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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.)

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    $\begingroup$ You need an example or a counter-example? Take any field, it's also a vector space over itself. That's what you're literally asking. Otherwise take any field $k$ and consider $k^2$ as a vector space over $k$: it's not a field. $\endgroup$ – Zavosh Oct 12 '14 at 5:31
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    $\begingroup$ The main difference in idea, put vaguely, is that fields are made of 'numbers' and vector spaces are made of 'collections of numbers' (vectors). You can multiply any two numbers together, and you can also take a collection of numbers and multiple them all with the same fixed number. $\endgroup$ – Zavosh Oct 12 '14 at 5:35
  • $\begingroup$ The complex numbers form a field. They are also a two-dimensional vector space over the field of real numbers and a one-dimensional vector space over the field of complex numbers (and an infinite dimensional vector space over the field of rational numbers). $\endgroup$ – Jyrki Lahtonen Oct 12 '14 at 13:09
  • $\begingroup$ Do you know what a field is? It looks like your question is for an example of a field since you're asking for a example of a vector space that's also a field and since all fields are vector spaces it follows that any field would do. $\endgroup$ – skyking Oct 6 '15 at 11:24
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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.

Edit As @hardmath points out, 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 over $\mathbb{F}$, and declare two 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 $\Bbb F$. This anyway gives an identification $$\mathbb{F}[x] / \langle f(x) \rangle \leftrightarrow \mathbb{F}^n.$$ On the other hand, this identification does depend on a choice of irreducible polynomial $f$ of degree $n$, so the field structure on $\mathbb{F}^n$ 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 $\{1, i\}$.

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    $\begingroup$ I've undone the edit made to this post, which didn't preserve the tone or the intent. I'm happy to discuss the content of the change further here in the comments, though. $\endgroup$ – Travis Willse Oct 12 '14 at 14:22
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    $\begingroup$ Given the OP's interest in special cases where field and vector space "coincide", it might be worth mentioning that field extensions provide examples of this. If field $\mathbb{F}$ is a subfield of (say) field $\mathbb{K}$, then $\mathbb{K}$ is both a field in itself and a vector space over the smaller field $\mathbb{F}$. We can always construct a multiplication rule for the Cartesian product $\mathbb{F}^n$ by treating it as an algebraic field extension of $\mathbb{F}$ (needs us to specify an irreducible polynomial of degree $n$ over $\mathbb{F}$). $\endgroup$ – hardmath Oct 19 '14 at 16:32
  • $\begingroup$ @hardmath Yes, that's a good point---perhaps you might like to promote your comment to a full answer? It certainly merits it. $\endgroup$ – Travis Willse Oct 19 '14 at 23:58
  • $\begingroup$ Thanks, Travis, but it perhaps merits another Question. Mindful of the OP's disclaimer of no mathematical background, I'm happy to draft behind your Answer, perhaps adding a little color for future Readers. $\endgroup$ – hardmath Oct 20 '14 at 0:29
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    $\begingroup$ In that case I added a short summary to the end of my answer explaining this in just a little more detail. Thanks for the very good comment! $\endgroup$ – Travis Willse Oct 20 '14 at 1:03
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A field is an algebraic structure allowing the four basic operations $+$, $-$, $\cdot$, and $:\,$, such that the usual rules of algebra hold, e.g., $(x+y)\cdot z=(x\cdot z) +(y\cdot z)$, etcetera, and division by $0$ is forbidden. The elements of a given field should be considered as "numbers". The systems ${\mathbb Q}$, ${\mathbb R}$, and ${\mathbb C}$ are fields, but there are many others, e.g., the field ${\mathbb F}_2$ consisting only of the two elements $0$, $1$ and satisfying (apart from the obvious relations) $1+1=0$.

