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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).

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    $\begingroup$ Instead of a humongous wall of text this question could just have one sentence. $\endgroup$
    – Christoph
    Commented Jul 5, 2018 at 9:49
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    $\begingroup$ I know but it took me ages to find a coherent set of definitions, so I put them here because I think having coherent definitions is 99% of what is needed to understand maths $\endgroup$ Commented Jul 5, 2018 at 9:50
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    $\begingroup$ Maybe you should rather ask which vector space properties/theorems remain valid for modules over division ring. Most of them do: e.g. we can have basis, well defined dimension, linear dependency, etc. $\endgroup$
    – Berci
    Commented Jul 5, 2018 at 10:23
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    $\begingroup$ @Christoph I thought you were exaggerating until I looked at rev #1. $\endgroup$ Commented Jul 5, 2018 at 11:46
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    $\begingroup$ Different people have different definitions. Just like some people see the existence of a $1$-element as a part of the definition of a ring, while others do not. As seen in the answer by Bernard, some people see a division ring (skew field) as a kind of field. What you ask, is just a matter of conventions. Your question already shows that different conventions exist for whether a module over a division ring $K$ should be called a $K$-vector space, or if that name should be reserved for "commutative fields". You will figure out from the context. Use the convention you and your peers like. $\endgroup$ Commented Jul 5, 2018 at 13:09

3 Answers 3

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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.

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  • $\begingroup$ I like this approach! $\endgroup$ Commented Jul 5, 2018 at 10:37
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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.

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  • $\begingroup$ Oh nice! I am still very new, and don't know too much about modules. Do you mean a unitary module (proofwiki.org/wiki/Definition:Unitary_Module_Axioms )? $\endgroup$ Commented Jul 5, 2018 at 10:38
  • $\begingroup$ "Most mathematicians": yes but, as @Bernard pointed out in his answer, it's not universal. For most French-speaking (or "bourbakists") mathematicians, a field is not necessarily commutative! $\endgroup$
    – paf
    Commented Jul 5, 2018 at 10:40
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    $\begingroup$ Yes. Modules over unitary rings are usually assumed to be unitary, unless otherwise stated. $\endgroup$
    – Berci
    Commented Jul 5, 2018 at 10:46
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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.

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  • $\begingroup$ IF F is a field, IF! That's the thing, can we consider a vector space to have commutativity or not, I wonder. In every vector space that I've worked on commutativity was a given, but now I'm so confused! Thanks anyway for your answer! $\endgroup$ Commented Jul 5, 2018 at 9:51
  • $\begingroup$ Commutativity is a property of the underlying division ring. When you define a vector space you always specify over what it is defined anyway. $\endgroup$ Commented Jul 5, 2018 at 9:54
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    $\begingroup$ Heh. What I mean is that "you have commutativity on your vector space only if you have it on your underlying division ring". $\endgroup$ Commented Jul 5, 2018 at 9:56
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    $\begingroup$ @Euler Welcome to the wide world of inconsistent definitions in mathematics. I have many books that assume field, and a couple that say "division ring". If you're writing, it's safest to say which convention you're using. $\endgroup$
    – Mark S.
    Commented Jul 5, 2018 at 10:01
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    $\begingroup$ @Euler_Salter There are many standards. And really, reals being cuts or cauchy sequences doesn't matter when we use reals, so the theory of reals doesn't need to be built up from quantifiers; foundations can often be swapped out without changing theories. Definitions are what make theorems interesting, and if new theories require definitions be bent, they get bent. $\endgroup$
    – Yakk
    Commented Jul 5, 2018 at 18:10

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