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An effective Cartier Divisor $D$ of a integer Scheme $X$ is associated to a set of tuple $\{(U_i, f_i)\}$ where $U_i$ prove a affine covering of $X$ and $f_i \in \Gamma(U_i, \mathcal{O}_X)$ are the regular local sections. These a defining in invertible ideal sheaf $\rho$ on $X$. Thus I can interpret the Cartier Divisor $D$ as support the quotient ideal $\mathcal{O}_X / \rho$.

My question is why has $D$ codimension $1$?

My ideas: As I know the codimenion of a subscheme is defined by $\operatorname{codim}(D,X) = \inf \{ \dim \mathcal{O}_{D,y} \mid y\in D \}$ (is it the right definition?)

Futhermore, because $D$ arises from $\mathcal{O}_X / \rho$ it's stalks are of the shape $\mathcal{O}_{D, y} =\mathcal{O}_{X, y} / (f)$ for a regular $f$. Applying Krull's Principle Ideal Theorem I get only the unequality $1 \ge dim(\mathcal{O}_{D,y})$, but how to get the equality?

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Your definition of codimension is wrong; you should be looking at $\mathcal{O}_{X,y}$, not $\mathcal{O}_{D,y}$. If $\mathcal{O}_{X,y}$ is $0$-dimensional, that means $y$ is the generic point of $X$, corresponding to the trivial ideal $\{0\}$ on any affine open subset. Since each $f_i$ is required to be nonzero, it cannot vanish at the generic point of $X$, so the generic point cannot be in $D$.

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  • $\begingroup$ Hi, thank you for the answer. I have a futher question: With your argument we get $codim(D,X) \ge 1$. Also, using Krull again, we get $1 \ge dim(\mathcal{O}_{X,y})-1 = dim(\mathcal{O}_{D,y})$. But this does provide the possiblities that $dim(\mathcal{O}_{X,y})$ equals $2$ or $1$, doesn't it? $\endgroup$
    – user267839
    Commented Oct 29, 2017 at 2:47
  • $\begingroup$ I don't know where you're getting $1 \ge dim(\mathcal{O}_{X,y})-1$ from. Why are you subtracting 1? $\endgroup$ Commented Oct 29, 2017 at 2:55
  • $\begingroup$ Oh I forgot to mention that because that $D$ is a support we have $y \in D$ is the same as $f_i \in m_y$. I used here a consequence of KPI Theorem that for every $f \in m_R$ we have $dim(R/{fR}) \ge dim(R)-1$ wich is a equality if $f$ isn't contained in any minimal prime ideal. Because $X$ integer and $f$ regular this automatically holds, doesn't it? $\endgroup$
    – user267839
    Commented Oct 29, 2017 at 3:04
  • $\begingroup$ OK, so $dim(\mathcal{O}_{X,y})-1 = dim(\mathcal{O}_{D,y})$. But I have no idea where you get the $1\geq$ at the start from. You want to minimize these dimensions, so you can choose $y$ such that $dim(\mathcal{O}_{D,y})=0$ (just take a minimal prime of the ring corresponding to $D$ on any affine open). $\endgroup$ Commented Oct 29, 2017 at 3:13
  • $\begingroup$ That means finding a prime ideal with property be minimal containing $f$... Can I find all time someone in this case? The problem is that the affine open sets in general don't correspond to Artinian rings ... $\endgroup$
    – user267839
    Commented Oct 29, 2017 at 3:30

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