description/proof of that for module, nonzero element nonzero-scalar multiplied is not necessarily nonzero
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 integers modulo natural number ring.
- The reader admits the proposition that any ring is canonically a module with a \(1\)-element basis.
Target Context
- The reader will have a description and a proof of the proposition that for a module, a nonzero element a-nonzero-scalar multiplied is not necessarily nonzero.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
not necessarily "\(\forall m \in M \setminus \{0\}, \forall r \in R \setminus \{0\} (r m \neq 0)\)"
//
2: Note
This is not because \(R\) is not commutative but because \(R\) is not any division ring: compare with the proposition that for any module over any division ring, each nonzero element each-nonzero-scalar multiplied is nonzero.
3: Proof
Whole Strategy: Step 1: see an example such that \(r m = 0\).
Step 1:
Let \(R = \mathbb{Z} / 6\), the integers modulo natural number ring, and \(M = R\), which is indeed a module, by the proposition that any ring is canonically a module with a \(1\)-element basis.
\(R = M = \{[0], ..., [5]\}\)
For \([3] \in M \setminus \{0\}\) and \([2] \in R \setminus \{0\}\), \([2] [3] = [6] = [0] = 0\).
So, for an \(m \in M \setminus \{0\}\) and an \(r \in R \setminus \{0\}\), \(r m \neq 0\) does not necessarily hold.
