Context. Apologies for the long post. As the title suggests, I am trying to compute a generic $Q_8$-polynomial in characteristic two.
Let $F$ be a field, and let $G$ be a finite group. A polynomial $f\in F(T_1,\ldots,T_n)[X]$ is called $G$-generic if the following conditions hold:
$f(T_1,\ldots,T_n,X)$ is irreducible, with Galois group $G$.
for any field extension $F'/F$, with $F'$ infinite, and every Galois extension $L'/F'$ with Galois group $G$, there exist $a'_1,\ldots,a'_n\in F'$ such that the specialized polynomial $f(a'_1,\ldots,a'_n,X)$ is irreducible in $F'[X]$ with splitting field $L'$.
Note that a $G$-generic polynomial does not necessarily exist. However, it does when Noether's problem has a positive answer. Indeed, if $V$ is a faithful representation of $G$ for which $F(V)^G/F$ is rational, then one may show that the minimal polynomial of a primitive element of $F(V)/F(V)^G$ is $G$-generic.
Claim. A $G$-generic polynomial always exists when $G$ is a $p$-group and $F$ has characteristic $p$.
Proof of the claim. A theorem of Miyata says that if $G\to \mathrm{GL}(V)$ is a triangularizable representation, then Noether's problem has a positive answer. Moreover, it is known that $p$-groups may be embedded into some $U_n(\mathbb{F}_p)$ (the $p$-Sylow subgroup of $\mathrm{GL}_n(\mathbb{F}_p)$ consisting of unipotent upper triangular matrices).
Hence, my strategy is:
to find injective morphism $Q_8\hookrightarrow U_n(F)$ for some $n\geq 1$, where $F$ is a given field of characteristic two
to compute algebraically independent elements $T_1,\ldots, T_n$ such that $F(T_1,\ldots,T_n)^{Q_8}$
to find a primitive element for $F(X_1,\ldots,X_n)/F(T_1,\ldots,T_n)$ and write up its minimal polynomial.
My results so far .
- After tedious computations, I found that $Q_8\simeq\langle I,J\rangle\subset U_4(F)$, where $I=\pmatrix{1 & 0 & 1 & 1\cr 0 & 1 & 0 & 1\cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1}$ and $J=\pmatrix{1 & 1 & 0 & 1\cr 0 & 1 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1}$.
The corresponding action (associated to the dual representation) on $F[X_1,\ldots,X_4]$ is then given by: $$I\cdot P=P(X_1,X_2,X_3+X_1,X_4+X_2+X_3), \ J\cdot P=P(X_1,X_2+X_1,X_3,X_4+X_2).$$
- I found $F[X_1,\ldots,X_4]^{Q_8}=F[T_1,T_2,T_3,T_4]$, with $$T_1=X_1, \ T_2=X_2(X_2+X_1), \ T_3=X_3(X_3+X_1), \ T_4=X_4^4+(X_2^2+X_2X_3+X_3^2)X_4^2+(X_2^2X_3+X_2X_3^2)X_4.$$
Note that $T_1,\ldots,T_4$ are algebraically independent over $F$.
- Replacing $X_i$ by $X_i/X_1$, we can safely specialize to $T_1=1$ without changing the generic property of the corresponding extension (this reflects the fact that the triangular representation of $Q_8$ factors to a faithful projective representation)
We then get a $Q_8$-extension $F(X_2,X_3,X_4)/F_0$, where $F_0=F(T_2,T_3,T_4)$.
Set $K=F_0(X_2,X_3)$, and let $g= X^4+(X_2^2+X_2X_3+X_3^2)X^2+(X_2^2X_3+X_2X_3^2)X+T_4\in K[X]$.
Then $X_4$ is a root of $g$, so $\mu_{X_4,K}\mid g$.
The field extension $K/F_0$ is a biquadratic extension of degree $4$. However, the polynomial $g$ cannot be irreducible over $K$, since $F(X_2,X_3,X_4)/F_0$ has degree $8$ by Artin's Lemma. But $g$ is indeed irreducible: specializing $X_2$ and $X_3$ to $T_4$ yields the polynomial $X^4+T_4^3X^2+T_4$, which is irreducible by Eisenstein's criterion.
