Berlekamp's Algorithm and Sage

We want to factor $f(x)=x^8+x^6+x^4+x^3+1$ over $\mathbb{F}_{2}$ using Berlekamp's algorithm. You can check up on the algorithm here: http://www.johnkerl.org/doc/iw2009/berlekamp.pdf It is easy to see that $f'(x) = x^2$ does not divide $f(x)$, but one can check this using the euclidean algorithm. In sage one checks this as follows:

 S.<x> = PolynomialRing(GF(2),'x')  
 f = x**8 + x**6 + x**4 + x**3 + 1; g = x**2  
 f.gcd(g)  

which will result in 1 when it is run. Next we compute $x^{iq}$ (mod $f(x)$) for $q=2, 0 <= i <= 7$. By using the Euclidean algorithm we get: $$x^0 \equiv 1$$ $$x^2 \equiv x^2$$ $$x^4 \equiv x^{4}$$ $$x^6 \equiv x^{6}$$ $$x^8 \equiv 1+x^3+x^4+x^6$$ $$x^{10} \equiv 1+x^2+x^4+x^5+x^6+x^7$$ $$x^{12} \equiv x^2+x^4+x^5+x^6+x^7$$ $$x^{14} \equiv 1+x+x^3+x^4+x^5$$ This gives us the matrix (now using sage again):

   
   
 [1 0 0 0 0 0 0 0]  
 [0 0 1 0 0 0 0 0]  
 [0 0 0 0 1 0 0 0]  
 [0 0 0 0 0 0 1 0]  
 [1 0 0 1 1 0 1 0]  
 [1 0 1 1 1 1 0 0]  
 [0 0 1 0 1 1 1 1]  
 [1 1 0 1 1 1 0 0]  
   
 

Then subtracting the identity matrix we get:

   
   
 [0 0 0 0 0 0 0 0]  
 [0 1 1 0 0 0 0 0]  
 [0 0 1 0 1 0 0 0]  
 [0 0 0 1 0 0 1 0]  
 [1 0 0 1 0 0 1 0]  
 [1 0 1 1 1 0 0 0]  
 [0 0 1 0 1 1 0 1]  
 [1 1 0 1 1 1 0 1]  
   

We can now use sage to find the basis of the null space of the above matrix as follows:

 M = MatrixSpace(GF(2), 8, 8)  
 A = M([0, 0, 0, 0, 0, 0, 0, 0,   
       0, 1, 1, 0, 0, 0, 0, 0,  
       0, 0, 1, 0, 1, 0, 0, 0,  
       0, 0, 0, 1, 0, 0, 1, 0,  
       1, 0, 0, 1, 0, 0, 1, 0,  
       1, 0, 1, 1, 1, 0, 0, 0,  
       0, 0, 1, 0, 1, 1, 0, 1,  
       1, 1, 0, 1, 1, 1, 0, 1])  
         
 A.kernel()  

Which will give the following result:

 Vector space of degree 8 and dimension 2 over Finite Field of size 2  
 Basis matrix:  
 [1 0 0 0 0 0 0 0]  
 [0 1 1 0 0 1 1 1]  
   

These correspond to the polynomials $g_{1}(x)=1$ and $g_{2}(x)=x+x^2+x^5+x^6+x^7$. Again using Euclidean algorithm we get $gcd(f(x),g_{2}(x))=x^6+x^5+x^4+x+1$ and $gcd(f(x),g_{2}(x)-1)=x^2+x+1$. Then, over $\mathbb{F_{2}}$, the canonical factorization is: $$f(x)=(x^6+x^5+x^4+x+1)(x^2+x+1)$$