Computing the Galois group of the splitting field of $X^{q+1} + X + T$ over the function field $\mathbb{F}_q(T)$ Let $k = \mathbb{F}_q$, $A=k[T]$ and $K=k(T)$. Let $f(X) := X^{q+1} + X + T \in K[X]$. Let $L$ be the splitting field of $f$ over $K$. (In other words, fixing an algebraic closure $K^{\mathrm{alg}}$ of $K$, let $L$ be the field generated by all the roots of $f$ in $K^{\mathrm{alg}}$ over $K$).
My question: How to compute the Galois group of $L/K$?

We explain the motivation for the above question in the remaining part of this post. But it seems that the question itself can be well-presented without the motivation below.
Motivation: The problem arises when I'm learning Drinfeld modules. Let $\phi: A \rightarrow \operatorname{End}(\mathbb{G}_a)$ be a Drinfeld module over $K$ of rank 2, defined by
$$
\phi_T := T + \tau + \tau^2.
$$
Then the absolute Galois group $\operatorname{Gal}_K$ acts on the $T$-torsion points $\phi[T]$ of $\phi$, which induces a Galois representation
$$
\pi_T: \operatorname{Gal}_K \rightarrow \operatorname{GL}_2(A/(T)) = \operatorname{GL}_2(\mathbb{F}_q).
$$
In paper here, the authors found that in general $\pi_T$ may not be surjective. My ultimate goal is to check that in this case, $\pi_T$ is indeed surjective.
Then by definition of $T$-torsion point, we have an injection of groups
$$
\overline{\pi_T}: \operatorname{Gal}(K(\phi[T])/K) \hookrightarrow \operatorname{GL}_2(\mathbb{F}_q),
$$
and $K(\phi[T])$ is merely the splitting field of $\Phi_T(X)$ over $K$, where $\Phi_T$ is the "untwisted version" of $\phi_T$:
$$
\Phi_T (X) = TX + X^{q} + X^{q^2} = X(T+ X^{q-1} + (X^{q-1})^{q+1}).
$$
So we break the extension $K(\phi[T])/K$ into the splitting field $L$ of $f(Y) := T + Y + Y^{q+1} \in K[Y]$ and an extension adjointing some "$(q-1)$-th" root. So to get my desired surjectivity of $\pi_T$, by counting $\#(\operatorname{GL}_2(\mathbb{F}_q))$, I'm hoping that at least we know the order of $L/K$ is $\geq q(q-1)(q+1)$.
But I got stuck here on how to calculating the Galois group, or at least the order of $L/K$.
Something more: It is more illustrating if one comes with a rank two Drinfeld module with nonsurjective $\pi_T$. I was told to try $\psi_T := T + \tau + T^q \tau^2$, for which $\pi_T$ has image $\operatorname{SL}_2(\mathbb{F}_q)$. This is even more chanllenging for me, as merely by counting may not be suffice to settle down the subgroup structure of $\operatorname{GL}_2(\mathbb{F}_q)$.
Sorry for such a long post and thank you all for help!

EDIT on July 13 after Jyrki Lahtonen's hint:
It seems that Gauss' lemma really works!  To show $f(X) \in K[X]=k(T)[X]$ is irreducible, suppose otherwise it is reducible in $k(T)[X]$, then by Gauss' lemma, it is reducible in $k[T][X]=k[T,X]$. Since $\deg_{T}(f)=1$, it factors as
$$
f(X,T)=f_1(X,T) f_2(X), 
$$
where $f_1(X,T) = T + g_1(X) \in k[X][T]$ and $g_1(X),f_2(X) \in k[X]$.
But now compare the coefficient of $T$ in $f \in k[X,T]$, we see that $f_2(X)=1$, showing that $f$ is irreducible in $k[X][T]$ and hence irreducible in $k(T)[X]$.
 A: Thanks Jyrki Lahtonen a lot for his help and inspiring comments. Now let me try to figure all these things out.
One of the key tool is the following proposition taken from Proposition 5.2 of this paper, quoted below:

