10

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 $\mathbb F_p$ ?

user26857
  • 52,094
  • Is that some $,p,$ or it can be any prime $,p,$ ? – DonAntonio Jun 23 '13 at 10:39
  • just $p \neq 2$. –  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. –  Jun 23 '13 at 11:00
  • So $X^4+1$ is irreducible in $\mathbb F_3$ ? –  Jun 23 '13 at 11:08
  • $x^4+1=x^4-2x^2+1-(-2)x^2=(x^2-1)^2-x^2=(x^2-1+x)(x^2-1-x)$ mod $3$. One thing is having a root another is being irreducible. – OR. Jun 23 '13 at 11:10

2 Answers2

27

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. $\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • 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 $\leq 2$. –  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 $\alpha^4 \neq 1$ ? –  Jun 23 '13 at 12:05
  • Yes...why? Because the equation $,x^2=1,$ has only two solutions on any field of characteristic$\neq 2;:;1,,,-1,$ , so if $,x=\alpha^4;$ ,then $,x^2=1;\wedge;x\neq 1\implies x=-1,$ – DonAntonio Jun 23 '13 at 12:16
  • 1
    Just for clarity. Thanks :) –  Jun 23 '13 at 12:17
  • @DonAntonio : I understand this except the idea of choosing $8$ :O could you please explain what made this $8$ to be the obvious choice.... –  Mar 09 '14 at 06:40
  • @PraphullaKoushik, the original question mentions the equation $;x^8-1=0;$ . Since the roots of unity of order $;2n;$ have arather close relation to those of order $;n;$ (can you see it?...:) ), I guess I tried that... – DonAntonio Mar 09 '14 at 11:48
  • @DonAntonio : I remember seeing somewhere relation between roots of unity of order 2n and of order n though could not actually state it... I would think for some more time.. :) If you have a copy of Dummit foote Abstract algebra I would request you to see that once.. It was mentioned with out any reason to consider $x^8-1$ (He might have mentioned but i could not realize it).. Corollary 16 of Chapter 14; Galois theory.. –  Mar 09 '14 at 12:19
  • Oh, no, no @PraphullaKoushik. There is a rather good, explained reason there to do that. Read it...it says that $;p^2-1=0\pmod 8\implies (x^8-1)\mid (x^{p^2-1}-1);$ and from here all rolls down. – DonAntonio Mar 09 '14 at 12:49
  • @DonAntonio : Oh yes yes.. I understand that except that it took so much time for me to realize... :) –  Mar 10 '14 at 02:24
  • This is the most beautiful proof I have read before of this problem. – iam_agf Nov 05 '16 at 01:33
16

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 $4$.

OR.
  • 5,941
  • We need $\left(\frac2p\right)=\left(\frac{-1}p\right)=1$. But,$$\left(\frac2p\right)=1\iff p\equiv1,7\pmod 8\text{ and }\left(\frac{-1}p\right)=1\iff p\equiv1\pmod 4$$ (http://www.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
  • 4
    +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