---
tag: K2N123JM35
kind: definition
name: Strong amalgamation property
msc: [08B25, 03C52]
defines: Q94Z7TB8B5
uses: [XSCSX39GEZ, SCRB74SWQN]
sources:
  - url: https://en.wikipedia.org/wiki/Amalgamation_property
---

## Definition

Fix a signature and let $\mathcal{K}$ be a class of algebras of this signature, closed under isomorphism.
The class $\mathcal{K}$ has the *strong amalgamation property* if the following holds.
For all $A, B, C \in \mathcal{K}$ and all [[embeddings|SCRB74SWQN]] $f \colon A \to B$ and $g \colon A \to C$, there are $D \in \mathcal{K}$ and embeddings $f' \colon B \to D$ and $g' \colon C \to D$ such that:

1. $f' \circ f = g' \circ g$;
2. $f'(B) \cap g'(C) = f'(f(A))$.

## Remarks

Clause 1 alone is the [[amalgamation property|XSCSX39GEZ]], so the strong amalgamation property implies the amalgamation property.
Clause 2 says that an element of $B$ and an element of $C$ have the same image in $D$ only if both come from the same element of $A$.

Example: the class of groups has the strong amalgamation property.
Take $D = B *_A C$, the free product of $B$ and $C$ amalgamated over $A$.
By the normal form theorem for amalgamated free products, $B$ and $C$ embed in $D$ and their images meet exactly in the image of $A$.

Non-example: the class of fields has the amalgamation property but fails clause 2.
Let $A = \Q$ and $B = C = \Q(\sqrt{2})$, with $f = g$ the inclusion.
In any field $D$ with embeddings $f', g'$ that agree on $\Q$, both $f'(\sqrt{2})$ and $g'(\sqrt{2})$ are roots of $x^2 - 2$, so $g'(\sqrt{2}) = \pm f'(\sqrt{2})$.
Hence $g'(\sqrt{2})$ lies in $f'(B) \cap g'(C)$ but not in the image of $\Q$.