Product of Irreducible Polynomials

Question

I am trying to show that the product of all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F}_2$ is $x^{16} - x$.

I know the irreducible factors have degree 16. The irreducible factors of degree 4 were found in this post: Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than 5. I notice that the roots of $x^{16} - x$ are a finite field of 16 elements and since each irreducible factor is distinct in the product there are 16 distinct roots of the product of all the polynomials of degrees 1,2 and 4.

How can I make use of the fact that splitting fields are unique (up to isomorphism) to help in this case?

Answer 1

$$ F_{2^4} = A \cup B \cup C $$

where $$ A = F_2 \\ B = F_{2^2} - F_2 \\ C = F_{2^4} - F_{2^2} $$

the two elements in $A$ are roots of two degree-1 polynomials the two elements in $B$ are roots of a single degree-2 polynomial the twelve elements in $C$ must therefore be the roots of three degree-four polynomials. in fact:

$$ x^{2^4} - x = x(x+1)(x^2+x+1)(x^4+x^3+1)(x^4+x^2+1)(x^4+x+1) $$

Answer 2

suppose $a$ is a root of $x^4+x +1 = 0$. then the other roots are the remaining three elements in the orbit of the Frobenius automorphism $x \to x^2$. these are $a^2, a^4 = a + 1$ and $a^8 = a^2 + 1$. you may find it a useful exercise to assign each of the other elements of $F_{2^4} \cong F(a)$ to their appropriate irreducible polynomials.

once you have the element $a$ the extension $F(a)$ contains the sixteen elements $0,1,a,a+1,a^2,a^2+1, a^2+a+1, a^3,a^3+1,a^3+a,a^3+a+1, a^3+a^2,a^3+a^2+1, a^3+a^2+a, a^3+a^2+a+1$. each of these must satisfy a polynomial of degree at most 4 because the dimension of $F_{2^4}$ as a vector space over $F_2$ is 4.