Free resolutions for group rings

[fieker]

Allows the following to work:

k, a= wildanger_field(3, 13)
k = normal_closure(k)
k = k[1]
I = idele_class_gmodule(k)
G = group(I.data[1])
ZG = group_algebra(ZZ, G)
fr = Oscar.GrpCoh.free_res(ZG; side = :right)
h = hom(fr, I.data[1])
quo(kernel(map(h, 2))[1], image(map(h, 1))[1])[1]

to compute (co)-homology like in the text book. Not yet perfect.

[thofma]

Do you want to add this as an example/test somewhere?

[ThomasBreuer]

When I try the example, idele_class_gmodule(k) throws an exception.

ERROR: AssertionError: all((in)(iU), u)
Stacktrace:
 [1] (::Oscar.GrpCoh.var"#ind_action#20"{…})(XX::GModule{…}, s::PermGroupElem)
   @ Oscar.GrpCoh /.../Oscar.jl/experimental/GModule/src/Cohomology.jl:440

(Sorry, I missed the info that dev versions of AbstractAlgebra and Hecke are needed. Will try again.)
(No, same error with dev versions of AbstractAlgebra and Hecke.)

[fieker]

this test is wrong - this is the mixup with the FundClass and the MeatAxe branches
I'll fix it and add an example

[fieker]

So, the interface is slowly emerging

• free_res extends however much you want (above, fr[5] will also work), thus H^4 can be computed
• there are nucleous functions to get proper 1 and 2 chains
• change_group should take a map U -> V of groups, resolutions of ZU and ZV and construct the chain map. here the interface is bad
• elements in the hom complex can be turned into homs and back

[ThomasBreuer]

Suggested change

	@test is_isomorphic(abelian_group([1]), quo(kernel(map(h, 1))[1], image(map(h, 1))[0])[1])
	@test is_isomorphic(abelian_group([1]), quo(kernel(map(h, 1))[1], image(map(h, 0))[1])[1])