Let $X$ be an algebraic variety (i.e. an integral $k$-scheme, such that $X \to \mathrm{Spec \;}k$ is separated and of finite type). Let $Y$ be a closed subscheme and $Y_1, \dots Y_n$ the irreducible components of $Y_{\mathrm{red}}$. Let $\xi_i$ be the generic point of $Y_i$. Since $Y$ is noetherian, then $\mathcal{O}_{Y,\xi_i}$ is a noetherian ring. We define the effective cycle associated to $Y$ as $<Y>:=\sum_{i=1}^nl_iY_i$, where $l_i$ is the length of the local ring $\mathcal{O}_{Y,\xi_i}$. In the book "3264 and all that" it is said that the ring $\mathcal{O}_{Y,\xi_i}$ has a finite composition series. I know that the length of $\mathcal{O}_{Y,\xi_i}$ is finite if and only if it is an artinian ring. But in this case i just see that it is noetherian. Why is it artinian too?

EDIT: I try to answer my own. Let $\xi$ be one of the $\xi_i$ and $Z$ the irreducible component associated to $\xi$. Choose $U$ to be an affine open subset of $Y$ such that $\xi \in U$. Then $Z \cap U$ is an irreducible component of $U$, with generic point $\xi$. If $U= \mathrm{Spec \;}A$, then $\xi=\mathfrak{p}$ is a minimal prime ideal and hence $\mathcal{O}_{Y,\xi}=\mathcal{O}_{U,\mathfrak{p}}$ is noetherian of dimension $0$, i.e. an artinian ring. Do you think it works?

  • 1
    $\begingroup$ Yes, it works $ $. $\endgroup$ – Georges Elencwajg Mar 28 at 9:23

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