Extension Theorem for Distributive Operations/Existence and Uniqueness
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.8$