2

As the title says: I want t prove or disprove the irreducibility of $X^4+1$ in $\frac{\mathbb{Z}}{p\mathbb{Z}}[X]$, $ p$ prime.

I have already proven that $X^4+1$ can't have linear factors in $\frac{\mathbb{Z}}{p\mathbb{Z}}[X]$, since it doesn't have roots in $\mathbb{Z}.$ By contradiction, such a root $k$ with $k^4=-1$ would have order $8$ which can't be since $8 \not| p$.

But that still leaves the possibility of a nontrivial decomposition with quadratic factors. So say $X^4+1 = (a_1x^2+q_1)(a_2X^2+q_2) \implies a_1a_2 = 1, a_1q_2+a_2q_1 = 0, q_1q_2 =0$. My hunch is that there is a contradiction somewhere in this equations, but I can't find it. Maybe there is none?

ghthorpe
  • 1,507
  • 1
    $X^4 + 1 = (X + 1)^4$ over $\Bbb Z/(2)$ – Edward Evans Dec 22 '17 at 14:28
  • 1
    $x^4+1$ is a square over $\mathbb{F}_2$, and for any odd prime $p$, the degree of the splitting field of $x^4+1=\Phi_8(x)$ over $\mathbb{F}_p$ is given by the least $k$ such that $8\mid (p^k-1)$. It follows that $x^4+1$ completely factors over $\mathbb{F}_p$ if $p\equiv 1\pmod{8}$ and factors as the product of two quadratic irreducible polynomials otherwise; it is never irreducible over $\mathbb{F}_p$, despite being irreducible over $\mathbb{Q}$. – Jack D'Aurizio Dec 22 '17 at 14:32

1 Answers1

3

For $p=2$ we have "freshman's dream" $$ x^4+1=(x+1)^4, $$ so it is certainly not irreducible. For $p=3$ we have $$ x^4+1=(x^2 + 2x + 2)(x^2 + x + 2). $$ In general see this duplicate:

Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime

Dietrich Burde
  • 130,978