There are several ways to generalize the notion of "algebraic closure" from fields to arbitrary commutative rings. A good overview is [On algebraic closures](https://www.jstor.org/stable/43737179) by R. Raphael. I am more interested in the notion suggested in [Stacks/0DCK](https://stacks.math.columbia.edu/tag/0DCK). In particular, I would like to know if there is a corresponding "algebraic closure" which is unique up to isomorphism - as in the case of fields.


A commutative ring $R$ is *absolutely integrally closed* if every monic polynomial over $R$ has a root in $R$. Equivalently, every monic polynomial over $R$ is a product of linear factors over $R$. (Notice that this differs from E. Enoch's notion of [totally integrally closed rings](https://www.ams.org/journals/proc/1968-019-03/S0002-9939-1968-0224600-9/S0002-9939-1968-0224600-9.pdf).)


It is easy to prove ([Stacks/0DCR](https://stacks.math.columbia.edu/tag/0DCR)) that every commutative ring $R$ has an integral extension $R \hookrightarrow R'$ such that $R'$ is absolutely integrally closed (and $R'$ is a free $R$-module, among other things). But clearly such an extension is not unique: Take any ideal $I \subseteq R'$ with $I \cap R = 0$, then $R \hookrightarrow R'/I$ has the same properties. So we have to make this extension tighter...


Recall that an extension $R \hookrightarrow R'$ is *tight* (aka *essential*) if for all ideals $I \subseteq R'$ we have $I \cap R = 0 \implies I =0$. If $R \hookrightarrow R'$ is any extension, there is always an ideal $J \subseteq R'$ which is maximal with the property $J \cap R = 0$ (Zorn's Lemma), and then $R \hookrightarrow R'/J$ is tight.


So for every commutative ring $R$ there is a tight integral ring extension $R \hookrightarrow R^{\text{aic}}$ such that $R^{\text{aic}}$ is absolutely integrally closed.


**Question.** Does this determine $R^{\text{aic}}$ up to isomorphism? If not, what properties do we need to add to make it unique?


The end result should be called the absolute integral closure of $R$, for short a.i.c. (Unfortunately, the name is already taken for [something else](https://arxiv.org/pdf/1409.0441.pdf) when $R$ is a domain.)


*Edit*. Actually the answer by Matthé proves that the a.i.c. in the case of a domain is what is already understood under this name, namely the integral closure inside the algebraic closure of the field of fractions. In this special case uniqueness holds.


*Edit.* Some simple facts:


(a) The a.i.c of a field is uniquely determined. It is the algebraic closure. (This is because every tight extension of a field is also a field.)


(b) If $R\_1,\dotsc,R\_n$ are commutative rings whose a.i.c. is uniquely determined, is straight forward to check that $R\_1 \times \cdots \times R\_n$ has a uniquely determined a.i.c. as well, with $(R\_1 \times \cdots \times R\_n)^{\text{aic}} = R\_1^{\text{aic}} \times \cdots \times R\_n^{\text{aic}}$.


(c) If $S \subseteq R$ is a multiplicative subset consisting of regular elements and $R^{\text{aic}}$ is some a.i.c. of $R$, then $S^{-1} R^{\text{aic}}$ is an a.i.c. of $S^{-1} R$. I'm not sure yet about uniqueness here, though.



Uniqueness of the "algebraic closure" of a commutative ring
Assume $R$ is reduced and $\mathrm{min}(R)$, the set of minimal prime ideals, is finite. Then $R$ has a *universal* aic, one in which every aic of $R$ can be embedded. We show this below.


Rings of this type have quite remarkable properties, and they include all Noetherian reduced rings. Domains are just the special case where $\mathrm{min}(R)$ is a one-point space. For if $\mathrm{min}(R)$ $=$ $\{\mathfrak{p}\}$, then $\mathfrak{p}$ $=$ $\bigcap\mathrm{min}(R)$ $=$ $\bigcap\mathrm{spec}(R)$ $=$ $\mathrm{nil}(R)$ $=$ $0$.


Let $R$ $\subseteq$ $S$ be a given tight integral extension, with $S$ an absolutely integrally closed ring. Then we have $\mathrm{nil}(S)\cap R$ $=$ $\mathrm{nil}(R)$ $=$ $0$, so by tightness $S$ must also be reduced.


