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Aug 24, 2010 at 18:44 vote accept TonyS
Aug 24, 2010 at 18:44 comment added TonyS Thanks. This looks better. I get the dual of the module you specified by transposing and then replace the $t$'s by $t^{-1}$? Then taking the determinant gives a polynomial in $t^{-1}$ with the desired pole order. So one does not need $\mathbb{Q}$-divisors at all, very good!
Aug 24, 2010 at 17:17 history answered Colin Ingalls CC BY-SA 2.5