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Show that if a field $F$ has characteristic $0$, then it is of infinite order.

Please someone help me. It is an unsolved exercise in my book, it is not homework.

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    $\begingroup$ If $n\times 1\neq 0$ for every $n$ then these elements are all different $\endgroup$ – user8268 Sep 15 '13 at 13:51
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    $\begingroup$ Well, suppose that F has finitely many elements. Then the subset {1, 1+1, 1+1+1,...} of F has only finitely many distinct elements. What would that imply? $\endgroup$ – Keshav Srinivasan Sep 15 '13 at 13:52
  • $\begingroup$ If it were finite, adding the multiplicative identity $1$ a number of times (say $p$) gets you the additive identity $0$. $\endgroup$ – mrk Sep 15 '13 at 13:55
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A ring $R$ has characteristic zero if there is a monomorphism $\phi : \mathbb{Z} \to R$. What does this imply about the relationship between $|\mathbb{Z}|$ and $|R|$?

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    $\begingroup$ Surely you meant to write $|R|$ at the end of your answer. $\endgroup$ – Asaf Karagila Sep 15 '13 at 14:02
  • $\begingroup$ Indeed. But $\mathbb{R}$ also has characteristic zero... $\endgroup$ – Elchanan Solomon Sep 15 '13 at 14:18
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    $\begingroup$ Beautiful solution. I can go to bed happy. $\endgroup$ – Daniel Montealegre Sep 18 '13 at 4:53
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A field with characteristic zero contains a subfield isomorphic to Q. The results follows by Q

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