Is a vector space over a ring or over a field?

Question

What is a vector space? I can see two different formulations, and between them there is one difference: commutativity.

DEFINITION 1 (See here)

Let $(F, +_F, \times_F)$ be a division ring. Let $(\mathcal{V}, +_\mathcal{V})$ be an abelian group. Let $(\mathcal{V}, +_\mathcal{V}, \cdot)_F$ be a unitary module over $F$. Then $(\mathcal{V}, +_\mathcal{V}, \cdot)_F$ is a vector space over $F$. That is, a vector space is a unitary module over a ring, whose ring is a division ring.

DEFINITION 2

Let $(F, +_F, \times_F)$ be a field. Let $(\mathcal{V}, +_\mathcal{V})$ be an abelian group. Let $\cdot: F\times \mathcal{V} \longrightarrow \mathcal{V}$ be a function. A vector space is $(\mathcal{V}, +_\mathcal{V}, \cdot)_F$ such that $\forall a,b, \in F$ and $\forall x,y \in \mathcal{V}$:

• $\cdot$ right distributive: $(a +_F b) \cdot x = (a\cdot x) +_\mathcal{V} (b\cdot x)$
• $\cdot$ left distributive: $\,\,\, a \cdot (x +_\mathcal{V} y) = (a\cdot x) +_\mathcal{V} (a\cdot y)$
• $\cdot$ compatible with $\times_F$: $(a\times_F b) \cdot x = a \cdot (b\cdot x)$
• $\times_F$ 's identity is $\cdot$'s identity: $1_F \cdot x = x$

There could also be other definitions,but for now it doesn't matter. What matter is that commutativity is not considered in the same way in both definitions! In the first definition, we ahve a division ring (not a commutative division ring, i.e. a field!), while in the second we have a field (i.e. a commutative division ring).

---

Notice that the key difference on which I am struggling is that on one side we have a division ring and on the other side a commutative division ring. The first is an abelian group $(R, +_R)$ under the $+_R$ binary operation, however $(R, \times_R)$ is only a group (i.e. not abelian, i.e. not commutative).

Answer 1

Usually, over a field.

On Wikipedia (I know, I know) I read that "Some authors use the term vector space to mean modules over a division ring" (cit.). That seems reasonable, as they are just extending the definition.

Note that in the division ring definition, if $F$ is a field, the two definitions become equivalent.

Answer 2

In $Bourbaki$, a field $F$ is not necessarily commutative, and they simply define left (resp. right) $F$-vector spaces as left (resp. right) $F$-modules.

Ref. N. Bourbaki, Algebra, ch.I, Algebraic Structures, §9 and ch. II, Linear Algebra, §1, n°1.

Answer 3

Usually, a vector space is an abelian group with a scalar multiplication with elements that come from a field.

It is true that most linear algebra keeps holding true if you drop the commutativity of the field (we are left with a division ring then), so that might be why the first definition calls it a vector space. Most mathematicians would call it a module over a division ring though.