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reducible polynomial modulo every prime

Ok so we have to prove the following: If $R=\mathbb{Z}_p$ for $p$ a prime, then, $x^4+1$ is reducible over $R[x]$. This is are my ideas (please let me know if there is an easier way to tackle this problem):

First if $p=2$ then we know that $x+1$ is a factor since $1^4+1=0$ and it follows that $x-1=x+1$ is a factor.

Hence from now on let us assume $p$ is odd. First if $x^4+1=(x^2+ax+b)(x^2+cx+d)$, I have that the leading coefficients in the factors is $1$ since I can always force this, and now I obtain the following:

$bd=1$

$ad+bc=0$

$d+ac+b=0$

$a+c=0$

Where the above are obtained by multiplying out the factors and setting the coefficients for the value they ought to be.

However, we can use the last equation and plug in the second equation, to obtain $a(b-d)=0$ and since $\mathbb{Z}_p$ is a field, then we know it has no zero divisors.

If $a=0$ then we have that $b=-d$ and hence $-d^2=1$, which means that $d^2=-1$. By using quadratic residues (here is where I worry since my class has not cover that) we have that if $p\equiv 1\pmod{4}$ then we know that (The Legendre Symbol) $(\frac{a}{p})=\bar{a}^{(p-1)/2}$, using $a=-1$, we have that $(\frac{a}{p})=1$, so $-1$ is not a quadratic residue, i.e., it is a square., hence

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

I am stuck in the case when $p\equiv 3\pmod{4}$. Any help, comments, hints, or thoughts would be appreciated!

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marked as duplicate by Jyrki Lahtonen, Nate Eldredge, JavaMan, Gerry Myerson, Asaf Karagila Feb 8 '12 at 11:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It would be best if you specify that $\mathbb{Z}_p$ stands for the integers modulo $p$. Be aware that, in this context, it is very (more?) common for $\mathbb{Z}_p$ to represent the $p$-adic integers, a completely different animal. $\mathbb{F}_p$ is more common for the field with $p$ elements. –  Arturo Magidin Feb 7 '12 at 15:53
    
Sorry. Im new here. I should have written $\mathbb{Z}/p\mathbb{Z}$. –  Daniel Montealegre Feb 7 '12 at 18:32
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No worries; $\mathbb{Z}_p$ is common shorthand for the field of $p$ elements, but it drives number theorists up a wall. (For that matter, a lot of group theorists object to its being used to denote the cyclic group of order $p$) –  Arturo Magidin Feb 7 '12 at 19:50
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I hate to spoil you all the fun, but this question is a dup. –  Jyrki Lahtonen Feb 7 '12 at 20:07

2 Answers 2

up vote 1 down vote accepted

If $p \equiv 1 \pmod{4}$, we have the factorization $$x^4 + 1 = (x^2 + i)(x^2 - i),$$ where $i \in \mathbb{F}_p$ is the element such that $i^2 = -1$.


When $p \equiv 3 \pmod{4}$, if $2$ is a square $\bmod{p}$ (say $a^2 = 2$), then

$$ x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1). $$

If $2$ is not a square, then $-2$ is a square $\bmod{p}$ (see my answer here for a proof). If, say $b^2 = -2$, then

$$ x^4 + 1 = (x^2 +bx -1)(x^2 -bx -1). $$

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Thanks! by the way, there is a small typo on your solution on the link. It says "-x is a quadratic nonresidue" It should read is a quadratic residue. –  Daniel Montealegre Feb 7 '12 at 7:53
    
@DanielMontealegre: In which line is the typo? –  JavaMan Feb 7 '12 at 8:09
    
On the other hand, suppose that p≡3(mod4) and −x is a quadratic nonresidue. Again, by definition, this implies that there exists an integer b such that b2=−x. However, this implies that –  Daniel Montealegre Feb 7 '12 at 18:30
    
@Daniel: Thanks for pointing out the typo. I have now fixed the proof. –  JavaMan Feb 7 '12 at 21:26

For any odd prime $p$, $p^2-1$ is divisible by $8$. Since its multiplicative group is cyclic, the finite field $\mathrm{GF}(p^2)$ contains a primitive eighth root of unity. Such an eighth root of unity satisfies $x^4+1=0$, so $x^4+1$ must have a quadratic or linear factor.

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