For $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, we can find a $\mathfrak{P}$ $\in$ $\mathrm{spec}(S)$ with $\mathfrak{P}\cap R$ $=$ $\mathfrak{p}$, because the extension $R$ $\subseteq$ $S$ is integral. Take a minimal prime ideal $\tilde{\mathfrak{p}}$ $\subseteq$ $\mathfrak{P}$ of $S$. Then $\tilde{\mathfrak{p}}\cap R$ $=$ $\mathfrak{p}$. If $I$ $=$ $\bigcap\_{\mathfrak{p}\in\mathrm{min}(R)}\tilde{\mathfrak{p}}$, then $I\cap R$ $=$ $\bigcap\mathrm{min}(R)$ $=$ $\mathrm{nil}(R)$ $=$ $0$, hence $I$ $=$ $0$. So if $\mathfrak{Q}$ $\in$ $\mathrm{min}(S)$, since $\mathrm{min}(R)$ is finite, the product $\prod\_{\mathfrak{p}\in\mathrm{min}(R)}\tilde{\mathfrak{p}}$ exists and is contained in $I$ $=$ $0$ $\subseteq$ $\mathfrak{Q}$. Thus $\mathfrak{Q}$ must be one of the $\tilde{\mathfrak{p}}$. Therefore, $S$ is also "semiglobal", that is to say, $\mathrm{min}(S)$ $=$ $\{\tilde{\mathfrak{p}}\mid\mathfrak{p}\in\mathrm{min}(R)\}$ is finite.


Let $K$ and $L$ be the total rings of fractions of $R$ and $S$, respectively. So $K$ $=$ $\mathrm{reg}(R)^{-1}R$, where $\mathrm{reg}(R)$ is the set of regular elements of $R$. The prime ideals of $K$ are of the form $\mathfrak{p}K$, where $\mathfrak{p}$ $\in$ $\mathrm{spec}(R)$ with $\mathfrak{p}\cap\mathrm{reg}(R)$ $=$ $\varnothing$, i.e. $\mathfrak{p}$ consists of zero divisors of $R$. These $\mathfrak{p}$ are precisely the minimal prime ideals of $R$. For if $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ and $r$ $\in$ $\mathfrak{p}$, then $r$ $\in$ $\mathfrak{p}R\_\mathfrak{p}$. Being a localisation of a reduced ring, $R\_\mathfrak{p}$ is again reduced. So $0$ $=$ $\mathrm{nil}(R\_\mathfrak{p})$ $=$ $\bigcap\mathrm{spec}(R\_\mathfrak{p})$ $=$ $\mathfrak{p}R\_\mathfrak{p}$, for $\mathfrak{p}R\_\mathfrak{p}$ is the only prime ideal of $R\_\mathfrak{p}$. (Note that $R\_\mathfrak{p}$ therefore must be a field.) So $r$ $=$ $0$ in $R\_\mathfrak{p}$, and hence there is an $r'$ $\in$ $R-\mathfrak{p}$ for which $rr'$ $=$ $0$ in $R$. So $r$ is a zero divisor of $R$. Conversely, if $rr'$ $=$ $0$ and $r'$ $\ne$ $0$, then there is a $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ with $r'$ $\notin$ $\mathfrak{p}$, because $\bigcap\mathrm{min}(R)$ $=$ $0$. Therefore, $r$ $\in$ $\mathfrak{p}$. So if a prime $\mathfrak{q}$ of $R$ contains only zero divisors, it is contained in $\bigcup\mathrm{min}(R)$. But this a finite union, and so by the [prime avoidance lemma](https://stacks.math.columbia.edu/tag/00DS) the ideal $\mathfrak{q}$ is a minimal prime of $R$.


So $\mathrm{spec}(K)$ $=$ $\{ \mathfrak{p}K\mid\mathfrak{p}\in\mathrm{min}(R)\}$. And $R$ $\subseteq$ $K$, for if $r$ $\in$ $R$ becomes zero in $K$, there is an $r'$ $\in$ $\mathrm{reg}(R)$ with $rr'$ $=$ $0$, hence $r$ $=$ $0$.


Note that $\mathfrak{p}K\cap R$ $=$ $\mathfrak{p}$ when $\mathfrak{p}$ is a minimal prime. Indeed, if $r=u/v$ in $K$ with $u$ $\in$ $\mathfrak{p}$ and $v$ $\in$ $R$ regular, then there is a $w$ $\in$ $\mathrm{reg}(R)$ with $w(vr-u)$ $=$ $0$ in $R$, so $vr$ $=$ $u$ $\in$ $\mathfrak{p}$. If $v$ $\in$ $\mathfrak{p}$, then by the above $v$ is a zero divisor of $R$, contradiction. So $r$ $\in$ $\mathfrak{p}$. As a result, $R/\mathfrak{p}$ $\subseteq$ $K/\mathfrak{p}K$.