Finally, I come to my questions:
Q1. Are my results 1) and 2) correct ?
Q2. Assuming that the answer to Q1 is YES, where did I go wrong in 3) ?
Remark. I strongly suspect that 2) is incorrect, but I did the computation several times, and ended up again and again with the same result. As far as I know (I've tried!) SAGE does not want to compute the ring of invariants if the characteristic of the base field divides the order of the group, so I cannot check my computations.
Side question. I found $I,J$ by trial and error, and this makes me extremely unhappy. Is there a clever way to find an embedding $Q_8\subset U_4(\mathbb{F}_2)$ ?
Note: This is not a full answer, but too long for a comment.
My computations seem to indicate so too.
You could check by hand, like so:
$$I\cdot T_4 = I\cdot(X_4^4+X_2^2X_4^2+X_2X_3X_4^2+X_3^2X_4^2+X_2^2X_3X_4+X_2X_3^2X_4)$$
Using action of $I$ gives $$I\cdot T_4 = (X_4+X_3+X_2)^4+X_2^2(X_4+X_3+X_2)^2+X_2(X_3+X_1)(X_4+X_3+X_2)^2+(X_3+X_1)^2(X_4+X_3+X_2)^2+X_2^2(X_3+X_1)(X_4+X_3+X_2)+X_2(X_3+X_1)^2(X_4+X_3+X_2)$$
Using $(a+b)^2=a^2+b^2$ gives $$I\cdot T_4=X_4^4+X_3^4+X_2^4+X_2^2X_4^2+X_2^2X_3^2+X_2^2X_2^2+X_2X_3X_4^2+X_2X_3X_3^2+X_2X_3X_2^2+X_2X_1X_4^2+X_2X_1X_3^2+X_2X_1X_2^2+X_3^2X_4^2+X_3^2X_3^2+X_3^2X_2^2+X_1^2X_4^2+X_1^2X_3^2+X_1^2X_2^2+X_2^2X_3X_4+X_2^2X_3X_3+X_2^2X_3X_2+X_2^2X_1X_4+X_2^2X_1X_3+X_2^2X_1X_2+X_2X_3^2X_4+X_2X_3^2X_3+X_2X_3^2X_2+X_2X_1^2X_4+X_2X_1^2X_3+X_2X_1^2X_2$$
Simplifying monomials gives $$I\cdot T_4=X_4^4+X_3^4+X_2^4+X_2^2X_4^2+X_2^2X_3^2+X_2^4+X_2X_3X_4^2+X_2X_3^3+X_2^3X_3+X_1X_2X_4^2+X_1X_2X_3^2+X_1X_2^3+X_3^2X_4^2+X_3^4+X_2^2X_3^2+X_1^2X_4^2+X_1^2X_3^2+X_1^2X_2^2+X_2^2X_3X_4+X_2^2X_3^2+X_2^3X_3+X_1X_2^2X_4+X_1X_2^2X_3+X_1X_2^3+X_2X_3^2X_4+X_2X_3^3+X_2^2X_3^2+X_1^2X_2X_4+X_1^2X_2X_3+X_1^2X_2^2$$
Collecting the same monomials gives
$$I\cdot T_4= X_1X_2X_3^2 +X_1X_2X_4^2 +X_1X_2^2X_3 +X_1X_2^2X_4 +2X_1X_2^3 +X_1^2X_2X_3 +X_1^2X_2X_4 +2X_1^2X_2^2 +X_1^2X_3^2 +X_1^2X_4^2 +X_2X_3X_4^2 +X_2X_3^2X_4 +2X_2X_3^3 +X_2^2X_3X_4 +4X_2^2X_3^2 +X_2^2X_4^2 +2X_2^3X_3 +2X_2^4 +X_3^2X_4^2 +2X_3^4 +X_4^4 $$
Using that $2=0$ gives
$$I\cdot T_4= X_1X_2X_3^2 +X_1X_2X_4^2 +X_1X_2^2X_3 +X_1X_2^2X_4 +X_1^2X_2X_3 +X_1^2X_2X_4 +X_1^2X_3^2 +X_1^2X_4^2 +X_2X_3X_4^2 +X_2X_3^2X_4 +X_2^2X_3X_4 +X_2^2X_4^2 +X_3^2X_4^2 +X_4^4 $$
Which is not the same as $$T_4=X_4^4+X_2^2X_4^2+X_2X_3X_4^2+X_3^2X_4^2+X_2^2X_3X_4+X_2X_3^2X_4$$
Because the terms $$I\cdot T_4-T_4=X_1X_2X_3^2 +X_1X_2X_4^2 +X_1X_2^2X_3 +X_1X_2^2X_4 +X_1^2X_2X_3 +X_1^2X_2X_4 +X_1^2X_3^2 +X_1^2X_4^2 $$ do not appear in $T_4$ but these do appear in $I\cdot T_4$.
