Question

Let $h:R\to S$ a homomorphism of rings and define a function $\overline{\mathbf{h}}\mathbf{:}\mathit{R}\left[x\right]\mathbf{\to }\mathit{S}\left[x\right]$ by the rule

$\overline{\mathbf{h}}\left(a_0+a_{1}x+\dots +a_n{x}^n\right)\mathbf{=}\mathit{h}\left(a_0\right)\mathbf{+}\mathit{h}\left(a_1\right)\mathit{x}\mathbf{+}\mathit{h}\left(a_2\right){\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathit{h}\left(a_n\right){\mathit{x}}^{\mathbf{n}}$

Prove that

(d) If $\mathit{R}\mathbf{\cong }\mathit{S}$, then $\mathit{R}\left[x\right]\mathbf{\cong }\mathit{S}\left[x\right]$.

Short answer

It is proved that if $R\cong S$, then $R\left[x\right]\cong S\left[x\right]$.

Step-by-step solution

Consider 

We know that an isomorphism is a bijective homomorphism.

Prove that  

From step 1, it implies that if $R\cong S$, then $R\left[x\right]\cong S\left[x\right]$.

Hence it is proved that if $R\cong S$, then $R\left[x\right]\cong S\left[x\right]$.