An example separating the amalgamation property from a weaker variant

Question

Let's consider an equational theory $T$ in an algebraic language language and the following statement (S): for every formula $\phi, \psi$, if $T \vdash \phi \to \psi$, then there exists a formula $\chi$ such that $T \vdash \phi \to \chi \land \chi \to \psi$. Here, $\phi$, $\psi$, $\chi$ are existential, universal, and quantifier-free formulas, respectively. One can show that (S) holds if and only if $T$ has the amalgamation property.

Now let's consider a similar statement (S'), which is just like (S) but $\phi, \psi, \psi$ are restricted to positive formulas. Basich (1974) showed that (S') corresponds to a weaker variant of the amalgamation property sometimes called the flat amalgamation property. Here, only diagrams of the form $B \leftarrow A \rightarrow A * F(X)$ are required to be amalgamated, where $*$ denotes the so-called free product, and $F(X)$ is the free $T$-algebra generated by $X$.

People's interest in both of these prima facie distinct principles suggest that they are actually different. I cannot, however, find an equational theory $T$ satisfying (S') but not (S). If I understand correctly, every variety of normal modal algebras satisfying (S') also satisfies (S), and these varieties are the only ones I can apply what little intuition I have regarding these things.

What $T$, if at all, satisfies (S') but not (S)?

Answer 1

The theory of rings (when I say "ring", I mean "commutative ring with unity") is an example, which I believe motivates the terminology "flat amalgamation property".

For any ring $A$ and any set $X$, the free product (coproduct) $A * F(X)$ is $A\otimes_{\mathbb{Z}} \mathbb{Z}[X] \cong A[X]$, the polynomial ring in variables $X$ with coefficients from $A$. Now for any embedding of rings $f\colon A\to B$, we get a canonical embedding of polynomial rings $A[X]\to B[X]$ (by applying $f$ to the coefficients), so $B[X]$ witnesses the flat amalgamation property.

The fancy way to say this is that the ring map $\mathbb{Z}\to \mathbb{Z}[X]$ is a flat morphism of rings, so taking the tensor product with $\mathbb{Z}[X]$ preserves injectivity. In particular, if $A\to B$ is injective, so is $A\otimes_ \mathbb{Z} \mathbb{Z}[X] \to B\otimes_\mathbb{Z} \mathbb{Z}[X]$, which is isomorphic to the canonical map $A[X]\to B[X]$.

On the other hand, the theory of rings does not have the amalgamation property (for embeddings). If there is a witness $D$ to the amalgamation property for embeddings $f\colon A\to B$ and $g\colon A\to C$, then this induces a map from the tensor product $B\otimes_A C$ (which is the pushout of $B$ and $C$ over $A$) to $D$. Since the maps $B\to D$ and $C\to D$ are injective, the maps $B\to B\otimes_A C$ and $C\to B\otimes_A C$ must also be injective. But this is not always the case - see here for an example where the map $A\to B\otimes_A C$ is not even injective.

Edit: If the example linked to above looks mysterious, here is a logician's point of view on what's happening there. Let $\varphi(a,b)$ be the formula $\exists x\, (x^2 = 0\land xb = a\land a\neq 0)$, and let $\psi(a,b)$ be the formula $\exists y\, (ya = b)$.

These two formulas are inconsistent, since together they imply $a = xb = xya = x^2yb = 0$ and $a\neq 0$. So the existential formula $\varphi(a,b)$ implies the universal formula $\lnot \psi(a,b)$. But there is no quantifier-free formula $\chi$ such that $\varphi(a,b)\rightarrow \chi(a,b)$ and $\chi(a,b)\rightarrow \lnot \psi(a,b)$.

This can be proven by finding a ring $A$ with distinguished elements $a$ and $b$ and extensions $A\subseteq B$ and $A\subseteq C$ such that $B\models \varphi(a,b)$ and $C\models \psi(a,b)$. If we had an interpolating $\chi(a,b)$, it would be true in $B$, hence true in $A$, and hence true in $C$ (since it is quantifier-free), but then $\lnot\psi(a,b)$ would be true in $C$, contradiction.

Well, the idea is to fix a field $k$ to work over and "freely adjoin" witnesses subject to the constraints: take $A = k[a,b]$, $B = A[x]/(x^2,a-bx)$, and $C = A[y]/(b-ay)$. But then this needs to be adjusted a bit to make the maps from $A$ to $B$ and $C$ injective, e.g. $a = bx$ and $x^2=0$ implies $a^2 = b^2x^2 = 0$, so we need to make $a^2 = 0$ in $A$. Continuing along these lines (and maybe making things more symmetrical for aesthetic reasons), we arrive at the example in the link.