Question

Show that the only irreducible quadratic in ${\mathit{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$is${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$. [Hint: List all the quadratics and use Corollary 4.19]

Short answer

It has been proved that the only irreducible quadratic in${\mathbb{Z}}_2\left[x\right]$ is $x^2+x+1$

Step-by-step solution

Theorem used

Corollary 4.19: Let F be a field and let$f\left(x\right)\in F\left[x\right]$ be a polynomial of degree 2 or 3. Then$f\left(x\right)$ is irreducible in$F\left[x\right]$ if and only if $f\left(x\right)$has no roots in F.

Proof

In ${\mathbb{Z}}_2\left[x\right]$the polynomials with degree 2 are:

$x^2,x^2+1,x^2+x+1$.

Now,$x^2=x\left(x\right)$

$x^2+1={\left(x+1\right)}^2$in ${\mathbb{Z}}_2\left[x\right]$

.$x^2+x=x\left(x+1\right)$

These three polynomials have a root in ${\mathbb{Z}}_2\left[x\right]$.

But$x^2+x+1$ has no root in ${\mathbb{Z}}_2\left[x\right]$.

Thus, it is the only irreducible quadratic in ${\mathbb{Z}}_2\left[x\right]$.