# When is the singularity of a semi-normal variety a double point singularity

Let $$X$$ be a semi-normal projective variety and $$p: \widetilde{X} \to X$$ be the normalization. Suppose that $$\widetilde{X}$$ is smooth and there exists two smooth divisors $$D_1, D_2 \subset \widetilde{X}$$ such that $$D_1 \cong D_2 \cong X_{\mathrm{sing}}$$ and $$p$$ induces as isomorphism between $$\widetilde{X} \backslash (D_1 \cup D_2)$$ and $$X\backslash X_{\mathrm{sing}}$$. Note that, the non-singular divisors $$D_1$$ and $$D_2$$ map to $$X_{\mathrm{sing}}$$ under the morphism $$p$$. Can we conclude that $$X$$ has double point singularities along $$X_{\mathrm{sing}}$$?

• What if transverse slices to $X_{\text{sing}}$ are tacnodal for some slices? I believe this can happen even in the semi-normal case. Jan 13, 2022 at 21:31
• Actually I looked at the examples I know where the transverse slices become tacnodal, and they actually are not semi-normal. So perhaps your conclusion is true. Jan 13, 2022 at 22:46

I think this is true if we assume that the isomorphisms $$D_1 \cong X_{\mathrm{sing}} \cong D_2$$ are also induced by $$p$$, at least in characteristic $$0$$ (though I think everything below is ok away from characteristic $$2$$). Edit: We also need to assume that the $$D_i$$ are disjoint in $$\widetilde{X}$$. If $$D_1 \cup D_2$$ is singular in $$\widetilde{X}$$ then we can also get worse singularities, see the edit below.
Let $$\tau : D_1 \cong D_2$$ denote the isomorphism induced by $$p$$ and let $$\bar{X}$$ be the quotient of $$\widetilde{X}$$ by the equivalence relation generated by $$x \sim \tau(x)$$. The quotient exists as a scheme in this case since $$D_i \subset X$$ are closed embeddings (this is an example of a pinching or Ferrand pushout, see e.g. this question and its answers). Then $$\bar{X}$$ is semi-normal with the required nodal singularities. Moreover, the map $$p$$ factors through a map $$q : \bar{X} \to X$$ by the universal property of quotients. By assumption $$q$$ is a bijection on points and isomorphism on residue fields so by semi-normality of $$X$$, $$q$$ is an isomorphism.
If we don't assume that the isomorphism $$D_1 \cong X_{\mathrm{sing}} \cong D_2$$ is induced by $$p$$, then this doesn't have to be true. For example, we can let $$\widetilde{X}$$ be two copies of $$\mathbb{P}^2$$ and $$D_i$$ be the conic $$x^2 + y^2 + z^2$$ the $$i^{th}$$ plane. Then we can consider the equivalence relation generated by identifying the two conics in the natural way as well as identifying $$(x,y,z) \in D_i$$ with $$(y,x,z) \in D_i$$. Then the quotient $$X$$ of $$\widetilde{X}$$ by this equivalence relation is semi-normal and $$D_i \cong \mathbb{P}^1 \cong X_{\mathrm{sing}}$$ but its singularities are not nodal. At a general point of $$X_{\mathrm{sing}}$$, the singularities will look like $$\mathbb{A}^1$$ times the union of the $$4$$ coordinate axes in $$\mathbb{A}^4$$ and $$p|_{D_i}$$ is $$2$$-to-$$1$$ onto $$X_{\mathrm{sing}}$$.
Edit: Here is an example when $$D_i$$ are smooth but $$D_1 \cup D_2$$ is not. Let $$D_i$$ be the coordinate axes in $$\mathbb{A}^2$$ and consider the equivalence relation that identifies the two axes by swapping them. The resulting semi-normal surface $$X$$ is $$\mathrm{Spec}$$ of the the subring $$\{f(x,y) \mid f(t,0) = f(0,t)\} \subset k[x,y].$$ I think this is isomorphic to $$u^3 - uvw + w^2 = 0$$ in $$\mathbb{A}^3$$ which is not nodal. The singularities get worse if $$D_1 \cup D_2$$ has worse singularities, e.g. if $$D_i$$ meet in a tacnode.