Extension Theorem for Distributive Operations/Existence and Uniqueness

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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 $*$.


Then:

There exists a unique operation $\circ'$ on $T$ which distributes over $*$ in $T$ and induces on $R$ the operation $\circ$.


Proof

We have by hypothesis that all the elements of $\struct {R, *}$ are cancellable.

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

So $\struct {T, *}$ is an abelian group.


Existence

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

$\forall x \in R: \map {\lambda_m} x = m \circ x$

Then:

\(\ds \map {\lambda_m} {x * y}\) \(=\) \(\ds m \circ \paren {x * y}\)
\(\ds \) \(=\) \(\ds \paren {m \circ x} * \paren {m \circ y}\) as $\circ$ distributes over $*$
\(\ds \) \(=\) \(\ds \map {\lambda_m} x * \map {\lambda_m} y\)

So $\lambda_m$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$.


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

This can be applied because $\struct {T, *}$ is abelian.

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:

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


By Homomorphism on Induced Structure to Commutative Semigroup:

$\lambda'_m * \lambda'_n$ is an endomorphism of $\struct {T, *}$ 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:

$\forall m \in R: \map {\rho_z} m = \map {\lambda'_m} z$


Then:

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


Therefore $\rho_z$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$.

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


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


By Homomorphism on Induced Structure to Commutative Semigroup:

$\rho'_y * \rho'_z$ is an endomorphism on $\struct {T, *}$ 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:

$\forall x, y \in T: x \circ' y = \map {\rho'_y} x$

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

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


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


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

\(\ds \paren {x * y} \circ' z\) \(=\) \(\ds \map {\rho'_z} {x * y}\)
\(\ds \) \(=\) \(\ds \map {\rho'_z} x * \map {\rho'_z} y\)
\(\ds \) \(=\) \(\ds x \circ' z * y \circ' z\)


\(\ds x \circ' \paren {y * z}\) \(=\) \(\ds \map {\rho'_{y * z} } x\)
\(\ds \) \(=\) \(\ds \map {\rho'_y} x * \map {\rho'_z} x\)
\(\ds \) \(=\) \(\ds x \circ' y * x \circ' z\)


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

$\blacksquare$


Uniqueness

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:

\(\ds y \mapsto m \circ_1 y\) \(,\) \(\ds y \in T\)
\(\ds y \mapsto m \circ' y\) \(,\) \(\ds y \in T\)

are endomorphisms of $\struct {T, *}$ 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:

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

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have just proved.

So these two mappings 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 $*$.

$\blacksquare$


Sources

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