6

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
  • 1
  • 13
  • 62
  • 125
baltazar
  • 1,427
  • 9
  • 15
  • If $A$ is noetherian, so is $A[X]$. – Olórin Jan 26 '15 at 15:45
  • 1
    So? How does that help prove that the prime is minimal over the ideal? – baltazar Jan 26 '15 at 15:47
  • 1
    This is fully proved in many books. (Can find an answer to your concrete question [here](https://books.google.ro/books?id=HYutxKuoIx8C&pg=PA170&lpg=PA170&dq=%22I+is+minimal+over+%28a1,+...+,+an%29&source=bl&ots=rn5AKhuRMs&sig=iK-oCZU1C8bfIA6qcZGW6OCIzh0&hl=en&sa=X&ei=eWjGVLnnOdLgaJn4gdAL&ved=0CCEQ6AEwAA#v=onepage&q&f=false), 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

2

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
  • 6,986
  • 8
  • 27
  • 47
  • also you can mark [this](http://math.stackexchange.com/questions/358423/a-short-proof-for-dimrt-dimr1) as favorite. – user 1 Jan 26 '15 at 16:24