Question

(a) By counting products of the form (x+a) (x+b) show that there are exactly $\left(p^2+p\right)/2$monic polynomials of degree 2 that are not irreducible in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

(b) Show that there are exactly$\left(p^2-p\right)/2$ monic irreducible polynomials of

degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

Short answer

1. $\frac{p^2-p}{2}+p=\frac{p^2+p}{2}$is the reducible monic polynomial of degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.
2. $p^2-\frac{p^2+p}{2}=\frac{p^2-p}{2}$is the irreducible monic polynomial of degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

Step-by-step solution

Obtain p2+p/2monic polynomials of degree 2 that are not irreducible in ℤpx

Let’s consider that, a=b.

Here, using unique factorization, the exact polynomial in the form of ${\left(x+a\right)}^2=x^2+2ax+a^2$.

By commutatively considering, if $a\ne b$then, $\left(x+a\right)\left(x+b\right)=\left(x+b\right)\left(x+a\right)$.

The polynomials in the form of $\left(\frac{p}{2}\right)=\frac{p^2-p}{2}$, adding these up would be$\frac{p^2-p}{2}+p=\frac{p^2+p}{2}$

is the reducible monic polynomial of degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

Obtain p2-p/2 monic polynomials of degree 2 that are not irreducible in ℤpx 

$x^2+a_{1}x+a_0$is the form of monic polynomials of degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

As polynomials are either reducible or irreducible, it can’t be both so there are,

$p^2-\frac{p^2+p}{2}=\frac{p^2-p}{2}$is the irreducible monic polynomial of degree 2 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.