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?

#### Attachments

CompuChip
Homework Helper
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).

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.

mathwonk