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

(b) $\overline{\mathbf{h}}$is injective if and only if $\mathit{h}$ is injective.

Short answer

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

Step-by-step solution

Prove that  h¯ is injective

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

$\begin{array}{rcl}\overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)& =& \sum _{i=0}^{m}h\left(a_i\right)x^i\\ & =& 0\\ & & \end{array}$

This implies that by the uniqueness of representation all $h\left(a_i\right)=0$, $a_i=0$and$\sum _{i=0}^m{a}_i{x}^i=0$.

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

Prove that  h is injective

It is given that $\overline{h}$ is injective, and we have to prove that h is injective as$\overline{h}$ is h when restricted to the copies of R and S in$R\left[x\right]$ and $S\left[x\right]$ respectively. Therefore, h is injective.