The Galois closure of $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$

Question

Consider the field extension $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ where $y$ is a root of the polynomial $g(T) = f(x,T) \in \mathbb{F}_2(x)[T]$, with

$$ \begin{array}{l} f(x,y) = x^{12} + x^{10}y^2 + x^{10}y + x^{10} + x^9y^2 + x^9y + x^8y^4 + x^8y^2 + x^8 \\ {}+ x^6y^6 + x^6y^5 + x^6y^4 + x^6y^3 + x^6y^2 + x^6y + x^6 + \\ x^5y^6 + x^5y^5 + x^5y^4 + x^5y^3 + x^5y^2 + x^5y + x^4y^8 + \\ x^4y^6 + x^4y^5 + x^4y^4 + x^4y^3 + x^4y^2 + x^4 + x^3y^6 + \\ x^3y^5 + x^3y^4 + x^3y^3 + x^2y^{10} + x^2y^9 + x^2y^8 + x^2y^6 + \\ x^2y^5 + x^2y^4 + x^2y^2 + x^2y + x^2 + xy^{10} + xy^9 + xy^6 + \\ xy^5 + xy^2 + xy + y^{12} + y^{10} + y^8 + y^6 + y^4 + y^2 + 1 \end{array} $$

I want to show that $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ is not a Galois extension.

We know that $f(x,y)$ is the unique non-linear factor of

$$ (x^{16}-x)(y^2-y)-(y^{16}-y)(x^2-x). $$

We have

$$ (x^{16}-x)(T^2-T)-(T^{16}-T)(x^2-x) = T(T+1)(T+x)(T+x+1)g(T). $$

In $\mathbb{F}_2(x,y)[T]$ we have

$$ g(T) = (T+y)(T+y+1)(T+y+x)(T+y+x+1)h(T), $$

with $h(T)$ of degree $8$. I want to find the Galois closure of $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$. Thus I can see that it is not Galois extension. Any help is appreciated.

Answer 1

Let $p(X,Y) = (X^{16}+X)(Y^2+Y)+(Y^{16}+Y)(X^2+X)$. If $p(x,y) = p(x,z) = 0$ then $p(x,y+z) = 0$ so for a given value of $x$, the roots of the corresponding polynomial form a $\Bbb F_2$-subspace, usually $4$-dimensional.

Notice that for any $x \in \overline{\Bbb F_2}$, $p(x,Y)$ has a double root $\iff$ $0$ is a double root of $p(x,Y)$ $\iff$ $x^{16}+x^2 = 0 \iff x^8+ x = 0 \iff x \in \Bbb F_8$ .

We already know that $p$ has $4$ roots $0,1,x,x+1 \in \overline {\Bbb F_2}(x)$.

Suppose there's another root $f \in \overline {\Bbb F_2}(X)$ such that $p(X,f(X)) = 0$.
By the rational root theorem, $f = a/b$ where $a$ divides $(X^8+X)^2$ and $b$ divides $(X^2+X)$. In particular, $f \in \Bbb F_8(X)$.
However there are lots of values of $x \in \Bbb F_{64}$ for which $p(x,Y)$ have only $4$ simple roots in $\Bbb F_{64}$, so this is impossible.

Next, let $y$ be a new root of $P$ and suppose that there's some $f \in \overline {\Bbb F_2}(X,y)$ such that the roots of $P(X,Y)$ are the subspace $\langle 1,X,y,f(X,y) \rangle$.
Then denoting $f^2 = Frob_2(f)$ (we square every coefficient of $f$), if $f \neq f^2$ then $f^2 + f \in \langle 1,X,y \rangle \subset \Bbb F_2(X,y)$. Hence $f^4 + f^2 = f^2 + f$ and $f^4 = f$ so that $f \in \Bbb F_4(X,y)$
However there are lots of values of $x \in \Bbb F_{2^{10}}$ for which $P(x,Y)$ have only $8$ simple roots in $\Bbb F_{2^{10}}$, so this is also impossible.

This shows that the splitting field of $P$ over$\Bbb F_2(X)$ has degree $12 \times 8 = 96$, and the Galois group is isomorphic to the matrix group

$G = \left\{ \begin{pmatrix} I_2 & A \\ 0 & B \end{pmatrix} \in GL_4(\Bbb F_2) \right\}$