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Let $R$ be a Noetherian ring, and let $\mathfrak p$ be a prime of $R$ of codimension $d$. Suppose that $P\subset R[X]$, $P$ prime, intersects $R$ in $\mathfrak p$. Prove that if $P\neq\mathfrak pR[X]$, then codim $P=d+1$ and that there are infinitely many $P$ like this.

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So, what's your question? –  Gerry Myerson Feb 9 '12 at 11:04
    
Proof that codim Q=d+1 and we have infinitely many Q like this.I forgot the word "proof". –  Strongart Feb 11 '12 at 11:02
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Then I suggest you edit the question accordingly. That will also have the effect of bringing it to the front page to increase its visibility on the site. –  Gerry Myerson Feb 11 '12 at 23:14
    
I see,thank you. –  Strongart Feb 12 '12 at 10:46

1 Answer 1

up vote 2 down vote accepted

For the part about codimension (= height) see Theorem 149 from Kaplansky, Commutative Rings, 1974.

The second part can be easily proved by reduction modulo $\mathfrak p$: reduce all by $\mathfrak p$, get an integral domain $R$ and try to prove that there are infinitely many primes in $R[X]$ lying over $(0)$. These are in a one-to-one correspondence with the prime ideals of $K[X]$, where $K$ is the field of fractions of $R$. Now try to prove that $K[X]$ has infinitely many primes, that is, there are infinitely many irreducible polynomials in $K[X]$.

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would the downvoter care to comment? –  robjohn Oct 17 '12 at 22:50

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