map.]

Short answer

It is proved that g is isomorphism

Step-by-step solution

Consider the given conditions

Since it is given that function $f:R\to S$is a isomorphism and function$g:S\to R$ is its inverse function, this implies that fand gare injective and surjective.

Therefore, we also know that$f\circ g$ is the identity map.

Show that g is a isomorphism  

Assume that x and y be arbitrary elements in S. As we know that f is surjective, this implies that there exits$x_1,y_1\in R$such that $f\left(x_1\right)=x$and $f\left(y_1\right)=y$.

Also, we have:

localid="1649231926138" $$\begin{gathered} x+y=f\left(x_1\right)+f\left(y_1\right) \\ =f\left(x_1+y_1\right) \\ xy=f\left(x_1\right)f\left(y_1\right) \\ =f\left(x_1{y}_1\right) \end{gathered}$$

Now prove that g is homomorphism:

$$\begin{gathered} g\left(x\right)+g\left(y\right)=g\left(f\left(x_1\right)+f\left(x_2\right)\right) \\ =x_1+y_1 \\ =g\left(f\left(x_1+y_1\right)\right) \\ =g\left(x+y\right) \\ g\left(x\right)g\left(y\right)=g\left(f\left(x_1\right)\right)g\left(f\left(y_1\right)\right) \\ =x_1{y}_1 \\ =g\left(f\left(x_1{y}_1\right)\right) \\ =g\left(xy\right) \end{gathered}$$

This implies that g is homomorphism. We already know that g is injective and surjective. This implies that g is isomorphism.

Hence it is proved that g is isomorphism.