How Do We Prove Something is a Submodule?

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To prove that a subset N is a submodule of a left R-module M, it must be shown that N is closed under scalar multiplication, contains the additive identity, is closed under addition, and includes the additive inverse for each element. The initial argument suggests that if N is closed under scalar multiplication, then setting r to 0 demonstrates the presence of the additive identity, and setting r to -1 shows the existence of inverses. However, it is clarified that this reasoning only holds if N is non-empty; otherwise, the additive identity may not be included. Therefore, it is essential to verify that N is non-empty before concluding that it satisfies all submodule properties. The discussion emphasizes the importance of addressing the non-emptiness condition in the proof.
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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.
 
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