Proposition: Let $K$ be a field and $f(X) \in K[X]$ be an irreducible polynomial of degree $n$. Let $K(\alpha_1, \ldots , \alpha_s)$ be the field obtained by adding $s$ roots of $f(X)$. Let $f_0(X) := f(X)$ and $f_s(X) \in  K(\alpha_1, \ldots, \alpha_s)[X]$ be the polynomial that is obtained by throwing away $s$ roots, so
$$
f_s(X) := \dfrac{p(X)}{(X − \alpha_1)(X − \alpha_2) \ldots (X − \alpha_s)}
$$
Then the Galois group of the splitting field of $f(X)$ over K contains at least $n(n − 1)\cdots(n − m)$ elements if $f_s(x)$ is irreducible in $K(\alpha_1, \ldots, \alpha_s)[X]$ for all $0 \leq s \leq m$.

So to compute the desired Galois group in our case, we need to throw out the roots of $f$ one by one.
First Step: Suppose $\lambda \in K^{\mathrm{alg}}$ such that $f(\lambda)=0$, i.e.
$$
\lambda + \lambda^{q+1} + T = 0 \text{ in } K^{\mathrm{alg}}.
$$
Hence $T = - \lambda - \lambda^{q+1}$, implying that $k(T)(\lambda) = k(\lambda)$. Hence $\lambda$ is transcendental over $k$, as otherwise $T$ is algebraic over $k$, which is absurd. So we substitute $T=- \lambda - \lambda^{q+1}$ into our polynomial as if it has coefficient in $k(\lambda)$.
Then throwing away the root $X=\lambda$, we ask: it the following polynomial $f_1(X)$ irreducible in $K(\lambda)[X]$?
$$
f_1(X) := \dfrac{- \lambda - \lambda^{q+1} + X + X^{q+1}}{X-\lambda} \in K(\lambda)[X] = k(\lambda)[X].
$$
We substitute $X \rightsquigarrow Y-1+\lambda$, it suffices to check the resulting polynomial $f_1(Y-1+\lambda) =: f_1^{\flat}(Y) \in k(\lambda)[X]$ is irreducible. We simplify
$$
f_1^{\flat}(Y) = f_1(Y+\lambda) = Y^q + \lambda Y^{q-1} + \lambda(\lambda^{q-1}-1).
$$
This is an $\lambda$-Eisenstein polynomial in $k[\lambda][Y]$, hence is irreducible in $k(\lambda)[Y]$.
Second Step: We further throw away another root of $f$, but now it is more convenient to work with $f_1^{\flat}$ instead. Starting with $f_1^{\flat}$, we make the substitution $ Y \rightsquigarrow \lambda Z$ to get
$$
f_1^{\flat\flat}(Z) := f_1^{\flat}(\lambda Z) = \lambda^q(Z^{q}+Z^{q-1}+1)+1.
$$
Since one can directly check that $\lambda \neq 0$, dividing $\lambda^q$ on both sides will cause no harm, so we consider
$$
f_1^{\flat\flat\flat}(Z) := \lambda^{-q} f_1^{\flat\flat}(Z) = Z^{q}+Z^{q-1}+(1+\lambda^{-q}).
$$
This is a good version of "$f_1$" that we're gonna use. Let $\mu \in K^{\mathrm{alg}}$ be a root of $f_1^{\flat\flat\flat}$, then
$$
\lambda^{q} = -\dfrac{1}{1+\mu^{q-1}+\mu^q} \in k(\mu).
$$
Hence $$K(\lambda, \mu) = k(\lambda, \mu) \supseteq K(\lambda^{q}, \mu)=k(\mu).$$ Now we regard $f_1^{\flat\flat\flat}$ as a polynomial with coefficient in $k(\lambda, \mu)$. Then
$$
f_1^{\flat\flat\flat}(Z) = Z^q + Z^{q-1} - \mu^{q-1}-\mu^{q} \in k(\lambda, \mu).
$$
Now we throw away the root $Z=\mu$, i.e. consider
$$
f_2(Z) := \dfrac{Z^q + Z^{q-1} - \mu^{q-1}-\mu^{q}}{Z-\mu} = \dfrac{Z^q -\mu^{q}}{Z-\mu} + \dfrac{Z^{q-1} - \mu^{q-1}}{Z-\mu}.
$$
To further simplify this, one notes the following identity: for any integer $\ell > 0$,
$$
Z^{\ell} - \mu^{\ell} = (Z-\mu)(Z^{\ell-1} + \mu Z^{\ell-2} + \cdots + \mu^{\ell-2} Z + \mu^{\ell-1}).
$$
Using the case $\ell=q, q-1$, we see
$$
f_2(Z) = Z^{q-1} + (\mu + 1)Z^{q-2} + \cdots + \mu^{q-3}(\mu+1)Z + \mu^{q-1}(\mu+1).
$$
This is an $(\mu+1)$-Eisenstein polynomial in $k[\lambda, \mu][Z]$, hence is irreducible in $k(\lambda, \mu)[Z]$.
Now it's time to use our quoted propsition above to see that the Galois group is at least of order $(q + 1)q(q - 1)$.
This is enough for me to get the corresponding result on the Galois representation associated to the Drinfeld module $\phi_T$, so we shall stop here.

