Question

Let $\mathit{h}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$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

(c) $\overline{h}$ is surjective if and only if h is surjective.

Short answer

It is proved that $\overline{h}$ is surjective if and only if h is surjective.

Step-by-step solution

Prove that  h¯ is surjective

It is given that h is surjective, and we have to prove that $\overline{h}$ is surjective. Since h is surjective, we have:

$\sum _{i=0}^m{s}_i{x}^i\in S\left[x\right]$

This implies that there are $r_i\in R$, such that $h\left(r_i\right)=s_i$. It implies that .

$\overline{h}\left(\sum _{i=0}^m{r}_i{x}^i\right)=\sum _{i=0}^m{s}_i{x}^i$

Therefore, $\overline{h}$ is surjective.

Prove that  h is surjective 

It is given that $\overline{h}$ is surjective. This implies that h is surjective.