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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.

Suppose $R \subseteq S$ a finite birational extension of reduced rings. Let $I = \text{Ann}_{K(R)}(S/R)$ (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. $\square$

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. $\square$

A number of people have asked me for this over the past couple years so maybe I should write it down (even though I think the experts already knew it). Let me write a slighter weaker statement (with excruciating details) first to get the idea. I'll try to add a more general statement in the next couple days. Regardless, please let me know about typos etc.

Lemma: Suppose first that $R \subseteq S$ is an extension of reduced Noetherian rings which is finite, birational and such that it is an isomorphism outside of a single closed point $\mathfrak{m} \in \text{Spec }R$. Then there an ideal $J \subseteq S$ with $J = J' S$ and a ring homomorphism $R/J' = T \to S/J$ such that $R$ is the pullback of $\big( S \to S/J \leftarrow T \big).$ is isomorphic to $R$ via the induced map.

Proof: Obviously I want to use $J' = \mathfrak{m}^n$ for some $n \gg 0$ and $J = J' S$. There's one slight problem, the map $R/J' \to S/J$ is not necessarily injective. So lets set $J'_n = (\mathfrak{m}^n S) \cap R$ and observe that then $J_n' S = ((\mathfrak{m}^n S) \cap R)S \subseteq \mathfrak{m}^n S$ hence $(J_n' S) \cap R = J_n'$. Note that the $J'_n$ are contained in the integral closure of $\mathfrak{m}^n$ and so are co-final with the $\mathfrak{m}^n$.

Next, for each $n$ set $R_n$ to be the pullback of $\big( S \to S/J_n \leftarrow R/J'_n \big)$. We have the maps $R \to R_n \to S$ (and these are injective because of the work above). The sequence $\{R_n / R\} \subseteq S/R$ is a sequence of descending Artinian modules, and so $R_n/R$ stabilizes for $n \gg 0$. But then $R_n$ stabilizes for $n \gg 0$ (by Nakayama). Hence for $n \gg 0$ and any $m \gg n$, $$(R_n/R)/(J'_m (R_n/R)) = (R_m/R)/(J'_m (R_m/R)) = (R_m/J_m) \big/ (R/J_m').$$ But now we claim that $R/J_m' = R_m/J_m$. To see this consider the map $R \to R_m/J_m$ and choose an element $\overline{(s, \overline{r})}$ in $R_m/J_m$, here we represent elements of the pullback as a pair with an element of $S$ and an element of $R/J_m'$. The claim is that $r \in R$ maps to it. We want to show that $s - r \in J_m$. But we already knew that was true since $(s, \overline{r})$ was in $R_m$. Hence $R \to R_m/J_m$ is surjective and the kernel is obviously $J_m'$. This proves the claim.

But now, claim in hand, we see that $(R_n/R)/J_m' = 0$ for all $m \gg 0$. Hence $R_n = R$ since the $J_m'$ are cofinal with the $\mathfrak{m}^m$. $\square$

That's great, and it immediately yields the weaker statement:

Proposition: Any non-normal reduced excellent Noetherian ring can be obtained as the final output of a finite sequence of pullbacks of rings $R_{i+1} = (R_i \to R_i/J_i \leftarrow R/J_i')$ where each $R \subseteq R_i$ is a subring of the normalization, $J_i' \subseteq R$ is an ideal, and $J_i = J_i' R_i$

Proof: Suppose that $R$ is non-normal with normalization $R' = R_0$. Let $Q_1, \ldots, Q_t \subseteq R$ be the minimal primes of the locus where $R \subseteq R_0$ is not an isomorphism. The lemma (and a bit of localization) guarantees that if we set $J_{0,n}' = ((Q_1 \cdots Q_t)^n R_0) \cap R$ for $n \gg 0$, we can find $R_1$ as the pullback of $\Big(R_0 \to R_0/J_{0,n} \leftarrow R/J_{0,n}' \Big)$ for $n \gg 0$ so that $R_1/R$ is supported on a locus strictly inside the support of $R_0/R$. Now do the same thing for $R \subseteq R_1$. Noetherian induction finishes it off. $\square$

But perhaps we wanted a single pullback and not a sequence of them. As I mentioned above, I'll try to write down the details in the next day or so, they should be very similar. (Or maybe there is an obvious way to combine a couple gluing diagrams into a single diagram).

