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Suppose $F$ is a field and there is a ring homomorphism from $\mathbb Z$ onto $F$. Show that $F \cong \mathbb Z_p$ for some prime $p$.

My try: If $F$ is of finite characteristics then it prime. So by first isomorphism theorem $F \cong \mathbb Z/p\mathbb Z$ and we know that $\mathbb Z/p\mathbb Z \cong \mathbb Z_p$ and hence the proof and on the other hand if $F$ is not of finite characteristics then its characteristics is zero and hence....

Is it correct explanation?? Please Help!!

Thank You!!

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Hint
Let $\sigma$ be a ring homomorphism from $\mathbb Z$ to $F.$ Then what is the kernel of $\sigma?$
And what does the kernel of $\sigma$ tell us about $F?$
A further explanationcan be added if needed, hope this helps.

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  • $\begingroup$ yes... $p\mathbb Z$ will be the kernel for a field with char. p $\endgroup$ – user8795 Jan 5 '15 at 2:11
  • $\begingroup$ And can the kernel of $\sigma$ be $(0)$? Namely, can $F$ have $0$ characteristic? $\endgroup$ – awllower Jan 5 '15 at 2:12
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    $\begingroup$ So, is $\mathbb Z$ a field? $\endgroup$ – awllower Jan 5 '15 at 2:14
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    $\begingroup$ Thus you have finished the proof, right? :) $\endgroup$ – awllower Jan 5 '15 at 2:16
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    $\begingroup$ @GPerez The condition says that the homomorphism is surjective. $\endgroup$ – awllower Jan 5 '15 at 4:20

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