# Finding a $\mathbb{C}$-balanced map from $\mathbb{C}^2\to\mathbb{R}^2$?

I was trying to show $\mathbb{C}\otimes_\mathbb{C}\mathbb{C}\cong\mathbb{R}^2$ as $\mathbb{R}$-modules. At one point I would like to prove the existence of an $\mathbb{R}$-module homomorphism from $\mathbb{C}\otimes_\mathbb{C}\mathbb{C}\to\mathbb{R}^2$ by showing there exists a $\mathbb{C}$-balanced map $\mathbb{C}\times\mathbb{C}\to\mathbb{R}^2$, and using the universal property.

One such map is $\psi(a+bi,c+di)=(ac-bd,ad+bc)$. My question is, how would one come up with this map without just trial and error? The term $ac-bd$ looks suspiciously like a determinant, so there must be some reason that this would be a go-to map.

-
Isn't it just because $(a+bi)\cdot(c+di)=(ac-bd)+(ad+bc)i$? – Raskolnikov Mar 8 '12 at 8:03

Perhaps it's clearer if you decompose the map into 2 parts.

First consider the map

$\mathbb{C} \otimes_\mathbb{C} \mathbb{C} \to \mathbb{C}$ sending $u \otimes v \mapsto uv$ (which is obviously $\mathbb{C}$-balanced)

Then follow it by the map

$\mathbb{C} \to \mathbb{R}^2$ sending $x + yi \mapsto (x,y)$ (which is obviously a map of $\mathbb{R}$-vector spaces)

So their composition is an $\mathbb{R}$-balanced map.

-
Thanks Ted, that makes it clear. – Hailie Mathieson Mar 11 '12 at 3:22