Question

Let R be a commutative ring and let $\mathit{k}\left(x\right)$ be a fixed polynomial in $R\left[x\right]$. Prove that there exists a unique homomorphism $\phi :R\left[x\right]\to R\left[x\right]$ such that

$\mathit{\phi }\left(r\right)\mathbf{=}\mathit{r}\mathbf{}\mathbf{\text{for}}\mathbf{}\mathbf{\text{all}}\mathbf{}\mathit{r}\mathbf{\in }\mathit{R}$and $\mathit{\phi }\left(x\right)\mathbf{=}\mathit{k}\left(x\right)$.

Short answer

It is proved that there exists a unique homomorphism $\phi :R\left[x\right]\to R\left[x\right]$ such that $\phi \left(r\right)=r\text{for}\text{all}r\in R$ and $\phi \left(x\right)=k\left(x\right)$.

Step-by-step solution

Define the mapping 

Define the mapping $\phi :R\left[x\right]\to R\left[x\right]$as $\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)=\sum _{i=0}^m{a}_{i}k{\left(x\right)}^i$. First, prove that $\phi$is a homomorphism as:

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

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

Since both conditions are satisfied, this implies that $\phi$ is a homomorphism.

Show the result

We have already proved that $\phi$ is a homomorphism. Now show that there exists a unique homomorphism $\phi :R\left[x\right]\to R\left[x\right]$ such that $\phi \left(r\right)=r\text{for}\text{all}r\in R$ and $\phi \left(x\right)=k\left(x\right)$.

Observe that for all $\phi \left(r\right)=r$ and $\phi \left(x\right)=k\left(x\right)$. From uniqueness, it follows that for any ring homomorphism,$\phi :R\left[x\right]\to S$ ; so we have:

data-custom-editor="chemistry" $\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)=\sum _{i=0}^m\phi \left(a_i\right)\phi {\left(x\right)}^i$

This implies that $\phi$ can be completely determined by the coefficient terms.

Hence it is proved that there exists a unique homomorphism $\phi :R\left[x\right]\to R\left[x\right]$ such that $\phi \left(r\right)=r\text{for}\text{all}r\in R$ and $\phi \left(x\right)=k\left(x\right)$.