EDIT: A quick speculation, if $R \subseteq S$ is finite birational. Consider the conductor ideal $I = \text{Ann}_R(S/R)$. This is an ideal of both $R$ and $S$. Then consider the pullback, $A = \{ S \to S/I \leftarrow R/I \}$. Is $A$ equal to $R$? It's true for seminormal $R$ since the conductor is radical in both $R$ and $S$.

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.

Suppose $R \subseteq S$ a finite birational extension of reduced rings. Let $I = \text{Ann}_{K(R)}(S/R)$ (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. $\square$

A number of people have asked me for this over the past couple years so maybe I should write it down (even though I think the experts already knew it). Let me write a slighter weaker statement (with excruciating details) first to get the idea. I'll try to add a more general statement in the next couple days. Regardless, please let me know about typos etc.

Lemma: Suppose first that $R \subseteq S$ is an extension of reduced excellent Noetherian rings which is finite, birational and such that it is an isomorphism outside of a single closed point $\mathfrak{m} \in \text{Spec }R$. Then there an ideal $J \subseteq S$ with $J = J' S$ and a ring homomorphism $R/J' = T \to S/J$ such that $R$ is the pullback of $\big( S \to S/J \leftarrow T \big).$ is isomorphic to $R$ via the induced map.

Proof: Obviously I want to use $J' = \mathfrak{m}^n$ for some $n \gg 0$ and $J = J' S$. There's one slight problem, the map $R/J' \to S/J$ is not necessarily injective. So lets set $J'_n = (\mathfrak{m}^n S) \cap R$ and observe that then $J_n' S = ((\mathfrak{m}^n S) \cap R)S \subseteq \mathfrak{m}^n S$ hence $(J_n' S) \cap R = J_n'$. Note that the $J'_n$ are contained in the integral closure of $\mathfrak{m}^n$ and so are co-final with the $\mathfrak{m}^n$.

Next, for each $n$ set $R_n$ to be the pullback of $\big( S \to S/J_n \leftarrow R/J'_n \big)$. We have the maps $R \to R_n \to S$ (and these are injective because of the work above). The sequence $\{R_n / R\} \subseteq S/R$ is a sequence of descending Artinian modules, and so $R_n/R$ stabilizes for $n \gg 0$. But then $R_n$ stabilizes for $n \gg 0$ (by Nakayama). Hence for $n \gg 0$ and any $m \gg n$, $$(R_n/R)/(J'_m (R_n/R)) = (R_m/R)/(J'_m (R_m/R)) = (R_m/J_m) \big/ (R/J_m').$$ But now we claim that $R/J_m' = R_m/J_m$. To see this consider the map $R \to R_m/J_m$ and choose an element $\overline{(s, \overline{r})}$ in $R_m/J_m$, here we represent elements of the pullback as a pair with an element of $S$ and an element of $R/J_m'$. The claim is that $r \in R$ maps to it. We want to show that $s - r \in J_m$. But we already knew that was true since $(s, \overline{r})$ was in $R_m$. Hence $R \to R_m/J_m$ is surjective and the kernel is obviously $J_m'$. This proves the claim.

But now, claim in hand, we see that $(R_n/R)/J_m' = 0$ for all $m \gg 0$. Hence $R_n = R$ since the $J_m'$ are cofinal with the $\mathfrak{m}^m$. $\square$

That's great, and it immediately yields the weaker statement:

Proposition: Any non-normal reduced excellent Noetherian ring can be obtained as the final output of a finite sequence of pullbacks of rings $R_{i+1} = (R_i \to R_i/J_i \leftarrow R/J_i')$ where each $R \subseteq R_i$ is a subring of the normalization, $J_i' \subseteq R$ is an ideal, and $J_i = J_i' R_i$

Proof: Suppose that $R$ is non-normal with normalization $R' = R_0$. Let $Q_1, \ldots, Q_t \subseteq R$ be the minimal primes of the locus where $R \subseteq R_0$ is not an isomorphism. The lemma (and a bit of localization) guarantees that if we set $J_{0,n}' = ((Q_1 \cdots Q_t)^n R_0) \cap R$ for $n \gg 0$, we can find $R_1$ as the pullback of $\Big(R_0 \to R_0/J_{0,n} \leftarrow R/J_{0,n}' \Big)$ for $n \gg 0$ so that $R_1/R$ is supported on a locus strictly inside the support of $R_0/R$. Now do the same thing for $R \subseteq R_1$. Noetherian induction finishes it off. $\square$

But perhaps we wanted a single pullback and not a sequence of them. As I mentioned above, I'll try to write down the details in the next day or so, they should be very similar. (Or maybe there is an obvious way to combine a couple gluing diagrams into a single diagram).

