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I am trying to solve the exercise in Atiyah, that $\dim(A[X]) = \dim (A) + 1$ for $A$ noetherian.
The very beginning poses a problem, he states in the hint that:

for a prime of height $m$ we can choose $m$ elements in that prime such that the prime is a minimal prime over the ideal generated by those $m$ elements.

How might one prove that first statement? And is there an alternative approach to proving this?

user26857
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baltazar
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  • If $A$ is noetherian, so is $A[X]$. – Olórin Jan 26 '15 at 15:45
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    So? How does that help prove that the prime is minimal over the ideal? – baltazar Jan 26 '15 at 15:47
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    This is fully proved in many books. (Can find an answer to your concrete question here, Theorem 10.1.2(3).) And I don't know if there is an alternative approach, since what you asked for shows a crucial thing: the height of a prime $p$ equals the height of $p[X]$. – user26857 Jan 26 '15 at 16:21
  • @user26857 the references helped. Thank you – baltazar Jan 26 '15 at 21:23
  • By (11.13) there are $a_1,...,a_m\in\mathfrak p$ such that $(a_1/1,...,a_m/1)$ is $\mathfrak p_{\mathfrak p}$-primary. Then $\mathfrak p$ is minimal over $(a_1,...,a_m)$. – Pierre-Yves Gaillard Oct 30 '18 at 18:28

1 Answers1

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for your first Question; the book Steps in Commutative Algebra (bySharp) Theorem 15.13:

Let $R$ be a commutative Noetherian ring and let $P \in Spec(R)$; suppose that $htP = n$. Then there exists an ideal $I$ of $R$ which can be generated by $n$ elements, has $ht\ I = n$, and is such that $I \subseteq P$
note that $ht\ I=ht\ P$, so $P$ is minimal prime ideal of $I$.

user 1
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