Krull's height theorem states that in a Noetherian, local ring $(A,\mathfrak m)$, for any $f \in \mathfrak m$, the minimal prime ideal containing $(f)$ is at most height $1$.

This is a very geometric statement and is essentially saying that hypersurfaces can cut down the dimension by at most one. However, I have never seen a geometrically motivated proof. All the proofs I have seen essentially muck around with symbolic powers of prime ideals and this seems very ad-hoc/unmotivated to me.

Surely a geometrical statement should have a geometric proof!

Does someone have a geometric way of seeing why this theorem should be true or what is going on? Or what is going on geometrically with the standard proof and symbolic powers?

  • 2
    $\begingroup$ The proof in the Stacks Project here stacks.math.columbia.edu/tag/00KD uses the interplay between the notion of ideal of definition, Hilbert polynomial, and Krull dimension. I think using the Hilbert polynomial is pretty geometric. But of course it isn't a completely geometric proof. $\endgroup$ – Samir Canning Jul 13 '18 at 20:03
  • 1
    $\begingroup$ Commutative algebra is to algebraic geometry as calculus to differential geometry. Some 'geometric' statements in AG rely on CA, just like some 'geometric' statements in DG rely on calculus. $\endgroup$ – R. van Dobben de Bruyn Jul 13 '18 at 22:48
  • 3
    $\begingroup$ Moreover, if the Noetherian hypothesis is dropped, a minimal prime over $(f)$ can actually have infinite Krull dimension! I'm not sure what this says about the geometry behind the proof... $\endgroup$ – R. van Dobben de Bruyn Jul 13 '18 at 22:49
  • $\begingroup$ @R.vanDobbendeBruyn I think of that result as "anti-geometry" in the sense that it tells you Noether normalization can't be generalized too far from the usual geometric thing about finiteness of generic projections. $\endgroup$ – Samir Canning Jul 14 '18 at 17:44

Anon's answer gives a beautiful geometric proof when $A$ is a variety. Below I am trying to give some geometric interpretation of the usual algebraic proof.

First a disclaimer: I'm not an algebraist, so the explanation below will be a learner's perspective, probably from an analytic perspective, and thus may seem idiosyncratic to experts.

I am going to start by interpreting two key ingredients used in the proof.

1. Symbolic power

Let $\frak p$ be a prime ideal of $A$. I think of the localization $A_{\frak p}$ as capturing the behavior of functions on a neighborhood of the generic point of $V({\frak p})$. To see what I mean, take for example $A=k[x,y]/(xy,y^2)$ and ${\frak p}=(y)$. Geometrically $Spec A$ is the $x$-axis plus some fuzz of order 2 at the origin, and $V({\frak p})$ is just the $x$-axis. Now look at $y\in A$. We have $y$ is nonzero in $A$ but becomes zero in $A_{\frak p}$ (where $y=xy/x=0$.) The geometric explanation is that $y$ is indeed zero on a neighborhood of $(x_0,0)$ for any $x_0\neq0$, because there is no fuzz around that point. The only reason for $y\neq0$ in $A$ is that it vanishes only to order 1 near the origin, which is captured by the fuzz (of order 2) there. But this happens at a single point in $V({\frak p})$ (a so called embedded prime), so that behavior is not generic (in the colloquial sense of the word), so we can still say that $y$ vanishes on a neighborhood of the generic point of $V({\frak p})$, which explains why it vanishes in $A_{\frak p}$.

Generalizing this example a bit, if we take $A=k[x,y]/(xy^n,y^{n+1})$, we see that $Spec A$ is the $x$-axis with multiplicity $n$ (in other words, with fuzz of order $n$ in the $y$ direction), plus some fuzz of order $n+1$ in the $y$ direction at the origin. Again let ${\frak p}=(y)$. Then ${\frak p}^{(m)}$ (the symbolic power) consists of functions that vanish to order $m$ at the generic point of $V({\frak p})$. Thus for $m<n$, ${\frak p}^{(m)}=(y^m)$, and for all $m\ge n$, ${\frak p}^{(m)}=(y^n)$. The fact that $y^n$ only vanishes to order $n$ near the origin does not matter, because again the origin is only a single point, not generic enough for $V({\frak p})$.

2. Nakayama's Lemma

The OP doesn't ask about this but since this lemma is used often in the proof I will also try to interpret it geometrically. Let $(A,{\frak m})$ be a local ring and $M$ be a finitely generated $R$ module. I think of $A$ as the germ of holomorphic functions on a neighborhood of the origin and $M$ as some sort of holomorphic vector bundle over that neighborhood. The condition of Nakayama's Lemma says that $M={\frak m}M$. Iterating this we get $M={\frak m}^nM$ for any $n\in\mathbb N$. This means that all sections of $M$ vanishes to arbitrarily high orders near the origin. By the holomorphic heuristics, all sections of $M$ vanish identically, so $M=0$.

