Is $\mathbb R^2$ a field? 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
 A: 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
$$
\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2
$$
A: 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?
A: 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.
A: 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.
