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The commutative field $K$ can be considered a vector space in itself.

I define a series of properties common to fields and vector spaces and then define each of these separately and guide how I am understanding both concepts. Some properties of algebraic structures common to vector spaces and fields are:

${\color{red}{\text{- Internal binary operation:}}}$

$A×A→A$ $(a,b)→a⋆b$

${\color{red}{\text{- External operation:}}}$

$A×B \to A$ $A×B \to C$ $B×A \to A$ $A×A \to B$

${\color{red}{\text{- Neutral element: a number within the set in use that, operated with any other number in the set, does not alter it.}}}$

$\exists~0 \in V: \forall~v\in V: v+0=0+v=v~~{\text{for all}}~~ v∈V. ~{\text{Neutral element}}$.

${\color{red}{\text{- Commutative property: the order of the values does not alter the result.}}}$

$∀~u,v \in V:u+v=v+u$

${\color{red}{\text{- Associative property: the order of operation does not alter the result.}}}$

$∀~u,v,w ∈V,(u+v)+w=u+(v+w)$

${\color{red}{\text{- Opposite element: is a number of the set that, operated on another number of the set, results in the neutral element.}}}$

$∀~u∈V \exists~-u∈V: u+(-u)=(-u) +u=\overrightarrow 0 $

${\color{red}{\text{- Distributive property: given 2 operations (for example: + , * ∈ A) the first will be distributive with respect to the second, if it is distributive to the left and to the right for any 3 elements: a, b, c ∈ A.}}}$

$∀~a,b,c ∈ A,(a\ast b)\circ c=(a\circ c)\ast (b\circ c), c\circ (a\ast b)=(c\circ a)\ast (c\ast b)$

Vector space.

Is an algebraic structure created from a non-empty set, an internal operation (called sum, defined for the elements of the set) and an external operation (called product by a scalar, defined between said set and another set, with body structure).

A vector space always has a field K on which it is defined, being in itself a triple $(V,+,\bullet)$ that satisfies:

• There is an “addition” operation that takes values $u~$ and $~v$ that belong to $V$ and the addition operation gives $u+v$ that belongs to $V$.

• There is a “multiplication” operation that takes values $\lambda$ that belongs to $K$ and $u$ that belongs to V, giving as a result $\lambda u$ that also belongs to $V$.

In both cases the operations “addition” and “multiplication” are not those referred to real numbers, in this case the notation used + and ∙ refer specifically to the operations applied to the vector space $V$ and relations with a field $K$. The elements of the vector space are called vectors and are represented as $u,v,w$,..., while the elements of the field will be called scalars and are represented as Greek letters $\alpha,\beta,\lambda,\mu$. Therefore, vectors can be added and subtracted from each other and can also be multiplied by scalars, complying with the usual algebraic properties. Vector spaces also have the following fundamental properties:

• Sum (+):

  • Neutral element.
  • Commutative property.
  • Associative property.
  • Opposite element.
  • Distributive property.

• Product (∙):

  • Associative property.
  • Distributive property.
  • Neutral element.

Field.

Algebraic structures that satisfy the form (B,⋆,△) where B is the set for which the two operations are defined (⋆,△). Fulfilling for .

(⋆) with:

  • Commutative property.
  • Symmetric element.
  • Neutral element.
  • Associative property.
  • Internal binary operation.

(△) with:

  • Internal binary operation.
  • Associative property.

With both operations the distributive property is verified. I imagine a vector space as a set of vectors and a field, intuitively, as a set of numbers, if so and in both cases (vector space and body) respectively fulfilling the properties described for each one.

If my previous statement is true: how is it possible to consider a field as a vector space? Could it be due to the coincidence between properties? or is it a particular case of some vector space? In that last case, if a practical example were provided it would be of great help.

CyclotomicField
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    I assume you're familiar with $\mathbb R^n$. What would $\mathbb R^1$ look like? – Erick Wong Feb 15 '24 at 00:50
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    Every field is a one dimensional vector space over itself, as you say. – lulu Feb 15 '24 at 00:52
  • I like to separate vectors and scalars notationally for this reason. For example take $\alpha , x \in K$ with $\alpha$ a scalar and $x$ a vector. It makes it a little easier to see why this is true. – CyclotomicField Feb 15 '24 at 01:32

1 Answers1

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You seem to have put it together already, but just for concreteness:

A field has defined multiplication and addition, with both being commutative and associative, having inverses and identities, and with multiplication distributing over addition. A vector space can be thought of as an abelian group with defined multiplication by a field which distributed over the group operation. If we take the additive group of a field as our abelian group, we can use multiplication by the field as our vector space scalar multiplication.

So it might be termed a "coincidence", but it's more just that fields with addition and multiplication already satisfy the requisite commutativity and distributive axioms.