Question

Show that$x^3-3$ is irreducible in ${\mathrm{\mathbb{Z}}}_7\left[x\right]$.

Short answer

It is proved that $x^3-3$is irreducible.

Step-by-step solution

Determine x3-3

Consider the given function, $x^3-3$in ${\mathrm{\mathbb{Z}}}_7\left[x\right]$, this is done by contradiction.

Now, using Theorem 4.11, $x^3-3$is written as a product of two polynomials of degree less than 3. Thus, the possibilities are:

$x^3-3=\left(ax+b\right)\left(c{x}^2+dx+e\right)or{x}^3-3=\left(ax+b\right)\left(cx+d\right)\left(ex+f\right)$or

Here, are real numbers because,

n	0 1 2 3 4 5 6
$n^3-3$mod 7	4 5 5 3 5 3 3

Result

Here,$x^3-3$ has no roots in${\mathrm{\mathbb{Z}}}_7$ and thus, factorization does not contain the linear factors contradiction.

Therefore, $x^3-3$is irreducible.

Hence, it is proved.