Now we turn to the actual proof of Krull's principal ideal theorem, which can found, for example, here.

By standard reduction, we can assume that $(A,\frak m)$ is a local domain, $f\neq0$ with a minimal prime ideal $\frak m$. Assume that there is a prime ideal $\frak p$ properly contained in $\frak m$, and our aim is to show that ${\frak p}=(0)$.

Consider $V(f)$. Since $\frak m$ is the maximal ideal of $A$ while at the same time a minimal prime ideal of $f$, $V(f)$ contains a single (scheme-theoretic) point, namely $\frak m$ itself (in algebraic terms, $A/(f)$ is an Artinian local ring), plus a finite order of fuzz around that point. Consider the symbolic power ${\frak p}^{(n)}$, that is, the ideal of functions that vanish to at least of order $n$ at a generic point of $V({\frak p})$. Then ${\frak p}^{(n)}|_{V(f)}$ (algebraically this is the ideal ${\frak p}^{(n)}+(f)/(f)$ in $A/(f)$) will include a finite order of fuzz near the unique point $\frak m$ of $V(f)$. The larger $n$ is, the more fuzz it can possibly include. Since the total order of fuzz around $\frak m$ is finite, for $n\gg1$ the order of fuzz included in ${\frak p}^{(n)}|_{V(f)}$ will not change (this is the DCC property for Artinian rings.) Written out algebraically, this amounts to ${\frak p}^{(n)}+(f)={\frak p}^{(n+1)}+(f)=\cdots$.

Now take $x\in{\frak p}^{(n)}$. Then $x$ vanishes at least to order $n$ generically on $V({\frak p})$, but we can write $x=y+fr$, where $y$ vanishes at least to order $n+1$ generically on $V({\frak p})$, so $fr$ vanishes at least to order $n$ generically on $V({\frak p})$. But $\frak p$ is not a point in $V(f)$ (whose only point is $\frak m$), so $f|_{V({\frak p})}\neq0$. Since $V({\frak p})$ is irreducible, $f$ does not vanish to any order generically on $V({\frak p})$. Hence $r$ vanishes at least to order $n$ generically on $V({\frak p})$. Translating back to algebra, we have ${\frak p}^{(n)}={\frak p}^{(n+1)}+f{\frak p}^{(n)}$.

Now we consider the module ${\frak p}^{(n)}/{\frak p}^{(n+1)}$. The above identity shows that every element in this module is a multiple of $f$. Iterating this we know that every element is a multiple of $f^m$ for all $m\in\mathbb N$. Since $f$ vanishes at $\frak m$, every element in ${\frak p}^{(n)}/{\frak p}^{(n+1)}$ vanishes to arbitrarily high order at $\frak m$. By Nakayama's Lemma, the module vanishes identically, so ${\frak p}^{(n)}={\frak p}^{(n+1)}$.

Now pass to the localization $A_{\frak p}$, that is, we forget about the behavior of functions at specific points of $V({\frak p})$, and only considers its behavior on a neighborhood of the generic point of $V({\frak p})$. Since the coordinate ring takes the generic behavior into account, it's unnecessary to restate it for the localization of the symbolic power. Thus ${\frak p}^{(n)}A_{\frak p}={\frak p}^nA_{\frak p}$, and it's stationary for $n\gg1$. Then any element in ${\frak p}^nA_{\frak p}$ vanishes to arbitrarily high order near $\frak p$. By Nakayama's Lemma again, ${\frak p}^nA_{\frak p}=(0)$.

To wrap up, I have to use some algebra (my previous geometric argument somewhat contradicts my earlier points.) Since $A$ is a domain, the localization $A\to A_{\frak p}$ is injective, so ${\frak p}^n=(0)$ in $A$. Again because $A$ is a domain, ${\frak p}=(0)$.

| cite | improve this answer | |

For a general Noetherian local ring, probably not, but in a "geometric situation" (algebraic varieties) there is a geometric proof in Mumford's Red Book, I, Section 7 (which he credits to Tate).

| cite | improve this answer | |
  • 1
    $\begingroup$ The argument is very nice and quite simple: The basic idea is that for varieties, we can use Noether Normalization to define a finite, surjective map $X \to \mathbb A^d$ where $d = \dim X$. Then, the image of the hypersurface $V(f)$ under this map is $V(Norm(f))$ and then you prove in the case of $\mathbb A^n$ that the principal ideal theorem holds+finite surjective maps preserve dimension. $\endgroup$ – Asvin Jul 16 '18 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.