Uniqueness of the "algebraic closure" of a commutative ring There are several ways to generalize the notion of "algebraic closure" from fields to arbitrary commutative rings. A good overview is On algebraic closures by R. Raphael. I am more interested in the notion suggested in Stacks/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.)
It is easy to prove (Stacks/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 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.
 A: Does this work for a counterexample? [EDIT: NO, IT DOES NOT. See comments.]
Let $\bar{\mathbb Z}$ be the ring of all algebraic integers in a fixed algebraic closure of $\mathbb Q$. (Surely this is an a.i.c. of $\mathbb Z$ in your sense.) Let $R$ be the fiber product ${\mathbb Z}\times_{\mathbb Z/p}{\mathbb Z}$ for some prime number $p$.
I think that $\bar{\mathbb Z}\times_{\bar{\mathbb Z}/p}\bar{\mathbb Z}$ is then an a.i.c. of $R$.
On the other hand, you can make a "twisted" version $\tilde R$ of this, by choosing an isomorphism $\varphi: \bar{\mathbb Z}/p\cong \bar{\mathbb Z}/p$ that is not the mod $p$ reduction of any isomorphism $\bar{\mathbb Z}\cong \bar{\mathbb Z}$ and then letting $\tilde R$ be the subring of $\bar{\mathbb Z} \times \bar{\mathbb Z}$ consisting of pairs $(x,y)$ such that the class of $y$ in $\bar{\mathbb Z}/p$ is $\varphi$ of the class of $x$.
To make such a map $\varphi$ (for some $p$), you can first choose a cubic Galois extension $K$ of $\mathbb Q$ and then choose a $p$ that splits in $K$. Then if $\mathcal O_K$ is the maximal order, $\mathcal O_K/p$ is the product of three copies of $\mathbb Z/p$. Choose an automorphism of this product, using a permutation (a $2$-cycle) that is not in the Galois group. This can be extended to an automorphism of $\bar{\mathbb Z}/p$.
A: 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 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.
