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 $X\_{\mathrm{sing}}$?



When is the singularity of a semi-normal variety a double point singularity
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](https://mathoverflow.net/questions/64294/gluing-along-closed-subschemes?rq=1)). 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 $X\_{\mathrm{sing}}$.


You might also want to take a look at Sections 5.1, 9.1 and 10.2 in Kollár's book *Singularities of the minimal model program*


**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.


