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First of all, sorry for my bad English.

I am looking for a general method to know for which primes $p$ a specific polynomial factorize in $Z_p[x]$.

For example my teacher considers $(x^3-1)=(x-1)(x^2+x+1)$ and wants to factorize $x^2+x+1$. He considers $p=5$ and replaces all elements of $Z_5$ instead of $x$; $x^2+x+1$ is never zero, so does not factorize. For $p=13$ he obtains zero, so in $Z_{13}$ the polynomial factorizes. Then he proves that it is possible to factorize $x^2+x+1$ in $Z_p$ if and only if $p \equiv 1 \pmod{3}$. In the first direction of the proof (polynomial factorize in $Z_p$ implies $p \equiv 1 \pmod{3}$) he takes a $z$ in $Z_p$ that is a zero for $x^2+x+1$ and implies that $z^3 \equiv 1 \pmod{p}$. Why?

If I have a different polynomial in $Z_p$ there is a general method to determine the equivalence class of $p$ like in the example? It maybe involves the polynomial's degree?

Thanks for your help!

Jean Marie
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Pterodattilo Frigidaire
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  • Is the problem that you don't understand how $z^3 \equiv 1 \pmod p$ is obtained? – Stefan4024 Jan 28 '17 at 19:02
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    We have $x^3-1=(x^2+x+1)(x-1)=0$, so $x^3=1$ in $\mathbb{F}_p$ for this $z$. – Dietrich Burde Jan 28 '17 at 19:06
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    To learn about algorithms see the Wikipedia article Factorization of polynomials over finite fields, and see any book on computer algebra for details, e.g. Kaltofen, Zippel, etc. – Bill Dubuque Jan 28 '17 at 19:08
  • I once posted a toy example of a run of Cantor-Zassenhaus algorithm. The goal there was to find zeros. Finding higher degree factors can also be done, but judging from your post this may be a bit advaced for you at the moment (try those again a bit later!). You can handle cubics over small fields by lookig for zeros, and quadratics modulo a larger prime using the law of quadratic reciprocity. With other specific polynomials I usually look for a trick first (like those on cyclotomic polynomials). If that fails I fire up my copy of Mathematica. – Jyrki Lahtonen Jan 28 '17 at 21:32

2 Answers2

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In general, the Berlekamp algorithm can give us a factorization of a given polynomial in $\mathbb{F}_p[x]$ for a prime $p$. If you want results for all $p$, there is no easy general congruence condition in general. For cyclotomic polynomials however see the discussion in this question. Sometimes we obtain easy conditions, e.g., the polynomial $x^4+1$ is for no prime $p$ irreducible over $\mathbb{F}_p$, see here.

Community
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Dietrich Burde
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  • Thank you. So he determines that the polynomial factorizes only for p=1(mod3) using the algorithm? For example if the polynomial is x^2+1 I think that p must be p=1(mod4), but I only understood that using numers, do I have to prove the double implication every time? Ps: i only need to know if the polynomial factorize, don't need to know the factors. – Pterodattilo Frigidaire Jan 28 '17 at 19:19
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There is a general method for prime $p$ and $$ a x^2 + bx + c $$ given by repeating the derivation of the Quadratic Formula. For primes $p \neq 2$ and $$ b^2 - 4 a c \neq 0 \pmod p, $$ we can factor when there is an ordinary integer $t$ such that $$ t^2 \equiv b^2 - 4 a c \pmod p. $$ Notice that, if $p | a,$ we may then take $t=b,$ but the thing is no longer quadratic anyway.

I would guess you will learn the Legendre symbol and quadratic reciprocity next.

After quadratics, it gets very hard, even for cubic polynomials. Unless you want a computer to do the factoring for you, which will work for one prime at a time, you may be dealing with class field theory, advanced material. For example $$ x^3 - x + 1 $$ is irreducible for primes $$ p = 2 u^2 + uv + 3 v^2, $$ for some integers $u,v.$ I don't recall what happens when $p=23.$ The cubic factors into three distinct linear factors for $$ p = u^2 + uv + 6 v^2. $$ Then $ x^3 - x + 1 $ factors as a linear times a quadratic when Legendre symbol $(-23|p) = -1.$

There are just a few examples of cubics where one may use reciprocity laws without resorting to class field theory. Ireland and Rosen do this one, proved by Gauss: $$ x^3 - 2 $$ is irreducible for primes $$ p = 4 u^2 + 2uv + 7 v^2, $$ for some integers $u,v.$ The cubic factors into three distinct linear factors for $$ p = u^2 + 27 v^2. $$ Then $ x^3 - 2 $ factors as a linear times a quadratic when Legendre symbol $(-3|p) = -1.$

Jacobi proved: $$ x^3 - 3 $$ is irreducible for primes $$ p = 7 u^2 + 3uv + 9 v^2, $$ for some integers $u,v.$ The cubic factors into three distinct linear factors for $$ p = u^2 + uv + 61 v^2. $$ Then $ x^3 - 3 $ factors as a linear times a quadratic when Legendre symbol $(-3|p) = -1.$

Will Jagy
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