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 $\{1, i\}$.
Best Answer
A $1$-dimensional vector space is not a field for the same reason that no vector space is a field. Multiplication of vectors with other vectors is not defined over vector spaces. So vector spaces are not even rings.
That said, given a $1$-D vector space $V$ over $F$ you can artificially define a multiplication function $\circ : V \times V \to V$ over $V$ thus. Since $V$ is $1$-dimensional over $F$, it is generated by a single vector $e$ of $V$. So any two vectors $x, y$ can be written uniquely as $x = \alpha \cdot e$ and $y = \beta \cdot e$ for scalars $\alpha, \beta$ (here $\cdot$ is the scalar action of $F$ on $V$).
So define $x \circ y := (\alpha \beta) \cdot e$. You can check that this is well-defined, commutative, has an identity and inverses, distributes over vector addition etc. To understand the motivation of this definition, notice that we can derive $x \circ e = e \circ x = x$ from it; so essentially what we are doing is that we are looking at the single basis element $e$ that generates $V$ as the multiplicative identity $1$ in $F$. Looked at this way, $V$ is isomorphic to $F$ itself.
However the reason we do not say a $1$-dimensional vector space $V$ is a field is because in general there is no unique way to turn some such $V$ into a field. For if $e$ generates $V$, then every (non-zero) scalar multiple of $e$ also generates $V$. Hence there are in general many different ways to define a multiplication function on $V$.
E.g. if you have the $1$-D vector space which is the line $y = x$ in $\Bbb R^2$, then you can look at it as being isomorphic to $\Bbb R$ in infinitely many ways. You can for instance get one isomorphism to $\Bbb R$ by identifying the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ with $1 \in \Bbb R$. Or alternatively, you can identify $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ with $1 \in \Bbb R$. Or you can identify $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$ with $1 \in \Bbb R$. So on and so forth. And each such identification provides a different way to turn the line $y = x$ into a field.