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Show that additive group of field of characteristic 0 is not cyclic. If it is so then the additive group will be isomorphic to $\Bbb Z$ from here how do I proceed.

I have seen Why must a field with a cyclic group of units be finite?, but here they deal with cyclic units.

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    $\begingroup$ Extended hints: - A field of characteristic zero has a copy of $\Bbb{Q}$ as a subfield, hence as a subgroup of the additive group. - A subgroup of a cyclic group is itself cyclic. - Can you show that $(\Bbb{Q},+)$ is not cyclic? $\endgroup$ – Jyrki Lahtonen Dec 1 '14 at 19:44
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As suggested in the hint, since a field of characteristic 0 contains a subring isomorphic to $\mathbb{Z}$ it also contains the field of all of its quotients, which is isomorphic to $\mathbb{Q}$. Every subgroup of a cyclic group is cyclic, but $\mathbb Q$ is not cyclic. To see this, note that in a cyclic group we have that for any nonzero element $x$ there is a maximum value of $n\in \mathbb Z$ such that there exists a $y$ in the group with $ny=x$ (such $n$ and $y$ essentially give a factorization of $x$, of which there are finitely many). However, in $\mathbb Q$ we have on the contrary that for any $x$ and any nonzero integer $n$ there exists a $y$ such that $ny=x$ (take $y=x/n$, identifying $n$ with an element of the field), so in particular there is no maximum $n$. Thus the additive group of the rational numbers is not cyclic.

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Suppose, on contrary, $\mathbb{F}$ is a field of characteristic zero and $(\mathbb{F},+)$ is cyclic generated by $\xi$, i.e. $\mathbb{F}=\langle\xi\rangle$.

NOTE: This is only possible when $\xi$ is the smallest positive element (or the biggest negative element) in $\mathbb{F}$ (why?)

Food for thought: Is there such an element in $\mathbb{F}$ considering the fact that $\mathbb{F}$ has characteristic zero (contains $(\mathbb{Q},+)$ as a subgroup)?

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