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 of Cancellable Semigroup, $$\left({T, *}\right)$$ is an abelian group.

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

So if $$\phi$$ and $$\psi$$ are two endomorphisms of $$T$$ coinciding on $$R$$ (i.e. whose restrictions to $$R$$ are the same mapping), then $$\phi = \psi$$.


 * For each $$m \in R$$, we define $$\lambda_m: R \to T$$ as:

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

As $$\circ$$ distributes over $$*$$, $$\lambda_m$$ is a homomorphism from $$\left({R, *}\right)$$ into $$\left({T, *}\right)$$.

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 $$*$$.


 * 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 $$*$$.


 * 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.

Similar arguments prove the rest.


 * 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 $$\circ'$$.