TITLE: Atiyah and Macdonald, exercise 11.7
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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 $I$.