How Do We Prove Something is a Submodule?

[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?

Thanks in advance

 

[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).

 

[Artusartos]

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]

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.

 

[blue_raver22]

I would like to clarify that the proof provided in the link is indeed complete. However, it may not explicitly mention all four conditions because they can be inferred from the given information.

The first condition, closure under scalar multiplication, is shown by the statement "rn \in N whenever n \in N and r \in R". This implies that N is closed under scalar multiplication.

The second condition, the additive identity being in N, is implied by the fact that N is a subgroup of M. This means that it must contain the identity element of M, which is also the additive identity.

The third condition, closure under addition, is shown by the statement "x+y \in N whenever x,y \in N". This implies that N is closed under addition.

The fourth condition, closure under inverses, is not explicitly mentioned because it is not necessary for a submodule to contain the inverse of every element. As long as N is closed under addition and scalar multiplication, it is considered a submodule. The inverse of an element can always be obtained by multiplying it with -1.

In conclusion, the proof provided in the link is complete and the four conditions for a submodule are implicitly shown. However, it is always important to be aware of all the necessary conditions and make sure they are met when proving something is a submodule.