Talk:Cohomological dimension

The article states that the cohomological dimension of G with coeffs in a ring R is 0 if and only if R[G] is semi simple. But

$\backslash mathbb\{Z\}$[trivial group] is isomorphic to $\backslash mathbb\{Z\}$, which is not semi-simple as a module over itself. Seamus Stoke (talk) 19:05, 22 January 2009 (UTC)

:I confirm this is a mistake in the article, but I seem to remember something extremely similar is actually true. In particular, the part about the group order being a unit is right for cd 0, I am pretty sure. So for instance Z[x,y]/(1-2x,y^2-1) is not semi-simple but is a group ring of a cyclic group of order 2 where the group has cd 0. Maybe the word semi-simple is just out of place. I used Cartan–Eilenberg as a reference (which only covers finite groups) and some articles of Swan, so I haven't looked through the references in this article any time recently. Here is the {{diff|Cohomological_dimension|diff=next|oldid=192463274|label=edit}} where it was made. It is probably true over fields. I think the blocks should be semi-simple, but I'm not sure how to phrase that for general rings. I think Jacobson semi-simple also does not work, since something like the 2-adics (as a group ring ovr the 2-adics of the trivial group) are not Jacobson semi-simple, but the trivial group does have cd 0. JackSchmidt (talk) 19:42, 22 January 2009 (UTC)

The definition for a group isn't quite right, since the modules are unrestricted, and so this is defining strict cd. They should be discrete torsion modules. This is from Serre I.3. I will expand this myself when I get a moment. Spectral sequence (talk) 06:23, 16 May 2013 (UTC)

I don't think that the claim " the field of laurent series $k((t))$ over an algebraically closed field k of non-zero characteristic also has absolute Galois group isomorphic to $\backslash hat\{\backslash mathbf\{Z\}\}$", is true, c.f. [https://mathoverflow.net/questions/133148/galois-cohomology-of-the-field-of-laurent-series](MO-post) (because there can be non-trivial Artin-Schreier coverings).

One way or another, the cited source only states it for algebraically closed fields of characteristic zero. Nurnochgeist (talk) 13:24, 12 September 2021 (UTC)