It follows that $K$ is a zero-dimensional reduced ring, that is, a von Neumann regular ring. For if $\mathfrak{p}$ and $\mathfrak{q}$ are minimals of $R$ with $\mathfrak{p}K$ $\subseteq$ $\mathfrak{q}K$, then $\mathfrak{p}$ $\subseteq$ $\mathfrak{q}K\cap R$ $=$ $\mathfrak{q}$, hence $\mathfrak{p}$ $=$ $\mathfrak{q}$. So if $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, we have $\mathfrak{p}K$ $\in$ $\mathrm{max}(K)$, and so $K/\mathfrak{p}K$ is a field. It is in fact the quotient field of $R/\mathfrak{p}$. The ring $L$ is also VNR, and it contains $S$ as a subring.


$R\subseteq S\subseteq L=\mathrm{reg}(S)^{-1}S$, and $\mathrm{reg}(R)$ $\subseteq$ $\mathrm{reg}(S)$. For if $rs$ $=$ $0$ with $r$ $\in$ $R$ and $s$ $\in$ $S-\{0\}$, then $r$ is in a minimal prime $\tilde{\mathfrak{p}}$ of $S$. But then $r$ $\in$ $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, so $r$ is a zero divisor in $R$. Thus $K$ $=$ $\mathrm{reg}(R)^{-1}R$ $\subseteq$ $\mathrm{reg}(R)^{-1}S$ (since localizations are flat) $\subseteq$ $\mathrm{reg}(S)^{-1}S$ $=$ $L$. Hence $K/\mathfrak{p}K$ is a subfield of $L/\tilde{\mathfrak{p}}L$. And the extension $K/\mathfrak{p}K$ $\subseteq$ $L/\tilde{\mathfrak{p}}L$ is algebraic, because $S$ is integral over $R$.


The natural map $K\to\prod\_{\mathfrak{p}\in\mathrm{min}(R)}K/\mathfrak{p}K$ is injective, for the kernel is the intersection of all prime ideals of $K$, and $K$ is reduced. As $\mathrm{dim}(K)$ $=$ $0$, by the CRT this is actually an isomorphism, and $K$ is a finite product of fields. It is easy to see that in fact $K/\mathfrak{p}K$ $\cong$ $K\_{\mathfrak{p}K}$ $\cong$ $R\_\mathfrak{p}$.


Let $\overline{K}$ $=$ $\prod\_{\mathfrak{p}\in\mathrm{min}(R)}C\_\mathfrak{p}$, where $C\_\mathfrak{p}$ is the algebraic closure of the field $K/\mathfrak{p}K$ (and of $L/\tilde{\mathfrak{p}}L$). This $\overline{K}$ may be regarded as the "algebraic closure" of $R$ (or, equally, of $K$, $S$ or $L$).


I claim that the integral closure $T$ of $R$ in $\overline{K}$ is tight over $R$. For let $I$ be a nonzero ideal of $T$, and take $0$ $\neq$ $t$ $\in$ $I$. For $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, denote the $\mathfrak{p}$-th coordinate of $t$ by $t\_\mathfrak{p}$, and pick a $\mathfrak{p}$ with $t\_\mathfrak{p}$ $\in$ $C\_\mathfrak{p}$ nonzero. Take a monic $f$ $\in$ $R[X]$ with $f(t)$ $=$ $0$ in $T$. If the constant term $f(0)$ is in $\mathfrak{p}$, it becomes zero in $K/\mathfrak{p}K$, hence in $C\_\mathfrak{p}$. As $t\_\mathfrak{p}$ is a root of the image of $f$ in $C\_\mathfrak{p}[X]$ and $C\_\mathfrak{p}$ is a field, $t\_\mathfrak{p}$ is also a root of (the image of) $g$ $=$ $(f-f(0))/X$ $\in$ $R[X]$. We then replace $f$ by $g$. If the new $f(0)$ is in $\mathfrak{p}$ again, repeat the process until $f(0)$ $\notin$ $\mathfrak{p}$.


