Question

Prove that${\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$ is irreducible in ${\mathit{\mathbb{Z}}}_{\mathbf{2}}\left[x\right]$. [Hint: Exercise 2 and Theorem 4.16]

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

Theorem 4.16: Let F be a field and let$f\left(x\right)\in F\left[x\right]$ and $a\in F$. Then a is the root of the polynomial if and only if$x-a$ is a factor of$f\left(x\right)$ in $F\left[x\right]$.

Proof

Here,${\mathbb{Z}}_2$ is a field.

And $x^4+x+1\in {\mathbb{Z}}_2\left[x\right]$.

Now, the only irreducible quadratic in${\mathbb{Z}}_2\left[x\right]$ is $x^2+x+1$(As proved in question 2).

Also, any in${\mathbb{Z}}_2$ is not a root of the polynomial.

Thus, according to the above theorem, there is no factor of polynomial in ${\mathbb{Z}}_2\left[x\right]$.

Since, there is no other factor, hence the polynomial is irreducible in ${\mathbb{Z}}_2\left[x\right]$.