This is a question about the hint in exercise 11.3.I part (b) of Vakil's FOAG notes, which is to prove the existence of a system of parameters for a Noetherian local ring. The statement of the full exercise is:
Suppose $(A,\mathfrak m)$ is a Noetherian local ring.
(a) Use Krull's Height Theorem to prove that if there are $g_1,\dots ,g_l$ such that $V(g_1,\dots ,g_l) = \{ \mathfrak m \}$, then $\dim A \leq l$.
(b) Let $d = \dim A$. Show that there exist $g_1,\dots ,g_d\in A$ such that $V(g_1,\dots ,g_d) = \{ \mathfrak m \}$. Hint: use induction on $d$. Find an equation $g_d$ knocking the dimension down by 1, i.e. $\dim A/(g_d) = d - 1$. Suppose $\mathfrak p_1, \dots ,\mathfrak p_n$ correspond to the irreducible components of ${\rm Spec} A$ of dimension $d$, and $\mathfrak q_i \supset \mathfrak p_i$ are prime ideals corresponding to irreducible closed subsets of codimension 1 and dimension $d-1$. Use prime avoidance to find $h_i \in \mathfrak q_i\setminus \cup_{j=1}^n \mathfrak p_j$. Let $g_d = \Pi_{i=1}^n h_i$.
Proceeding with the inductive step as in the bolded hint, we can pick out the $\mathfrak q_i$ of both codimension 1 and dimension $d-1$ because we can just take them to be the second prime ideal in every chain of length $d$ starting with the minimal prime $\mathfrak p_i$. Because $\mathfrak q_i$ has height 1, it cannot be contained in any of the $\mathfrak p_j$, hence the prime avoidance lemma applies and we can find an element $h_i$ as stated.
The point I don't understand is this: why do we need to take the product of the $h_i$ at the end? It seems the purpose of doing that is to ensure that $g_d$ is contained in every $\mathfrak q_i$. But isn't it already true that setting $g_d = h_1$ gives a quotient $A/(g_d)$ which is a Noetherian local ring of dimension $d-1$, since we're avoiding all the minimal primes? What am I missing?
If you were to take $g_d = h_1$, then indeed $\text{Spec }A/(g_d)$ has an irreducible component (corresponding to $q_1$) of dimension $d - 1$. But this does not ensure that $A/(g_d)$ is pure of dimension $d - 1$. We could now have irreducible components of $A/(g_d)$ of dimension smaller than $d-1$, not corresponding to any of the $\mathfrak{q}_i$.