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    • CommentRowNumber1.
    • CommentAuthorZhen Lin
    • CommentTimeFeb 10th 2015

    Created Beck module, mentioned it (once) on the tangent category page.

    • CommentRowNumber2.
    • CommentAuthorTim_Porter
    • CommentTimeFeb 11th 2015

    There was work done on the ‘derived module’ of a group homomorphism in

    R. H. Crowell, The derived module of a homomorphism, Advances in Math., 5, (1971), 210 – 238.

    This is related to both Beck modules and to Fox derivatives. (i discussed this in the Menagerie, and it relates to the linearisation of crossed complexes in Ronnie Brown’s work.)

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeFeb 11th 2015
    • (edited Feb 11th 2015)

    Very nice! I have added a few more pointers to this from other relevant entries.

    I have also added pointer to Beck 67.

    • CommentRowNumber4.
    • CommentAuthorTim_Porter
    • CommentTimeAug 28th 2018
    • (edited Aug 28th 2018)

    Added a reference to a preprint by Markus Szymik relating Beck modules of quandles to the detection of the unknot. NB: Richard, he is at Trondheim.

    diff, v3, current

    • CommentRowNumber5.
    • CommentAuthorRichard Williamson
    • CommentTimeAug 28th 2018
    • (edited Aug 28th 2018)

    Hehe, yes, Markus and I were colleagues for a while, and meet occasionally :-). He is a very nice chap! I believe is on sabbatical at the moment, in Cambridge I think.

    • CommentRowNumber6.
    • CommentAuthorn.mertes
    • CommentTimeMay 17th 2020

    Let AA be a commutative ring, and let MM be an AA-module. Then any square-zero extension BAB\to A with kernel MM can be viewed as a torsor in CRing/A\text{CRing}/A for the abelian group object AMAA\oplus M\to A corresponding to MM. The group action (AM)× ABB(A\oplus M)\times_A B\to B is such that ((a,m),b)((a, m), b) is mapped to b+mb+m.

    Is every torsor for AMAA\oplus M\to A necessarily a square-zero extension? More specifically, is the category of torsors for AMAA\oplus M\to A equivalent to the category of square-zero extensions of AA by MM?

    • CommentRowNumber7.
    • CommentAuthorn.mertes
    • CommentTimeJun 9th 2020
    I answered my question listed above in this article I also attempted to generalize this situation to the setting of schemes.