Can anyone help me to find a method for calculating the splitting field for a polynomial over a function field?
I think this feature is not currently supported..
Splitting field in math software.
math-softwaresplitting-field
Related Solutions
There is a typo in the statement of Lemma 67 in your source. The $n$th roots of unity are in $\Bbb{F}_p$ only if $n\mid p-1$ or, iff $p\equiv1\pmod n$ (not $n\equiv1\pmod p$ as is written there). Therefore that Lemma does not apply.
In fact, the polynomial $x^{15}-2$ is NOT irreducible in $\Bbb{F}_7[x]$. This follows trivially from the fact that $3^5=243\equiv-2\pmod 7$. Therefore $$ x^{15}-2=(x^3)^5+3^5=(x^3+3)(x^{12}-3x^9+3^2x^6-3^3x^3+3^4). $$
We immediately see that $x^3+3$ has no zeros in $\Bbb{F}_7$ (the cubes in that field are $0,\pm1$), so it is irreducible. Therefore the polynomial has a zero $\alpha$ in $\Bbb{F}_{7^3}$.
To get the splitting field of $x^{15}-2$ we need, as you observed, the primitive 15th roots of unity. We easily see that $$ 7^4=2401\equiv1\pmod{15}. $$ The multiplicative group of the field $\Bbb{F}_{7^4}$ is cyclic of order $7^4-1$, and thus it contains a primitive 15th root of unity $\zeta$.
A consequence of all this is that the splitting field of this polynomial is $$ \Bbb{F}_7[\alpha,\zeta]=\Bbb{F}_{7^{12}}. $$
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)
Best Answer
You can find the splitting field with the Isprime and PrimaryDecomposition function
thus $$T^3+T(x^3+y)+1 = (T-z)(T-w)(T-(u+z+w)) \\\in \Bbb{Q}[x,y,z,w]/J [T]$$ where $$J=(x^2+y^3+y , \qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad \qquad z^3+z(x^3+y)+1, \qquad \qquad \qquad\\ \qquad \qquad w^2 + z w + y^9 z + 3 y^7 z + 3 y^5 z + y^3 z + y^2 z + 2 y z^3 + 2 y + z^5 + 2 z^2) $$