Extension Theorem for Distributive Operations

Theorem
Let $\left({R, *}\right)$ be a commutative semigroup, all of whose elements are cancellable.

Let $\left({T, *}\right)$ be an inverse completion of $\left({R, *}\right)$.

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

Then:
 * 1) There is a unique operation $\circ'$ on $T$ which distributes over $*$ in $T$ and induces on $R$ the composition $\circ$
 * 2) If $\circ$ is associative, then so is $\circ'$
 * 3) If $\circ$ is commutative, then so is $\circ'$
 * 4) If $e$ is an identity for $\circ$, then $e$ is also an identity for $\circ'$
 * 5) Every element cancellable for $\circ$ is also cancellable for $\circ'$.

Proof
By Inverse Completion an Abelian Group, $\left({T, *}\right)$ is an abelian group.

Existence of Distributive Operation
For each $m \in R$, we define $\lambda_m: R \to T$ as:


 * $\forall x \in R: \lambda_m \left({x}\right) = m \circ x$

Then:

So $\lambda_m$ is a homomorphism from $\left({R, *}\right)$ into $\left({T, *}\right)$.

Now, by the Extension Theorem for Homomorphisms, every homomorphism from $\left({R, *}\right)$ into $\left({T, *}\right)$ is the restriction to $R$ of a unique endomorphism of $\left({T, *}\right)$.

We have just shown that $\lambda_m$ is such a homomorphism.

Therefore there exists a unique endomorphism $\lambda'_m: T \to T$ which extends $\lambda_m$.

Now:

By Homomorphism on Induced Structure, $\lambda'_m * \lambda'_n$ is an endomorphism of $\left({T, *}\right)$ that, as we have just seen, coincides on $R$ with $\lambda'_{m * n}$.

Hence $\lambda'_{m * n} = \lambda'_m * \lambda'_n$.

Similarly, for each $z \in T$, we define $\rho_z: R \to T$ as:


 * $\forall m \in R: \rho_z \left({m}\right) = \lambda'_m \left({z}\right)$

Then:

Therefore $\rho_z$ is a homomorphism from $\left({R, *}\right)$ into $\left({T, *}\right)$.

Consequently there exists a unique endomorphism $\rho'_z: T \to T$ extending $\rho_z$.

By Homomorphism on Induced Structure, $\rho'_y * \rho'_z$ is an endomorphism on $\left({T, *}\right)$ that coincides (as we have just seen) with $\rho'_{y * z}$ on $R$.

Hence $\rho'_{y * z} = \rho'_y * \rho'_z$.

Now we define an operation $\circ'$ on $T$ by:


 * $\forall x, y \in T: x \circ' y = \rho'_y \left({x}\right)$

Now suppose $x, y \in R$. Then:

so $\circ'$ is an extension of $\circ$.

Next, let $x, y, z \in T$. Then:

So $\circ'$ is distributive over $*$.

Uniqueness of Distributive Operation
To show that $\circ'$ is unique, let $\circ_1$ be any operation on $T$ distributive over $*$ that induces $\circ$ on $R$.

Since $\circ'$ and $\circ_1$ both distribute over $*$, for every $m \in R$, the mappings:

are endomorphisms of $\left({T, *}\right)$ that coincide on $R$ so must be the same mapping.

Therefore $\forall m \in R, y \in T: m \circ_1 y = m \circ' y$.

Similarly, for every $y \in T$, the mappings:

are endomorphisms of $\left({T, *}\right)$ that coincide on $R$ by what we have just proved, so must be the same mapping.

Hence $\forall x, y \in T: x \circ_1 y = x \circ' y$.

Thus $\circ'$ is the only operation on $T$ which extends $\circ$ and distributes over $*$.

Proof of Associativity
Suppose $\circ$ is associative.

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

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

Therefore $\forall x \in T, n, p \in R: \left({x \circ' n}\right) \circ' p = x \circ' \left({n \circ' p}\right)$.

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

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

Therefore $\forall x, y \in T, p \in R: \left({x \circ' y}\right) \circ' p = x \circ' \left({y \circ' p}\right)$.

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

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

Therefore $\circ'$ is associative.

Proof of Commutativity
Suppose $\circ$ is commutative.

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

are endomorphisms of $\left({T, *}\right)$ that coincide on $R$ by the commutativity of $\circ$ and hence are the same mapping.

Therefore $\forall x \in T, n \in R: x \circ' n = n \circ' x$.

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

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

Therefore $\circ'$ is commutative.

Proof of Identity
Let $e$ be the identity element of $R$.

Then the restrictions to $R$ of the endomorphisms $\lambda_e: x \to e \circ' x$ and $\rho_e: x \to x \circ' e$ of $\left({T, *}\right)$ are monomorphisms.

But then $\lambda_e$ and $\rho_e$ are monomorphisms by the Extension Theorem for Homomorphisms, so $e$ is the identity element of $T$.

Proof of Cancellability
To prove that every element of $R$ cancellable for $\circ$ is also cancellable for $\circ'$:

Let $a$ be an element of $R$ cancellable for $\circ$.

Then the restrictions to $R$ of the endomorphisms $\lambda_a: x \to a \circ' x$ and $\rho_a: x \to x \circ' a$ of $\left({T, *}\right)$ are monomorphisms.

But then $\lambda_a$ and $\rho_a$ are monomorphisms by the Extension Theorem for Homomorphisms, so $a$ is cancellable for $\circ'$.