1. Jan 14, 2013

### Artusartos

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?

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2. Jan 14, 2013

### CompuChip

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).

3. Jan 14, 2013

### Artusartos

Thanks.

4. Jan 15, 2013

### mathwonk

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: Jan 15, 2013