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.
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.