description/proof of that kernel of linear map is submodule of domain
Topics
About: module
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that the kernel of any linear map is a submodule of the domain.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
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Statements:
\(Ker (f) \in \{\text{ the submodules of } M_1\}\)
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2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is closed under linear combination; Step 2: conclude the proposition.
Step 1:
Let us see that \(Ker (f)\) is closed under linear combination.
Let \(m, m' \in Ker (f)\) and \(r, r' \in R\) be any.
\(f (r m + r' m') = r f (m) + r' f (m') = r 0 + r' 0 = 0\).
So, \(r m + r' m' \in Ker (f)\).
\(0 \in Ker (f)\).
Step 2:
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, \(Ker (f)\) is a submodule of \(M_1\).
