Question

If$\mathit{f}\left(x\right)\mathbf{=}{\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{0}}\mathbf{ }\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathit{g}\left(x\right)\mathbf{=}{\mathit{b}}_{\mathbf{r}}{\mathit{x}}^{\mathbf{r}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathit{b}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{b}}_{\mathbf{0}}\mathbf{,}\mathbf{ }$ and$\mathbf{ }\mathit{h}\left(x\right)\mathbf{=}{\mathit{c}}_{\mathbf{s}}{\mathit{x}}^{\mathbf{s}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathit{c}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{0}}$ are polynomials in $\mathrm{\mathbb{Z}}\left[x\right]$such that$\mathit{f}\left(x\right)\mathbf{=}\mathit{g}\left(x\right)$,$\mathbf{ }\mathit{h}\left(x\right)$

, show that in ${\mathrm{\mathbb{Z}}}_n\left[x\right]$, $\overline{\mathbf{f}}\left(x\right)\mathbf{=}\overline{\mathbf{g}}\left(x\right)\overline{\mathbf{h}}\left(x\right)$.Also, see Exercise 19 in Section 4.1.

Short answer

We proved that, $\overline{f}\left(x\right)=\overline{g}\left(x\right)\overline{h}\left(x\right)$.

Step-by-step solution

Defining the given polynomials and writing in summation form

Let $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0, g\left(x\right)=b_r{x}^r+\cdots +b_{1}x+b_0$and $h\left(x\right)=c_s{x}^s+\cdots +c_{1}x+c_0$be the given polynomials.

Then suppose $f\left(x\right)=n, g\left(x\right)=r$and $h\left(x\right)=s$.

If $f\left(x\right)=g\left(x\right)h\left(x\right)$, then we have,

$n=r+s$ and $a_k=\sum _{i+j=k}{b}_i{c}_j$in$\mathrm{\mathbb{Z}}$ .

To prove f¯x=g¯xh¯x

Now, in ${\mathrm{\mathbb{Z}}}_n$, we have,

.$\left[a_k\right]=\left[\sum _{i+j=k}{b}_i{c}_j\right]=\sum _{i+j=k}\left[b_i\right]\left[c_j\right]$

Thus, we have,

$\begin{array}{rcl}\overline{f}\left(x\right)& =& \sum _{k=0}^n\left[a_k\right]x^k\\ & =& \sum _{k=0}^{r+s}\left(\sum _{i+j=k}\left[b_i\right]\left[c_j\right]\right)x^k\\ & =& \left(\sum _{i=0}^r\left[b_i\right]x^i\right)\left(\sum _{j=0}^s\left[c_j\right]x^j\right)\\ & =& \overline{g}\left(x\right)\overline{h}\left(x\right)\\ & & \end{array}$

$\therefore \overline{f}\left(x\right)=\overline{g}\left(x\right)\overline{h}\left(x\right)$

Hence, we proved that $\overline{f}\left(x\right)=\overline{g}\left(x\right)\overline{h}\left(x\right)$.