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Dec 6, 2017 at 12:29 vote accept Ali Taghavi
Nov 29, 2017 at 11:29 comment added მამუკა ჯიბლაძე @YCor Again, I have to agree. Althoug I see at least one interesting topology, coming from the compact-open topology on the space of all continuous maps from $\mathbf C$ to $\mathbf{CP}^1$, indeed any topology on the semigroup itself was never mentioned
Nov 29, 2017 at 11:21 comment added YCor @მამუკაჯიბლაძე A group isomorphism is trivial. I was not claiming more since I'm not aware of any canonical topology, so saying that "topologies match" does not make any sense to me. No topology was even specified on the semigroup!
Nov 29, 2017 at 10:45 comment added მამუკა ჯიბლაძე @YCor You are right. At least, a bijection is indeed trivial. Whether topologies match I don't see so quickly.
Nov 29, 2017 at 10:35 comment added YCor @მამუკაჯიბლაძე it's trivially the case, isn't it?
Nov 29, 2017 at 7:10 comment added მამუკა ჯიბლაძე This is really nice! I wonder if this Grothendieck group can be in turn naturally identified with the group of rational functions with zeroes and poles confined to the unit circle
S Nov 29, 2017 at 5:37 history suggested CommunityBot CC BY-SA 3.0
Added paragraph breaks.
Nov 29, 2017 at 5:07 review Suggested edits
S Nov 29, 2017 at 5:37
Nov 29, 2017 at 5:06 comment added Julian Rosen Since the question asks about all polynomials, not just the monic ones, it seems an extra factor of $\mathbb{C}^\times$ is needed.
Nov 28, 2017 at 22:54 history answered Anton Fetisov CC BY-SA 3.0