# Is every $1$-dimensional vector space a field?

We say that every field $$F$$ "is" a $$1$$-D vector space over itself. By this we mean that if we consider the elements of $$F$$ as both vectors and scalars, then we get a vector space by using the addition and multiplication from $$F$$.

It seems just as easy to go in the other direction and interpret any $$1$$-D vector space as a field. But I've never seen it written that every $$1$$-D vector space "is" a field.

Why?

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.

• Thanks very much! I had something like this lurking in the back of my mind, but kept using $\mathbb{R}$ as my example. Since it has a canonical basis, it's obviously a bad example. One point of clarification though. You say "Multiplication of vectors with other vectors is not defined over vector spaces." But of course, multiplication with vectors is not defined over fields and yet we are fine saying they "are" vector spaces. So it seems the point is not whether it's defined, but whether there's a natural choice to use, right? – A_P Aug 11 at 17:33
• @A_P But it is defined. For something to be a vector space, only scalar multiplication with vectors have to be defined. In the case of fields, this is just the usual field multiplication. You just have to view the field elements as both the scalars and the vectors simultaneously. – 0XLR Aug 11 at 23:21
• Hmm but you said it yourself: "(has) to be defined." "Multiplication with vectors" had no meaning until we chose to view the elements as vectors, and made a choice to use the usual multiplication. So it still seems to me that the point is that this choice is natural, no? In the other direction, if we have a standard basis for our vector space (e.g., $\mathbb{R}$), can't we choose to say it "is" a field because of the natural choice of basis? This seems to be what Chris Culter's answer is indicating. – A_P Aug 12 at 2:20
• What I am saying is that the choice of the usual multiplication as the scalar multiplication is so natural that mathematicians already assume that the choice has been made when they defined a field. It is in this sense that they say that a field is a vector space: the choice has already been made implicitly. And yes, the motivation for this implicit choice is the "naturality". – 0XLR Aug 12 at 4:08
• In the reverse direction, however, the point is that we don't in general have a standard basis for most vector spaces (even $1$-D). Like in my example above of the line $L: y = x$ in $\Bbb R^2$. Each of the choices $(1,1), (2, 2), (3, 3)$ are equally good basis vectors for that vector space. What Chris Culter is saying is that if we chose to fix one of those basis vectors, say $(1, 1)$, then you could say that the pair $(L, (1, 1))$ is canonically isomorphic to $\Bbb R$. This pair $(L, (1, 1))$ is a pointed vector space and strictly speaking not the same mathematical object as $L$. – 0XLR Aug 12 at 4:18

If you want the interpretation to be natural, or canonical, or unique, then the answer is generally no.

Consider the most familiar case: real vector spaces. Suppose $$V$$ is a $$1$$-dimensional field over $$\mathbb R$$. Does that mean there's a natural way to identify $$V$$ with $$\mathbb R$$? Not really. For example, $$V$$ doesn't know which side is supposed to be positive and which side is negative. It doesn't know which of its elements are supposed to be integers. It doesn't know how to multiply or divide. In fact, for any $$v\neq0$$ in $$V$$, there is a vector space isomorphism $$V\to\mathbb R^1$$ that sends $$v\mapsto1$$.

On the other hand, a $$1$$-dimensional vector space with a distinguished basis is canonically isomorphic to its base field. That's kind of a silly statement, but, well, there it is. Choosing a basis adds a lot of information.

There is also the special case of a $$1$$-dimensional vector space $$V$$ over the field with $$2$$ elements, $$\mathbb F_2$$. There is a unique vector space isomorphism between $$V$$ and $$\mathbb F_2$$, because at least $$V$$ knows where $$0$$ is and where it isn't.