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I am asked to prove that every prime ideal P of a ring R can be obtained as the kernel of a homomorphism to a field.

I know that the kernel of a homomorphism is an ideal. I need to start from an arbitrary prime ideal and show that it is the kernel of a homomorphism. So it would seem that this would not help, since it would be proving something (that it is an ideal) that I am already assuming (the prime ideal).

Could someone please give ideals how to prove this?

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Hints:

The projection $R\to R/P$ is a homomorphism of $R$ to a domain, and this map's kernel is $P$.

Integral domains embed in their field of fractions.

Can you take it from here?

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  • $\begingroup$ I see how R/P is an integral domain, but I do not understand its embedding into the field of fractions. $\endgroup$ – math1234567 Mar 29 '14 at 13:48
  • $\begingroup$ @violin.lover Then you should read about it $\endgroup$ – rschwieb Mar 29 '14 at 20:24

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