What is the main difference between a vector space and a field?

It is true that vector spaces and fields both have operations we often call multiplication, but these operations are fundamentally different, and, like you say, we sometimes call the operation on vector spaces scalar multiplication for emphasis.

The operations on a field $\mathbb{F}$ are

• $+$: $\mathbb{F} \times \mathbb{F} \to \mathbb{F}$
• $\times$: $\mathbb{F} \times \mathbb{F} \to \mathbb{F}$

The operations on a vector space $\mathbb{V}$ over a field $\mathbb{F}$ are

• $+$: $\mathbb{V} \times \mathbb{V} \to \mathbb{V}$
• $\,\cdot\,$: $\mathbb{F} \times \mathbb{V} \to \mathbb{V}$

One of the field axioms says that any nonzero element $c \in \mathbb{F}$ has a multiplicative inverse, namely an element $c^{-1} \in \mathbb{F}$ such that $c \times c^{-1} = 1 = c^{-1} \times c$. There is no corresponding property among the vector space axioms.

It's an important example---and possibly the source of the confusion between these objects---that any field $\mathbb{F}$ is a vector space over itself, and in this special case the operations $\cdot$ and $\times$ coincide.

On the other hand, for any field $\mathbb{F}$, the Cartesian product $\mathbb{F}^n := \mathbb{F} \times \cdots \times \mathbb{F}$ has a natural vector space structure over $\mathbb{F}$, but for $n > 1$ it does not in general have a natural multiplication rule satisfying the field axioms, and hence does not have a natural field structure.

Remark As @hardmath points out in the below comments, one can often realize a finite-dimensional vector space $\mathbb{F}^n$ over a field $\mathbb{F}$ as a field in its own right if one makes additional choices. If $f$ is a polynomial irreducible over $\mathbb{F}$, say with $n := \deg f$, then we can form the set $$\mathbb{F}[x] / \langle f(x) \rangle$$ over $\mathbb{F}$: This just means that we consider the vector space of polynomials with coefficients in $\mathbb{F}$ and declare two polynomials to be equivalent if their difference is some multiple of $f$. Now, polynomial addition and multiplication determine operations $+$ and $\times$ on this set, and it turns out that because $f$ is irreducible, these operations give the set the structure of a field. If we denote by $\alpha$ the image of $x$ under the map $\mathbb{F}[x] \to \mathbb{F}[x] / \langle f(x) \rangle$ (since we identify $f$ with $0$, we can think of $\alpha$ as a root of $f$), then by construction $\{1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}\}$ is a basis of (the underlying vector space of) $\mathbb{F}[x] / \langle f \rangle$; in particular, we can identify the span of $1$ with $\Bbb F$, which we may hence regard as a subfield of $\mathbb{F}[x] / \langle f(x) \rangle$; we thus call the latter a field extension of $\Bbb F$. In particular, this basis defines a vector space isomorphism $$\mathbb{F}^n \to \mathbb{F}[x] / \langle f(x) \rangle, \qquad (p_0, \ldots, p_{n - 1}) \mapsto p_0 + p_1 \alpha + \ldots + p_{n - 1} \alpha^{n - 1}.$$ Since $\alpha$ depends on $f$, this isomorphism does depend on a choice of irreducible polynomial $f$ of degree $n$, so the field structure defined on $\mathbb{F}^n$ by declaring the vector space isomorphism to be a field isomorphism is not natural.

Example Taking $\Bbb F := \mathbb{R}$ and $f(x) := x^2 + 1 \in \mathbb{R}[x]$ gives a field $$\mathbb{C} := \mathbb{R}[x] / \langle x^2 + 1 \rangle.$$ In this case, the image of $x$ under the canonical quotient map $\mathbb{R}[x] \to \mathbb{R}[x] / \langle x^2 + 1 \rangle$ is usually denoted $i$, and this field is exactly the complex numbers, which we have realized as a (real) vector space of dimension $2$ over $\mathbb{R}$ with basis $\{1, i\}$.

