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Let R be an integral domain. Then how to show that intersection of non zero prime ideals of R[x] is zero.

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Hint $\ $ If $\,f\,$ is in every prime ideal $\ne 0\,$ then $\,1+xf\,$ is in no maximal ideal so is a unit, so $\,f = 0.$

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  • $\begingroup$ Can we prove this using localisation of rings $\endgroup$ – user156736 Jun 12 '14 at 14:34
  • $\begingroup$ @user156736 You could, but why use more complex methods when such a simple proof exists? $\endgroup$ – Bill Dubuque Jun 12 '14 at 15:01
  • $\begingroup$ Dear BillDubuque : I don't know, I think that's a legitimate question that shouldn't have been invalidated. Some students get out of control with such questions, but this particular one looks fine. Perhaps you would consider withdrawing your hasty dismissal of the student's curiosity? @user156736 Regards $\endgroup$ – rschwieb Jun 12 '14 at 17:04
  • $\begingroup$ Dear @BillDubuque : If you don't recognize that Why do that when you have my proof? dismisses the poser's request, then yes, you speak a very different language than I do, and there is little else that can be said. Hopefully the poster pursues the line of thought despite your admonition. Regards $\endgroup$ – rschwieb Jun 13 '14 at 13:28
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    $\begingroup$ @rschwieb To clarify, my question to the OP was meant sincerely. Namely, I believe that understanding the motivation for seeking alternative proofs can be very useful when teaching. Please try to view things from a more positive standpoint when there is ambiguity. $\endgroup$ – Bill Dubuque Jun 13 '14 at 14:02
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If $p(x) \in R[x]$ is nonzero with leading coefficient $a$, then $p(x)^n\neq 0$ since the leading coefficient of $p(x)^n$ is $a^n$ and $a^n \neq $ as $R$ is an integral domain. Assume for the moment that the constant coefficient of $p(x)$ is nonzero, we have that $p(x) \not \in (x)$ and by a well known application of Zorn's lemma gives a prime ideal extending $(x)$ not containing $p(x)$. We can of course factor $x$ out of $p(x)$ to reduce to the last case, this leaves only the possibility that $p(x)=ax^k$ but then $p(x)\not \in (x-1)$, and the same argument proves that there is a prime ideal not containing $p(x)$

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  • $\begingroup$ Easier: any prime ideal containing the nonunit $\,1+x p(x),\,$ does not contain $\,p(x)\ne 0,\,$ see my answer. $\endgroup$ – Bill Dubuque Jun 12 '14 at 15:47
  • $\begingroup$ Sir as I am begineer in this paper so would you simplify a bit for me.. $\endgroup$ – user156736 Jun 15 '14 at 9:00

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