# Is vector space a field? Or more than that?

As you all know, vector space is closed under scalar multiplication, scalar product, vector product and addition. If I take scalar product, vector space is a field, but if i take vector product, vector space is a group. Is there any term designating this kind of set? Plus scalar multiplication is a binary operation from two same sets to different set. What kind of operation is this?

-
A vector space is not even a ring and there is no such a thing as vector product in general. – azarel Jul 3 '12 at 1:48
You're right that scalar multiplication is an operation from two different sets to one of them. This is why your statement that "If I take scalar product, vector space is a field" is incorrect. You should look up group actions to see a similar definition in general. – Matt Jul 3 '12 at 1:48
The closest thing to what you are describing is an algebra over a field (that is, a vector space equipped with a bilinear vector product). – F M Jul 3 '12 at 1:50
"Is there any term designating this kind of set?" Yes, "vector space". – Alex Becker Jul 3 '12 at 1:56
Scalar multiplication is a binary operation from sets that are not the same in general. A scalar product is an action $\cdot:F\times K\rightarrow K$ where $F$ is a field and $K$ is an abelian group. – F M Jul 3 '12 at 1:56

## 2 Answers

Let $K$ be a field. A vector space over $K$ is a set $V$, together with an operation $+\colon V\times V\to V$, and a function $\cdot\colon K\times V\to V$ (or alternatively, a family of functions $\lambda_c\colon V\to V$, indexed by elements $c\in K$) subject to certain identities and conditions.

The set $V$ does not normally contain $K$, so it makes no sense to talk about the vector space $V$ as being "closed under scalar multiplication". In addition, there is in general no "product" of vectors, so it does not make sense to talk about "vector multiplication."

Even in the case where we have a cross product of vectors (e.g., $\mathbb{R}^3$), this operation does not make $V$ into a group (the product is not associative).

Scalar multiplication is not an operation; an operation is always a function from a cartesian power of a set to the set. However, you can use currying to view the scalar multiplication as a family of unary operations on $V$, as indicated above, where for each $c\in K$ and each $\mathbf{v}\in V$, we define $\lambda_c(\mathbf{v})=c\mathbf{v}$.

(Added. Once you take into account the vector space axioms that relate to the scalar multiplication, it turns out that the scalar multiplication $K\times V\to V$ is actually an action of the field $K$ on the additive group $(V,+)$. But to truly call it an "action", you need the function to satisfy certain properties; just having a function $K\times V\to V$ does not make it an action, whereas any function $S\times S\to S$ is a binary operation on $S$, regardless of its properties.)

So pretty much all of your questions are based on misstatements of fact. Seems hard to answer them accurately.

Now, there is a situation in which some of what you say might make sense: if $F$ and $K$ are fields, and $F\subseteq K$, then we can view $K$ as a vector space over $F$ by "forgetting" about multiplication inside of $K$ when neither factor is in $F$. But here you really are going the other way: you already have a field, and you are obtaining a vector space by restricting the multiplication $\cdot\colon K\times K\to K$ to $F\times K$.

There is another situation in which you may have a product between "vectors": when you have an algebra. If $F$ is a field, a $K$ algebra is a ring with (a copy of) $F$ in the center of $K$. But again, what you have is a richer structure that, by "forgetting" part of the structure, yields a vector space. Much like you can have a ring and, by forgetting the product, obtain a group.

-
Arturo, isn't a function from $K\times V$ to $V$ usually called an action? i.e. multiplication by scalars is an action of $K$ on $V$ (maybe it is worth mentioning it has a name) – Pedro Tamaroff Jul 3 '12 at 2:31
@Peter: Well, the fact that it's an action is due to the axioms of a vector space; a plain function would not qualify as an action (whereas any function $V\times V\to V$ is automatically a binary operation on $V$, regardless of what properties like associativity, etc. it might have). But it may be worth mentioning. – Arturo Magidin Jul 3 '12 at 2:33
@PeterTamaroff You wouldn't typically call it "an action from $K\times V\to V$". If you were to refer to it as an action, you'd probably say "an action of $K$ on $V$". – Alex Becker Jul 3 '12 at 2:33
@AlexBecker Yes, you're right, my mistake. Fixed. Could I find you in the chat? – Pedro Tamaroff Jul 3 '12 at 2:35
@tomasz: Didn't I say so already in the penultimate paragraph? – Arturo Magidin Jul 3 '12 at 2:41

A vector space over a field is a set with its own definition

A field is a set with its own definition

But later on you will find that the field axioms agree with vector space axioms

Therefore, a field over itself is a vector space but a vector space does not imply a field For example, polynomials are vector space but also a ring, not a field

-
Downvoting: their definitions will not coincide later on. – Mark Fantini Jul 16 '14 at 2:57