Is $x^4 + 4$ irreducible in $\mathbb{Z}_5$?

Question

Well, I'm having doubts, isnt that $\mathbb{Z}_5$ has no zero divisors, and now you cant factor $x^4 + 4$ ?

Answer 1

Obviously $x-1$ is a factor since $1^4+4=0$.

Answer 2

\begin{align} x^4+4&=x^4-1\\ &=(x^2-1)(x^2+1)\\ &=(x-1)(x+1)(x^2-4)\\ &=(x-1)(x+1)(x+2)(x-2)\\ &=(x-1)(x-2)(x-3)(x-4). \end{align}

Answer 3

You've already been answered, but observe it is easier than it appears on this case since $\;4=-1\pmod 5\;$ , so doing arithmetic modulo $\;5\;$ all along we get

$$x^4+4=x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x-2)(x-3)$$

The last step follows easily if we know that $\;-1\;$ is a quadratic residue as $\;5=1\pmod4\;$, for example. Or else directly trying the well known formula for the roots of a quadratic.

Answer 4

Note that $x^4+4$ is not irreducible over any field, since $$x^4+4=(x^2+2x+2)(x^2-2x+2).$$