Extension Theorem for Distributive Operations

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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:

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


Proof

By hypothesis, all the elements of $\left({R, *}\right)$ are cancellable.

Thus Inverse Completion of Commutative Semigroup is Abelian Group can be applied.

So $\left({T, *}\right)$ 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:

\(\displaystyle \lambda_m \left({x * y}\right)\) \(=\) \(\displaystyle m \circ \left({x * y}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({m \circ x}\right) * \left({m \circ y}\right)\) as $\circ$ distributes over $*$
\(\displaystyle \) \(=\) \(\displaystyle \lambda_m \left({x}\right) * \lambda_m \left({y}\right)\)

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:

\(\, \displaystyle \forall m, n, z \in R: \, \) \(\displaystyle \lambda_{m * n} \left({z}\right)\) \(=\) \(\displaystyle \left({m * n}\right) \circ z\) Definition of $\lambda$
\(\displaystyle \) \(=\) \(\displaystyle \left({m \circ z}\right) * \left({n \circ z}\right)\) Distributivity of $\circ$ over $*$
\(\displaystyle \) \(=\) \(\displaystyle \lambda_m \left({z}\right) * \lambda_n \left({z}\right)\) Definition of $\lambda$
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda_m * \lambda_n}\right) \left({z}\right)\) Here $*$ is the operation induced on $T^T$ by $*$


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:

\(\displaystyle \rho_z \left({m * n}\right)\) \(=\) \(\displaystyle \lambda'_{m * n} \left({z}\right)\) Definition of $\rho_z$
\(\displaystyle \) \(=\) \(\displaystyle \lambda'_m \left({z}\right) * \lambda'_n \left({z}\right)\) Behaviour of $\lambda'_{m * n}$
\(\displaystyle \) \(=\) \(\displaystyle \rho_z \left({m}\right) * \rho_z \left({n}\right)\) Definition of $\rho_z$


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 $\rho_z$.


\(\, \displaystyle \forall y, z \in T, m \in R: \, \) \(\displaystyle \rho_{y * z} \left({m}\right)\) \(=\) \(\displaystyle \lambda'_m \left({y * z}\right)\) Definition of $\rho_{y * z}$
\(\displaystyle \) \(=\) \(\displaystyle \lambda'_m \left({y}\right) * \lambda'_m \left({z}\right)\) $\lambda'_m$ is a homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \left({\rho_y * \rho_z}\right) \left({m}\right)\) Definition of $\rho_y$ and $\rho_z$


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:

\(\displaystyle x \circ' y\) \(=\) \(\displaystyle \rho_y \left({x}\right)\) as $\rho'_y = \rho_y$ on $R$
\(\displaystyle \) \(=\) \(\displaystyle \lambda_x \left({y}\right)\) Definition of $\rho_x$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y\) Definition of $\lambda_x$


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


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

\(\displaystyle \left({x * y}\right) \circ' z\) \(=\) \(\displaystyle \rho'_z \left({x * y}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \rho'_z \left({x}\right) * \rho'_z \left({y}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle x \circ' z * y \circ' z\)


\(\displaystyle x \circ' \left({y * z}\right)\) \(=\) \(\displaystyle \rho'_{y * z} \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \rho'_y \left({x}\right) * \rho'_z \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle x \circ' y * x \circ' z\)


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

$\blacksquare$


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:

\(\displaystyle y \mapsto m \circ_1 y\) \(,\) \(\displaystyle y \in R\)
\(\displaystyle y \mapsto m \circ' y\) \(,\) \(\displaystyle y \in R\)

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:

\(\displaystyle x \mapsto x \circ_1 y\) \(,\) \(\displaystyle x \in T\)
\(\displaystyle x \mapsto x \circ' y\) \(,\) \(\displaystyle x \in T\)

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 $*$.

$\blacksquare$


Proof of Associativity

Suppose $\circ$ is associative.

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

\(\displaystyle x \mapsto \left({x \circ' n}\right) \circ' p\) \(,\) \(\displaystyle x \in T\)
\(\displaystyle x \mapsto x \circ' \left({n \circ' p}\right)\) \(,\) \(\displaystyle x \in T\)

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:

\(\displaystyle y \mapsto \left({x \circ' y}\right) \circ' p\) \(,\) \(\displaystyle y \in T\)
\(\displaystyle y \mapsto x \circ' \left({y \circ' p}\right)\) \(,\) \(\displaystyle y \in T\)

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:

\(\displaystyle z \mapsto \left({x \circ' y}\right) \circ' z\) \(,\) \(\displaystyle z \in T\)
\(\displaystyle z \mapsto x \circ' \left({y \circ' z}\right)\) \(,\) \(\displaystyle z \in T\)

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.

$\blacksquare$


Proof of Commutativity

Suppose $\circ$ is commutative.

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

\(\displaystyle x \mapsto x \circ' n\) \(,\) \(\displaystyle x \in T\)
\(\displaystyle x \mapsto n \circ' x\) \(,\) \(\displaystyle x \in T\)

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:

\(\displaystyle z \mapsto z \circ' y\) \(,\) \(\displaystyle z \in T\)
\(\displaystyle z \mapsto y \circ' z\) \(,\) \(\displaystyle z \in T\)

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.

$\blacksquare$


Proof of Identity

Let $e$ be the identity element of $\left({R, \circ}\right)$.

Then the restrictions to $R$ of the endomorphisms $\lambda_e: x \mapsto e \circ' x$ and $\rho_e: x \mapsto x \circ' e$ 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 $T$.


$\blacksquare$


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 $\lambda_a: x \mapsto a \circ' x$ and $\rho_a: x \mapsto x \circ' a$ 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 $\circ'$.

$\blacksquare$


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