The general case of Krull’s height theorem is obtained by induction on the number \(r\) of generators. The case \(r = 1\) is the hardest part of the proof, and we state it separately.
Theorem 6.21. (Krull’s principal ideal theorem / Hauptidealsatz) Let \(A\) be a Noetherian ring, let \(a \in A\) be a non-unit, and let \(\mathfrak {p}\) be a prime ideal which is minimal among the prime ideals containing \(a\). Then \(h(\mathfrak {p}) \leq 1\).
Proof. We may suppose that \(A\) is a local ring with maximal ideal \(\mathfrak {p}\). Indeed, by Corollary 6.15(ii), the prime ideals of \(A_\mathfrak {p}\) containing \(\frac {a}{1}\) correspond to the prime ideals of \(A\) contained in \(\mathfrak {p}\) and containing \(a\), so \(\mathfrak {p}A_\mathfrak {p}\) is minimal among the prime ideals of \(A_\mathfrak {p}\) containing \(\frac {a}{1}\), and \(h(\mathfrak {p}) = \dim A_\mathfrak {p} = h(\mathfrak {p}A_\mathfrak {p})\).
So let \(A\) be local with maximal ideal \(\mathfrak {p}\), and let \(\mathfrak {q} \subsetneq \mathfrak {p}\) be a prime ideal. We want to show that \(h(\mathfrak {q}) = 0\), that is, that \(\mathfrak {q}A_\mathfrak {q}\) is the only prime ideal of the local ring \(A_\mathfrak {q}\). Since \(\sqrt {(0)}\) is the intersection of all prime ideals, it suffices to show that the maximal ideal \(\mathfrak {q}A_\mathfrak {q}\) is nilpotent. Note that \(a \notin \mathfrak {q}\), as \(\mathfrak {p}\) is minimal among the prime ideals containing \(a\), so \(\frac {a}{1}\) is a unit in \(A_\mathfrak {q}\).
Let \(\iota : A \to A_\mathfrak {q}\) be the localisation map, and let \[ \mathfrak {q}^{(n)} \coloneqq \iota ^{-1}(\mathfrak {q}^n A_\mathfrak {q}) = \{ x \in A : xs \in \mathfrak {q}^n \text { for some } s \in A \setminus \mathfrak {q} \} \] be the \(n\)-th symbolic power of \(\mathfrak {q}\), see the exercise in Lecture 19. These form a chain of ideals \(\mathfrak {q} = \mathfrak {q}^{(1)} \supset \mathfrak {q}^{(2)} \supset \ldots \) in \(A\), and by Lemma 6.14(i) we have \(\mathfrak {q}^{(n)} A_\mathfrak {q} = \mathfrak {q}^n A_\mathfrak {q}\).
Step 1: \((a) + \mathfrak {q}^{(n)} = (a) + \mathfrak {q}^{(n+1)}\) for some \(n\). The prime ideals of \(A/(a)\) correspond to the prime ideals of \(A\) containing \(a\), and \(\mathfrak {p}\) is the only one of these, as \(A\) is local and \(\mathfrak {p}\) is minimal over \(a\). So \(\dim A/(a) = 0\), and \(A/(a)\) is Artinian by Theorem 6.8. Hence the chain of images of \((a) + \mathfrak {q}^{(1)} \supset (a) + \mathfrak {q}^{(2)} \supset \ldots \) in \(A/(a)\) becomes stationary, which gives the claim.
Step 2: \(\mathfrak {q}^{(n)} = \mathfrak {q}^{(n+1)}\). Let \(x \in \mathfrak {q}^{(n)}\). By Step 1, \(x = ra + y\) with \(r \in A\) and \(y \in \mathfrak {q}^{(n+1)}\). Then \(ra = x - y \in \mathfrak {q}^{(n)}\), so \(\frac {r}{1} \frac {a}{1} \in \mathfrak {q}^nA_\mathfrak {q}\), and since \(\frac {a}{1}\) is a unit, \(\frac {r}{1} \in \mathfrak {q}^n A_\mathfrak {q}\), that is, \(r \in \mathfrak {q}^{(n)}\). This shows that \[ \mathfrak {q}^{(n)} = a\mathfrak {q}^{(n)} + \mathfrak {q}^{(n+1)} \subset \mathfrak {p}\mathfrak {q}^{(n)} + \mathfrak {q}^{(n+1)}. \]As \(A\) is Noetherian, \(\mathfrak {q}^{(n)}\) is a finitely generated \(A\)-module, and Nakayama’s lemma for the submodule \(\mathfrak {q}^{(n+1)} \subset \mathfrak {q}^{(n)}\) gives \(\mathfrak {q}^{(n)} = \mathfrak {q}^{(n+1)}\).