Ok, after supper, let me fill up the following up arguments.
So now, we see the Galois group of the splitting field $L$ of $f$ over $K$ has degree $\geq (q + 1)q(q - 1)$. Now for the extension $K(\phi[T])/K$, we break it into $K(\phi[T]) / L$ and $L/K$. The remaining mysterious extension is $K(\phi[T]) / L$, yet this is merely adding the $(q-1)$-th root of the roots of $f$ in $K^{\mathrm{alg}}$, so $[K(\phi[T]) : L] \geq q-1$. By the tower formula and the Galois correspondence, we see
$$
\operatorname{Gal}(K(\phi[T])/K) = [K(\phi[T]): K] \geq (q + 1)q(q - 1)^2.
$$
Now comes the essential role that Drinfeld module plays: it provides us with the injection
$$
\overline{\pi_T}: \operatorname{Gal}(K(\phi[T])/K) \hookrightarrow \operatorname{GL}_2(\mathbb{F}_q).
$$
It is then elementary to see that $\#(\operatorname{GL}_2(\mathbb{F}_q))=(q + 1)q(q - 1)^2$. So the monomorphism $\overline{\pi_T}$ is an isomorphism, as desired!

Remarks:

*

*There is still some questions that remains to be answered: the structure of the Galois group $L/K$? Though by the above arguments, we see that $\operatorname{Gal}(L/K)$ is a subgroup of $\operatorname{GL}_2(\mathbb{F}_q)$ of order $(q + 1)q(q - 1)$. But can we settle down the explicit group structure of $\operatorname{Gal}(L/K)$ by merely knowing its order? So may be we can ask: is the subgroup lattice of $\operatorname{GL}_n(\mathbb{F}_q)$ available for us now?


*Here we have an example showing that $\pi_T$ is surjective. But it seems that verifying an example of nonsurjective $\psi_T$ (for example, given in the post in the question part) is even more chanllenging. The main obstruction for me is that all the above arguments relies on counting, but for non-surjective $\pi_T$, getting a lower bound of $[L:K]$ that is strictly less than $\#(\operatorname{GL}_2(\mathbb{F}_q))$ does NOT help!
So I still get stuck here for the case $\psi_T$. ....
So maybe I should post another post in this site for the above questions in the remark, i.e.

*

*Describe the subgroup lattice of $\operatorname{GL}_n(\mathbb{F}_q)$ (or even for other reductive groups over finite fields)?

*The nonsurjective $\pi_T$ given by the Drinfeld module $\psi_T$.

Thank you all for paying attention to this post and providing encouraging helps!! :)