A number of people have asked me for this over the past couple years so maybe I should write it down (even though I think the experts already knew it). Let me write a slighter weaker statement (with excruciating details) first to get the idea. I'll try to add a more general statement in the next couple days. Regardless, please let me know about typos etc.

Lemma: Suppose first that $R \subseteq S$ is an extension of reduced Noetherian rings which is finite, birational and such that it is an isomorphism outside of a single closed point $\mathfrak{m} \in \text{Spec }R$. Then there an ideal $J \subseteq S$ with $J = J' S$ and a ring homomorphism $R/J' = T \to S/J$ such that $R$ is the pullback of $\big( S \to S/J \leftarrow T \big).$ is isomorphic to $R$ via the induced map.

Proof: Obviously I want to use $J' = \mathfrak{m}^n$ for some $n \gg 0$ and $J = J' S$. There's one slight problem, the map $R/J' \to S/J$ is not necessarily injective. So lets set $J'_n = (\mathfrak{m}^n S) \cap R$ and observe that then $J_n' S = ((\mathfrak{m}^n S) \cap R)S \subseteq \mathfrak{m}^n S$ hence $(J_n' S) \cap R = J_n'$. Note that the $J'_n$ are contained in the integral closure of $\mathfrak{m}^n$ and so are co-final with the $\mathfrak{m}^n$.

Next, for each $n$ set $R_n$ to be the pullback of $\big( S \to S/J_n \leftarrow R/J'_n \big)$. We have the maps $R \to R_n \to S$ (and these are injective because of the work above). The sequence $\{R_n / R\} \subseteq S/R$ is a sequence of descending Artinian modules, and so $R_n/R$ stabilizes for $n \gg 0$. But then $R_n$ stabilizes for $n \gg 0$ (by Nakayama). Hence for $n \gg 0$ and any $m \gg n$, $$(R_n/R)/(J'_m (R_n/R)) = (R_m/R)/(J'_m (R_m/R)) = (R_m/J_m) \big/ (R/J_m').$$ But now we claim that $R/J_m' = R_m/J_m$. To see this consider the map $R \to R_m/J_m$ and choose an element $\overline{(s, \overline{r})}$ in $R_m/J_m$, here we represent elements of the pullback as a pair with an element of $S$ and an element of $R/J_m'$. The claim is that $r \in R$ maps to it. We want to show that $s - r \in J_m$. But we already knew that was true since $(s, \overline{r})$ was in $R_m$. Hence $R \to R_m/J_m$ is surjective and the kernel is obviously $J_m'$. This proves the claim.

But now, claim in hand, we see that $(R_n/R)/J_m' = 0$ for all $m \gg 0$. Hence $R_n = R$ since the $J_m'$ are cofinal with the $\mathfrak{m}^m$. $\square$

That's great, and it immediately yields the weaker statement:

Proposition: Any non-normal reduced excellent Noetherian ring can be obtained as the final output of a finite sequence of pullbacks of rings $R_{i+1} = (R_i \to R_i/J_i \leftarrow R/J_i')$ where each $R \subseteq R_i$ is a subring of the normalization, $J_i' \subseteq R$ is an ideal, and $J_i = J_i' R_i$

Proof: Suppose that $R$ is non-normal with normalization $R' = R_0$. Let $Q_1, \ldots, Q_t \subseteq R$ be the minimal primes of the locus where $R \subseteq R_0$ is not an isomorphism. The lemma (and a bit of localization) guarantees that if we set $J_{0,n}' = ((Q_1 \cdots Q_t)^n R_0) \cap R$ for $n \gg 0$, we can find $R_1$ as the pullback of $\Big(R_0 \to R_0/J_{0,n} \leftarrow R/J_{0,n}' \Big)$ for $n \gg 0$ so that $R_1/R$ is supported on a locus strictly inside the support of $R_0/R$. Now do the same thing for $R \subseteq R_1$. Noetherian induction finishes it off. $\square$

But perhaps we wanted a single pullback and not a sequence of them. As I mentioned above, I'll try to write down the details in the next day or so, they should be very similar. (Or maybe there is an obvious way to combine a couple gluing diagrams into a single diagram).

EDIT: A quick speculation, if $R \subseteq S$ is finite birational. Consider the conductor ideal $I = \text{Ann}_R(S/R)$. This is an ideal of both $R$ and $S$. Then consider the pullback, $A = \{ S \to S/I \leftarrow R/I \}$. Is $A$ equal to $R$? It's true for seminormal $R$ since the conductor is radical in both $R$ and $S$.