Is $x^4+x^2+1$ an irreducible polynomial over $\Bbb Z/2\Bbb Z$?

Question

Is $x^4+x^2+1$ an irreducible polynomial over $\Bbb Z/2\Bbb Z$?

According to this site the answer is no. But I can't find the factors. Can you?

Answer 1

Over any ring:

$$x^4+x^2+1=(x^4+2x^2+1)-x^2=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)$$

In the case of $\mathbb Z/2\mathbb Z$, $2x^2=0,$ and $x^2-x+1=x^2+x+1,$ but this polynomial is reducible in any ring.

Another way to see this is that:

$$x^4+x^2+1 = \frac{x^6-1}{x^2-1}$$

And $x^6-1$ factors as $(x^3-1)(x^3+1)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$.

Alternatively, over $\mathbb Z/2\mathbb Z$, $(a+b)^2=a^2+b^2$, so any polynomial over $\mathbb Z_2$ with only even exponents is a square. For example:

$$x^2+1=(x+1)^2\\ x^6+x^4+1=(x^3+x^2+1)^2\\ x^{12}+x^4+1=(x^6+x^2+1)^2$$

Answer 2

$(x^{2}+x+1)^{2}=x^{4}+2\cdot x^{3}+3\cdot x^{2}+2\cdot x+1=x^{4}+x^{2}+1$ in $\mathbb{Z}_{2}$.