Step 3: \(\mathfrak {q}A_\mathfrak {q}\) is nilpotent. By Step 2, \(\mathfrak {q}^nA_\mathfrak {q} = \mathfrak {q}^{(n)}A_\mathfrak {q} = \mathfrak {q}^{(n+1)}A_\mathfrak {q} = \mathfrak {q}^{n+1}A_\mathfrak {q} = (\mathfrak {q}A_\mathfrak {q})(\mathfrak {q}^nA_\mathfrak {q})\). The ring \(A_\mathfrak {q}\) is Noetherian by Lemma 6.14(i), so Nakayama’s lemma for the \(A_\mathfrak {q}\)-module \(\mathfrak {q}^nA_\mathfrak {q}\) gives \(\mathfrak {q}^nA_\mathfrak {q} = \{0\}\). □
\(\blacktriangleright \) Geometrically, let \(f \in k[X_1, \ldots , X_n]\) be non-constant, with \(k\) algebraically closed. By the dictionary of Lecture 14, the irreducible components of the hypersurface \(\mathcal {V}(f)\) are the sets \(\mathcal {V}(\mathfrak {p})\), where \(\mathfrak {p}\) runs over the minimal prime ideals containing \(f\). The theorem says that each of these has \(h(\mathfrak {p}) \leq 1\), and since \((0) \subsetneq \mathfrak {p}\), in fact \(h(\mathfrak {p}) = 1\). Thus every irreducible component of a hypersurface has codimension one.
\(\sphericalangle \) Carefully examine the proof to see why we needed to use the symbolic powers \(\mathfrak {q}^{(n)}\) rather than the ordinary powers \(\mathfrak {q}^n\) of the ideal \(\mathfrak {q}\).
\(\sphericalangle \) Let \(\mathfrak {p} = (X^4 - YZ, Y^2 - XZ, X^3Y - Z^2) \subset k[X,Y,Z]\). This is the kernel of the homomorphism \(k[X,Y,Z] \to k[T]\) sending \(X, Y, Z\) to \(T^3, T^5, T^7\), hence a prime ideal. Show that \(\mathfrak {p}^{(2)} \neq \mathfrak {p}^2\).
Before we go on to the proof of Krull’s height theorem, let us see an application of the principal ideal theorem.
Theorem 6.22. Let \(\mathfrak {p} \subset \mathfrak {q}\) be prime ideals in a Noetherian ring \(A\). If there exists a prime ideal strictly between them, then there are infinitely many.
Proof. The prime ideals between \(\mathfrak {p}\) and \(\mathfrak {q}\) correspond to the prime ideals between \((0)\) and \(\mathfrak {q}/\mathfrak {p}\) in the integral domain \(A/\mathfrak {p}\), so we may suppose that \(A\) is an integral domain and \(\mathfrak {p} = (0)\). Suppose for a contradiction that \(\mathfrak {p}_1, \ldots , \mathfrak {p}_n\), with \(n \geq 1\), are the only prime ideals strictly between \((0)\) and \(\mathfrak {q}\). By prime avoidance (Proposition 2.9), \(\mathfrak {q} \not \subset \mathfrak {p}_1 \cup \ldots \cup \mathfrak {p}_n\), so we can pick \(a \in \mathfrak {q}\) with \(a \notin \mathfrak {p}_i\) for all \(i\). Then \(\mathfrak {q}\) is minimal among the prime ideals containing \(a\): such a prime ideal contained in \(\mathfrak {q}\) is \((0)\), one of the \(\mathfrak {p}_i\), or \(\mathfrak {q}\), and the first two do not contain \(a\). By Theorem 6.21, \(h(\mathfrak {q}) \leq 1\), contradicting the chain \((0) \subsetneq \mathfrak {p}_1 \subsetneq \mathfrak {q}\). □
\(\blacktriangleright \) In \(k[X,Y]\), the prime ideals strictly between \((0)\) and \((X,Y)\) are the ideals \((f)\) with \(f\) irreducible and \(f(0,0) = 0\), that is, the irreducible curves through the origin, see Figure 15.
\(\sphericalangle \) Let \(A\) be a Noetherian ring with minimal prime ideals \(\mathfrak {p}_1, \ldots , \mathfrak {p}_s\), so that \(\sqrt {(0)} = \mathfrak {p}_1 \cap \ldots \cap \mathfrak {p}_s\) by Proposition 6.4. Show that every element of \(\bigcup _{i=1}^s \mathfrak {p}_i\) is a zero-divisor. Deduce that, with the assumptions of Theorem 6.21, \(h(\mathfrak {p}) = 1\) if \(a\) is not a zero-divisor.