What is the main difference between a vector space and a field?

In my opinion both are almost same. However there should be some differenes like any two elements can be multiplied in a field but it is not allowed in vector space as only scalar multiplication is allowed where scalars are from the field.

Could anyone give me atleast one counter- example where field and vector space are both same.
Every field is a vector space but not every vectorspace is a field.
I need an example for which a vector space is also a field.

Thanks in advance. (I’m not from mathematical background.)

Answer

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\}.

Attribution
Source : Link , Question Author : Vineel Kumar Veludandi , Answer Author : Travis Willse

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