# divisorial part of an ideal

hello everybody! Let $X$ a smooth variety and $I$ an ideal sheaf of $\mathcal{O}_X$ . I have read that if $dim(X)\geq 2$ an ideal cannot be written as product of prime ideals, but the codimension 1 primes can be separeted from the rest. That is, there is a unique largest effective divisor $Div(I)$ such that $I\subset\mathcal{O}_X(-Div(I))$. but i don't understand what are the codimensions primes. can someone could show me an example? thanks

Codimension=height.

To find that divisor and the corresponding ideal look at the ideal of the codimension $1$ part of the subscheme $\big(\mathrm{supp}(\mathscr O_X/I), \mathscr O_X/I\big)$.

• thanks! but what do you mean by the support of $\mathcal{O}_X /I$? Mar 29, 2011 at 7:54

Suppose that $X$ is affine: $X=\mathrm{Spec}(A)$. Then $I$ is an ideal of the noetherian, integrally closed ring $A$ (this property in fact is all you need, smoothness is not required). The ring $A$ can be written as

$A=\bigcap\limits_{p\in\mathrm{Spec} (A):\mathrm{height}(p)=1}A_p$.

The divisorial hull of $I$ is defined as

$\widehat{I}=\bigcap\limits_{p\in\mathrm{Spec} (A):\mathrm{height}(p)=1}IA_p$.

The ideal $\widehat{I}$ then defines a Weil divisor $D$ on $X$ in the usual way: namely

$v_p(D):=v_p(f_p)$ for $p\in\mathrm{Spec} (A):\mathrm{height}(p)=1$,

where $v_p$ is the discrete valuation associated to the valuation ring $A_p$ and $\widehat{I}A_p=f_pA_p$. This divisor is the one you are searching for.