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This question is motivated by an attempt to understand what is going on in Tom's post from a certain point of view.

Let $\mathbb N^+$ be the free semigroup on one generator (so the positive natural numbers with addition). Is it true that the underlying semigroup of the pro-algebraic completion of $\mathbb N^+$ can be described as $\mathbb N^+\sqcup G$ where $G$ is the pro-algebraic completion of the additive group of integers, where adding an element of $\mathbb N^+$ to an element of $G$ is done by first identifying an element of $\mathbb N^+$ with its copy in $\mathbb Z\subseteq G$?

Assume the ground field is algebraically closed if you like.

Remark: If you replace pro-algebraic by profinite, then it is true.

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  • $\begingroup$ Also the analogue for free monogenic compact semigroups is true. $\endgroup$ Dec 18, 2011 at 5:09
  • $\begingroup$ I have made a very small style edit --- if I have trod on your style, I apologize, and you should rever the edit. I notices that there was unneeded TeX in the title, which slows load times for some users, so I removed that, and in the process I made a similar change in the body. $\endgroup$ Dec 18, 2011 at 5:54
  • $\begingroup$ (Incidentally, "rever" = "reverse", of course, and apparently I hit a keyboard neighbor when I meant to hit a "d", among other typos in my previous comment.) $\endgroup$ Dec 20, 2011 at 5:26
  • $\begingroup$ Could you give a link to a definition of the pro-algebraic completeion of a semigroup (or give a definition yourself)? I'm having trouble finding it. Thanks. $\endgroup$ Dec 21, 2011 at 18:32
  • $\begingroup$ The definition has only been made for groups on the literature but makes sense for semigroups. A pro-algebraic semigroup is a projective limit of (linear) algebraic semigroups. The pro-algebraic completion is the left adjoint of the forgetful functor from pro-algebraic semigroups to semigroups. $\endgroup$ Jan 6, 2012 at 1:49

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