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

I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$.

Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$?

Has $f$ a zero in $\mathbb F_p$ ?

• Is that some $\,p\,$ or it can be any prime $\,p\,$ ? – DonAntonio Jun 23 '13 at 10:39
• just $p \neq 2$. – user42761 Jun 23 '13 at 10:51
• Then I'm not sure what exactly is the question here: the equation $\, x^8=1\,$ has always solution in $\,\Bbb F_{p^2}^*\,$ since this last is an abelian group of even order. Are you arguing then that $\,x^4+1\,$ is always reducible modulo any odd prime $\,p\,$ ? – DonAntonio Jun 23 '13 at 10:58
• Yes, thats what the question states. – user42761 Jun 23 '13 at 11:00
• So $X^4+1$ is irreducible in $\mathbb F_3$ ? – user42761 Jun 23 '13 at 11:08

I think I know, finally!, what you meant with this exercise. Be sure you can follow and prove all the following:

Claim: $\,x^4+1\in\Bbb F_p[x]\,$ is reducible for any prime $\,p\,$

Proof : For $\,p=2\,$ the claim follows from $\,x^4+1=(x+1)^4\bmod 2\,$ , so let us suppose $\,p\,$ is odd.

Note that for any such prime, $\,p^2-1=0\bmod 8\,$ (why?) , so that the cyclic multiplicative group $\,\Bbb F_{p^2}^*\,$ has a cyclic subgroup of order $\;8\;$ , and from here that you have an element of order $\,8\,$ in $\,\Bbb F_{p^2}^*\;$ (and not merely an element s.t. $\,\alpha^8=1\,$ , which is completely trivial). Since

$$x^8-1=(x^4+1)(x^4-1)$$

over any field, all the roots of the right hand side are contained in $\;\Bbb F_{p^2}^*\;$ , and since this is an extension of degree two over the prime field $\,\Bbb F_p\,$ (and observe that $\,x^4+1\,$ is always a polynomial over this prime field!), it cannot be this polynomial is irreducible over the prime field since then any of its roots would form an extension of $\,\Bbb F_p\,$ of degree $\;4\;$ , which is absurd. $\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$

• Hmmm...would it be too much to ask for a little explanation about the downvote here? Thanks. – DonAntonio Jun 23 '13 at 11:46
• We have that $\alpha$ is a zero of $X^4+1$. If $X^4+1$ would be irreducible it would be the minimum polynomial of $\alpha$ over $\mathbb F_p$ and thus $[\mathbb F_p(\alpha) : \mathbb F_p] = 4$ which cannot be true since $\mathbb F_{p^2} : \mathbb F_p(\alpha)$ is a field extension and thus this degree must be $\leq 2$. – user42761 Jun 23 '13 at 11:57
• Exactly @Andre. I wonder whether the downvoter thought something of what I wrote is wrong or else he was upset for seeing the answer...oh, well. – DonAntonio Jun 23 '13 at 12:03
• I don't know. Your answer helped me the most. Thanks. One remark: Is it the true $\alpha^4 = -1$ since $(\alpha^4)^2 = 1$ and $\alpha^4 \neq 1$ ? – user42761 Jun 23 '13 at 12:05
• Just for clarity. Thanks :) – user42761 Jun 23 '13 at 12:17

We can use three different factorizations:

If $2$ is a square mod $p$. (which occurs for $p=\pm1$ mod $8$)

$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$ Assume $q^2=2$ mod $p$. Then $x^4+1=(x^2+1+qx)(x^2+1-qx)$.

If $-2$ is a square mod $8$. (which occurs for $p=1$ or $3$ mod $8$)

$x^4+1=x^4-2x^2+1+2x^2=(x^2-1)^2-(-2)x^2=(x^2-1+rx)(x^2-1-rx)$, where $r^2=-2$ mod $8$.

Finally, if $-1$ is a square mod $8$. (which happens for $p=1$ mod $4$)

$x^4+1=x^4-(-1)=(x^2+r)(x^2-r)$, where $r^2=-1$ mod $p$.

Checking if all the cases are covered.

All (odd) primes are congruent to either $1,3,5,7$ mod $8$. The cases $1$ and $7$ are covered by the first factorization. The case $3$ by the second. The case $5$ by the third, since $p=5$ mod $8$ implies $p=1$ mod $4$.

• We need $\left(\frac2p\right)=\left(\frac{-1}p\right)=1$. But,$$\left(\frac2p\right)=1\iff p\equiv1,7\pmod 8\text{ and }\left(\frac{-1}p\right)=1\iff p\equiv1\pmod 4$$ (mathreference.com/num-mod,qr12.html), but what about $p\equiv3,5\pmod 8?$ – lab bhattacharjee Jun 23 '13 at 11:08
• $p=5$ mod $8$ is inside $p=1$ mod $4$. The case $p=3$ mod $8$ is covered by the factorization that uses that $-2$ is a square mod $p$. – OR. Jun 23 '13 at 11:13
• @frankin, got you point. We need $\left(\frac2p\right)=1\text{ or } \left(\frac{-2}p\right)=1$ or both – lab bhattacharjee Jun 23 '13 at 11:17
• +1. We don't even need to know specifically for which primes $2, -2, -1$ are squares. If neither $2$ nor $-1$ is a square, their product certainly is (uses only that the multiplicative group is cyclic). The claim doesn't even require $p\ge 3$ as in fact $X^4+1=(X^2+1)^2$ over $\mathbb F_2$. – Hagen von Eitzen Jun 23 '13 at 11:27