# Square-free factorization of univariate polynomials over Galois Fields

I am looking at an algorithm in the book, 'Algorithms for Computer Algebra' by Geddes/Czapor/Labahn, and the algorithm on p. 345, 'Finite Field Square-Free Factorization', which square -free factorizes univariate polynomials over Galois Fields, of order '$q$'. Where, '$q$' here is a prime, or a power of a prime, so that $q = p^m$, where '$p$' is prime, and '$m$' is a positive integer.

On page 344, there is a theorem that states that if $a(x) = a_0 + a_1x + \ldots + a_n x^n$ be a polynomial of degree '$n$' in $GF(q)[x]$ satisfying $a'(x) = 0$. Then $a(x) = b(x)^p$, for some polynomial b(x), and that each coefficient of $b(x)$, is given by :-

$$b_i = a_{ip}^{1/p} = a_{ip}^{p^{m-1}}$$

Now, I have myself produced an example, to test this theorem, so I have my $a(x)$ as :-

$$a(x) = 2 x^8 + 4 x^4 + 8 x^2 + 7$$

where here, I have chosen $q = 16, p = 2, m = 4$

Now, $a'(x) = 0$ (Hope you agree with that?)

So, if I am interpreting this theorem correctly, then I make my corresponding coefficients for $b(x)$, as:-

$b_0 = a_0^8 = 7^8 = 1$, $b_1 = a_2^8 = 8^8 = 0$, $b_2 = a_4^8 = 4^8 = 0,$ $b_3 = a_6^8 = 0^8 = 0,$ $b_4 = a_8^8 = 2^8 = 0$

All the other coefficients will be zero, so there's no point going any further - so my final result for $b(x)$ is:- $b(x) = 1$

Yet this cannot possibly be correct, as $b(x)^p$ does not equal $a(x)$.

Does anyone else have this book, or know this theorem. Can you see what's going wrong here please?

Thanks Jeremy

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I took the liberty of TeXifying your post. I'm relatively sure that I got everything correctly, but please check. – Jyrki Lahtonen Dec 14 '13 at 12:29
And the example does work out right. The field $GF(16)$ is of characteristic two, so $2=0$, when they are treated as elements of $GF(16)$. So inside the polynomial ring $GF(16)[x]$ you do have $$a(x)=2x^8+4x^4+8x^2+7=0x^8+0x^4+0x^2+1=1.$$ May be you have a wrong idea of what $GF(16)$ looks like? The bottom half of this answer lists all the (non-zero) elements of $GF(16)$ as well as shows some example of how to do arithmetic in that field. – Jyrki Lahtonen Dec 14 '13 at 12:33
Hi Jyrki, Thanks for the response. I was under the impression that with Galois Fields, the characteristic was the 'q' value, so 16 in this case. But it is the 'p' value you use ie. '2' in this case? – Jeremy Watts Dec 14 '13 at 13:36
That's correct. The characteristic of a field is always a prime number (or zero). The arithmetic operations in $GF(16)$ are different from those of $\Bbb{Z}_{16}$. – Jyrki Lahtonen Dec 14 '13 at 14:00