Let X be a non-projective non-reduced scheme and let D be an effective Cartier divisor on X. Why is D disjoint from $Ass(\mathcal{O}_X)$? In other words, why can't reduced points lie in the support of any effectice Cartier divisor?

The question comes from trying to understand Kleimann's example of a non-projective, non-reduced scheme for which there is not a bijective correspondence between Cartier divisors and line bundles.

Thanks in advance for any insight.


If $x$ is point in the support of $D$, then $D$ is defined locally at $x$ by a regular element in the maximal ideal $m_x$ of $O_{X,x}$. If $x$ is an associated point, then, by definition, $m_x$ doesn't contain any regular element, contradiction.

  • $\begingroup$ The nice answer of Georges reminds me add the hypothesis $X$ is (locally) noetherian. $\endgroup$ – user18119 Sep 4 '12 at 22:23

We can reduce to the definition of an effective Cartier divisor $D$ on the scheme $X$, which is a closed subscheme $D\subset X$.
Such a divisor is locally defined on an affine open subset $U= Spec(A)\subset X$ by $U\cap D= V(f)$ where $f\in A$ is not a zerodivisor .
To conclude you have to remember that if the ring $A$ is noetherian its zero divisors are exactly the union of its associated primes (Atiyah-Macdonald Prop.4.7) : $$Zdiv(A)=\bigcup_{\mathfrak p\in Ass(A)} \mathfrak p=\bigcup_{\mathfrak p\; \text {a minimal prime}} \mathfrak p$$

This definition of Cartier divisor has many advantages.
In particular we have $dim(D)\lt dim(X)\:$ (if $dim(X)\lt \infty $) and this would be violated with a less stringent definition of "divisor":
For example take for $X$ the cross $X=V(xy)\subset k[x,y]=\mathbb A^2_k$ ($k$ a field).
If you allowed $D=V(\bar x)\subset X$ as a Cartier divisor, you would have $dim(D)=dim(X)$: this is the punishment for allowing a zero divisor like $\bar x$ (with $\bar x \bar y=0)$ to define a Cartier divisor.

  • $\begingroup$ $\dim(D)<\dim(X)$ if $\dim(X)$ is finite :). $\endgroup$ – user18119 Sep 4 '12 at 11:23
  • $\begingroup$ Dear QiL, you are of course right: corrected. $\endgroup$ – Georges Elencwajg Sep 4 '12 at 12:10

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