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Dec 18, 2018 at 19:43 vote accept 352506
Dec 17, 2018 at 20:23 answer added Will Sawin timeline score: 1
Dec 10, 2018 at 18:49 comment added KConrad I don't know what proof you have in mind for the second version, but all you're doing to define $g$ in terms of $f$ is adding values of $f$, hence only the additive structure is relevant. Does your second proof really involve multiplying elements of $R$ together where neither term is an integer?
Dec 10, 2018 at 14:38 comment added Andreas Blass The convolution doesn't seem to use the multiplicative part of the ring structure. You're convolving a sequence $g$ of elements of $R$ with a sequence $\mu$ of natural numbers, and that makes sense in any additive abelian group.
Dec 10, 2018 at 14:16 comment added 352506 Perhaps I should reformulate my question in terms of the proofs of the formulas. The proof of the second version uses the multiplicative structure of $R$ (via the convolution). Is there an analogous proof of the first version?
Dec 10, 2018 at 2:35 comment added KConrad The second version never uses the ring structure in an essential way. It is formulated only with the additive group structure of ${\mathcal A}(R)$, so version 2 is really a special case of version 1. The "common generalization" is just version 1.
Dec 10, 2018 at 2:01 history asked 352506 CC BY-SA 4.0