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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?

Sil
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Mike
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2 Answers2

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

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:

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

David Holden
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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.

David Holden
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  • But how do the other roots relate to the the root of $x^4 + x + 1$? – Mike Dec 19 '19 at 03:36
  • Is it maybe because all the roots are contained in $\mathbb{F}{16}$ and we know they are distinct? Specifically, each of the irreducible polynomials splitting field is contained in $\mathbb{F}{16}$ and by the separability of the product of the polynomials we know that they comprise 16 distinct elements. – Mike Dec 19 '19 at 03:39
  • i've added a bit to my auxiliary answer to expand your point – David Holden Dec 19 '19 at 04:31
  • Ok, but is the reason that we can say that the roots of each of these polynomials are the ones you listed above is for the reason I said before? – Mike Dec 19 '19 at 04:35