Extension Theorem for Distributive Operations/Associativity

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {R, *}$ be a commutative semigroup, all of whose elements are cancellable.

Let $\struct {T, *}$ be an inverse completion of $\struct {R, *}$.

Let $\circ$ be an operation on $R$ which distributes over $*$.


Let $\circ'$ be the unique operation on $T$ which distributes over $*$ in $T$ and induces on $R$ the operation $\circ$.


Then:

If $\circ$ is associative, then so is $\circ'$.


Proof

We have that $\circ'$ exists and is unique by Extension Theorem for Distributive Operations: Existence and Uniqueness.


Suppose $\circ$ is associative.

As $\circ'$ distributes over $*$, for all $n, p \in R$, the mappings:

\(\ds x \mapsto \paren {x \circ' n} \circ' p\) \(,\) \(\ds x \in T\)
\(\ds x \mapsto x \circ' \paren {n \circ' p}\) \(,\) \(\ds x \in T\)

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the associativity of $\circ$ and hence are the same mapping.

Therefore:

$\forall x \in T, n, p \in R: \paren {x \circ' n} \circ' p = x \circ' \paren {n \circ' p}$


Similarly, for all $x \in T, p \in R$, the mappings:

\(\ds y \mapsto \paren {x \circ' y} \circ' p\) \(,\) \(\ds y \in T\)
\(\ds y \mapsto x \circ' \paren {y \circ' p}\) \(,\) \(\ds y \in T\)

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping.

Therefore:

$\forall x, y \in T, p \in R: \paren {x \circ' y} \circ' p = x \circ' \paren {y \circ' p}$


Finally, for all $x, y \in T$, the mappings:

\(\ds z \mapsto \paren {x \circ' y} \circ' z\) \(,\) \(\ds z \in T\)
\(\ds z \mapsto x \circ' \paren {y \circ' z}\) \(,\) \(\ds z \in T\)

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping.

Therefore $\circ'$ is associative.

$\blacksquare$


Sources