EDIT: You could find an alternative for $T_4$ that is certainly invariant under $Q_8$ by considering the $Q_8$-orbit of $X_4$:
And then taking the product of all these images:
$T_4=X_4(I\cdot X_4)(J\cdot X_4)(IJ\cdot X_4)(I^2\cdot X_4)(I^3\cdot X_4)(J^3\cdot X_4)(IJ^3\cdot X_4)=(X_4+X_3+X_2+X_1)(X_4+X_3+X_2)(X_4+X_3+X_1)(X_4+X_3)(X_4+X_2+X_1)(X_4+X_2)(X_4+X_1)X_4$ $$=X_4^8 + X_4^4\left(X_3^4+X_2^4+X_1^4+X_3^2X_2^2+X_3^2X_1^2+X_2^2X_1^2+X_3^2X_2X_1+X_3X_2^2X_1+X_3X_2X_1^2\right) + X_4^2\left(X_3^2X_2^2X_1^2+X_3^4X_2X_1+X_3X_2^4X_1+X_3X_2X_1^4+X_3^4X_2^2+X_3^4X_1^2+X_3^2X_2^4+X_3^2X_1^4+X_2^4X_1^2+X_2^2X_1^4\right) + X_4\left(X_3^4X_2^2X_1+X_3^4X_2X_1^2+X_3^2X_2^4X_1+X_3^2X_2X_1^4+X_3X_2^4X_1^2+X_3X_2^2X_1^4\right) $$
Is certainly invariant (because group-multiplication permutes the group). Note that $T_1,T_2,T_3$ are also of this form. But in this case it seems to be quite a stupid degree-$8$ polynomial.
Maybe we can do something smarter with something of the form $T_4=(q_1\cdot X_4)(q_2\cdot X_4)(q_3\cdot X_4)(q_4\cdot X_4)+(q_5\cdot X_4)(q_6\cdot X_4)(q_7\cdot X_4)(q_8\cdot X_4)$. The problem is that in characteristic $2$, the $X_4^4$ term will cancel out in this setting. Maybe we can take $q_1\dots q_8$ such that some lower-degree term containing $X_4$ does not cancel out.
EDIT 2:
Unfortunately we cannot: If we assume WLOG that $q_1=e$ and $q_2\in\{I,J,K\}$, then multiplying by $q_2$ gives that the term with $q_2$ must also contain $q_2^2=\overline{e}$. But then it contains $e$ and $\overline{e}$, so it must also contain $\overline{e}q_2=\overline{q_2}$. Therefore, we have something of the form $$(e\cdot X_4)(I\cdot X_4)(I^2\cdot X_4)(I^3\cdot X_4)+(J\cdot X_4)(J^3\cdot X_4)(K\cdot X_4)(K^3\cdot X_4)$$ or with $I,J,K$ permuted. Checking these expressions shows that all non-constant terms (i.e. with a factor $X_4$) cancel out.