# Vector spaces - "over a field"

1. What do we mean by saying “vector space over a field”? I read many posts and listened to lectures but it seems to me that people have different definitions/explanations for the terminology “over a field” – in one lecture, the professor said that when we say a vector space "over a field" we mean that the scalars with which we multiply the elements of the vector space are taken from some field F. In another lecture, a professor said that “over a field” means that the components of the elements in the vector space are from some field F. After hearing their explanations, I got confused – does the terminology “over a field” refers to the elements of the vector space themselves or to the scalars we multiply the vectors in the V.S. with?

I would appreciate it if you could explain the terminology of “over a field” in more detail than those professors did.

1. When talking about a polynomial over a field F, we mean that the coefficients of the polynomial are taken from some field F. My question is the following: do the roots of the polynomial are from the field F because the coefficients belong there too, or in other words – does the choice of a field of the coefficients determines the field to which the roots of the polynomial belong?

Sorry about the bad English (I'm not a native speaker) and the silly question, I just started learning linear algebra so it’s still new and hard for me…

Thank you very much for reading my question!

• 1. Field: en.wikipedia.org/wiki/Field_(mathematics) . Note that e.g. $\mathbb Q, \mathbb R, \mathbb C$ are fields. For the start, it is a good idea to just pretend that your field is $\mathbb R$ and follow that intuition. 2. If coefficients belong to a field, the roots might belong to that field or to a "bigger" field. For example, the roots $\pm\sqrt{2}\not\in\mathbb Q$ even if the coefficients of $x^2-2$ belong to $\mathbb Q$. The roots $\pm i\not\in\mathbb R$ even if the coefficients of $x^2+1$ belong to $\mathbb R$. Sep 27 '21 at 9:49
• Every vector space is over a field. The field is usually $\Bbb R$ when you first encounter vector spaces. It turns out any finite dimensional vector space over $\Bbb F$ is isomorphic to $\Bbb F^n$ for some $n$. but in general, you should consider vectors and scalars to be different things. Sep 27 '21 at 9:55
• More generally, we have modules over rings. This is defined in basically the same way, but where the ring $R$ of scalars can be something like the integers, $\Bbb Z$. In the special case where $R$ is a field, a module is just a vector space over $\Bbb R$. Sep 27 '21 at 9:58
• And yes - it is the scalars (for scalar multiplication) come from a field. The vectors are abstract entities, they don't "have" components per se. To have components ("co-ordinates") - you first need to have a "co-ordinate system" (i.e. a basis), which is a tuple of linearly-independent vectors that span the whole space. Then, if the base is $(v_1,v_2,\ldots,v_n)$, for a vector $v$ there is a unique representation of $v=a_1v_1+a_2v_2+\ldots+a_nv_n$, where $a_i$ are the scalars (from the field), and then you can call those scalars the "components" of the vector. Sep 27 '21 at 10:01

This circles back to the definition of a vector space: a vector space $$V$$ over a field $$F$$ is a set with two operations $$+ : V \times V \to V$$ and $$\cdot : F \times V \to V$$ such that
1. $$(V,+)$$ is an abelian group
2. $$+, \cdot$$ are compatible with the field operations: $$1 \cdot v = v$$, $$(ab) \cdot v = a \cdot (b \cdot v)$$, $$a \cdot (u + v) = a \cdot v + a \cdot v$$, $$(a + b) \cdot v = a \cdot v + b \cdot v$$
So this "over some field" is baked right into the definition - it just means "the elements we can multiply our vectors by". For example if you take the vector space $$\mathbb{R}^2$$ with the usual operations you can multiply each element by real numbers and thus it's a vector space over $$\mathbb{R}$$, but you could also multiply them by rational numbers which means it's also a vector space over $$\mathbb{Q}$$.
1. When we say that $$V$$ is a vector space over a field $$F$$, what that means is your first option: that the scalars are taken from $$F$$. For instance $$\Bbb R$$ is a $$1$$-dimensional vector space over $$\Bbb R$$, and an infinite-dimensional vector space over $$\Bbb Q$$.
2. It needs context. If we say that $$P(x)$$ has no roots, it is a good idea to add where it has no roots. For instance $$x^2+1\in\Bbb R[x]$$ and it has no roots in $$\Bbb R$$, but it has roots in $$\Bbb C$$.