The definition of effective Cartier divisor that I'm using is: a closed immersion whose corresponding quasicoherent sheaf of ideals is an invertible sheaf.

Let $X$ be a scheme that is dimension 1 and locally of finite type over a field $k$ (but can be singular or have multiple components). Let $p$ be a closed point on $X$. Then why is the canonical morphism $f : Spec(k(p)) \rightarrow X$ is an effective Cartier divisor?

This is my attempt so far:

Since $p$ is a closed point, $f : Spec(k(p)) \rightarrow X$ is a homeomorphism onto its range and the range is closed. Also, the pullback morphism $f^\# : O_X \rightarrow f_* O_{Spec(k(x))}$ is surjective (as on an affine open neighbourhood $U = Spec(A)$ of $p$, we have $A \rightarrow Quot(A/p) = A/p$ since $p$ is a closed point so it's a maximal ideal, so pullback is surjective). Therefore it's a closed immersion.

Let $\mathcal{I}$ be the corresponding quasicoherent sheaf of ideals. For open sets U that don't contain $p$, $\mathcal{I} \sim O_X|_{U}$ so is an invertible sheaf on there. But now I'm stuck on the part where open sets $U$ that contain $p$.

Is the statement even true? If so, how do I prove $\mathcal{I}$ is invertible on some open set $U$ that contains $p$?

  • 1
    $\begingroup$ the ideal sheaf is functions that vanish at $P$. picking a uniformizer at $P$, dividing by that uniformizer give an iso (in some neighborhood of $P$) from functions that vanish at $P$ to all functions. $\endgroup$
    – hunter
    Dec 20, 2020 at 7:12

1 Answer 1


This statement is true precisely when $p$ is a regular point of $X$. If $p$ is regular, then $\mathcal{O}_{X,p}$ is a regular local ring of dimension one and hence a DVR, so we can select a uniformizer which gives a generator on some neighborhood of $p$. Conversely, if the ideal sheaf of $X$ is locally principal then its stalk at $p$ must be a principal ideal of $\mathcal{O}_{X,p}$. This suffices to show that $\mathcal{O}_{X,p}$ is a DVR, which is a regular local ring.


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