A number of people have asked me for a reference since I wrote down that answer in the other question so I'll try to write a reference here (I'm sure some experts knew it before though). I originally wrote a complicated Noetherian induction in this answer *but* I just realized this is really easy.

**Main point:** There is a *canonical* way to find the thickened subschemes that glue to give the non-normal variety. Indeed, the non-normal locus already has a canonical scheme structure. Let's use that!

Suppose $R \subseteq S$ a finite birational extension of reduced rings. Let $I = R :_{K(R)} S = \{z \in K(R)\;|\; zS \subseteq R \}$ (a fractional ideal -- usually called the conductor). Note that $I$ is an ideal of $R$ (since in particular it multiplies $R$ back into $R$) but it is also an ideal of $S$. Indeed, take $s \in S$ and $x \in I$, then $xs$ also still multiplies other $s' \in S$ into $R$ as well.

**Theorem:** *With notation as above if $A$ is the pullback of $\big( S \to S/I \leftarrow R/I \big)$ then $A = R$.*

*Proof:* Obviously we have $R \subseteq A \subseteq S$. We want to show that $R \to A$ surjects as well. Take an element $(s, \overline{r})$ in $A$ (this is a pair of elements $s \in S$ and $\overline{r} \in R/I$ with common image in $S/I$). Let $r \in R$ be any representative for $\overline{r}$. Consider $s - r \in S$. Obviously $s-r$ is sent to zero in $S/I$ and so $s - r \in I \subseteq R$. But then $s \in R$ as well.

The map $R \to A$ sends $x$ to $(x, \overline{x})$. Therefore the map $R \to A$ surjects as claimed.
$\square$