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In page 4 of Lang's Introduction to Algebraic Geometry, Proposition 1 states that if $a^n = 1$ then $a = 1$ or $n = 0$ and decudes that the map $a \mapsto a^n$ is an isomorphism from $\Gamma \rightarrow \Gamma$. Here $\Gamma$ is a commutative multiplicative group with ordering, i.e. $\exists S$ such that $S$ is a semigroup and $\Gamma = S \cup 1 \cup S^{-1}$ is a disjoint union.

However, I think this is false because we can take $\Gamma$ to be the set of positive rational numbers and the semigroup $S$ to be the set of rationals less than 1. The power map clearly cannot be surjective (2 is not the square of a rational for example).

I'm wondering if this is just a typo or rather we are implicitly making a connection between $\Gamma$ and the multiplicative group of some algebraically closed field? The section talks about valuations and places within the context of algebraically closed fields so that would make sense. Any help is appreciated, thanks.

green frog
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  • Torsion-freeness implies that the map is injective, but not necessarily surjective. – Geoffrey Trang Jul 07 '21 at 01:23
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    Lang doesn't deduce that the power map is an isomorphism from $\Gamma$ to $\Gamma$. He deduces that it is an isomorphism from $\Gamma$ into itself. These don't mean the same thing. An isomorphism from a set into itself is an isomorphism of the set with its image under the map. That is, Lang says it's an isomorphism of $\Gamma$ with the range of the power map, not necessarily with the codomain of the power map. – Gerry Myerson Jul 07 '21 at 04:53
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    @GerryMyerson I see. That makes perfect sense. Thank you! – green frog Jul 07 '21 at 05:43
  • Homomorphism:https://math.stackexchange.com/q/1132336/688539 – Clemens Bartholdy Mar 10 '22 at 00:46

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