Problems with proof of Krull's height theorem

Question

I want to understand the proof of next

Theorem. Let $A$ a Noetherian ring and $\mathfrak a=(a_1,...,a_n)$ a proper ideal of $A$. Let $\mathfrak p\in\mathrm{Spec}(A)$ a minimal ideal over $\mathfrak a$. Then $\operatorname {ht}(\mathfrak p) \leq n$.

I don't understand the affirmation: because $\mathfrak p A_{\mathfrak p}$ is minimal over $\mathfrak aA_{\mathfrak p}$ is enough to prove that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak m$, $\mathfrak a=(a_1,...,a_n)$ and $\mathfrak m=\sqrt{\mathfrak a}$ then $\operatorname {ht}\mathfrak m \leq n$. Can you help me? Thank you!

Answer 1

Do you know these facts?

• $A_p$ is local ring with $p A_p$, maximal.
• $aA_p$ is an ideal of $A_p$ which can be generated by $n$ elements.
• $p A_p$ is minimal over $aA_p$.
• $\operatorname {ht} pA_p= \operatorname {ht} p.$
• In a local ring $(R,m)$, if $m$ is minimal over an ideal $J$, then $\sqrt J = m$.

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Now in order to prove that $\operatorname {ht} p \leq n$ one can prove $\operatorname {ht} pA_p \leq n.$ But $p A_p$ is the maximal ideal of local ring $A_p$ and is minimal over an ideal generated with $n$ elements. So it is enough to show that $\operatorname {ht} \leq n$ under the additional hypotheses that the ring is local with $p$ maximal.