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Let $F$ be a infinite field of characteristic 2 whose multiplicative group $F^*$ is torsion free. I would like to conclude that $F^*$ is decomposable or find an example where $F^*$ is indecomposable.

If $F^*$ is (direct-sum) indecomposable, then:

  • $F$ cannot be purely transcendental over $F_2$
  • $F^*$ has infinite rank.

I don't know anything beyond that.

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  • $\begingroup$ FWIW, my instinct would be to try looking at something like the field of Hahn series over $F_2$ with value group $\mathbb{Q}$. But I'm not sure how to calculate the multiplicative group, and it's getting late where I am. $\endgroup$
    – Todd Trimble
    Commented Jun 14, 2014 at 4:33
  • $\begingroup$ @ToddTrimble: It seems to me that this field is perfect and henselian, hence every element has $n$th roots for every $n$ (since this trivially holds for the residue field). So, the multiplicative group is just a vector space over $\mathbb Q$. I think there might be more hope with a discrete value group. $\endgroup$
    – Emil Jeřábek
    Commented Jun 14, 2014 at 11:10
  • $\begingroup$ And I’m talking nonsense with discrete valuations, because if the value group is free, $K^\times$ splits as $O^\times\oplus\Gamma$. $\endgroup$
    – Emil Jeřábek
    Commented Jun 14, 2014 at 12:38
  • $\begingroup$ Hi Sunil. I don't really have any ideas, but I wondered if you knew any examples of fields (apart from finite fields of order one more than a prime power) where the multiplicative group is indecomposable? $\endgroup$
    – Jeremy Rickard
    Commented Jun 18, 2014 at 15:47
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    $\begingroup$ Thanks for the reference! First counterexample published by my mathematical grandfather in the year of my birth: I should probably have known about that! $\endgroup$
    – Jeremy Rickard
    Commented Jun 18, 2014 at 18:09

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