Of course, to argue by passage to char. 0 you have to assume $\ell$ is not the characteristic (so restrict to abelian connected Galois covers with degree not divisible by the characteristic).

This is true over any separably closed field with any $\ell$. The key is that it is true with coefficients in any finite abelian group: every abelian connected finite Galois cover of $X$ arises by $\phi$-pullback from a unique such cover of the Jacobian $J$. Milne discusses this with references in his article on Jacobians in the book "Arithmetic Geometry". Or use deformation of curves and smooth/proper base change to reduce to char. 0, and then to $\mathbf{C}$ where it reduces to determining the analytification of $(J, \phi)$ in terms of that of $X$.

@Rupert: For any connected reductive group $H$ (such as $Z_G(S)$) and any $h \in \mathscr{D}(H)$, we know its conjugacy class in $H$ is the same as its conjugacy class in $\mathscr{D}(H)$. In what sense is your question is not an instance of that?

I think there are also some E$_6$ counterexamples as well (over suitable fields). For "most" types it is OK by inspection of classification theorems.

No. Assume $G$ is semisimple (else silly counterexamples via non-split central tori). For a maximal $k$-torus $T\supset S$ and basis $\Delta$ of the set $\Phi$ of absolute roots, a basis of absolute roots of $Z_G(S)$ is $\Delta_0=\ker(\Delta\rightarrow{\rm{X}}(S))$. You want $\Delta_0$ of size $\dim(T)-\dim(S)$; i.e., $D:=\Delta -\Delta_0$ of size $\dim(S)$ $(\dim(T)=\#\Delta$ since $G$ is semisimple). Restriction $D\rightarrow{\rm{X}}(S)$ maps onto a basis of relative roots, so you want it injective; i.e., $\#D$ is the $k$-rank. False for many non-quasi-split unitary groups over fields.

@user43326: I know almost nothing about modern homotopy theory, but I've seen other settings where a global cohomological computation encounters complications because certain signs are merely "locally constant", and so I wondered if the disconnectedness of orthogonal groups might be contributing discrepancies governed by local signs and not a single overall global sign. But this is just pure speculation, so feel free to ignore it (or maybe Oscar R-W sees something about it).

@Menny: In your own comment about you wrote things adelically with SO(q) instead of O(q); I was just pointing out that you must have meant to use O(q) there.

@Menny: You meant to include "equivalent to $f$" in some places in your definitions (otherwise you aren't using $f$), to use orthogonal groups rather than special orthogonal groups, and to assume $f$ is non-degenerate over $\mathbf{Q}$. Your ${\rm{gen}}_S(f)$ is the kernel (in the sense of pointed sets) of the map ${\rm{gen}}(f) \rightarrow {\rm{gen}}(f_S)$, where $f_S$ denotes the associated quadratic module over $\mathbf{Z}_S$ (over which there is the evident notion of "genus" as over any Dedekind domain). Clearly $S=\emptyset$ answers your question, so did you want $S$ not empty?

@Ian: Just as ${\rm{GL}}_n(\mathbf{R}) \rightarrow {\rm{PGL}}_n(\mathbf{R})$ is a nontrivial $\mathbf{R}^{\times}$-torsor, for number fields $F$, ${\rm{GL}}_n(O_F) \rightarrow {\rm{PGL}}_n(O_F)$ has cokernel ${\rm{Pic}}(O_F)[n]$, so this is nontrivial when $\gcd(h_F,n) \ne 1$. The $\mathbf{Z}$-group scheme PGL$_n$ represents the automorphism functor of $\mathbf{P}^{n-1}_{\mathbf{Z}}$, and is the affine open subscheme complementary to $\{\det=0\}$ in the projective space (over $\mathbf{Z}$) of $n\times n$ matrices (coordinate ring is the degree-0 part of $\mathbf{Z}[x_{ij}][1/\det]$).

@Ian: I should have clarified that the algebraic group PSL$_n$ is the same as PGL$_n$; its group of $K$-points for a field $K$ is ${\rm{GL}}_n(K)/K^{\times}$ and not ${\rm{SL}}_n(K)/\mu_n(K)$. The latter does not satisfy Galois descent in $K$ since ${\rm{H}}^1(K, \mu_n) = K^{\times}/(K^{\times})^n$ is usually nontrivial, whereas the former does satisfy Galois descent in $K$ due to Hilbert 90.

@Niels: When I read SGA1 as a student I worked out the above as a translation of those constructions in terms which can be expressed "at finite level", and I found the process of connecting up those two points of view to be very helpful both for understanding proofs there and when reading SGA4, Deligne's Weil II (with its Weil sheaves), etc. I'm sorry to hear that you think it is "too involved".

To clarify the end of Jim's answer, for a number field $F$ and $n\ge 2$ there are many finite-index non-congruence subgroups $\Gamma_2\subset {\rm{PGL}}_n(O_F)$, yet the cokernel ${\rm{H}}^1(O_F,\mu_n)$ of $f:{\rm{SL}}_n(O_F) \rightarrow {\rm{PGL}}_n(O_F)$ is finite, so replacing $\Gamma_2$ with a finite-index subgroup ensures $\Gamma_2=f(\Gamma_1)$ for $\Gamma_1=f^{-1}(\Gamma_2)$. But $\Gamma_1$ has finite index in ${\rm{SL}}_n(O_F)$, so it is a congruence subgroup if $n \ge 3$ and $F$ has a real place. Then $\Gamma_1$ is congruence with image containing no congruence subgroup.

@Ian: It is a bit delicate to make examples for $G_1 = {\rm{SL}}_2$ and $G_2 = {\rm{PGL}}_2$ over $\mathbf{Q}$ since ${\rm{SL}}_2$ doesn't have the congruence subgroup property. For $n > 2$ life is easier: there are lots of non-congruence subgroups of ${\rm{PGL}}_n(\mathbf{Z})$ yet their preimages in ${\rm{SL}}_n(\mathbf{Z})$ are finite-index and hence congruence (as $n > 2$!), so that provides a bonanza of congruence subgroups in ${\rm{SL}}_n(\mathbf{Z})$ whose image in ${\rm{PGL}}_n(\mathbf{Z})$ is non-congruence (due to being contained in a non-congruence subgroup).

@Neils: Where is the first place that is not understandable, and is there an alternative method to make it explicit (including addressing the role of the base point) if one were to take the more concrete viewpoint of $\pi_1$ in terms of automorphisms of connected finite etale covers (as that is closer to the spirit of Galois theory, which is not generally taught in terms of automorphisms of fiber functors)?

@Tom: Every congruence subgroup of $G_2(\mathbf{Q})$ contains the image of a congruence subgroup of $G_1(\mathbf{Q})$ since its preimage in $G_1(\mathbf{Q})$ is such a subgroup. Indeed, this reduces to checking that $G_1(\mathbf{A}^{\infty}) \rightarrow G_2(\mathbf{A}^{\infty})$ is topologically proper (so preimage of a compact open subgroup is a compact open subgroup), which holds because $G_1 \rightarrow G_2$ is a finite morphism.

@anon: is your implicit definition of "congruence subgroup" $G(\mathbf{Q}) \cap K$ for a compact open subgroup $K$ of $G(\mathbf{A}^{\infty})$?

Do whatever you wish. If you alert me via comment that you make such an edit then I will delete my (then moot) comments.

For any noetherian ring $R$ it suffices to test against prime ideals $I$. Indeed, by direct limit considerations we just need ${\rm{Tor}}_1(N,M)=0$ for finitely generated $N$, and any such $N$ has a finite composition series whose successive quotients are $R/P$ for prime ideals $P$ since $R$ is noetherian.