By the finiteness of the minimal spectrum of $R$, the product of the minimal primes $\mathfrak{q}$ $\ne$ $\mathfrak{p}$ cannot be contained in $\mathfrak{p}$, so there exists a $c$ $\in$ $R$ that is in all minimals of $R$ except $\mathfrak{p}$. Put $h$ $:=$ $cf$. Then $h(t\_\mathfrak{p})$ $=$ $0$, and for $\mathfrak{p}$ $\ne$ $\mathfrak{q}$ $\in$ $\mathrm{min}(R)$, we have $c$ $=$ $0$ in $K/\mathfrak{q}K$, so $h$ $=$ $0$ in $C\_\mathfrak{q}[X]$. Hence $h(t)$ $=$ $0$ in $T$. But $h(0)$ $=$ $cf(0)$ $\notin$ $\mathfrak{p}$, and therefore $h(0)$ is a nonzero element of $tT$ $\subseteq$ $I$ that is in $R$. This settles tightness.


Since $T$ is integral over $R$ and clearly absolutely integrally closed, it is an absolute integral closure of $R$. The image of $S$ in $\overline{K}$ under the composition map $S\subseteq L\rightarrowtail\prod\_{\mathfrak{p}\in\mathrm{min}(R)}L/\tilde{\mathfrak{p}}L$ $\subseteq$ $\overline{L}$ $=$ $\overline{K}$ is integral over $R$, so it is contained in $T$.


An $e\in\overline{K}$ is idempotent iff for all $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ its $\mathfrak{p}$-th coordinate is either $0$ or $1$. Clearly, the idempotents are integral over $R$, hence they are in $T$, and there are $2^{|\mathrm{min}(R)|}$ of them. Denote the one with $e\_\mathfrak{p}$ $=$ $1$ and $0$ elsewhere by $e\_{(\mathfrak{p})}$. Then every idempotent is the sum of the elements of a subset of $\{e\_{(\mathfrak{p})}\mid\mathfrak{p}\in\mathrm{min}(R)\}$. And $e\_{(\mathfrak{p})}e\_{(\mathfrak{q})}$ $=$ $0$ when $\mathfrak{p}$ $\ne$ $\mathfrak{q}$. So the $e\_{(\mathfrak{p})}$ form what is known as a fundamental system of orthogonal idempotents.


Then $S=T$ iff the $e\_{(\mathfrak{p})}$ are all in $S$. For if they are, and $t$ $\in$ $T$, fix a $\mathfrak{p}$, and let $f$ $\in$ $R[X]$ be monic with $f(t)$ $=$ $0$. Then $f(t\_\mathfrak{p})$ $=$ $0$ in $C\_\mathfrak{p}$. Write $f$ $=$ $\prod\_{1\leq i\leq n}(X-s\_i)$ in $S[X]$, so that we have $f$ $=$ $\prod\_{1\leq i\leq n}(X-(s\_i)\_{\mathfrak{p}})$ in $C\_\mathfrak{p}[X]$. But then $t\_\mathfrak{p}$ $=$ $(s\_i)\_{\mathfrak{p}}$ for some $i$, because $C\_\mathfrak{p}$ is a field. Put $s$ $=$ $s\_i$. Since $e\_{(\mathfrak{p})}$ is in $S$, so is $e\_{(\mathfrak{p})}s$. Its $\mathfrak{p}$-th component is $t\_\mathfrak{p}$, and it has zero for the other $\mathfrak{q}$. So $e\_{(\mathfrak{p})}s$ is actually equal to $e\_{(\mathfrak{p})}t$. The sum of the $e\_{(\mathfrak{p})}$, taken over the $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, is equal to $1$. So $t$, which is therefore the sum of the $e\_{(\mathfrak{p})}t$, must be in $S$.


Note that, by integrality, lying-over holds for $S/R$. So when $R$ is a domain, there's a prime $\mathfrak{q}$ of $S$ with $\mathfrak{q}\cap R$ $=$ $0$. As $S$ is tight over $R$, we have $\mathfrak{q}$ $=$ $0$, so $S$ is a domain. Hence it contains $2^{|\mathrm{min}(R)|}$ idempotents, namely just the trivial ones, $0$ and $1$.


