Is $\mathbb R^2$ a field?
Solution 1:
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?
Solution 2:
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.