Question

Let $\phi :R\left[x\right]\to R$ be the function that maps each polynomial in R[x] onto its constant term (an element of R). Show that $\phi$is a surjective homomorphism of rings.

Short answer

It is proved that $\phi$is a surjective homomorphism.

Step-by-step solution

Given part

It is given that$\phi :R\left[x\right]\to R$ is a function which maps each polynomial in R[x] onto its constant term.

Then we have,

$\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)=a_0$

Proof part

Now, we need to prove that $\phi$is a homomorphism.

Consider that a_i,b_j=0 for . Then, we have:

$i>m,j>n$

$\begin{array}{rcl}\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)+\left(\sum _{j=0}^n{b}_j{x}^i\right)& =& \phi \left(\sum _{j=0}^{max\left(m,n\right)}\left(a_i+b_j\right)x^i\right)\\ & =& a_0+b_0\\ & =& \phi \left(\sum _{i=0}^m{a}_i{x}^i\right)+\phi \left(\sum _{j=0}^n{b}_j{x}^i\right)\end{array}$

Now, find the product as follows:

$\begin{array}{rcl}\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)+\left(\sum _{j=0}^n{b}_j{x}^i\right)& =& \phi \left(\sum _{j=0}^{max\left(m,n\right)}\left(a_i+b_j\right)x^i\right)\\ & =& a_0+b_0\\ & =& \phi \left(\sum _{i=0}^m{a}_i{x}^i\right)+\phi \left(\sum _{j=0}^n{b}_j{x}^i\right)\end{array}$

Also, note that for all $a\in R$, then, we have $a=a{x}^0\in R\left[x\right]$and $\phi \left(a\right)=a$.

Hence proved,$\phi$ is a surjective homomorphism.