I'm taking a course in Complex Analysis and my instructor commented that Complex Analysis differs from Multivariable Real Analysis in its definition of differentiablity.
He said that Complex differentiation is much more stronger than Differentiability for a function from $\mathbb{R}^n$ to $\mathbb{R}^m$ and I'm able to see his point by looking at the definition of Complex differentiability and the definition of Real differentiability.
It is clear that the whole of the above discussion hinges on the new product structure defined on $\mathbb{R}^2$ as
$$(a,b).(x,y)=(ax-by,ay+bx)$$
which makes $\mathbb{R}^2$ into a field (With the usual addition). (And naturally defining differentiability here gives rise to holomorphic functions.)
My Question
Can we define a new product $\star$ on $\mathbb{R}^2$ such that it becomes a field? (With the usual addition.) Or all the field structures (on $\mathbb{R}^2$) isomorphic to the field structure of $\mathbb{C}$ ?
Can we do Analysis successfully in this new plane with the standard Topology.
Can we do the same thing in $\mathbb{R}^n$ $n\geq3$.
The only field extension of $\mathbb R$ which is $2$-dimensional as a $\mathbb R$-vector space is $\mathbb C$. And, no, we cannot do it in $\mathbb R^n$ with $n>2$. That is, there is no $n$-dimensional field extension of $\mathbb R$ with $n>2$; take a look at the Frobenius theorem). The best we have are the quaternions, which are a $4$-dimensional extension of $\mathbb R$, but which does not form a field; their multiplication is not commutative.