An interesting (yet curious) example is an ultraproduct of etale coomology with $\mathbb Z/\ell$-coefficients over all $\ell$ different from the characteristic. It was used by Gabber to show that $\ell$-adic cohomology is torision free for all but a fini for a finite number of $\ell$ (and for a fixed smooth and projective variety).

My comment was more referring to what I believe to be Grothendieck's motivation. Finding a natural way to interpret Weil cohomology theories as the cohomology with respect to some topology (and some sheaf in it) would be an a posteriori fact (and a wonderful one at that).

Well it is even worse I thought about the distinction for $K$ where I noted that it doesn't matter as $K$ is totally complex but forget that it matters for $\mathbb Q[\sqrt{19\cdot151}$. What is much worse I however is that when I asked Magma (and also Kant to doublecheck) about the class number I did for the field obtained by adjoining one root of the polynomial and not all of them. Hence I got class number (which is probably OK) but for the wrong field... Back to the drawing board.

Here is a specialisation of the question at the end. The polynomial (an example "dear to Artin" according to Lang, if I remember the comment in his book correctly) $x^5-x+1$ gives an unramified $A_5$-extension $K$ of $\mathbb Q[\sqrt{19\cdot151}]$. Is this the maximal unramified extension of $\mathbb Q[\sqrt{19\cdot151}]$? The class group of $K$ is trivial so a further solvable extension is out. On the other hand Odlyzko's bounds are not strong enough to give anything it seems.

The proof in for instance Milnor: Morse theory of the real case works without changes in the complex case.

As we always have $\mathrm{Ext}^2(A,M)=0$, the short exact sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/n\to0$ gives that $\mathrm{Ext}^1(A,\mathbb Z/n)=0$ for all $n$ if $\mathrm{Ext}^1(A,\mathbb Z)=0$.

Indeed,there is a wonderful proof that is short enough to fit into one of these comments... Consider a finite projective resolution of a module. Shifted it is also a finite resolution of the left hand side which is injective. As all but the last modules in that resolution are injective the last on must also be. Hence any module with a finite projective resolution must be injective and hence projective.

I think Odlyzko's bounds on the root discriminants (On conductors and discriminants) can be used to go a bit further as they are better than the Minkowski bounds.

Then you will no doubt enjoy learning about groups with periodic cohomology. They are completely classified and Swan proved that a finite group has periodic cohomology iff it admits a free action on a CW-complex homotopic to a sphere.

Very few finite groups have periodic cohomology. The conclusion that all non-trivial finite $G$ have cohomology in arbitrarily high dimension is of course still true.

That the lowest term of the Taylor expansion determines the singularity is not always true. It is true because the Hessian is non-singular in which case there is only one type as we are working over the complex numbers (it is over the reals that we have different indices).

@BCnrd: Characteristic $0$ comes in when one uses the fact that $exp(r)$ and $log(1+r)$ are defined when $r$ is nilpotent (this is actually the basis for the fact that group schemes are smooth so they are equally algebraic). This implies that $mR$ as an additive group and $1+mR$ as multiplicative group are isomorphic when $m$ is a nilpotent ideal in $R$. This is false when we are not in characteristic $0$, see for instance the case of $1+tF[[t]]$ and $tF[[t]]$ when $F$ is a finite field. Mumford in his book "Curves on surfaces" has a nice description in these terms of what happens for $p>0$.

The quaternion and dihedral groups of order $8$.

Don't get me wrong, I would only be too happy to have someone understand properly the non-unipotent part. I think that if one is to use Tannakian techniques a generalisation of Tannakian categories will be needed where the category is allowed to grow like a scheme under base extension and not just like a zero-dimensional scheme (or rather stack) that is the case with current Tannakian categories (a case in point is a pro-torus whose character group is the additive group of the base). The unipotent theory gives a complete and satisfying answer today but it is clearly not the whole story.

The Hessian (the matrix of second derivatives) is invertible so the result follows from the (complex version of) the Morse lemma.