This is an exercise in Ravi Vakil's AG notes. My question is about part b. Namely, in the induction step recommended in the hint, why can't I just choose $g_d$ to be any element outside $\cup_j \mathfrak{ p}_j$? Why do I need $g_d$ to be in all of the $\mathfrak{ q}_i$? Maybe the dimension of $A/(g_d)$ will go down by more than $1$, but that just means I would need fewer than $d$ generators in the statement of the proposition (and in fact that is prohibited by part a). It makes me think I am misunderstanding something fundamental.
Edit: I have typed the exercise below.
Let $(A, \frak m)$ be a Noetherian local ring.
b) Let $d= \dim A$. Show that there exist $g_1,…,g_d\in A$ such that $V(g_1,…,v_d) = \{[\frak m]\}$.
Hint: use induction on $d$. Find an equation $g_d$ knocking the dimension down by 1, i.e., $\dim A/(g_d) = \dim A -1$. Suppose $\frak p_1,…,p_n$ correspond to the irreducible components of $Spec A$ of dimension $d$, and $\frak p_i \subset q_i$ are prime ideals corresponding to irreducible closed susbsets of codimension $1$ and dimension $d-1$. Use prime avoidance to find $h_i \in \frak q_i- \cup_{j=1}^n p_j$. Let $g_d= \Pi_{i=1}^n h_i.$
Best Answer
Yes, your argument looks absolutely correct to me, and in fact you can use it to prove the following more general fact:
Lemma 1: If $A$ is a (commutative) Noetherian ring, and $\mathfrak{p}<A$ is a prime of height $d$, then there exist $g_1,\dots,g_d\in \mathfrak{p}$ such that $\mathfrak{p}$ is a minimal prime over $\langle g_1,\dots,g_d\rangle$.
Proof: By induction on $d$. The case $d=0$ is clear, since then $\mathfrak{p}$ is minimal over the zero ideal, which is generated by $\varnothing$. For the inductive step, suppose $d>0$. Now, since $A$ is Noetherian, it has finitely many minimal prime ideals; let $\mathfrak{q}_1,\dots,\mathfrak{q}_k$ be a complete list. Since $\operatorname{ht}\mathfrak{q}_i=0<d=\operatorname{ht}\mathfrak{p}$ for each $i$, we have $\mathfrak{p}\nsubseteq \mathfrak{q}_i$ for each $i$, and hence by prime avoidance we may find $g_d\in \mathfrak{p}\setminus\bigcup_{i=1}^k\mathfrak{q}_i$. Then the prime ideal $\mathfrak{p}\big/\langle g_d\rangle$ of the Noetherian ring $A\big/\langle g_d\rangle$ has height strictly smaller than $d$ – say $\operatorname{ht}\mathfrak{p}\big/\langle g_d\rangle=l$ – and now by the inductive hypothesis there exist $g_1,\dots,g_{l}\in A$ such that $\mathfrak{p}\big/\langle g_d\rangle$ is a minimal prime over $\langle \bar{g}_1,\dots,\bar{g}_{l}\rangle$. Now $\mathfrak{p}$ is a minimal prime over $\langle g_1,\dots,g_l,g_d\rangle$, so we are done. $\blacksquare$
Note that the height actually is knocked down by precisely one; in other words, in the proof of Lemma 1, we actually must have $l=d-1$. Why? Well, by Krull's height theorem, if $\mathfrak{p}$ is minimal over $\langle g_1,\dots,g_l,g_d\rangle$ then $\operatorname{ht}\mathfrak{p}\leqslant l+1$. But $l<d$, and $\operatorname{ht}\mathfrak{p}=d$, so this forces $l=d-1$, as desired. So your proof in fact subsumes Ravi's completely. Also, because of this fact, ie since $l=d-1$, we may actually use the same argument to prove a slightly stronger fact:
Lemma 2: If $A$ is a (commutative) Noetherian ring, and $\mathfrak{p}<A$ is a prime of height $d$, then there exist $g_1,\dots,g_d\in \mathfrak{p}$ such that $\langle g_1,\dots,g_d\rangle$ has height $d$. $\blacksquare$
The proof is essentially identical, although it does invoke Krull's height theorem, which is not needed in Lemma 1.