Trivial module over quotient ring

[User8764552]

Dear all,

I'm sorry for such a silly question, but it bugs me.

Given a graded connected commutative ring R over QQ with unity (actually, a quotient of polynomial ring), I want to define the trivial R-module having QQ in degree 0 and zero otherwise.

(Then I want to do homological algebra computations with this R-module.)

I try the following code just to see that the quotient keeps all degree 1 generators (not what needed).

A=QQ[v1,v2,v3]/(v1*v2*v3)

I=ideal {v1,v2,v3}

M=module A/I

betti M

                   0 1
o4 = total: 1 3
               0: 1 3

Do you know how to define the trivial module in M2?

Best regards, User

[David Eisenbud]

try

M = coker vars A.

M = A^1/I would also work.

Your A/I is a ring, not a module 

DE

--

David Eisenbud

Professor of Mathematics

University of California Berkeley

https://eisenbud.github.io/

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[User8764552]

Dear David,

Thank you for your answer.

Both of the suggested variants give the A-module with 5 generators in degree 1 (which is different from what I am looking for).

Another close variant that misses the target: "module A/A^1" gives 1 generator in degree 1 (in addition to 1 gen in degree 0, of course).

This problem looks specific for QuotientRing.

For polynomial rings there is no problem with defining the trivial module by similar commands to the above.

--

Best regards, User.

среда, 22 мая 2024 г. в 06:17:32 UTC+2, d...@berkeley.edu:

[jche...@gmail.com]

Dear User,

You may be misinterpreting the output of "betti" - the number in column index 1 is the number of relations, not degree 1 generators. The "1" you see in column 0, row 0 indicates a single generator in degree 0. See https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/Macaulay2Doc/html/_betti.html

Justin

[User8764552]

Dear Justin,

Thank you for your elaboration.

I am a happy fool now, problem never existed.

--

Best regards, User

среда, 22 мая 2024 г. в 18:43:17 UTC+2, jche...@gmail.com: