Vakil FOAG 11.3.B I am thinking about how to use Krull's PIT to prove this statement (11.3.B on Vakil's notes):
If $(A,m,k)$ is a Noetherian local ring with maximal ideal $m$, and $f \in m$, then $\dim  A/(f) \geq \dim A-1$.
What puzzles me is that Krull's PIT only gives us information about the codimension of $V(f)$. How can I know its dimension from its codimension?
 A: Since the asker mentions in a comment to the other answer that this exercise is done before developing much theory of dimension of local rings, here's a short, self-contained proof. In fact, this proof can be interpreted geometrically and used almost word for word to prove problem $11.3.C$ in Vakil, that in projective space, hyperplane intersections reduce dimension by at most one, amongst other things. I'll put the geometric interpretation at the end.
Take any maximal chain of primes in $A, \mathfrak{p}_0\lneq \dots \lneq \mathfrak{p}_n$ (so that $n = \dim(A)$ and $\mathfrak{p}_n = \mathfrak{m}$). Then since $f \in \mathfrak{m}$, there is some $k$ such that $f \in \mathfrak{p}_k \setminus \mathfrak{p}_{k-1}$ (or $k = 0$). If $k = 0$, then $\dim(A/(f)) \geq n$ so we're done. If not, then there is some prime $\mathfrak{p}_{k-1}+(f)\leq \mathfrak{q}_{k-1}\leq \mathfrak{p}_k$ minimal with respect to this property. Then, repeating this process, there is some prime $\mathfrak{p}_{k-2}+(f) \leq \mathfrak{q}_{k-2} \leq \mathfrak{q}_{k-1}$ minimal with this property, giving a chain of primes $\mathfrak{q}_0\leq \dots \leq \mathfrak{q}_{k-1} \leq \mathfrak{p}_k \lneq \dots \lneq \mathfrak{p}_n$ which all contain $f$. We may have that $\mathfrak{q}_{k-1} = \mathfrak{p}_k$, but I claim that all the $\mathfrak{q}_i$ are distinct, which proves the theorem.
Indeed, suppose $\mathfrak{q}_i = \mathfrak{q}_{i+1}$. Then by construction, $\mathfrak{q}_{i+1}$ is minimal over $\mathfrak {p}_i + (f)$. Then by Krull's theorem (note that this is the one and only time we apply Krull in the proof) $\mathfrak{q}_{i+1}$ has height $0$ or $1$ over $\mathfrak {p_i}$. But $\mathfrak {p_i} \lneq p_{i+1} \lneq \mathfrak{q}_{i+1}$ (where we have the second inequality by the fact that $f \in \mathfrak{q}_{i+1} \setminus \mathfrak{p}_{i+1}$) so in fact $\mathfrak{q}_{i+1}$ has height at least $2$ over $\mathfrak{p_i}$, a contradiction.
Geometric interpretation:
This is a purely algebraic statement, but the proof is, at it's heart, geometric (and this is the more natural way to come up with it). What this amounts to is showing that the hyperplane $V(f) \subset X=\rm{Spec}(A)$ has dimension at least $\dim(A)-1$ (although in fact it proves the slightly stronger statement that the intersection of any irreducible component with $V(f)$ has codimension at most one less).
In the proof, we take $X_0\lneq \dots \lneq X_n$ a maximal chain of irreducible closed subsets in $X$. We observe that since $X_0 \subset V(f)$, all of the intersections are non-empty, and take $r$ such that $X_r$ is contained in $V(f)$ and $X_{r+1}$ isn't, so that $X_0 \lneq \dots \lneq X_r$ remains a chain of irreducible closed subsets in $V(f)$. We then pick out irreducible components $Y_i$ of $X_i\cap V(f)$ iteratively, giving a chain $X_0 \lneq \dots \lneq X_r \leq Y_{r+1}\leq \dots \leq Y_n$. We then complete the proof by showing that all of the $Y_i$ are distinct, finishing the proof. I think it's interesting that the proof makes a lot more sense geometricallt, you're really just following your nose until it's time to show that the $Y_i$ are distinct, which seems very hard to do geometrically. Recall that in the algebraic proof, this was where we used Krull's principal ideal theorem and it was relatuvely easy, the hard part was coming up with the strategy, which turns out to be the "obvious" strategy geometrically.
Remark: We only actually used that $A$ was a local ring in one place, to show that the hyperplane intersections were non-empty (algebraically, to show that some prime contained $f$). The local property is really not the key thing for this argument, it's the non-empty intersection. In both algebraic and geometric versions we used this non-emptiness to find somewhere to start (the $\mathfrak{p}_i$ and the $X_i$) and this is what allows us to generalise the proof.
A: In order to prove Krull's principal ideal theorem, the usual route is to first prove that the dimension of a local ring $(A,\mathfrak{m})$ (defined as the max length of chains of primes) coincides with the minimal number of generators of an $\mathfrak{m}$-primary ideal. This is e.g. Theorem 11.14 of Atiyah-MacDonald (and it is sometimes called the Dimension Theorem). It is this fact that we will use.
The desired inequality appears in the proof of Corollary 11.18 of Atiyah-MacDonald, using the above result. For completeness, I will summarize it here: take elements $g_1,\ldots,g_{d} \in \mathfrak{m}$ such that (their images) generate an $\mathfrak{m}/(f)$-primary ideal in $A/(f)$, where $d=\dim A/(f)$. Then, the ideal $(f,g_1,\ldots,g_d)$ in $A$ is $\mathfrak{m}$-primary, hence $\dim A$ is less than or equal to $\dim A/(f) + 1$, the number of generators of this ideal.
