9
$\begingroup$

I was very confused by the proof of Artin–Rees / Krull intersection theorem when I was younger.

Now that I learnt about blow up— I saw the Rees algebra again and I want to now gain a better understanding of the proof motivated by blow-up.

I consider the universal property of blow-up from Vakil; it makes your subscheme Cartier.

I will explain how to prove the Krull intersection in this way, I haven't been able to do this for Artin–Rees and this is the content of my question; is it obvious when the ideal is Cartier, and does it follow from taking the blowup? Let me try to illustrate on Krull intersection:

Statement: Let $A$ be a Noetherian local ring with $m$ its maximal ideal, then $M = \bigcap m^n =0$.

Proof:

First suppose that $m = (f)$ for a non-zero divisor $f$. In this case I claim that $fM = M$. Indeed, given $x \in M$ there are $x_i$ with $f^i x_i = x$. Now we have $f x_1 = f^{i} x_{i}$ but $f$ is a non-zero divisor so $x_1 = f^{i-1} x_i$, this proves that $fM = M$ so by Nakayama $M=0$.

$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Bl{Bl}$Now for general $I$, we take the blowup (some notation; $Y=\Spec(A)$, $X = \Spec(A/m)$, $B = \Bl_X(Y)$, $E = E_X$. Sadly the blowup of a local ring is not local. Instead if the local ring is the stalk at a point $p$ in a scheme $Z$, then we can blowup $Z$ in $p$ and take the fibered product with the stalk (So it's like an infinitesimal neighborhood of the exceptional divisor). I don't know how to finish here; we can pullback $\bigcap m^n$ so that now we know it vanishes in many points, but we need to show at all points and why it implies $\bigcap m^n$ is 0.

So my question is how to deduce the general case from the Cartier case in Krull, and how to prove Artin–Rees using this strategy.

Thanks! (Not exactly research question but I didn't get any interaction in math exchange.)

Partial work

I think I can close down Krull intersection. I will assume the underlying schemes are irreducible and Noetherian.

Lemma 1: Krull intersection is true for a single element in this form; let $\Spec(A)$ be an irreducible Notherian scheme and $f$ a non-zero divisor, then $\bigcap f^n = 0$. Indeed suppose $f$ vanishes at $p$, then by the local version we know $\cap f^n = 0$ on some open neighborhood, but if it contained a nonzero $g$, then $g$ would also be nonzero on an open subset so they'd meet, a contradiction.

Lemma 2: Suppose $\Spec(A) \to \Spec(B)$ is a surjective map of schemes (on points), then a nonzero sheaf pulls back to a nonzero sheaf. Indeed if we have our $M$ with $M \otimes k(p) \neq 0$ for some $p \in Spec(B)$, then taking $q \in Spec(A)$ with $q \to p$ we see that $k(q) \otimes_B M/p \neq 0$.

General proof of Krull intersection in nice cases— an irreducible Noetherian local ring:

Taking the blowup we have $B \to Y$ surjective on points (since it's projective and contains the open set $Y - X$), so it's enough to show the pullback of $\bigcap I^n$ is $0$. Inside a given affine in $B$, we have the exceptional is generated by $f$ for a nonzero divisor $f$, and the pullback is certainly contained in $\bigcap f^n$ so the claim is proved.

Extending those arguments to Artin–Rees

I took the easiest case above. I think the general case is slightly hard because pullback doesn't work with intersection when the map is not flat. Could it be a miracle that the blowup of a local ring at an ideal (or at the very least the maximal ideal) is flat over it? It feels like Artins–Rees is saying there is some approximate flatness.

$\endgroup$
2
  • $\begingroup$ You say "I didn't get any interaction in math exchange", but I don't find any question on this topic from you there. Could you link to it? $\endgroup$
    – LSpice
    Sep 22, 2021 at 23:35
  • $\begingroup$ @LSpice I deleted it after a few days, think it's not good practice to keep a question active on both sites $\endgroup$
    – Andy
    Sep 23, 2021 at 14:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.