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Let $F$ be any field of zero characteristic, $F^{\ast}$ its multiplicative group and $T$ is the torsion group. Is it true that $T$ is a direct summand for $F^{\ast}$?

edited Aug 26 2011 at 10:37

Gjergji Zaimi
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asked Aug 26 2011 at 9:15

Konstantin
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His conjecture holds for any number field, since $F^* / T$ is free. – Hurkyl Aug 26 2011 at 10:12

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$F^*/T$ is certainly torsion-free, but for $F=\mathbb{C}$ it is (uniquely) divisible and therefore not free. In that case we still have a splitting because $T$ is also divisible and so is injective in the category of abelian groups. – Neil Strickland Aug 26 2011 at 10:28

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It's too bad that some people use "number field" to mean subfield of $\mathbb C$, while others use "number field" to mean finite extension of $\mathbb Q$, and probably others use it to mean algebraic extension of $\mathbb Q$. – Tom Goodwillie Aug 26 2011 at 18:31

Gjergji Zaimi · Accepted Answer · 2011-08-26 10:38:38Z UTC

This was a problem that was asked by Fuchs in his book "Abelian groups" (1958). It was first solved in negative by P. M. Cohn in "Eine Bemerkung uber die multiplikative Gruppe eines Korpers", Arch. Math. (Basel) 13 (1962) 344-48. (MR0146252). Later W. May gave a counterexample as an algebraic extension of $\mathbb Q$, in "Multiplicative groups of fields", Proc. London Math. Soc. (3) 24 (1972), 295–306. (MR0294490)

torsion group of the multiplicative group of a field

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