4

Let $(X,\mathcal O_X)$ be a Noetherian scheme. Let $\mathfrak I$ and $ \mathcal J$ be two quasi-coherent sheaf of ideals on $X$.

Is it true that for every $x\in X$, we have $(\mathfrak I \mathcal J)_x=\mathfrak I_x \mathcal J_x$ ?

If this is not true in general, do we at least have that for every $x \in X$ and every integer $n>0$, that $(\mathcal J_x)^n =(\mathcal J^n)_x $ ?

uno
  • 1,560
  • 1
    What definition precisely are you using for the product of quasi-coherent sheaves of ideals? I can think of a couple possibilities, one which would work for general sheaves of ideals, and another which would involve gluing $\tilde I \cdot \tilde J := (IJ) \tilde{~}$ on affine open subsets. – Daniel Schepler Jul 09 '19 at 21:14
  • @Daniel Schepler: I thought there was a standard definition for product of ideal sheaves , I mean Stacks project, Eisenbud's Geometry of Schemes book and other references uses product of ideal sheaf all the time ... isn't it just the sheaf associated to the presheaf $U \to \mathfrak I (U) \mathcal J(U) $ ... ? – uno Jul 10 '19 at 10:24

1 Answers1

2

Yes. Fix $f \in \mathscr{O}_{X,x}$. Then if $f \in (\mathfrak{I}\mathfrak{J})_x$, that means that for some open neighborhood $U$ of $x$, $f \in \mathfrak{I}(U)\mathfrak{J}(U)$. Suppose $f = \sum_{i=1}^n g_i h_i$ with $g_i \in \mathfrak{I}(U), h_i \in \mathfrak{J}(U)$; then each $g_i \in \mathfrak{I}_x$ and each $h_i \in \mathfrak{J}_x$ so $f \in \mathfrak{I}_x \mathfrak{J}_x$.

Conversely, if $f \in \mathfrak{I}_x \mathfrak{J}_x$, suppose $f = \sum_{i=1}^n g_i h_i$ with each $g_i \in \mathfrak{I}_x$ and $h_i \in \mathfrak{J}_x$. Since there is a finite number of terms, there is some open neighborhood $U$ of $x$ such that each $g_i \in \mathfrak{I}(U)$ and each $h_i \in \mathfrak{J}(U)$. It follows that $f \in \mathfrak{I}\mathfrak{J}(U)$, so $f \in (\mathfrak{I}\mathfrak{J})_x$.

Daniel Schepler
  • 20,364