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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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Also the analogue for free monogenic compact semigroups is true. –  Benjamin Steinberg Dec 18 '11 at 5:09
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. –  Theo Johnson-Freyd Dec 18 '11 at 5:54
Thanks Theo!!!! –  Benjamin Steinberg Dec 18 '11 at 13:20
(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.) –  Theo Johnson-Freyd Dec 20 '11 at 5:26
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. –  Tom Leinster Dec 21 '11 at 18:32
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