A vector space $X$ is in the first place an "additive structure" satisfying the rules we associate with such structures, e.g., $a+({-a})=0$, etc. In addition any vector space has associated with it a certain field $F$, the field of scalars for that vector space. The elements $x$, $y\in X$ cannot only be added and subtracted, but they can be as well scaled by "numbers" $\lambda\in F$. The vector $x$ scaled by the factor $\lambda$ is denoted by $\lambda x$. This scaling satisfies the laws we are accustomed to from the scaling of vectors in ${\mathbb R}^3$: $$\lambda(x+y)=\lambda x+\lambda y,\qquad (\lambda+\mu)x=\lambda x+\mu x\ .$$

Asking "What is the difference between a vector space and a field" is similar to asking "What is the difference between tension and charge" in electrodynamics. In both cases the simple answer would be: "They are different notions making sense in the same discipline".

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Simply speaking a field is a structure where you can add, subtract, multiply and divide with "normal" rules retained. For example $\mathbb Q$, $\mathbb R$ and $\mathbb C$ are fields (but not $\mathbb Z$ since you can't divide and still be in $\mathbb Z$).

A vector space on the other hand is a structure "above" a field where the normal vector space operations are defined and relates to the field (called the scalars) in the way one would expect. You should be able to add vectors, and you should be able to multiply them with a scalar with "normal" behavior (fx $0\overline u$ should be the null vector and $1\overline u=u$ and $(a+b)\overline u = a\overline u + b\overline u$ and so on). Note that scalar product to be defined is noting that is required to qualify as a vector space.

Normal vector spaces are $\mathbb F^n$ where $\mathbb F$ is a field (but there exists vector spaces that are not of finite dimensions). As a special case a field is also a vectorspace over itself (ie with $n=1$), because the rules of a vector space is just a reduction of the rules that applies to a field.

So the example you're asking for is fx $\mathbb R$.

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$\mathbb{R}$ and $\mathbb{C}$ are fields as well as vector spaces over $\mathbb{R}$. More generally any field is a vector space over its subfields. This is simple to prove.

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Once you view an object as a field you stop seeing it as a vector space on something smaller or over itself:

  1. Any field is a vector space over itself.
  2. If $\mathbb{K}$ then $\mathbb{K}[x]$ is the vector space of all the polynomials with coefficient in $\mathbb{K}$. This set is an algebra but not a field. Let $g$ be an irreducible polynomial in $\mathbb{K}[x]$ then we define the linear vector subspace $A_g=\{p(x)\in\mathbb{K}[x]:p(x)=g(x)q(x)\text{ for some }q\in\mathbb{K}[x]\}$ (which is an ideal) and then the quotient $\mathbb{K}/A_g$ is a field. (e.g. $\mathbb{C}$ is build this way over the polynomial $x^2+1$.)
  3. The space of all meromorphic functions over a Riemann surface is a field and a vector space over $\mathbb{C}$.
  4. The space of rational function over a field $\mathbb{K}$, noted as $\mathbb{K}(x)$ form a field. (this case coincides to the case in point 3 when the Riemann surface is a sphere.)

In general a vector space is the set of function from a set to a field. Let $A$ be any set and $\mathbb{K}$ a field, then: $$V=\{f:A\to\mathbb{K}\}$$ is a vector space with the operations induced by the field operations. While a field is the same set with an additional property of multiplication which must form a group when removing the zero vector.

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    $\begingroup$ Why the (-1)? Explain so that I can try to improve the answer. $\endgroup$ – Lolman Oct 12 '14 at 13:56
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A given set is said to be closed under certain operation, when any two elements of that set undergone that operations gives a result which lies back in the same set

A field is closed under addition, subtraction, multiplication and division Ex: R(set of all real numbers) or C(set of all complex numbers).

Where as vector space is a set of elements associated with scalar field (R or C) which satisfies this two conditions: 1 The set of elements is closed under addition 2 The set of elements is closed under scalar multiplication with respect to scalar field

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For the sake of intuition, we can think of a field as R or C, and we can think of a vector space as R$^n$ for any natural number $n$. More generally, for any field F, and for any $n\in$ N, F$^n$ is a vector space over F.

Note that a field is a type of vector space.

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