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As the title. Or rather, for any integer $m$ which is not the characteristic, does such an 'integer division' exist?

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  • $\begingroup$ Why the vote to close and downvote? This seems a reasonable question. $\endgroup$ – Noah Schweber Apr 13 '16 at 21:22
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The crucial fact is the following: if $k$ is a field of characteristic zero, there is a natural embedding $e$ of $\mathbb{Q}$ into $k$.

To prove this, first note that subring of $k$ generated by $1_k$ is isomorphic to $\mathbb{Z}$ (since the characteristic is zero), yielding an embedding $d$ of $\mathbb{Z}$ into $k$. Now we can extend this embedding of $\mathbb{Z}$ to an embedding $e$ of $\mathbb{Q}$ as desired: $e({p\over q})=d(p)\times d(q)^{-1}$.

Division by an integer $n$ in $k$ is then conducted by multiplying by $e({1\over n}$).

Exercise: by a similar argument, embed $\mathbb{Z}/p\mathbb{Z}$ into any field of characteristic $p$.

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As long as $m \neq 0$ it must have an inverse. The fact that $m$ is integer does not matter.

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  • $\begingroup$ But an integer is not necessarily in an arbitrary field. Here by mx, I mean n times' addition of x. $\endgroup$ – Piccaberini Apr 13 '16 at 19:34
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    $\begingroup$ All integers make sense in any ring. But they can be zero sometimes. $\endgroup$ – Captain Lama Apr 13 '16 at 19:38

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