Assume $u, v \in \mathcal{D}'(\mathbb{R}^n)$ are distributions with compact support. Denote by $\operatorname{WF}(\bullet) \subset T^*\mathbb{R}^n \setminus 0$ the wavefront set of a distribution $\bullet$. If $\operatorname{WF}(u) \cap \operatorname{WF}(v) = \emptyset$, then their product $uv$ is well defined. If $uv = 0$, does this imply that at an open and dense set of points one of the two distributions vanish, that is, $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c \subset \mathbb{R}^n$ is open and dense?

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For the background on the wavefront set and distribution theory, see Chapter 8 of Hörmander's book [The Analysis of Linear Partial Differential Operators I](https://doi.org/10.1007/978-3-642-61497-2). For products under the wavefront set condition, see Theorem 8.2.10 of the same book. I've also asked this question on [Mathematics StackExchange][2].

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Remarks:
- As noted below by Vinicius in the comments, if $u, v \in C^\infty_c(\mathbb{R})$ with $\operatorname{supp}(u) = [-1, 0]$ and $\operatorname{supp}(v) = [0, 1]$, then $uv = 0$ but $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c = \mathbb{R}\setminus 0$, so open and dense is the most we can hope for.

- The same question makes sense also if $u$ and $v$ don't have compact support. In that case, if for simplicity we set $n = 2$ and there are transversal smooth vector fields $X$ and $Y$ such that $X u = 0$ and $Y v = 0$, then the wavefront set condition is automatically satisfied. In fact, in a suitable local coordinate system adapted to $X$ and $Y$, one can show that $uv$ is locally a tensor product and so at each point either $u = 0$ or $v = 0$. One should keep in mind the trivial case when $X = \partial_{x_1}$ and $Y = \partial_{x_2}$.


  [1]: https://link.springer.com/book/10.1007/978-3-642-61497-2
  [2]: https://math.stackexchange.com/questions/4479812/product-of-distributions-under-wavefront-set-condition-is-zero