Why do we not need to show that $a-b \in W$ to verify that $W$ is a subspace?

Question

Question: What is the most general setting in which it suffices to check $a+b \in W$ instead of $a-b \in W$ when trying to verify a sub-algebraic-object? When that algebraic object (1) has a set of scalars which are closed under additive inverses and possesses a multiplicative identity and (2) satisfies the right distributive property?

I know that for a (commutative) group, for an ideal, and even for a (commutative) ring (with unit), if you want to show that a set $W$ is an additive subgroup (for the latter two cases as part of showing that it is a subideal, subring), given $a,b$ in $W$ it is necessary (e.g. here) to check that $$a-b \in W \iff \text{W is closed under addition, and additive inverses}.$$ In other words, showing that $W$ is closed under addition does not suffice to show that it is closed under additive inverses.

However, to show that $W$ is a subspace of a vector space, one usually only asks students (e.g. here) to verify that, given $a,b \in W$, one has $$a+b \in W.$$

Why? My guess would be that, for vector spaces $(-1)b=-b$ always, so the closure under additive inverses follows from checking closure under scalar multiplication. Is this correct?

Does this reasoning extend to modules as well? At least provided that the ring in question has a unit $1$ distinct from the additive identity $0$, and thus being closed under additive inverses has a $-1$ element as well for which $$0=(0)b=(-1+1)b=(-1)b+(1)b = (-1)b + b \implies (-1)b = -b.$$ And does it also hold for algebras, i.e. when checking that $a+b \in W, sa \in W, ab \in W$, it is the closure under scalar multiplication $sa \in W$ that guarantees us the closure under additive inverses?

If it holds for modules, however, then doesn't that contradict the claim that it doesn't hold for rings with unit? Isn't any ring with unit a module over itself?

Answer 1

My guess would be that, for vector spaces $(-1)b=-b$ always, so the closure under additive inverses follows from checking closure under scalar multiplication. Is this correct?

That is indeed correct, and generalises from vector spaces to modules $M$ over (unital, if that isn't part of the used definition) rings where the scalar multiplication satisfies $1\cdot m = m$ for all $m\in M$, since by the distributivity of the scalar multiplication we obtain $(-1)\cdot m = -m$ for all $m\in M$. (Note that the argument doesn't need $-1 \neq 1$.)

If it holds for modules, however, then doesn't that contradict the claim that it doesn't hold for rings with unit? Isn't any ring with unit a module over itself?

Yes, a ring is a module over itself, and thus the argument works for submodules, i.e. (left resp. right) ideals. It does not work for subrings, since a subset of a ring that is closed under addition and multiplication and contains $1$ need not contain $-1$ (e.g. $\mathbb{N}\subset \mathbb{Z}$).

Answer 2

If $W$ is a non empty subset of a vector space $V$ then the minimal "things" you need to check to ensure $W$ is a vector space in own right is that it is closed under sums of vectors AND also under scalar multiples of vectors. Just the former is not enough (take for example the set of all vectors of the form (k,0) in $\mathbb{R}^{2}$ for all integers $k$. Then this is closed under addition but (1/2)(1,0) is not in this set.

From the axioms of a vector space you can prove that the "minus" vector is the same vector as the scalar multiple of minus 1 with that vector. So the second axiom of a subspace ensures that if $w\in W$ then $-w\in W$ and then the first subspace axiom gives you $v,w\in W \Rightarrow v-w \in W$.

Hope this helps