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This question already has an answer here:

It is a standard result that finite fields have cyclic multiplicative groups (the nonzero elements with respect to field multiplication).

A recent discussion on this Question leads me to ask about the converse:

Is a field $F$ whose multiplicative group $F^*$ is cyclic necessarily finite?

Clearly the biggest a cyclic group can be is countable, and thus $F$ would also be at most countable.

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marked as duplicate by Jyrki Lahtonen Sep 15 '15 at 4:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Does the algebraic closure of $\mathbb F_p$ have cyclic multiplicative group? $\endgroup$ – William Stagner Sep 15 '15 at 1:43
  • $\begingroup$ @WilliamStagner: No, since any generator of that algebraic closure would lie in a simple finite algebraic extension of $\mathbb{F}_p$, and thus would generate only that many nonzero powers. $\endgroup$ – hardmath Sep 15 '15 at 1:48
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Yes. If $F^*$ is infinite cyclic, then $F^*$ is torsion-free, and in particular this means that $F$ has only one square root of $1$. Thus $F$ has characteristic $2$, and furthermore $F$ can contain no nontrivial finite extension of $\mathbb{F}_2$ (since any such finite extension would give torsion in $F^*$). Thus $F$ contains a subfield of the form $\mathbb{F}_2(x)$. But the group of units of $\mathbb{F}_2(x)$ is free of infinite rank (generated by the irreducible polynomials over $\mathbb{F}_2$), and in particular cannot be a subgroup of a cyclic group. This is a contradiction.

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