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 $ ?
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 $f \in (\mathfrak{I}\mathfrak{J})_x$.