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I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.

In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.

This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?

Thanks

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    $\begingroup$ A field has multiplication. How would you define multiplication on $\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't "obvious" until you realize that the resulting field is the complex numbers...) $\endgroup$ – Thomas Andrews Dec 4 '12 at 16:07
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    $\begingroup$ $\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. $\endgroup$ – Rankeya Dec 4 '12 at 16:08
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    $\begingroup$ But, if you want to try to do this for $\mathbb{R}^n$, for $n \geq 3$ such that $\mathbb{R}$ is naturally embedded in $\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. $\endgroup$ – Rankeya Dec 4 '12 at 16:11
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    $\begingroup$ $\mathbb R^3$ is a vector space. It turns out, there is no "good" multiplication that you can define on $\mathbb R^3$ that makes it a field. There is a multiplication on $\mathbb R^4$ that makes $\mathbb R^4$ almost a field, minus commutativity of multiplication. $\endgroup$ – Thomas Andrews Dec 4 '12 at 16:13
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    $\begingroup$ Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. $\endgroup$ – Adam Rubinson Dec 4 '12 at 16:37
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If you define:

$$(a,b)+(x,y):=(a+x,b+y)$$

$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$

then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?

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    $\begingroup$ Spoiler: take a look at one of the comments given. $\^\smile\^$ $\endgroup$ – Frenzy Li Dec 4 '12 at 16:10
  • $\begingroup$ complex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? $\endgroup$ – vondip Dec 4 '12 at 16:11
  • $\begingroup$ Yes, it would, because addition in $\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. $\endgroup$ – Rankeya Dec 4 '12 at 16:15
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    $\begingroup$ What do you mean "vector space"? Any field is a vector space over any subfield, so the field $\,\Bbb R^2\cong\Bbb C\,$ is a vector field ove an infinite number of subfields, say $\,\Bbb C\,,\,\Bbb R\,,\,\Bbb Q\,,\,\Bbb Q(i)\ldots\,$ , etc. $\endgroup$ – DonAntonio Dec 4 '12 at 16:15
  • $\begingroup$ So would that mean that I any vector space could be defined when F -being the field, as : F^n ? $\endgroup$ – vondip Dec 4 '12 at 16:17
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It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.

The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.

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Adding to the above answer. With the usual exterior multiplication of the $\mathbb{R}-\{0\}$ as a ring with the natural addition and multiplication you can not make a field out of $\mathbb{R}^{2}-(0,0)$ \ But there may exist other products such as the one in the answers which can make a field out of ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ \ According to one of the theorems of Field theory every field is an Integral domain. So by considering : ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ With the following natural product: $(A,B)*(C,D)=(AB,CD) $ We see that $(1,0)*(0,1)=(0,0) $ Which means that $\mathbb{R}^{2} $is not an integral domain and hence not a field.

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  • $\begingroup$ Why are you considering $\mathbb{R}^*\times\mathbb{R}^*$? $\endgroup$ – Tobias Kildetoft Feb 23 '15 at 10:16
  • $\begingroup$ Bcoz he is asking that if $\mathbb{R}^{2} $is a field ... And I think he meant with the exterior multiplication and exterior addition. Otherwise it is obvious we can consider it as field aince it is isomorphism to $\mathbb{C}$ and hence a field $\endgroup$ – Arvin Rasoulzadeh Feb 23 '15 at 10:21
  • $\begingroup$ But what does that have to do with this? You are removing way more elements than $0$ when you consider this (in fact, it is clear that the usual multiplication does turn this into a group). $\endgroup$ – Tobias Kildetoft Feb 23 '15 at 10:22
  • $\begingroup$ And since the natural multiplication is defined on $\mathbb{R}$ We have to consider $\mathbb{R}^{*}$ as the set which the natural multiplication works on $\endgroup$ – Arvin Rasoulzadeh Feb 23 '15 at 10:24
  • $\begingroup$ No, to be a field we would need $(\mathbb{R}\times\mathbb{R})\setminus \{0\}$ to be a group, not $\mathbb{R}^*\times\mathbb{R}^*$. $\endgroup$ – Tobias Kildetoft Feb 23 '15 at 10:25
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Usually in mathematics one defines these structures as tuples

A field is a triple $(K,+,\cdot)$ such that $K$ is a set and [...] and $\cdot:K \times K \rightarrow K$

A Vectorspace is a triple $(V,+,\cdot)$ such that $V$ is a set and [...] and $\cdot: K \times V \rightarrow V$

So your question is meaningless: A set (say $\mathbb R^2$) cannot be a field or a vectorspace or a group or anything - only if you add some additional structure (most of the time operations) you can ask this question.

For example $\mathbb R^2$ can be the set used in the definition of a field, as well as the underlying set used in the definition of a vectorspace. And we are happy, the addition operation $$ +:\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2 $$ is the same, and the multiplicaton for "the" vectorspace structure $$ \mathbb R \times \mathbb R^2 \rightarrow \mathbb R^2 $$ is "compatible" with the multiplication for "the" field structure $$ \mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2 $$

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