Are quartic minimal polynomials over $\mathbb{Q}$ always reducible over $\mathbb{F}_p$?

This situation arose while studying biquadratic extensions.

Let $\mathbb{Q}(\alpha)$ is some biquadratic extension, with $m(x)$ the minimal polynomial of $\alpha$. Suppose that $m(x)\in\mathbb{Z}[x]$. Basic field theory and simple definitions tells us that $\deg(m(x))=4$, and $m(x)$ is irreducible over $\mathbb{Q}$. However, is $m(x)$ always reducible over $\mathbb{F}_p$, for $p$ a prime?

The reason I'm interested is that I noticed that the quartic $x^4+1$, known to be irreducible over $\mathbb{Q}$ is actually reducible over $\mathbb{F}_p$ for the first few primes at least. For example, $x^4+1$ can be reduced as $(x+1)^4\pmod{2}$, $(x^2+x-1)(x^2-x-1)\pmod{3}$, $(x^2+2)(x^2-2)\pmod{5}$, $(x^2+3x+1)(x^2-3x+1)\pmod{7}$.

So I'm wondering if $m(x)$ is the minimal polynomial of some algebraic integer of degree $4$, is it actually reducible over $\mathbb{F}_p$ for any prime?

An elementary argument, without any explicit reference to Galois theory:

Suppose the biquadratic extension is $K = \mathbb{Q}(\sqrt{B}, \sqrt{C})$, where $B$ and $C$ are integers (and none of $B$, $C$, $BC$ a rational square). Then any element $\alpha$ of $K$ has the form

$\alpha = a + b\sqrt{B} + c\sqrt{C} + d \sqrt{BC}$

for some $a, b, c, d \in \mathbb{Q}$. This $\alpha$ satisfies the polynomial

$$\begin{align*}f(X) =& (X - a - b\sqrt{B} - c\sqrt{C} - d\sqrt{BC})\cdot(X - a + b\sqrt{B} + c\sqrt{C} - d\sqrt{BC})\\ &\qquad{}\cdot (X - a + b\sqrt{B} - c\sqrt{C} + d\sqrt{BC})\cdot(X - a - b\sqrt{B} + c\sqrt{C} + d\sqrt{BC}).\end{align*}$$

Convince yourself that $f(X)$ is in $\mathbb{Q}[X]$. (This is where a bit of Galois theory helps.) If $\alpha$ is integral over $\mathbb Q$, then in fact $f(X) \in \mathbb{Z}[X]$. If $\alpha$ generates $K$ over $\mathbb{Q}$, then $f(X)$ is irreducible and is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. We want to show that if $\alpha$ is integral then $f(X)$ will always factor in $\mathbb{F}_p[X]$.

The key observation is that at least one of $B$, $C$, and $BC$ will be a square modulo $p$. Convince yourself that this is true: the product of two nonsquares modulo $p$ is a square modulo $p$. (One way to do this is note that a nonsquare has to be an odd power of a generator of $\mathbb{F}_p^\times$.)

So let's suppose that $BC \equiv S^2 \pmod{p}$. (The argument will work essentially the same way if we assume that $B$ or $C$ is a square modulo $p$ instead; convince yourself of this at the end.)

So we have

$\begin{align*}f(X) =& \big((X - a - d\sqrt{BC}) - (b\sqrt{B} + c\sqrt{C})\big)\big((X - a - d\sqrt{BC}) + (b\sqrt{B} + c\sqrt{C})\big)\\ &\qquad {} \cdot \big((X - a + d\sqrt{BC}) + (b\sqrt{B} - c\sqrt{C})\big)\big((X - a + d\sqrt{BC}) - (b\sqrt{B} - c\sqrt{C})\big)\\ =& \big((X - a - d\sqrt{BC})^2 - (b\sqrt{B} + c\sqrt{C})^2\big)\\ &\qquad {} \cdot \big((X - a + d\sqrt{BC})^2 - (b\sqrt{B} - c\sqrt{C})^2\big)\\ =& \big((X - a)^2 + d^2BC - 2d(X-a)\sqrt{BC} - b^2B - c^2C - 2bc \sqrt{BC}\big)\\ &\qquad {} \cdot \big((X - a)^2 + d^2BC + 2d(X-a)\sqrt{BC} - b^2B - c^2C + 2bc \sqrt{BC}\big)\\ \equiv& \big((X - a)^2 + d^2BC - 2d(X-a)S - b^2B - c^2C - 2bc S\big)\\ &\qquad {} \cdot \big((X - a)^2 + d^2BC + 2d(X-a)S - b^2B - c^2C + 2bcS\big) \pmod{p},\end{align*}$ which is a product of two quadratics in $\mathbb{F}_p[X]$.

In the case where all three of $B$, $C$, $BC$ are squares modulo $p$, the quadratic factors of $f(X)$ will split further into linear factors in $\mathbb{F}_p[X]$.

As to your other question -- or maybe it was just a typo, but in any case -- no, it's not true that every irreducible monic quartic polynomial in $\mathbb{Z}[X]$ factors over every $\mathbb{F}_p$. For example, $g(X) = X^4 + X + 1$ is irreducible in $\mathbb{F}_2[X]$. To see this, first plug $X= 0$ and $X =1$ into $g(X)$ to see that there are no mod $2$ roots. To see that $g(X)$ doesn't factor into a product of two irreducible quadratics, note that $X^2 + X + 1$ is the only monic irreducible quadratic of $\mathbb{F}_2$ (write them all down and check for roots), and $(X^2 + X + 1)^2 = X^4 + 2X^3 + 3X^2 + 2X + 1 \equiv X^4 + X^2 + 1 \not\equiv g(X)$. Since $g(X)$ is not divisible by any irreducible linears or quadratics, it has to be irreducible.

In fact, you can show that every $\mathbb{F}_p[X]$ has irreducible polynomials of every degree.

You may find of interest the paper below, which refers to the pertinent theory and also presents an algorithm to compute the splitting type of quartics over $\:\mathbb F_p\:.\:$

Adams, William W. Splitting of Quartic Polynomials, Math. Comp. 43 167 (1984) 329-343.