description/proof of that for module with finite basis, module is 'modules - linear morphisms' isomorphic to components module
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
- The reader knows a definition of product module.
- The reader admits the proposition that any ring is canonically a module with a \(1\)-element basis.
- The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
- The reader admits the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any module with any finite basis, the module is 'modules - linear morphisms' isomorphic to the components module.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\), also as the canonical module mentioned in the proposition that any ring is canonically a module with a \(1\)-element basis
\(J\): \(\in \{\text{ the finite index sets }\}\), such that \(\vert J \vert = d\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\), with any finite basis, \(B = \{b^j \vert j \in J\}\)
\(R^d\): \(= \text{ the product module }\)
\(f\): \(: M \to R^d, m_{J_l} b^{J_l} \mapsto (m_{J_1}, ..., m_{J_d})\)
//
Statements:
\(f \in \{\text{ the 'modules - linear morphisms' isomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) is valid; Step 2: see that \(f\) is linear; Step 3: see that \(f\) is a bijection; Step 4: conclude the proposition.
Step 1:
\(f\) is valid, because for each \(m \in M\), \((m_{J_1}, ..., m_{J_d})\) is uniquely determined, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: when \(m_{J_l} b^{J_l}\) does not contain a basis element, the component is defined to be \(0\).
Step 2:
Let us see that \(f\) is linear.
Let \(m, m' \in M\) and \(r, r' \in R\) be any.
\(m = m_{J_l} b^{J_l}\) and \(m' = m'_{J_l} b^{J_l}\).
\(f (r m + r' m') = f (r m_{J_l} b^{J_l} + r' m'_{J_l} b^{J_l}) = f ((r m_{J_l} + r' m'_{J_l}) b^{J_l}) = (r m_{J_1} + r' m'_{J_1}, ..., r m_{J_d} + r' m'_{J_d}) = r (m_{J_1}, ..., m_{J_d}) + r' (m'_{J_1}, ..., m'_{J_d}) = r f (m) + r' f (m')\).
So, \(f\) is linear.
Step 3:
\(f\) is an injection, because for each \(m = m_{J_l} b^{J_l}, m' = m'_{J_l} b^{J_l} \in M\) such that \(m \neq m'\), \((m_{J_1}, ..., m_{J_d}) \neq (m'_{J_1}, ..., m'_{J_d})\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(f\) is a surjection, because for each \((r_{J_1}, ..., r_{J_d}) \in R^d\), \(r_{J_l} b^{J_l} \in M\) and \(f (r_{J_l} b^{J_l}) = (r_{J_1}, ..., r_{J_d})\).
So, \(f\) is a bijection.
Step 4:
By the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism, \(f\) is a 'modules - linear morphisms' isomorphism.
