description/proof of that range of linear map between modules is submodule of codomain
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% module.
- The reader knows a definition of linear map.
- The reader admits the proposition that for any module, any nonempty subset of the module is a submodule if and only if the subset is closed under linear combination.
Target Context
- The reader will have a description and a proof of the proposition that the range of any linear map between any modules is a submodule of the codomain.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(f (M_1)\):
//
Statements:
\(f (M_1) \in \{\text{ the submodules of } M_2\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f (M_1)\) is closed under linear combination.
Step 1:
Let us see that \(f (M_1)\) is closed under linear combination.
Let \(f (m), f (m') \in f (M_1)\) and \(r, r' \in R\) be any.
\(r f (m) + r' f (m') = f (r m + r' m') \in f (M_1)\), because \(f\) is linear.
By the proposition that for any module, any nonempty subset of the module is a submodule if and only if the subset is closed under linear combination, \(f (M_1)\) is a submodule of \(M_2\).
