Epimorphism preserves Modules

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.

Let $\phi: G \to H$ be an epimorphism.


Then $H$ is an $R$-module.


Corollary

Let $\struct {G, +_G, \circ}_R$ be an unitary $R$-module.

Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.

Let $\phi: G \to H$ be an epimorphism.


Then $H$ is a unitary $R$-module.


If $G$ is a unitary $R$-module, then so is $H$.


Proof

If $\struct {G, +_G, \circ}_R$ is an $R$-module, then:

$\forall x, y, \in G, \forall \lambda, \mu \in R$:

$(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
$(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$


If $\phi: G \to H$ is an epimorphism, then:

$\forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$
$\forall x \in S: \forall \lambda \in R: \map \phi {\lambda \circ x} = \lambda \circ \map \phi x$
$\forall y \in H: \exists x \in G: y = \map \phi x$


As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$.

If $\phi$ were not an epimorphism, that is not surjective, we would have no way of knowing the behaviour of elements of $H$ outside of the image of $G$.

Hence the specification that $\phi$ needs to be an epimorphism.


Now we check the module axioms in turn.


Module Axiom $\text M 1$: Distributivity over Module Addition

\(\ds \lambda \circ \paren {\map \phi x +_H \map \phi y}\) \(=\) \(\ds \lambda \circ \paren {\map \phi {x +_G y} }\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ \paren {x +_G y} }\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\lambda \circ x} +_G \paren {\lambda \circ y} }\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ x} +_H \map \phi {\lambda \circ y}\)

Thus Module Axiom $\text M 1$: Distributivity over Module Addition is shown to hold for $H$.


Module Axiom $\text M 2$: Distributivity over Scalar Addition

\(\ds \paren {\lambda +_R \mu} \circ \map \phi x\) \(=\) \(\ds \map \phi {\paren {\lambda +_R \mu} \circ x}\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {\lambda \circ x} +_G \paren {\mu \circ x} }\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ x} +_H \map \phi {\mu \circ x}\)
\(\ds \) \(=\) \(\ds \lambda \circ \map \phi x +_H \mu \circ \map \phi x\)

Thus Module Axiom $\text M 2$: Distributivity over Scalar Addition is shown to hold for $H$.


Module Axiom $\text M 3$: Associativity

\(\ds \paren {\lambda \times_R \mu} \circ \map \phi x\) \(=\) \(\ds \map \phi {\paren {\lambda \times_R \mu} \circ x}\)
\(\ds \) \(=\) \(\ds \map \phi {\lambda \circ \paren {\mu \circ x} }\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\map \phi {\mu \circ x} }\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\mu \circ \map \phi x}\)

Thus Module Axiom $\text M 3$: Associativity is shown to hold for $H$.


So all the module axioms for $H$ are satisfied.

$\blacksquare$


Also see


Sources