SO since I have a field F I know the only ideals of F would be F itself and 0. I'm not sure how to tie this into the fact I have an onto ring homomorphism. i f you can let me know if I'm on the right track that'd be awesome

  • $\begingroup$ A homomorphism out of a field is either injective or the 0 homomorphism. $\endgroup$ – Yeldarbskich Mar 14 '15 at 17:49

By what is given, $R\cong F/\ker\Phi$ and either $\ker\Phi=0$ or $\ker \Phi=F$.

  • $\begingroup$ How do you know R is isomorphic to the factor ring F/kerΦ ? $\endgroup$ – Nicole Mar 14 '15 at 17:48
  • $\begingroup$ 1st Isomorphism Theorem. $\endgroup$ – Nishant Mar 14 '15 at 17:48
  • $\begingroup$ oh cool thanks! So I am missing where the onto ring homomorphism comes in to play $\endgroup$ – Nicole Mar 14 '15 at 17:49
  • $\begingroup$ The first isomorphism theorem says that for $\phi : R\to S$ a ring homomorphism, $R/\ker\phi\cong im(\phi)$, but when $\phi$ is surjective, the image is all of $S$. $\endgroup$ – Stahl Mar 14 '15 at 17:52
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    $\begingroup$ okay, so if I added Since F-->R is an onto ring homomorphism, we know kerΦ is an ideal of F. And then I continue my proof of how the only ideals of F are 0 and itself $\endgroup$ – Nicole Mar 14 '15 at 17:55

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