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Theorem: If $R$ is an integral domain then the intersection of all nonzero prime ideals of $R[x]$ is zero.

The proof of this theorem(Kaplansky) in the paper by J Peter May but it is proved by using concepts of R-algebra and few more lemmas on it. I was wondering if it can be done in more simpler(elementary) way as alternative so that a beginner can understand it. Can any one please figure it out... Thank you

Bill Dubuque
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Tamang Rigen
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    Funnily enough, I was talking to Dr. May as I read this problem, and I read him the question. He said, "if there was a remarkably simpler way to prove this, don't you think I would have proved it that way?" – Rushabh Mehta Jun 26 '21 at 15:11
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    Are you missing some assumptions? For example, for the ring $R = \mathbb{R}[x] / \langle x^2 \rangle$, the intersection of all prime ideals (all of which are nonzero) is $\langle x \rangle$. – Daniel Schepler Jun 26 '21 at 15:12
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    @Don Thousand -- Mathematics is full of theorems whose original proofs were far more complicated than necessary. – uniquesolution Jun 26 '21 at 15:13
  • @Danial Schepler u can visit the page https://www.google.com/search?q=munshi%27s+proof+of+the+nullstellensatz&oq=munsi+proof+&aqs=chrome.1.69i57j0i13i30.5189j0j4&client=ms-android-motorola-rev2&sourceid=chrome-mobile&ie=UTF-8 – Tamang Rigen Jun 26 '21 at 15:33
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    @uniquesolution I know, it was a joke on his part. – Rushabh Mehta Jun 26 '21 at 15:37
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    If $R$ has nonzero nilpotents, this is false. – Mohan Jun 26 '21 at 15:42
  • You forgot the crucial hypothesis (in paragraph $2)$ "the letter $R$ will always stand for an integral domain". The proof is essentially the same as in Theorems $18-21$ in Kaplansky's Commutative Rings. All the proofs are very easy - a few lines using only basic ring theory, and all obvious (except possibly his Theorem $1$ (see this answer), which is also a few lines of simple algebra, along with Zorn's Lemma - a basic result for localization). If this is not "simple" to you then please explain why, so we can help you past any obstacles. – Bill Dubuque Jun 26 '21 at 21:11
  • See here for one simpler way, a la Euclid's classical proof: $ $ if $,f,$ is "diviible by" (i.e. in) all max primes, then $1+xf,$ is divisible by no max primes, so it is a unit, so $,f = 0,$ by $R$ a domain. $\ \ $ – Bill Dubuque Jun 27 '21 at 02:47
  • @ Bill Dubuque can u please elaborate a bit about Euclid classical proof I m not getting it – Tamang Rigen Jun 27 '21 at 04:59

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