Krull's principal ideal theorem

Let ${\displaystyle A}$ be a Noetherian ring, x an element of it and ${\displaystyle {\mathfrak {p}}}$ a minimal prime over x. Replacing A by the localization ${\displaystyle A_{\mathfrak {p}}}$, we can assume ${\displaystyle A}$ is local with the maximal ideal ${\displaystyle {\mathfrak {p}}}$. Let ${\displaystyle {\mathfrak {q}}\subsetneq {\mathfrak {p}}}$ be a strictly smaller prime ideal and let ${\displaystyle {\mathfrak {q}}^{(n)}={\mathfrak {q}}^{n}A_{\mathfrak {q}}\cap A}$, which is a ${\displaystyle {\mathfrak {q}}}$-primary ideal called the n-th symbolic power of ${\displaystyle {\mathfrak {q}}}$. It forms a descending chain of ideals ${\displaystyle A\supset {\mathfrak {q}}\supset {\mathfrak {q}}^{(2)}\supset {\mathfrak {q}}^{(3)}\supset \cdots }$. Thus, there is the descending chain of ideals ${\displaystyle {\mathfrak {q}}^{(n)}+(x)/(x)}$ in the ring ${\displaystyle {\overline {A}}=A/(x)}$. Now, the radical ${\displaystyle {\sqrt {(x)}}}$ is the intersection of all minimal prime ideals containg ${\displaystyle x}$; ${\displaystyle {\mathfrak {p}}}$ is among them. But ${\displaystyle {\mathfrak {p}}}$ is a unique maximal ideal and thus ${\displaystyle {\sqrt {(x)}}={\mathfrak {p}}}$. Since ${\displaystyle (x)}$ contains some power of its radical, it follows that ${\displaystyle {\overline {A}}}$ is an Artinian ring and thus the chain ${\displaystyle {\mathfrak {q}}^{(n)}+(x)/(x)}$ stabilizes and so there is some n such that ${\displaystyle {\mathfrak {q}}^{(n)}+(x)={\mathfrak {q}}^{(n+1)}+(x)}$. It implies:

${\displaystyle {\mathfrak {q}}^{(n)}={\mathfrak {q}}^{(n+1)}+x\,{\mathfrak {q}}^{(n)}}$,

from the fact ${\displaystyle {\mathfrak {q}}^{(n)}}$ is ${\displaystyle {\mathfrak {q}}}$-primary (if ${\displaystyle y}$ is in ${\displaystyle {\mathfrak {q}}^{(n)}}$, then ${\displaystyle y=z+ax}$ with ${\displaystyle z\in {\mathfrak {q}}^{(n+1)}}$ and ${\displaystyle a\in A}$. Since ${\displaystyle {\mathfrak {p}}}$ is minimal over ${\displaystyle x}$, ${\displaystyle x\not \in {\mathfrak {q}}}$ and so ${\displaystyle ax\in {\mathfrak {q}}^{(n)}}$ implies ${\displaystyle a}$ is in ${\displaystyle {\mathfrak {q}}^{(n)}}$.) Now, quotienting out both sides by ${\displaystyle {\mathfrak {q}}^{(n+1)}}$ yields ${\displaystyle {\mathfrak {q}}^{(n)}/{\mathfrak {q}}^{(n+1)}=(x){\mathfrak {q}}^{(n)}/{\mathfrak {q}}^{(n+1)}}$. Then, by Nakayama's lemma (which says a finitely generated module M is zero if ${\displaystyle M=IM}$ for some ideal I contained in the radical), we get ${\displaystyle M={\mathfrak {q}}^{(n)}/{\mathfrak {q}}^{(n+1)}=0}$; i.e., ${\displaystyle {\mathfrak {q}}^{(n)}={\mathfrak {q}}^{(n+1)}}$ and thus ${\displaystyle {\mathfrak {q}}^{n}A_{\mathfrak {q}}={\mathfrak {q}}^{n+1}A_{\mathfrak {q}}}$. Using Nakayama's lemma again, ${\displaystyle {\mathfrak {q}}^{n}A_{\mathfrak {q}}=0}$ and ${\displaystyle A_{\mathfrak {q}}}$ is an Artinian ring; thus, the height of ${\displaystyle {\mathfrak {q}}}$ is zero. ${\displaystyle \square }$

Krull’s height theorem can be proved as a consequence of the principal ideal theorem by induction on the number of elements. Let ${\displaystyle x_{1},\dots ,x_{n}}$ be elements in ${\displaystyle A}$, ${\displaystyle {\mathfrak {p}}}$ a minimal prime over ${\displaystyle (x_{1},\dots ,x_{n})}$ and ${\displaystyle {\mathfrak {q}}\subsetneq {\mathfrak {p}}}$ a prime ideal such that there is no prime strictly between them. Replacing ${\displaystyle A}$ by the localization ${\displaystyle A_{\mathfrak {p}}}$ we can assume ${\displaystyle (A,{\mathfrak {p}})}$ is a local ring; note we then have ${\displaystyle {\mathfrak {p}}={\sqrt {(x_{1},\dots ,x_{n})}}}$. By minimality, ${\displaystyle {\mathfrak {q}}}$ cannot contain all the ${\displaystyle x_{i}}$; relabeling the subscripts, say, ${\displaystyle x_{1}\not \in {\mathfrak {q}}}$. Since every prime ideal containing ${\displaystyle {\mathfrak {q}}+(x_{1})}$ is between ${\displaystyle {\mathfrak {q}}}$ and ${\displaystyle {\mathfrak {p}}}$, ${\displaystyle {\sqrt {{\mathfrak {q}}+(x_{1})}}={\mathfrak {p}}}$ and thus we can write for each ${\displaystyle i\geq 2}$,

${\displaystyle x_{i}^{r_{i}}=y_{i}+a_{i}x_{1}}$

with ${\displaystyle y_{i}\in {\mathfrak {q}}}$ and ${\displaystyle a_{i}\in A}$. Now we consider the ring ${\displaystyle {\overline {A}}=A/(y_{2},\dots ,y_{n})}$ and the corresponding chain ${\displaystyle {\overline {\mathfrak {q}}}\subset {\overline {\mathfrak {p}}}}$ in it. If ${\displaystyle {\overline {\mathfrak {r}}}}$ is a minimal prime over ${\displaystyle {\overline {x_{1}}}}$, then ${\displaystyle {\mathfrak {r}}}$ contains ${\displaystyle x_{1},x_{2}^{r_{2}},\dots ,x_{n}^{r_{n}}}$ and thus ${\displaystyle {\mathfrak {r}}={\mathfrak {p}}}$; that is to say, ${\displaystyle {\overline {\mathfrak {p}}}}$ is a minimal prime over ${\displaystyle {\overline {x_{1}}}}$ and so, by Krull’s principal ideal theorem, ${\displaystyle {\overline {\mathfrak {q}}}}$ is a minimal prime (over zero); ${\displaystyle {\mathfrak {q}}}$ is a minimal prime over ${\displaystyle (y_{2},\dots ,y_{n})}$. By inductive hypothesis, ${\displaystyle \operatorname {ht} ({\mathfrak {q}})\leq n-1}$ and thus ${\displaystyle \operatorname {ht} ({\mathfrak {p}})\leq n}$. ${\displaystyle \square }$