Splitting Field of $x^4 + x^3 + 1$ over $\mathbb{F}_{32}$

abstract-algebrafinite-fieldssplitting-field

I'm trying to find the splitting field described in the title. I believe I have figured it out, but my method seems a bit involved and I'm wondering if there is any simpler way to obtain the result.

My method:

This polynomial is actually contained in $\mathbb{F}_2[x]\subset\mathbb{F}_{32}[x]$ so it should suffice to find the splitting field $K$ of the polynomial over $\mathbb{F}_2$ and get the composite $K\mathbb{F}_{32}$. Note that over $\mathbb{F}_2$ this polynomial is irreducible since it has no roots and the only irreducible polynomial of degree $2$ over $\mathbb{F}_2$ is $x^2 + x + 1$ which does not square to our polynomial. Since Gal($K/\mathbb{F}_2)\leq S_4$, and contains the degree $4$ extension $E = \mathbb{F}_2[x]/(x^4 +x^3 + 1)$, we have that $[K:\mathbb{F}_2]= 4$ or $8$. Over $E$, a tedious computation shows that our polynomial splits as: $$ x^4 + x^3 + 1 = (x+\bar{x})(x+\bar{x}^2)(x^2 + (1 + \bar{x} + \bar{x}^2)x + (\bar{x} + 1)).$$ Now the polynomial $ x^2 + ( 1 + \bar{x} + \bar{x}^2)x + (\bar{x} + 1))$ becomes the irreducible $x^2 +x + 1$ over the quotient $E/(\bar{x})\cong \mathbb{F}_2$, so it must be irreducible in $E[x]$. Thus $E$ is not the splitting field of our polynomial over $\mathbb{F}_2$, implying the splitting field is the unique extension of degree $8$, namely $\mathbb{F}_{2^8}$. Since this field contains $\mathbb{F}_{32}$, we have that this is also the splitting field of this polynomial over $\mathbb{F}_{32}$.

Edit: Woops, $\mathbb{F}_{2^5}$ is definitely not contained in $\mathbb{F}_{2^8}$ so this last part should say the splitting field is $\mathbb{F}_{2^8}\mathbb{F}_{2^5} = \mathbb{F}_{2^{40}}$.

Best Answer

Note that the lattice of fields of the shape $$\Bbb F_{\displaystyle 2^r}$$ corresponds to the lattice of the $r$-values w.r.t. division. The field $$\Bbb F_{32}=\Bbb F_{2^5}$$ intersects (in a common embedding) the fields $\Bbb F_{2^k}$ for $k=1,2,3,4$ only in $\Bbb F_2$, in the prime field.

The polynomial $$ f=X^4+X^3+1\in \Bbb F_2[X]$$ is irreducible.

To see these, note that there is no root of it in $\Bbb F_2$. The only possibility to factor it would be as a product of two irreducible polynomials of degree two. But there is only one such irreducible polynomial, it is reciprocal, $X^2+X+1$, its square is reciprocal, $X^4+X^2+1$, but it is not our polynomial.

Form here, the splitting field of $f$ over $\Bbb F_2$ is $\Bbb F_{2^4}\cong \Bbb F_2[X]/(f)$.

The minimal field containing $\Bbb F_{2^4}$ and $\Bbb F_{2^5}$ is $$\Bbb F_{\displaystyle 2^{4\cdot 5}} = \Bbb F_{\displaystyle 2^{20}} \ , $$
which is the splitting field of $f$ considered as a polynomial over $\Bbb F_5$.

Later EDIT:

Let us split the polynomial $T^4 + T^3 +1 \in F[T]$ over the field $F=\Bbb F_2[X]/(f)=\Bbb F_2(a)$, where $a=X$ modulo $(f)$ is the generator of $F$, and the minimal relation over the prime field is $a^4+a^3+1=0$.

  • First of all, $a$ is a root in $F$ of $T^4 + T^3 +1$.

  • The multiplicative order of $a$ is $2^4-1=15$, it generates the cyclic multiplicative group $F_{16}^\times$.

  • The Frobenius morphism ($u\to u^2$) applied on the relation $a^4+a^3+1=0$ then gives: $$ \begin{aligned} 0 &=a^4+a^3+1\\ 0 &=(a^2)^4+(a^2)^3+1\\ 0 &=(a^4)^4+(a^4)^3+1\\ 0 &=(a^8)^4+(a^8)^3+1\ . \end{aligned} $$ So we have $T^4+T^3+1=(T-a)(T-a^2)(T-a^4)(T-a^8)$.

Computer checks:

sage: var('x');
sage: F.<a> = GF(2^4, modulus=x^4+x^3+1)
sage: F
Finite Field in a of size 2^4
sage: a.minpoly()
x^4 + x^3 + 1
sage: R.<T> = PolynomialRing(F)
sage: (T-a) * (T-a^2) * (T-a^4) * (T-a^8)
T^4 + T^3 + 1
sage: factor(T^4+T^3+1)
(T + a) * (T + a^2) * (T + a^3 + 1) * (T + a^3 + a^2 + a)
sage: a, a^2, a^4, a^8
(a, a^2, a^3 + 1, a^3 + a^2 + a)
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