# Irreducible polynomial which is reducible modulo every prime

How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?

For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$ gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?

• Dec 1, 2016 at 12:58
• The answers show how to prove that $f=x^4+1$ is reducible modulo every prime $p$. If you want to show that $f$ is irreducible in $\mathbb{Z}[x]$, one way is to apply Eisenstein's criterion with $p=2$ to $f(x+1)=(x+1)^4+1=x^4+4x^3+6x^2+4x+2$. Feb 27, 2018 at 19:20
• What happens is that the splitting field of $x^4+1$ is $\Bbb Q(\zeta_8) / \Bbb Q$, which is not a cyclic extension, so there are no inert primes in that extension. Apr 6, 2018 at 13:24
• Note the factorisations $(x^2+i)(x^2-1)$ and $x^4+1=x^4+2x^2+1-2x^2=(x^2+1+x\sqrt 2)(x^2+1-x\sqrt 2)$ and $x^4+1=x^4-2x^2+1+2x^2=(x^2-1+x\sqrt {-2})(x^2-1-x\sqrt{ -2})$ so if your field contains square roots of $-1$ or $\pm 2$ you can realise one of these factorisations. Sometimes a handy perspective if you know or spot a real factorisation and want to find one in a finite field. Oct 19, 2020 at 14:06

If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+1=X^4-a^2=(X^2+a)(X^2-a).$$ If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we have $$X^4+1=(X^2+1)^2-(bX)^2=(X^2+bX+1)(X^2-bX+1)$$ If $p$ is odd and neither $-1$ nor $2$ is a square, then their product $-2$ is a square, say $-2=c^2$. (Without using anything even remotely as deep as quadratic reciprocity, this follows immediately from the fact that $\Bbb F_p^\times$ is a cyclic group of even order). Then we have $$X^4+1=(X^2-1)^2-(cX)^2=(X^2-cX-1)(X^2+cX-1)$$

• Good job! :-) ${}$ Jun 8, 2016 at 18:52
• I feel this is a really simple and beautiful proof. Jan 22, 2017 at 21:22
• Above is the most elegant solution to this problem, at least amongst the ones I know. Feb 17, 2019 at 11:35
• This construction generalizes; for example, the minimal polynomial of $\sqrt2+\sqrt3$ over $\mathbb Q$ (which happens to be $x^4-10x^2+1$) is another irreducible polynomial that is reducible mod every prime $p$. Apr 20 at 3:36

For every odd prime $p$ we have $8\mid p^2-1$. The multiplicative group of the finite field $F=GF(p^2)$ is cyclic of order $p^2-1$. Putting these two bits together tells us that there is a primitive root $u$ of order $8$ in $F$. We must have $u^4=-1$, because $-1$ is the only element of multiplicative order two. Because $F$ is a quadratic extension of $\mathbf{Z}/p\mathbf{Z}$, the minimal polynomial of $u$ is of degree $\le 2$. That minimal polynomial is then a factor of $$x^4+1=(x-u)(x-u^3)(x-u^5)(x-u^7)=(x-u)(x-u^3)(x+u)(x+u^3).$$

====================

Edit: Here's an idea for finding the factorization. I split it into cases according to the residue class of $p$ modulo 8. Assume first that $p\equiv 1\pmod 4$ (or $p$ equivalent to $1$ or $5$ modulo 8). In that case all we need is a square root $i$ of $-1$ modulo $p$. IIRC there is an algorithm for finding two integers $x,y$ such that $p=x^2+y^2$, and then $i=x*y^{-1}$ is the desired square root in the prime field $F_p=GF(p)$. A factorization is then $$x^4+1=(x^2+i)(x^2-i).$$ Observe that if $p\equiv1\pmod8$ then both quadratic factors will split further.

If $p\equiv 3\pmod 8$, then $u$ is not in the prime field, and its conjugate is $u^p=u^3$. Therefore the minimal polynomial is $$m(x)=(x-u)(x-u^p)=(x-u)(x-u^3)=x^2-[u+u^3]x + u^4= x^2-ax-1,$$ where $a$ is some unknown element of the prime field. Because $u^5=-u$ and $u^7=-u^3$, the other factor of $x^4+1$ must be $m(-x)=x^2+ax-1$. We need to find the coefficient $a$. Let's multiply $$(x^2-ax-1)(x^2+ax-1)=(x^2-1)^2-a^2x^2=x^4-(2+a^2)x^2+1.$$ We see that we have found the factorization, if we can find $a=\sqrt{-2}$. It is well known that when $p\equiv 3\pmod 8$, then $-2$ is a quadratic residue modulo $p$ confirming our finding.

