The Algebraic Hartog’s Lemma

The “Algebraic Hartog’s Lemma” is Lemma 13.5.11 in Ravi’s book. Slightly restated, it says the following:

Let $R$ be an integrally closed Noetherian integral domain and let $K = \text{Frac}(R)$ . Let $\theta \in K$ . The following are equivalent:

(1) $\theta$ is in $R$ .

(2) For all dvrs $D$ with $R \subseteq D \subseteq K$ , we have $\theta \in D$ .

(3) For all codimension one primes $\mathfrak{p}$ , we have $\theta \in R_{\mathfrak{p}}$ .

The implications $(1) \implies (2) \implies (3)$ are straightforward, but I find $(3) \implies (1)$ quite confusing. And I have to say that Ravi’s proof doesn’t help me; it feels like a lot of pushing around modules that I don’t follow. So I put in a fair bit of time trying to find my own route and wound up frustrated: I thought I was making a fair bit of progress, but then I get stuck at the end.

I’m very glad to hear about other proofs that people like!

Okay, we want to prove $(3) \implies (1)$ . We’ll attack the contrapositive: Given $\theta \not\in R$ , we want to find a codimension one prime $\mathfrak{p}$ such that $\theta \not\in R_{\mathfrak{p}}$ .

Put $R' = R[\theta^{-1}]$ and put $J' = (\theta^{-1}) R'$ .

Claim: $J'$ is not the unit ideal of $R'$ .

Proof: If $J'$ is $(1)$ , then $1 = \theta^{-1}(a_0 + a_1 \theta^{-1} + \cdots + a_n \theta^{-n})$ for some $a_i$ in $R$ . Then $\theta^{n+1} = a_0 \theta^n + a_1 \theta^{n-1} + \cdots + a_n$ , so $\theta$ is integral over $R$ , which we assumed not to be the case. $\square$

Thus, there is a prime ideal $\mathfrak{p}'$ of $R'$ with $\mathfrak{p}' \supseteq J'$ , and thus $\theta^{-1} \in \mathfrak{p}'$ . So $\theta^{-1} \in \mathfrak{p} R'_{\mathfrak{p}'}$ , and therefore $\theta \not \in R'_{\mathfrak{p}'}$ . Moreover, since we assumed $R$ to be Noetherian, by Krull’s principal ideal theorem, we can assume that $\mathfrak{p}'$ is a codimension one prime of $R'$ .

So far, we seem really close to the desired goal. And it feels to me like what I have done so far is not that difficult, and is reasonably well motivated. The trouble is that we have $\theta \not \in R'_{\mathfrak{p}'}$ and not $\theta \not \in R_{\mathfrak{p}}$ for some codimension one prime of $\mathfrak{p}$ .

I’ll now explain my failure to finish the proof. It is quite possible, though, that you should close your browser now, get out a notepad, and start thinking on your own, because I am not convinced that anything that follows is helpful. I also hope, at some point, to write a blogpost explaining how this all works for non-Noetherian things, since it is much more straightforward than I realized.

The most direct thing to do, in my opinion, is to put $\mathfrak{p} = \mathfrak{p}' \cap R$ . Fortunately, $\mathfrak{p}$ is prime. But is it codimension one?

Let’s do an example. Let $R = k[x,y]$ and let $\theta = x y^{-1}$ . Then $R' = k[x, x^{-1} y]$ and the ideal $J' = (x^{-1} y) R'$ is prime, so we take $\mathfrak{p}' = (x^{-1} y)$ . We have $\mathfrak{p} = R \cap \mathfrak{p}' = (y)$ . So, in this case, we win! The ideal $(y)$ is, indeed, codimension one in $R$ .

On the other hand, the map $\text{Spec}(R') \to \text{Spec}(R)$ is not dimension preserving even in this very simple example. Look at the prime ideal $x R'$ in $R'$ , and note that $y = x (x^{-1} y) \in x R'$ . So $x R' \cap R = \langle x,y \rangle$ . The ideal $x R'$ is codimension one in $R'$ , and $\langle x,y \rangle$ is codimension two in $R$ . So we need to use something specific to the prime $\mathfrak{p}'$ .

My first thought is to localize to $\mathfrak{p}'$ . We have an inclusion $R_{\mathfrak{p}} \subset R'_{\mathfrak{p}'}$ . If $R'_{\mathfrak{p}'}$ is finite over $R_{\mathfrak{p}}$ , then we are done. In our example, in fact have equality! $R_{\mathfrak{p}} = k[x,y]_{(y)} = k[x, x^{-1} y]_{(x^{-1} y)} = R'_{\mathfrak{p}'}$ , because $x$ is a unit in $k[x,y]_{(y)}$ .

