How Do We Prove Something is a Submodule?

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Discussion Overview

The discussion revolves around the criteria for proving that a subset is a submodule of a left R-module. Participants explore the necessary conditions for a subset to qualify as a submodule, referencing both textbook definitions and specific proofs.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant outlines the textbook definition of a submodule, listing four conditions that must be satisfied: closure under scalar multiplication, inclusion of the additive identity, closure under addition, and inclusion of additive inverses.
  • Another participant suggests that if closure under scalar multiplication holds, one can derive the presence of the additive identity and inverses by setting specific values for the scalar.
  • A different participant reiterates the previous point, emphasizing the need to demonstrate that the additive identity and inverses can be derived from the closure property.
  • One participant counters that closure under scalar multiplication does not imply the presence of the additive identity unless the subset is non-empty, indicating a potential gap in the argument presented by others.

Areas of Agreement / Disagreement

Participants express differing views on whether the closure under scalar multiplication alone is sufficient to establish the presence of the additive identity and inverses, indicating that the discussion remains unresolved.

Contextual Notes

There is an assumption that the subset being considered is non-empty, which affects the implications drawn from the closure property. The discussion does not resolve whether the proof is complete without addressing this assumption.

Artusartos
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My textbook says that...

If M is a left R-module, then a submodule N of M...is an additive subgoup N of M closed under scalar multiplication: rn \in N whenever n \in N and r \in R.

So if we want to prove that something is a submodule, we need to show that...

1) It closed under scalar multiplication
2) The additive idenitity is in N
3) N is closed under additition
4) If x is in N, then so is its inverse

Right?

But, in the link that I attached, it only shows 1) and 3), right? Can anybody tell me why? Is the proof still considered complete?

Thanks in advance
 

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Isn't it as simple as: If (1) holds, set r = 0 to get (2) and r = -1 to get (4)?
(OK, you might want to show that 0n is the additive identity for any n and that -n is the additive inverse of any n).
 
CompuChip said:
Isn't it as simple as: If (1) holds, set r = 0 to get (2) and r = -1 to get (4)?
(OK, you might want to show that 0n is the additive identity for any n and that -n is the additive inverse of any n).

Thanks.
 
1 does not imply 2, unless the subset considered is non empty. i.e. 1 implies that IF the subset contains anything, then it also contains 0.
 
Last edited:

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