Proving that $\mathbb{C}$ has a natural definition I would like to prove a theorem in complex analysis which states:

Let $K$ be a commutative field. We suppose that $L$ is a sub-field of $K$ and that $K$ is thus a vector space of finite dimension $n$ over $L$.
Then, either $n=1$ and $K=L$, or $n=2$ and $K$ is isomorphic to $\mathbb{C}$. If there is no restriction on $K$ being commutative, you obtain the field of quaternions for $n=4$.


*

*Beyond the definition of the complex field as such, I can't understand why it can't be possible to define a complex field isomorphic to $\mathbb{R}^n$ for all $n$ if one defines a multiplication law for each $n$ and thus build a corresponding ring and then a field for each $n$. Why is it not possible to define such a multiplication rule?


*Can somebody give me some hints how to prove the theorem?
Thanks for any comment.
 A: As written, strictly speaking, your theorem is false. You need to add the additional requirement that $L = \Bbb R$. Your statement then becomes:
What field extensions of finite degree of the real numbers can we obtain?
The usual approach is to do this:
1) Show that if $K$ is a finite extension of $\Bbb R$ it is an algebraic extension of $\Bbb R$.
2) Show that any finite field extension of $\Bbb R$ can be realized as a (ring) isomorph of a quotient $\Bbb R[x]/\langle f(x)\rangle$, where $f$ is an irreducible polynomial in $\Bbb R[x]$.
3) Use the Fundamental Theorem of Algebra (this is where complex analysis typically comes into play-in proving the FTOA, using Liouville's Theorem) to show that the only irreducible polynomials in $\Bbb R[x]$ are of degree $1$ or $2$.
4) Show that if $[K:\Bbb R] = 2$, then $K$ contains a root $j$ of $x^2 + 1$ (hint: use the quadratic formula to produce this root from the discriminant of your irreducible quadratic).
To answer your other question:
Sir William Rowan Hamilton searched for a long time for a way to turn $\Bbb R^3$ into a division algebra over $\Bbb R$. He finally turned instead to $\Bbb R^4$, and discovered the quaternions. In 1877, Ferdinand Georg Frobenius proved Hamilton's search for such a $3$-dimensional division algebra was indeed pointless, which you can read about here.
A "hand-wavy" argument as to "why" only powers of two work (up to a point) is that multiplication is desired to be bilinear, and bilinear functions play well with squares, but not other powers. The truth of the matter is a wee bit more complicated, but also beautiful.
