Question

Show that there are ${\mathit{n}}^{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{-}{\mathit{n}}^{\mathbf{k}}$polynomials of degree K in${\mathbf{\mathbb{Z}}}_{\mathbf{n}}\left[x\right]$ .

Short answer

It is proved that $n^{k+1}-n^k$, polynomials of degree K in${\mathrm{\mathbb{Z}}}_n\left[x\right]$

Step-by-step solution

Step 1:To find a polynomial in ℤnx

Let us see the definitions of Reducible polynomial and Irreducible polynomial,

A nonconstant polynomial $f\left(x\right)\in F\left[x\right]$, where F is a field, is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials(units).

A nonconstant polynomial that is not irreducible is called reducible polynomial.

Consider a polynomial $p\left(x\right)\in {\mathrm{\mathbb{Z}}}_n\left[x\right]$of degree K such that $p\left(x\right)=a_0+a_{1}x+a_2{x}^2+\cdots +a_k{x}^k$with $a_k\ne 0$.

Thus, we have n possible values for $a_i$with $0⩽i⩽k$and $\left(n-1\right)$possible values for $a_k$.

To show there  nk+1-nkare polynomials

We have to show that there are$n^{k+1}-n^k$ polynomials of degree k in ${\mathrm{\mathbb{Z}}}_n\left[x\right]$.

From step 1, we can find the number of polynomials in ${\mathrm{\mathbb{Z}}}_n\left[x\right]$.

Therefore,

$\begin{array}{rcl}\left(n-1\right)n^k& =& n{n}^k-n^k\\ & =& n^{k+1}-n^k\\ & & \end{array}$

Therefore $\left(n-1\right)n^k=n^{k+1}-n^k$.

Hence, there $n^{k+1}-n^k$are polynomials of degree k in ${\mathrm{\mathbb{Z}}}_n\left[x\right]$