Factoring $X^{16}+X$ over $\mathbb{F}_2$

I just asked wolframalpha to factor $X^{16}+X$ over $\mathbb{F}_2$. The normal factorization is $$ X(X+1)(X^2-X+1)(X^4-X^3+X^2-X+1)(X^8+X^7-X^5-X^4-X^3+X+1) $$ and over $GF(2)$ it is $$ X(X+1)(X^2+X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1). $$ Does the second form follow from the first, or is there a different way to factor over $\mathbb{F}_2$? I noticed that simply replacing the $-$ signs with $+$ signs in the first factorization doesn't yield the second one.

Solution 1:

It shouldn't be a surprise that switching from integer (or rational) coefficients to modular coefficients allows for further factorization. The simplest example is probably the factorization $$ x^2+1=x^2+2x+1=(x+1)^2 $$ over $F_2$.

Here the key difference between the two factorizations is that the polynomial of degree 8 splits into a product of two quartic polynomials over $F_2$: $$(x^4+x+1)(x^4+x^3+1)=x^8+x^7+x^5+x^4+x^3+x+1$$ that is equal to (up to sign changes) your last factor.

Once you start on finite fields in your studies you will immediately learn that all the elements of $GF(16)$ are roots of the polynomial $x^{16}+x$. As that field is a degree 4 extension of $F_2$, all those elements have minimal polynomials of degree a factor of 4. Furthermore, all such irreducible polynomials appear as factors of $x^{16}+x$. Now, we easily see that both $x^4+x+1$ and its reciprocal polynomial $x^4+x^3+1$ are both irreducible in the ring $F_2[x]$, so they must appear as factors.

Solution 2:

The second form does follow from the first. Note that over $\mathbb F_2$, the pair of polynomials $$ X^2 - X + 1 \quad \text{ and } \quad X^2 + X + 1 $$ and the other pair $$ X^4 - X^3 + X^2 - X + 1 \quad \text{ and } \quad X^4 + X^3 + X^2 + X + 1 $$ are the same. The only difference that comes up in $\mathbb F_2$ is that the polynomial of degree $8$ factors.

Hope that helps,