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

(a) $\overline{\mathbf{h}}$ is a homomorphism of rings

Short answer

It is proved that $\overline{h}$ is a homomorphism

Step-by-step solution

Define the map 

Since we have the map $\overline{h}:R\left[x\right]\to S\left[x\right]$ such that $\overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)=\sum _{i=0}^{m}h\left(a_i\right)x^i$,

We have to show that $\overline{h}$ is a homomorphism

Prove that h¯ is a homomorphism 

Prove that $\overline{h}\left(\left(\sum _{i=0}^m{a}_i{x}^i+\sum _{j=0}^n{b}_j{x}^j\right)\right)=\overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)+\overline{h}\left(\sum _{j=0}^n{b}_j{x}^j\right)$ as:

$\begin{array}{rcl}\overline{h}\left(\left(\sum _{i=0}^m{a}_i{x}^i+\sum _{j=0}^n{b}_j{x}^j\right)\right)& =& \overline{h}\left(\sum _{i=0}^{\max \left(m,n\right)}\left(a_i+b_i\right)x^i\right)\\ & =& \sum _{i=0}^{\max \left(m,n\right)}h\left(a_i+b_i\right)x^i\\ & =& \sum _{i=0}^{\max \left(m,n\right)}\left(h\left(a_i\right)+h\left(b_i\right)\right)x^i\\ & =& \left(\sum _{i=0}^{m}h\left(a_i\right)x^i\right)+\left(\sum _{j=0}^{n}h\left(b_j\right)x^j\right)\\ & =& \overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)+\overline{h}\left(\sum _{j=0}^n{b}_j{x}^j\right)\\ & & \end{array}$

Prove that $\overline{h}\left(\left(\left(\sum _{i=0}^m{a}_i{x}^i\right)\left(\sum _{j=0}^n{b}_j{x}^j\right)\right)\right)=\overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)\overline{h}\left(\sum _{j=0}^n{b}_j{x}^j\right)$.

$\begin{array}{rcl}\overline{h}\left(\left(\left(\sum _{i=0}^m{a}_i{x}^i\right)\left(\sum _{j=0}^n{b}_j{x}^j\right)\right)\right)& =& \overline{h}\left(\sum _{k=0}^{m+n}\left(\sum _{i+j=k}{a}_i{b}_j\right)x^k\right)\\ & =& \sum _{k=0}^{m+n}h\left(\sum _{i+j=k}{a}_i{b}_j\right)x^k\\ & =& \sum _{k=0}^{m+n}\left(\sum _{i+j=k}h\left(a_i\right)h\left(b_j\right)\right)x^k\\ & =& \left(\sum _{i=0}^{m}h\left(a_i\right)x^i\right)\left(\sum _{j=0}^{n}h\left(b_j\right)x^j\right)\\ & =& \overline{h}\left(\sum _{i=0}^m{a}_i{x}^i\right)\overline{h}\left(\sum _{j=0}^n{b}_j{x}^j\right)\\ & & \end{array}$

Hence it is proved that $\overline{h}$ is a homomorphism.