Singular

[quote="noncommuting"]My apologies for the possibly confusing context of the question.

The answer I am getting is the answer that I wanted, that the homology vanishes. The example is too complicated to do by hand. However, I will say that the ring is (almost) a Weyl-algebra.

I want a "proof" that this homology vanishes -- which is why I want to check if anything about the homology command is critically assuming the ring to be commutative.

Because homolog_lib is listed as a commutative algebra library, I am a bit hesitant.[/quote]

There is an ongoing work on the nchomalg.lib, which implements homological computations for G-algebras (which Singular:Plural supports). Contact me or post your computation here for being 100% sure. Indeed there is a possibility that some computations can be incorrect. homolog.lib can only work by chance in non-commutative ring.

Regards,
Viktor

AFAIK 'homology' only uses 'modulo' command and module-concatination which will work fine in any non-commutative setting.

Therefore if you are content with the idea behind it (section Compute: of its manual entry) it should work for you.

ps: Happy New Year!

My apologies for the possibly confusing context of the question.

The answer I am getting is the answer that I wanted, that the homology vanishes. The example is too complicated to do by hand. However, I will say that the ring is (almost) a Weyl-algebra.

I want a "proof" that this homology vanishes -- which is why I want to check if anything about the homology command is critically assuming the ring to be commutative.

Because homolog_lib is listed as a commutative algebra library, I am a bit hesitant.

If you have the impression that some results are not correct

-- and if you want to get an answer here --

then present the comptuation you have done.

I have been using the homology command with a non-commutative base ring. Is there any reason that the answer I am getting would not be correct?

Thanks in advance.