Krull's principal ideal theorem

In commutative algebra, Krull's principal ideal theorem, named after Wolfgang Krull (1899–1971), gives a bound on the height of a principal ideal in a commutative Noetherian ring. The theorem is sometimes referred to by its German name, Krulls Hauptidealsatz (Satz meaning "proposition" or "theorem").

Precisely, if R is a Noetherian ring and I is a principal, proper ideal of R, then each minimal prime ideal over I has height at most one.

This theorem can be generalized to ideals that are not principal, and the result is often called Krull's height theorem. This says that if R is a Noetherian ring and I is a proper ideal generated by n elements of R, then each minimal prime over I has height at most n. The converse is also true: if a prime ideal has height n, then it is a minimal prime ideal over an ideal generated by n elements.[1]

The principal ideal theorem and the generalization, the height theorem, both follow from the fundamental theorem of dimension theory in commutative algebra (see also below for the direct proofs). Bourbaki's Commutative Algebra gives a direct proof. Kaplansky's Commutative ring includes a proof due to David Rees.

Proofs

Proof of the 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, letting ${\displaystyle I=(x)}$, we get that both sides are zero and ${\displaystyle {\mathfrak {q}}^{(n)}={\mathfrak {q}}^{(n+1)}}$, 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. ${\displaystyle \square }$

Proof of the height theorem

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 ${\displaystyle \operatorname {ht} ({\mathfrak {p}})\leq n}$. ${\displaystyle \square }$

References

1. Eisenbud, Corollary 10.5. harvnb error: no target: CITEREFEisenbud (help)

• Eisenbud, David (1995). Commutative Algebra with a View Toward Algebraic Geometry. Graduate Texts in Mathematics. 150. Springer-Verlag. doi:10.1007/978-1-4612-5350-1. ISBN 0-387-94268-8.
• Matsumura, Hideyuki (1970), Commutative Algebra, New York: Benjamin, see in particular section (12.I), p. 77
• http://www.math.lsa.umich.edu/~hochster/615W10/supDim.pdf