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    • CommentRowNumber1.
    • CommentAuthorDavidRoberts
    • CommentTimeDec 13th 2010
    • (edited Dec 13th 2010)

    At Galois theory we have the following:

    If AA is not the zero ring and NN is free with basis (w i) iinI(w_i)_{i\inI}, then the cardinality #I#I only depends on NN, and not on the choice of basis

    is this right, considering the discussion at the thread on rank?

    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeDec 13th 2010
    • (edited Dec 13th 2010)

    Only if AA is assumed to be commutative or noetherian or something, which doesn’t seem to be the case.

    I think that I’ve edited the page so that its claims about rank are correct. These claims are far from complete, but they cover what seems to be used (and the complete version is at rank).

    • CommentRowNumber3.
    • CommentAuthorDavidRoberts
    • CommentTimeDec 13th 2010

    Thanks for that.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeDec 13th 2010

    I’ll check later today with the lecture notes. I think we need to assume commutative rings pretty much throughout. But let me check

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeDec 13th 2010

    Commutative is enough ?

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeDec 13th 2010

    I wrote:

    I’ll check later today with the lecture notes.

    Ah, silly me, I thought I’d had to go home to see my hardcopy of Lenstra’s lecture there, but we have the online copy

    All rings in his convention are commutative and unital. I added the adjective “commutative” to where the first ring that is not a field appears at Galois theory. What is typed there is pretty much directly from Lenstra’s notes. So I think it should be right now. But you can all check.

    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeDec 13th 2010

    OK, thanks.

    • CommentRowNumber8.
    • CommentAuthorTobyBartels
    • CommentTimeDec 14th 2010

    Commutative is enough ?

    According to Wikipedia, yes, although there is no citation or proof.