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In his answer to this MO question, Karl Schwede claimed that every non-normal variety can be obtained by an appropriate pushout diagram, as sketched in that answer. This would give substance to the heuristics according to which "a normal variety is a variety that has no undue gluing of subvarieties or tangent spaces" (again, see K.Schwede's answer and the examples therein).

Q. Has a proof of that fact been written down somewhere since then? If yes, where? And if not, could anybody who knows it sketch it here on MO?

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  • $\begingroup$ Just a comment. One might hope for the following: given a variety $X$ and a proper birational $f:X'\to X$, let $Z\subseteq X$ be the locus where $f$ is not a local isomorphism (with reduced subscheme structure). For $n\geq 1$, let $Z_n$ denote the $n$-th infinitesimal thickening of $Z$ in $X$. Then for $n\gg 0$, $X$ is the pushout of $Z_n \leftarrow f^{-1}(Z_n)\to X'$. To answer your question, we would apply this to the normalization map (as hinted by Karl Schwede). $\endgroup$ Nov 7, 2014 at 1:16
  • $\begingroup$ I think the claim above could follow from some results in Artin's paper "Algebraization of Formal Moduli II", where instead of $Z_n$ we consider their limit (the formal completion). The trick seems to be to choose an appropriate $n\gg 0$. $\endgroup$ Nov 7, 2014 at 1:18

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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. $\square$

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