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Let $F$ be a field. When happens that the additive group of $F$ is isomorphic to the multiplicative group?

It is easily to work out that $F$ must have characteristic $0$, but then what?

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  • $\begingroup$ Consider the case of finite fields. Every finite field is of odd order, so there does not exist a non-trivial element $x$ such that $2x = 0$. But there will always exist a non-trivial $y$ such that $y \cdot y = 1$. $\endgroup$ – Myridium Apr 5 '15 at 22:28
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    $\begingroup$ Not every finite field is of odd order...in fact, there are infinite finite fields of even order. $\endgroup$ – Timbuc Apr 5 '15 at 22:36
  • $\begingroup$ It is meant every finite field of characteristic not $2$ is of odd order. If the characteristic is $2$ then this argument fails to be valid since every element is of order $2$ in the additive group. However, clearly there is no group isomorphism between $\mathbb{F}_{2^n}$ and $\mathbb{F}_{2^n}^\times$ for any $n$ as they have different orders as groups. So the argument is valid. $\endgroup$ – Eoin Apr 5 '15 at 22:50
  • $\begingroup$ @Eoin Thank you for that, I assumed what Myridium meant, I just remarked a careless phrase in that comment, that's all. $\endgroup$ – Timbuc Apr 6 '15 at 8:06
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If you already have that char$\,\Bbb F=0\;$ then what does $\;-1\;$ map to? This is an element of order two in $\;\Bbb F^*\;$ so it must map to an element of order two in $\;\Bbb F\;$ .

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  • $\begingroup$ It was easy... Thanks! $\endgroup$ – W4cc0 Apr 5 '15 at 22:40

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