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?

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    $\begingroup$ 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. $\endgroup$ Jul 13, 2018 at 20:03
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    $\begingroup$ 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. $\endgroup$ Jul 13, 2018 at 22:48
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    $\begingroup$ 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... $\endgroup$ Jul 13, 2018 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. $\endgroup$ Jul 14, 2018 at 17:44

2 Answers 2


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, ${\frak p}=(0)$.


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).

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    $\begingroup$ 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, 2018 at 19:57

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