To sum up:


**1) $\boldsymbol{T}$ is the universal aic of $\boldsymbol{R}$. That is to say, it contains all other aic's.**


**2) An aic of $\boldsymbol{R}$ is isomorphic to $\boldsymbol{T}$ iff it contains $\boldsymbol{2^{|\mathrm{min}(R)|}}$ idempotents.**


**3) Domains have uniquely determined aic's.**


*Edit.* To this sum we can now add:


**4) If $\boldsymbol{k}$ is a field, then the ring $\boldsymbol{R=k[X,Y]/(XY)}$ has non-isomorphic aic's.**


Here, $\mathrm{min}(R)$ $=$ $\{\mathfrak{p},\mathfrak{q}\}$, with $\mathfrak{p}$ $=$ $(X)$ and $\mathfrak{q}$ $=$ $(Y)$, if we denote the images of $X$ and $Y$ in $R$ by the same symbols. And $K/\mathfrak{p}K$ $=$ $k(Y)$, with algebraic closure $C\_\mathfrak{p}$. The image of $R$ in $C\_\mathfrak{p}$ is the polynomial ring $k[Y]$. Let $T\_\mathfrak{p}$ be the integral closure of $k[Y]$ in $C\_\mathfrak{p}$, and $T\_\mathfrak{q}$ the integral closure of $k[X]$ in $C\_\mathfrak{q}$ $\cong\_k$ $C\_\mathfrak{p}$ (under $X\mapsto Y$). We then have $T$ $=$ $T\_\mathfrak{p}\times T\_\mathfrak{q}$ in $C\_\mathfrak{p}\times C\_\mathfrak{q}$ $=$ $\overline{K}$.


If $\overline{k}$ is the algebraic closure of $k$, there are ring homomorphisms $T\_\mathfrak{p}
\xrightarrow{\varphi}\overline{k}
\xleftarrow{\psi}T\_\mathfrak{q}$ extending the maps $\varphi:k[Y]\to \overline{k}\gets k[X]:\psi$ defined by $Y\mapsto$$\,0\,$↤$\,X$. For, the algebraic closure $C\_\mathfrak{p}$ results from the field $k(Y)$ by a transfinite series of field extensions, indexed by some ordinal number $\alpha$ (which we can take to be $\omega$ $=$ $\aleph\_0$ or the cardinal number of $k$, whichever is the largest of the two, for that must be the cardinality of the field $C\_\mathfrak{p}$). When $\gamma$ $\le$ $\alpha$ is a limit ordinal, the $\gamma$-th step simply consists of taking the union of the fields constructed in the earlier steps, plus taking the union of the maps $\varphi$ constructed. And when $\gamma$ $=$ $\beta+1$ is a successor ordinal, this step is the adjunction of a new algebraic element to the current field $F\_\beta$. By the induction hypothesis, we then have $\varphi:I\_\beta\to\overline{k}$, where $I\_\beta$ is the set of integers of $F\_\beta$ over $k[Y]$, and $F\_\gamma$ $=$ $F\_\beta[Z]/(g)$ for some monic irreducible $g$ $=$ $g(Z)$ $\in$ $F\_\beta[Z]$. If $u$ $\in$ $I\_\gamma-I\_\beta$ is a new integer and $f$ $\in$ $F\_\beta[Z]$ is the minimal polynomial of $u$ over $F\_\beta$, its coefficients are in $I\_\beta$. (They are elementary symmetric functions in the conjugates of $u$ over $F\_\beta$, and each of the latter is integral over $I\_\beta$.) Then $u$ can be identified with $Z\text{ mod }f$ $\in$ $I\_\beta[Z]/(f)$. Take a $v$ in $\overline{k}$ that is a zero of the image of $f$ in $\overline{k}[Z]$ under $\varphi$. Note that the homomorphism $\varphi:I\_\beta[Z]/(f)\to\overline{k}$, for which $Z\text{ mod }f\mapsto v$ and $\varphi\upharpoonright I\_\beta:I\_\beta\to\overline{k}$ is the map already built, is well defined. It is now clear that $\varphi$ extends to $\varphi:I\_\gamma\to\overline{k}$. Ultimately, for $\gamma$ $=$ $\alpha$, we obtain the desired $\varphi:T\_\mathfrak{p}$ $=$ $I\_\alpha\to\overline{k}$. And we can let $\psi:T\_\mathfrak{q}\to\overline{k}$ be the composition $T\_\mathfrak{q}\overset{\sim}{\underset{\text{nat}}{\to}}T\_\mathfrak{p}\underset{\varphi}{\to}\overline{k}$.