In every example I have tried, this is what happened: I always found that $R_{\mathfrak{p}}= R'_{\mathfrak{p}'}$ . I don’t know how to prove this. We want to show that $\theta^{-1} \in R_{\mathfrak{p}}$ , so we want to write $\theta= \alpha/\beta$ for $\alpha$ , $\beta \in R$ and $\beta \not\in \mathfrak{p}$ . I’m not sure how you’d do that …

If we knew that $R'_{\mathfrak{p}'}$ was a finitely generated $\mathfrak{R}_{\mathfrak{p}}$ module, we could hope to apply Nakayama’s lemma to show that $R_{\mathfrak{p}} \to R'_{\mathfrak{p}'}$ is surjective. But, if we knew that $R'_{\mathfrak{p}'}$ was a finitely generated $\mathfrak{R}_{\mathfrak{p}}$ module, then we would already know that the map $\text{Spec}(R'_{\mathfrak{p}'}) \to \text{Spec}(R_{\mathfrak{p}})$ was dimension preserving. And it seems odd to aim for proving that $R'_{\mathfrak{p}'}$ is finite over $\mathfrak{R}_{\mathfrak{p}}$ when the evidence suggests the much stronger claim that $R_{\mathfrak{p}}= R'_{\mathfrak{p}'}$ .

Here is a lemma which seems like it should help:

Lemma: The natural map $R/\mathfrak{p} \to R'/\mathfrak{p}'$ is an isomorphism.

Proof: First of all, let’s understand where this map comes from. The kernel of the composite $R \to R' \to R'/\mathfrak{p}'$ is, by definition, $R \cap \mathfrak{p}' = \mathfrak{p}$ . So we get a natural injection $R/\mathfrak{p} \hookrightarrow R'/\mathfrak{p}'$ . It remains to prove surjectivity. Since $R' = R[\theta^{-1}]$ , the ring $R'/\mathfrak{p}'$ is generated over $R/\mathfrak{p}$ by $\theta^{-1}$ . So all we need to show is that $\theta^{-1}$ is in the image of our injection. But $\theta^{-1} \in \mathfrak{p}'$ so $\theta^{-1} = 0$ in $R'/\mathfrak{p}'$ , and this is obvious. $\square$

This lets us prove that $\mathfrak{p}$ is codimension one in $R$ in a very large number of cases:

Proposition: Let $R$ have finite Krull dimension. Then $\mathfrak{p}$ is codimension one in $R$ .

Proof: Let $d$ be the Krull dimension of $R$ . Let $\theta = a/b$ for $a$ , $b \in R$ . Then $\dim R[x] = d+1$ . By the Krull principal ideal theorem, each irreducible component of $\text{Spec} R[x]/(bx-a)$ is $d$ -dimensional. One of these components is $\text{Spec}(R')$ , so $\dim R' = d$ . Then $R'/\mathfrak{p}' = R/\mathfrak{p}$ has dimension $d-1$ . So $\mathfrak{p}$ must be codimension one in $R$ . $\square$

In a very different direction, I considered trying to show that $R'_{\mathfrak{p}'}$ is a dvr, or at least that it is contained in some dvr $D$ with $\theta \not\in D$ . In that way, we could at least show that (3) implies (2). Since $\mathfrak{p}'$ is codimension one in $R'$ , this follows immediately if we know that $R'$ is integrally closed.

But $R'$ has no reason to be integrally closed: Take $R = k[x,y]$ and $\theta = x^2 y^{-2}$ . Then $\theta^{-1} = (x^{-1} y)^2 \in R'$ but $x^{-1} y \not\in R'$ .

So I considered replacing $R'$ by its normalization $\tilde{R'}$ . But the normalization of a Noetherian ring is not always Noetherian! So I’m not sure where to go from here …

One thought on “The Algebraic Hartog’s Lemma”

Hao XIAO on May 15, 2024 at 1:41 pm said:

It seems that, in principle, a proof of 3=>1 fails to avoid using associated primes/primary ideals/primary decomposition.

Besides Stacks Project Tag 031T, see for example Siegfried Bosch’s book on algebraic geometry. It’s Proposition 5 on Page 434.

In fact, Vakil’s proof uses primary ideals implicitly.

I was looking for a proof independent of primary ideals, but nothing interesting was found. 🙁

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