In the last case $p\equiv 7\pmod8$ the minimal polynomial of $u$ over $F_p$ is $$m(x)=(x-u)(x-u^p)=(x-u)(x-u^7)=x^2-[u+u^7]x+u^8=x^2-bx+1$$ for some $b\in F_p$. Again the other factor is $m(-x)$, and a similar calculation shows that we need $b=\sqrt{2}$. Again this fits together with the known fact that in this case $2$ is a quadratic residue modulo $p$.

==================

Edit(2): TonyK described the following methods for finding the square roots. They depend on the fact that if $p$ is an odd prime, and $gcd(a,p)=1$, then $a^{(p-1)/2}\equiv\pm1\pmod p$. Here we have the plus sign, if and only if $a$ is a quadratic residue (=QR) modulo $p$.

If $p\equiv 3\pmod8$, then we know that $2$ is not a QR modulo $p$. Therefore $2^{(p-1)/2}\equiv -1\pmod p$. Hence $2^{(p+1)/2}\equiv -2\pmod p$. But here $(p+1)/2$ is an even integer, so writing $z=2^{(p+1)/4}$ we get $z^2\equiv 2^{(p+1)/2}\equiv -2$, and we have found a square root of $-2$.

Similarly, if $p\equiv 7\pmod 8$, we know that $2$ is a quadratic residue modulo $p$. This time $2^{(p+1)/2}\equiv 2$, and the same calculation shows that $z=2^{(p+1)/4}$ is a square root of $2$ in $F_p$.

If $p\equiv 5\pmod 8$, then again $2$ is not a QR modulo $p$, so $2^{(p-1)/2}\equiv -1\pmod p$ and $(p-1)/2$ is even. Thus $z=2^{(p-1)/4}$ is a square root of $-1$. If $p\equiv 1\pmod 8$, then we cannot use $2$ (but could use any non-QR in its place, or the method mentioned earlier).

• why $[F:\mathbf{Z}/p\mathbf{Z}]=2$? Oct 30, 2011 at 12:51
• If $p\equiv 3\pmod 4$, then the square root of $2$ or $-2$ (whichever one exists) is simply $2^{(p+1)/4}$. Oct 30, 2011 at 14:32
• @PraphullaKoushik: Looking at the eighth roots of unity here is a consequence of the fact that $x^4+1\mid x^8-1$. Thus all the zeros of $x^4+1$ have multiplicative order eight. Also, surprisingly many of the questions about finite fields on this site can be resolved by using the cyclicity of their multiplicative groups. That and the fact that the Frobenius automorphism generates the Galois groups. Mar 9, 2014 at 9:12
• So when thinking about a problem related to finite fields one of the first things I look at is whether I can deduce the multiplicative order of the unknown elements (here the zeros of $x^4+1$). Knowing that order is sufficient information in determining the field that the said element generates. The rest is an easy application of Galois theory. Mar 9, 2014 at 9:17
• @PraphullaKoushik: So to get the splitting field we want to adjoin the eighth roots of unity to the prime field (unless they are already there). That will never take higher than a quadratic extension, because $8\mid p^2-1$. If you prefer a Galois theoretic point of view it goes like: over the rationals adjoining the eighth roots (powers of $(1+i)/\sqrt2$) gives us all $i$, $\sqrt2$ and $\sqrt{-2}$. Over $\Bbb{F}_p$ there is only one quadratic extension, so it does not matter whether we need to adjoin one or more of $i$, $\sqrt2$ or $\sqrt{-2}$ we get all three for the price of one. Mar 9, 2014 at 15:02

When $p=2$ then just note $x^4+1=(x+1)^4$.
Now if $p$ is odd then $8\mid p^2-1 \implies x^4+1\mid x^{p^2-1}-1\mid x^{p^2}-x$. Let $a$ be a root of $x^4+1$ in some extension of $\mathbb F_p$. So, $[\mathbb F_p(a):\mathbb F_p]=4$ if we let $x^4+1$ is irreducible over $\mathbb F_p$. But from $x^4+1\mid x^{p^2}-x$ we can say $a\in\mathbb F_{p^2} \implies [\mathbb F_p(a):\mathbb F_p]\leq 2$, a contradiction.

• Why from the divisibility fact you can say that the root $a$ of $x^{4}+1$ is in $\mathbb{F}_{p^2}$? Nov 5, 2016 at 1:03
• What does it mean to be root of $x^{p^2}-x$ ? Nov 5, 2016 at 3:53