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I recently learned about vector spaces, and had two questions:

  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!

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    $\begingroup$ 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 $\mathbb R$. $\endgroup$ Sep 27 '21 at 9:49
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    $\begingroup$ 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. $\endgroup$
    – Elliot G
    Sep 27 '21 at 9:55
  • $\begingroup$ 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 $\Bbb R$. $\endgroup$
    – Elliot G
    Sep 27 '21 at 9:58
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    $\begingroup$ 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. $\endgroup$ Sep 27 '21 at 10:01
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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 $\mathbb{Q}$.

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  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 $\Bbb C$.
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