Repeated Factor of a Polynomial

Question

I am told that the polynomial $x^9+x^8+x^6+x^5+x^4+x^3+1\in \mathbb F_2[x]$ has a repeated factor. I know that there is some theorem relating repeated factors to the derivative of the polynomial, but I'm unable to find the specific result. What is this result I'm looking for? Is it even helpful here? If not, what will be helpful?

Edit 1: I have found the following theorem in a textbook:

Let $f$ and $P$ be polynomials with coefficients in a field $k$. On one hand, if $P^2$ divides $f$, then $P$ divides $\gcd(f,f')$. On the other hand, if the characteristic $p$ of $k$ is positive, if $P$ is irreducible, if every element $a\in k$ has a $p$th root in $k$, and if $P$ divides $\gcd(f,f')$, then $P^2$ divides $f$.

I find that if we set $f(x)=x^9+x^8+x^6+x^5+x^4+x^3+1, \gcd(f,f')=1$ so since we can't have a polynomial dividing this constant, we're at a dead end. Am I misunderstanding this theorem? Or (more likely) have I computed the $\gcd$ incorrectly?

Edit 2: From this post Derivative criterion for repeated factors it must be the case that $f$ is not a product of linear factors, assuming my $\gcd$ is correct.

So how can I search for the factors? Is there an "elegant" way via a quick application of some theorem?

I don’t know, what will be helpful when you want to find a repeated factor of a polynomial? Surely finding a single factor wouldn’t help. — Divide1918, Dec 13 '23 at 09:18
Do you mean a statement like this: "If $(x-a)^k$ is a factor of the polynomial, then $(x-a)^{k-1}$ is a factor of its derivative."? — DominikS, Dec 13 '23 at 09:25
https://math.stackexchange.com/questions/505510/derivative-criterion-for-repeated-factors — Exodd, Dec 13 '23 at 09:38
I am sceptical whether your calculation (which is not visible) of $\gcd(f,f')$ is correct. If (according to the answer) $(x^3+x+1)^2$ is a factor, then $(x^3+x+1)$ should divide $f'$ as well. — DominikS, Dec 13 '23 at 10:47
@DominikS Indeed, I have added the computation of $gcd(p,p')$ to my answer and it is not equal to 1 as said in the question. — Jean Marie, Dec 13 '23 at 12:25

Jean Marie · Accepted Answer · 2023-12-15T17:47:04.503

Edit : The idea of computing $\gcd(p,p')$ with

$$p=x^9+x^8+x^6+x^5+x^4+x^3+1 \ \text{and} \ p'=x^8+x^4+x^2$$

is a good one but the result is not $1$ ; it is

$$\gcd(p,p')=x^6+x^2+1$$

Indeed, applying the Euclidean algorithm based on recursive property :

$$\gcd(a,b)=\gcd(b,r) \ \text{for Euclidean division} \ a=bq+r$$

we get :

$$\gcd(p,p')=\gcd(p',r_1)=\gcd(r_1,r_2)=\gcd(r_2,r_3)$$

with the successive remainders :

$$\begin{cases}r_1&=&x^8+x^6+x^4+1\\r_2&=&x^6+x^2+1\\ r_3&=&x^6+x^2+1\end{cases}$$

The process stops there ("last nonzero remainder" criteria) because the next remainder is $0$.

Let us check it with the following little SAGEmath program :

B.<x>=GF(2)[]
p=x^9+x^8+x^6+x^5+x^4+x^3+1;
q=p.factor();q
#
pp=p.derivative();pp
r=gcd(p,pp);r

giving the following answers :

$$q=(x^3 + x^2 + 1)(x^3 + x + 1)^2$$

$$pp=p'=x^8+x^4+x^2$$

$$r=\gcd(p,p')=x^6 + x^2 + 1=(x^3+x+1)^2$$

Remark : using SAGEmath online is very easy ; just call it in this way :

https://sagecell.sagemath.org/

type your program, then hit the button "Evaluate'.

score 3 · Answer 2 · answered Dec 13 '23 at 15:35

We all do things our own way. This is what I did.

First clearly $f(0)\ne 0$ and $f(1)\ne 0$ so $f$ has no linear factors.

Now a quick calculation gives $$ f'(x)=x^8+x^4+x^2=x^2(x^6+x^2+1)=x^2(x^3+x+1)^2. $$

Hence any common factor (if there is one) involves only powers of the two irreducible factors of $f'$, namely $x$ and $x^3+x+1$. We've already eliminated $x$ as a factor.

Hence if $f$ has a repeated factor it is $f(x)=(x^3+x+1)^2 g(x)$, where $g(x)$ is cubic with no linear factor. Hence $g(x)$ is one of the two possible irreducible cubics, namely $x^3+x+1$ or $x^3+x^2+1$.

It is very easy to multiply out and see that indeed $f(x)=(x^3+x+1)^2 (x^3+x^2+1)$.

Repeated Factor of a Polynomial

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