Now let $T\_0=\{(u,v)\in T=T\_\mathfrak{p}\times T\_\mathfrak{q}\mid\varphi(u)=\psi(v)\}$ be the pullback. It contains the image of $R$ in $T$. Indeed, every $r$ $\in$ $R$ is of the form $\lambda+Xf(X)+Yg(Y)$ with $\lambda$ $\in$ $k$ and $f$ and $g$ are univariate polynomials. This maps to $(\lambda+Yg(Y),\lambda+Xf(X))$ in $R/\mathfrak{p}\times R/\mathfrak{q}$ $=$ $k[Y]\times k[X]$ $\subseteq$ $T$, and thence to $(\lambda,\lambda)$ in $\overline{k}\times\overline{k}$ under $\varphi\times\psi$, since $\varphi(Y)$ $=$ $0$ $=$ $\psi(X)$. And $T\_0$ is integral and tight over $R$. For let $t\_0$ $=$ $(u,v)$ $\in$ $T\_0-R$. Say $u$ $\ne$ $0$. Then $Yt\_0$ $=$ $(Yu,0)$ is in $T\_0-\{0\}$, and we have $f(u)$ $=$ $0$ for some monic $f$ $=$ $f(Z)$ $\in$ $k[Y][Z]$ with $f(0)$ $\in$ $k[Y]$ non-zero, for example for the minimal polynomial $f$ of $u$ over $k(Y)$. If $f^\bullet$ $\in$ $R[X]$ denotes the same polynomial $f$ $\in$ $k[Y][Z]$ $\subseteq$ $R[Z]$, then $g$ $:=$ $Y^{\mathrm{deg}(f)}f^\bullet(Y^{-1}Z)$ $\in$ $R[Z]$ has $g(Yu)$ $=$ $0$ in $T\_\mathfrak{p}$ and $g(0)$ = $Y^{\mathrm{deg}(f)}f(0)$ $\ne$ $0$. But $Y$ $=$ $0$ in $T\_\mathfrak{q}$, so $g(0)$ vanishes in $T\_\mathfrak{q}$, hence $g(Yt\_0)$ $=$ $0$ in $T\_0$. This shows that $0$ $\ne$ $g(0)$ is in $Yt\_0T\_0\cap R$ $\subseteq$ $t\_0T\_0\cap R$.


Finally, $T\_0$ is an absolutely integrally closed ring. For let $f$ $\in$ $T\_0[Z]$ be monic, of degree $n$, say. Then $f$ $=$ $\prod\_{1\leq i\leq n}(Z-(u\_i,v\_i))$ in $T[Z]$ for suitable $u\_i$ $\in$ $T\_\mathfrak{p}$ and $v\_i$ $\in$ $T\_\mathfrak{q}$, as $T$ is absolutely integrally closed. But then $f$ $=$ $\prod\_{1\leq i\leq n}(Z-(u\_i,v\_{\pi(i)}))$ in $T[Z]$ $=$ $(T\_\mathfrak{p}\times T\_\mathfrak{q})[Z]$ for every permutation $\pi$ of the indices $1,\cdots,n$. For $i$ $<$ $n$, let $(a\_i,b\_i)$ $\in$ $T\_0$ be the coefficient of $Z^i$ in $f$. So $(a\_0,b\_0)$ $=$ $(-1)^n\prod\_{1\leq i\leq n}(u\_i,v\_i)$, and so on, up to $(a\_{n-1},b\_{n-1})$ $=$ $-\sum\_{1\leq i\leq n}(u\_i,v\_i)$. Then we have $(-1)^n\prod\_{1\leq i\leq n}\varphi(u\_i)$ $=$ $\varphi(a\_0)$ $=$ $\psi(b\_0)$ $=$ $(-1)^n\prod\_{1\leq i\leq n}\psi(v\_i)$, and so forth, up to $-\sum\_{1\leq i\leq n}\varphi(u\_i)$ $=$ $\varphi(a\_{n-1})$ $=$ $\psi(b\_{n-1})$ $=$ $-\sum\_{1\leq i\leq n}\psi(v\_i)$. It follows that in the ring $\overline{k}[Z]$ we have the equality $\prod\_{1\leq i\leq n}(Z-\varphi(u\_i))$ $=$ $\prod\_{1\leq i\leq n}(Z-\psi(v\_i))$, and hence $\varphi(u\_i)$ $=$ $\psi(v\_{\pi(i)})$ for all $i$, for some permutation $\pi$.


So $T\_0$ is an aic of $R$. But since $e\_{(\mathfrak{p})}$ $=$ $(1,0)$ is not in $T\_0$, the rings $T\_0$ and $T$ cannot be isomorphic. One is connected, while the other is not. Quod demonstrandum erat, et nunc demonstratum *est*.


**Edit** A paper based on the answer is now on [arXiv](https://arxiv.org/abs/2212.06738).


