So for a set F where addition and multiplication are closed and the axioms are satisfied, F is a field.

For V to be a vector space, vector addition and scalar multiplication must be closed as well right?

Now for V to be a vector space over F, does that mean that V must first of all be closed under vector addition and scalar multiplication. Then all the elements of V must be elements of F as well. Is that right? The vector space over F must be contained in or equal to F?

• Have a look at a simple example to check your thinking. $V:=\mathbb R^2$ is a vectorspace over field $F:=\mathbb R$. Also have a look in a book-script introducing vectorspaces. Mar 27, 2015 at 11:08
• No. A simple exemple: $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Mar 27, 2015 at 11:12
• @EmilioNovati That is a nice example.
– mvw
Mar 27, 2015 at 11:53

So for a set F where addition and multiplication are closed and the axioms are satisfied, F is a field.

There should be a $1 \in \mathbb{F}$ and subtraction and division should be closed as well, but that probably is an implication from the closedness of addition and multiplication and the axioms.

For V to be a vector space, vector addition and scalar multiplication must be closed as well right?

That is true the axioms should say that $V$ is an additive group regarding the addition (AEDDA) and an commutative multiplicative group regarding the scalar multiplication (ANIC).

Now for V to be a vector space over F, does that mean that V must first of all be closed under vector addition and scalar multiplication.

We had that. The results must be in $V$.

Then all the elements of V must be elements of F as well.

The scalars are taken from $\mathbb{F}$ and $V$ must be compatible with it. An 1-dimensional Vector space is probably isomorph to $\mathbb{F}$, if not identical.

$$V = \mathbb{R}, v = (x) \in V \wedge x \in \mathbb{R}$$

But for example $$V = \mathbb{R}^2 = \{ \left. v = (x, y)^T \,\right|\, x, y \in \mathbb{R} \}$$ seems different from the field $\mathbb{R}$ to me.

Is that right? The vector space over F must be contained in or equal to F?

Not in general. If they were always identical it would be a field and the vector space definition would be superfluous.

The scalar multiplication is not a multiplication between vectors, but between a field element (scalar) and a vector giving a vector. In general there is no vector multiplication giving a vector and allowing a division of vectors.

Or said differently: A field is a vector space but not every vector space is a field.

Sorry for my comment about coordinate spaces, I am too used to them. So a coordinate space is a vector space. But not every vector space is a coordinate space.

E.g. $$V = C^1 = \{ \left. f: \mathbb{R} \to \mathbb{R} \,\right|\, f \mbox{ is differentiable on } \mathbb{R} \}$$ is a vector space but no coordinate space (at least no finite dimensional one). I can add differentiable functions and get a differentiable function. I can also take any scalar $\alpha$ from $\mathbb{R}$ and use that to scalar multiply some $f \in V$ and get a $(\alpha f)\in V$.

• I still don't really understand what it means for a vector space to be "over" a field. Mar 27, 2015 at 11:20
• so scalars that are coordinates of the vector space are also elements of the field but even if these scalars are now used to create vector spaces in different dimensions then the individual coordinates are still elements of the field. So this is why say $\mathbb R^2$ is a vector space over $\mathbb R$ Mar 27, 2015 at 11:27
• @stephancasey Actually, saying "over a field" means the scalars/coefficients come from the field you are over. It doesn't necessarily mean the coordinates/components of the vectors are over the same field. For example, since $\Bbb C$ is a field, we could look at the example of $\Bbb R^{5}$ over $\Bbb C$. Here, the coordinates of the vectors are from $\Bbb R$. This is not a vector space, but it is a good example of what "over a field" means. When we say "over $\Bbb C$", we mean: since vectors are closed under scalar multiplication, for all $\alpha \in \Bbb C$, $v \in \Bbb R^{5}$.. Mar 27, 2015 at 11:30
• I think I see what you're saying. So you have to take a scalar element of the field and a vector from the vector space. Then when you take multiplication of the scalar and the vector then the components of the new vectors created must also be part of the original vector space which is $\mathbb R^5$ in this case Mar 27, 2015 at 11:41
• @StephanCasey Yes, you are right! And now you know what it means to say "over a field". It means the scalars (the "numbers" we use as the coefficients when we want to take "multiples" of the vectors, i.e., scalar multiples) come from the specified field. Mar 27, 2015 at 12:27

Note three out of eight axioms of vector space are related to scalar fields.

• $0\vec v=\vec 0$, where $0$ is additive identity of field $F$
• $1\vec v=\vec v$, where $1$ is multiplicative identity of field $F$
• $s(t\vec v)=(st)\vec v$, where $s,t\in F$ requires multipilcation closure $st\in F$

Generally speaking, $V$ and $F$ don't have set relation. Field $F$ provides a group action on $V$, called scalar multiplication $f:F\times V\to V$ with $(c,\vec v)\mapsto c\vec v$ satisfying axioms of vector space.

The abstract approach in mathematics should be motivated by concrete examples but should not be grounded in it. It should stand alone.

Let $V=\{x,y\}$ and define a binary operation $+$ on it satisfying the necessary properties. This will work if you let $x+y=y+x=y$ and $x+x=y+y=x$.

Next take a field $F=\{a,b\}$ where $a$ and $b$ are the additive and multiplicative identity respectively. Denote scalar multiplication by a $\odot$ and define it as follows: $a\odot x = a\odot y=b\odot x=x$ and $b\odot y=y$.

This algebraic structure qualifies as a vector space $V$ over the field $F$.

It is your own definition of scalar multiplication which is forcing the result to be in $V$. Otherwise $V$ and $F$ are two different sets. We may have $V=\{x=\mbox{apple},y=\mbox{banana}\}$ and $f=\{a=\mbox{giraffe},b=\mbox{tiger}\}$.