Question

Show that, under congruence modulo ${\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{}\mathit{i}\mathit{n}\mathbf{}{\mathbf{\mathbb{Z}}}_{\mathbf{3}}\left[x\right]$ in , there are exactly 27 distinct congruence classes.

Short answer

It is proved that there are 27 distinct congruence classes.

Step-by-step solution

Corollary statement

Assume that p (x)is a non-zero polynomial of degree n in F [x] and congruence modulo p (x).

1. When f(x)is divided by p (x), we have $\left[f\left[x\right]\right]=\left[r\left(x\right)\right]$, where r (x) is the remainder and $f\left(x\right)\in F\left[x\right]$.
2. Let S be the set of all polynomials whose degrees are less than the degree of p (x), then each congruence class modulo p (x) is also the class of some polynomial in S, where each class is distinct.

Proof part

The given expression is in $x^3+2x+1in{\mathrm{\mathbb{Z}}}_3\left[\mathrm{x}\right]$.

Here, n is 3, which is a prime. And, the degree of p (x)is 3.

The formula of number of congruence classes is $n^{\text{degree of}p\left(x\right)}$.

So, for the given expression, we have 3³ = 27.

Hence, there are 27 congruence classes.