description/proof of that for module, nonempty subset is submodule iff subset is closed under linear combination
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% module.
Target Context
- The reader will have a description and a proof of 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.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(S\): \(\subseteq M\), \(\neq \emptyset\)
//
Statements:
\(S \in \{\text{ the submodules of } M\}\)
\(\iff\)
\(\forall s_1, s_2 \in S, \forall r_1, r_2 \in R (r_1 s_1 + r_2 s_2 \in S)\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(r_1 s_1 + r_2 s_2 \in S\); Step 2: see that \(S\) is a submodule of \(M\); Step 3: suppose that \(S\) is a submodule of \(M\); Step 4: see that \(r_1 s_1 + r_2 s_2 \in S\).
Step 1:
Let us suppose that \(\forall s_1, s_2 \in S, \forall r_1, r_2 \in R (r_1 s_1 + r_2 s_2 \in S)\).
Let us see that \(S\) satisfies the conditions for being a module.
1) \(\forall s_1, s_2 \in S (s_1 + s_2 \in S)\) (closed-ness under addition): \(s_1 + s_2 = 1 s_1 + 1 s_2 \in S\).
2) \(\forall s_1, s_2 \in S (s_1 + s_2 = s_2 + s_1)\) (commutativity of addition): it holds because it holds on the ambient \(M\).
3) \(\forall s_1, s_2, s_3 \in S ((s_1 + s_2) + s_3 = s_1 + (s_2 + s_3))\) (associativity of additions): it holds because it holds on the ambient \(M\).
4) \(\exists 0 \in S (\forall s \in S (s + 0 = s))\) (existence of 0 element): there is an \(s \in S\), because \(S \neq \emptyset\), and \(0 = 0 s + 0 s \in S\), and \(s + 0 = s\), because it holds on the ambient \(M\).
5) \(\forall s \in S (\exists s' \in S (s' + s = 0))\) (existence of inverse element): \(-1 s + 0 s = -1 s \in S\), and \(-1 s + s = 0\), because it holds on the ambient \(M\).
6) \(\forall s \in S, \forall r \in R (r . s \in S)\) (closed-ness under scalar multiplication): \(r . s = r . s + 0 . s \in S\).
7) \(\forall s \in S, \forall r_1, r_2 \in R ((r_1 + r_2) . s = r_1 . s + r_2 . s)\) (scalar multiplication distributability for scalars addition): it holds because it holds on the ambient \(M\).
8) \(\forall s_1, s_2 \in S, \forall r \in R (r . (s_1 + s_2) = r . s_1 + r . s_2)\) (scalar multiplication distributability for elements addition): it holds because it holds on the ambient \(M\).
9) \(\forall s \in S, \forall r_1, r_2 \in R ((r_1 r_2) . s = r_1 . (r_2 . s))\) (associativity of scalar multiplications): it holds because it holds on the ambient \(M\).
10) \(\forall s \in S (1 . s = s)\) (identity of 1 multiplication): it holds because it holds on the ambient \(M\).
Step 3:
Let us suppose that \(S\) is a submodule of \(M\).
Step 4:
\(\forall s_1, s_2 \in S, \forall r_1, r_2 \in R (r_1 s_1 + r_2 s_2 \in S)\) holds, because the conditions for being a module requires that.
