Question

Let$\phi :\mathrm{\mathbb{Z}}\left[x\right]\to {\mathrm{\mathbb{Z}}}_n\left[x\right]$ be the function that maps the polynomial$a_0+a_{1}x+\dots +a_k{x}^k$in $\mathrm{\mathbb{Z}}\left[x\right]$onto the polynomial $\left[a_0\right]+\left[a_1\right]x+\dots +\left[a_k\right]x^k$, where$\left[a\right]$ denotes the class of the integer a in ${\mathrm{\mathbb{Z}}}_n$. Show that $\phi$is a surjective homomorphism of rings.

Short answer

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

Step-by-step solution

Define the mapping 

Define the mapping $\phi :\mathrm{\mathbb{Z}}\left[x\right]\to {\mathrm{\mathbb{Z}}}_n\left[x\right]$as $\phi \left(\sum _{i=0}^m{a}_i{x}^i\right)=\sum _{i=0}^m\left[a_i\right]x^i$. First Prove that $\phi$is a homomorphism as:

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

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

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

Show that is a surjective homomorphism of rings

Since we have already proved that$\phi$ is a homomorphism. Now show that $\phi$is a surjective.

Observe that for all $\sum _{i=0}^m\left[a_i\right]x^j\in {\mathrm{\mathbb{Z}}}_n$, we have $\sum _{i=0}^m{a}_i{x}^j\in \mathrm{\mathbb{Z}}$and $\phi \left(\sum _{i=0}^m{a}_i{x}^j\right)=\sum _{i=0}^m\left[a_i\right]x^j$, this implies that$\phi$ is a surjective.

Hence it is proved that $\phi$is a surjective homomorphism.