A field is an algebraic structure allowing the four basic operations $+$, $-$, $\cdot$, and $:\,$, such that the usual rules of algebra hold, e.g., $(x+y)\cdot z=(x\cdot z) +(y\cdot z)$, etcetera, and division by $0$ is forbidden. The elements of a given field should be considered as "numbers". The systems ${\mathbb Q}$, ${\mathbb R}$, and ${\mathbb C}$ are fields, but there are many others, e.g., the field ${\mathbb F}_2$ consisting only of the two elements $0$, $1$ and satisfying (apart from the obvious relations) $1+1=0$.

A vector space $X$ is in the first place an "additive structure" satisfying the rules we associate with such structures, e.g., $a+({-a})=0$, etc. In addition any vector space has associated with it a certain field $F$, the field of scalars for that vector space. The elements $x$, $y\in X$ cannot only be added and subtracted, but they can be as well scaled by "numbers" $\lambda\in F$. The vector $x$ scaled by the factor $\lambda$ is denoted by $\lambda x$. This scaling satisfies the laws we are accustomed to from the scaling of vectors in ${\mathbb R}^3$: $$\lambda(x+y)=\lambda x+\lambda y,\qquad (\lambda+\mu)x=\lambda x+\mu x\ .$$

Asking "What is the difference between a vector space and a field" is similar to asking "What is the difference between tension and charge" in electrodynamics. In both cases the simple answer would be: "They are different notions making sense in the same discipline".

Simply speaking a field is a structure where you can add, subtract, multiply and divide with "normal" rules retained. For example $\mathbb Q$, $\mathbb R$ and $\mathbb C$ are fields (but not $\mathbb Z$ since you can't divide and still be in $\mathbb Z$).

A vector space on the other hand is a structure "above" a field where the normal vector space operations are defined and relates to the field (called the scalars) in the way one would expect. You should be able to add vectors, and you should be able to multiply them with a scalar with "normal" behavior (fx $0\overline u$ should be the null vector and $1\overline u=u$ and $(a+b)\overline u = a\overline u + b\overline u$ and so on). Note that scalar product to be defined is noting that is required to qualify as a vector space.

Normal vector spaces are $\mathbb F^n$ where $\mathbb F$ is a field (but there exists vector spaces that are not of finite dimensions). As a special case a field is also a vectorspace over itself (ie with $n=1$), because the rules of a vector space is just a reduction of the rules that applies to a field.

So the example you're asking for is fx $\mathbb R$.

Once you view an object as a field you stop seeing it as a vector space on something smaller or over itself:

1. Any field is a vector space over itself.
2. If $\mathbb{K}$ then $\mathbb{K}[x]$ is the vector space of all the polynomials with coefficient in $\mathbb{K}$. This set is an algebra but not a field. Let $g$ be an irreducible polynomial in $\mathbb{K}[x]$ then we define the linear vector subspace $A_g=\{p(x)\in\mathbb{K}[x]:p(x)=g(x)q(x)\text{ for some }q\in\mathbb{K}[x]\}$ (which is an ideal) and then the quotient $\mathbb{K}/A_g$ is a field. (e.g. $\mathbb{C}$ is build this way over the polynomial $x^2+1$.)
3. The space of all meromorphic functions over a Riemann surface is a field and a vector space over $\mathbb{C}$.
4. The space of rational function over a field $\mathbb{K}$, noted as $\mathbb{K}(x)$ form a field. (this case coincides to the case in point 3 when the Riemann surface is a sphere.)

In general a vector space is the set of function from a set to a field. Let $A$ be any set and $\mathbb{K}$ a field, then: $$V=\{f:A\to\mathbb{K}\}$$ is a vector space with the operations induced by the field operations. While a field is the same set with an additional property of multiplication which must form a group when removing the zero vector.

A given set is said to be closed under certain operation, when any two elements of that set undergone that operations gives a result which lies back in the same set

A field is closed under addition, subtraction, multiplication and division Ex: R(set of all real numbers) or C(set of all complex numbers).

Where as vector space is a set of elements associated with scalar field (R or C) which satisfies this two conditions: 1 The set of elements is closed under addition 2 The set of elements is closed under scalar multiplication with respect to scalar field