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Consider the following algorithm from page 240 of this pdf:

Irreducibility-Test(f)
1 $n ← \deg(f)$
2 if $X^{p^n} \not\equiv X (\mod f)$
3 $\quad$ then return "no"
4 for the prime divisors $r$ of $n$
5 $\quad$ do if $X^{p^{n/r}} \equiv X (\mod f)$
6 $\quad\quad$ then return "no"
7 return "yes"

Now, this algorithm basically checks if $\deg(f)$ is the smallest positive integer $d$ such that $ f \,\mid\, X^{p^d} - X$. The author says, that this implies irreducibility of $f$ and the algorithm relies on this (lines 4 to 7).

However, the assumed implication isn't right, see the answer to my question.

Is this algorithm simply wrong, or did I get something mixed up? If it is wrong, (how) can it be fixed? If it is correct, where is my misunderstanding?

polynomial_donut
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    The linked source has non-congruence in line 2 (which makes more sense). – Andreas Blass Nov 21 '18 at 15:12
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    Perhaps replacing line 5 by "do if $gcd(x^{p^{n/r}}-x,f)\neq 1$" would suffice ? Just a silly thought... – user120527 Nov 21 '18 at 15:22
  • @user120527 I think I don't completely get the reasoning behind your suggestion (completely on me, probably). From my understanding, lines 4-6 work to sort out reducible polynomials, but not necessarily all of them - with your line change, irreducible polynomials still 'reach the end', but I don't quite see how all reducible ones would be sorted out. – polynomial_donut Nov 21 '18 at 19:05
  • This algorithm gives the wrong answer if and only if $f$ is a square-free product of lower degree irreducible polynomials $f(x)=p_1(x)p_2(x)\cdots p_k(x)$, $p_i$ irreducible for all $i$, such that $n$ is the least common multiple of the degree os the factors $p_i(x)$. – Jyrki Lahtonen Nov 22 '18 at 06:28
  • (cont'd) If there are repetitions among the irreducible factors $p_i$, then this will be observed as $f$ fails to be a factor of $x^{p^n}-x$. Given that $f$ is square-free (after passing the test in step 2) $f$ will be a factor of $x^{p^{n/r}}-x$ if and only if all the factors of $f$ have degrees that are factors of $n/r$. But if $n$ is the l.c.m. of the numbers $\deg p_i$, then none of the tests involving $x^{p^{n/r}}-x$ will catch the factors. – Jyrki Lahtonen Nov 22 '18 at 06:34
  • I didn't check the entire passage, but more often than not a factorization algorithm (and also an irreducibility test) of a polynomial in $\Bbb{F}_q[x]$ first verifies that $f$ is square-free (by calculating $\gcd(f,f')$), and then looks for "equal degree parts". After all, $\gcd(f,x^{p^m}-x)$ gives the product of irreducible factors of degrees dividing $m$, so doing this for factors of $m$ will give us that. At that point we can then run Cantor-Zassenhaus to find those factors. – Jyrki Lahtonen Nov 22 '18 at 06:40
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    Anyway, as Eric Wofsey's answer explains, any product of a linear, an irreducible quadratic and an irreducible cubic will pass this test $n=3+2+1=6=lcm{3,2,1}$. Similarly a product of distinct irreducible polynomials of degrees $5,2,2,1$ will pass the test as $5+2+2+1=10=lcm{5,2,2,1}$. I don't see a remedy other than to calculate $\gcd(f,x^{p^{n/r}}-x)$ in step 5 rather than simply check non-divisibility. – Jyrki Lahtonen Nov 22 '18 at 06:45
  • As finding the factorization makes such gcd-calculations necessary anyway, that should not deter us. – Jyrki Lahtonen Nov 22 '18 at 06:46
  • Anyway, calculating $gcd(f,x^{p^{n/r}}-x)$, and observing that result is not equal to $1$ will catch the presence of any factor of degree dividing $n/r$. With $n=6$, $r=2$, you will catch the $3+2+1=6$ case for not being irreducible etc. – Jyrki Lahtonen Nov 22 '18 at 06:56
  • @JyrkiLahtonen Thank you for your comments. I'm a little lazy,
    'But if $n$ is the l.c.m. of the numbers $\deg (p_i)$, then none of the tests involving $X^{p^{n/r}} - X$ will catch the factors.'

    Can you explain this a little?

    Also, I'm not sure I understand why you wrote 'As finding the factorization makes such gcd-calculations necessary anyway, that should not deter us.' (why would I be deterred from what?)

    Only if you have the time...

     – polynomial_donut Nov 22 '18 at 15:40

1 Answers1

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The algorithm should work out with the change that user120527 suggested in a comment on this question.

If $f$ is irreducible, then if we have $f \,\mid\, P_d:=X^{q^d} - X$ for some $d > 0$, we know that $\deg(f) \,\mid\, d$ since $P_d$ is the squarefree product of all irreducible polynomials of $\mathbb{F}_q[X]$ whose degree divides $d$. Hence, $\deg (f) \leq d$.
This implies that $f \,\not\mid\, P_{n/r}$ for all prime factors $r$ of $n = \deg (f)$, because $\deg(f) > \deg(f)/r = n/r$, meaning that $f$ will pass through to the end in the algorithm above (note that $f \,\not\mid\, P_{n/r} \overset{f\, \text{prime}}\iff \gcd\left(f, P_{n/r}) \neq 1 \right)$.

Conversely, if $f$ is reducible, then we have $f = gh$ where $g$ is irreducible and $h$ is nontrivial. Assuming that the algorithm didn't cancel at line 2, we hence know that $g \,\mid\, f \,\mid\, P_{\deg(f)}$, so that $$ \deg (g) \,\mid\, \deg (f) $$ Now, since $h$ is nontrivial we know $\deg (g) < \deg (f)$, so that with the above, there is a prime factor $r$ of $\deg(f)$ such that $\deg (g) \,\mid\, \deg(f)/r = n/r$ and then, $$ g \,\mid\, P_{n/r} \iff \gcd\left(g, P_{n/r}\right) \neq 1 \implies \gcd\left(f, P_{n/r}\right) \neq 1 $$ and $f$ will fail the test in line 5 of the changed algorithm here:

Irreducibility-Test(f)
1 $n ← \deg(f)$
2 if $X^{p^n} \not\equiv X (\mod f)$
3 $\quad$ then return "no"
4 for the prime divisors $r$ of $n$
5 $\quad$ do if $\;\mathbf{\,\textbf{gcd}\left(f,\, X^{p^{n/r}} - X \right) \neq 1}$
6 $\quad\quad$ then return "no"
7 return "yes"

polynomial_donut
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