Why is $i = (0,1)$?

Question

My question concerns the construction of the complex numbers, why complex multiplication is the way it is, and why $i$ must be $(0,1)$. All the arguments I have seen tend to rely on the fact that we can represent each complex number uniquely as $a + bi$, which seems to assume that $i = (0,1)$, or they assume that we know the formula for complex multiplication already. I am wondering whether “$i = (0,1)$”, and the formula for complex multiplication, can be deduced.

Suppose the following: There exists a way to turn $\mathbb{R}^2$ into a field. Call this field $\mathbb{F}$, with addition and multiplication given by $+$ and $\times$, respectively.

• The addition in $\mathbb{F}$ is the usual vector addition: $(a,b) + (c,d) = (a + c, b + d)$.

• The multiplication in $\mathbb{F}$ is not yet known explicitly. (But since $\mathbb{F}$ is a field, the multiplication must obey all the usual laws like distributivity, associativity, etc.)

• We require that $(a,b) \times (1,0) = (a,b)$ for all $(a,b)$. (This is reasonable: If $\mathbb{F}$ is to “extend” $\mathbb{R}$, where $\mathbb{R}$ is thought of as lying on the $x$-axis, then the multiplicative identity should remain the same.)

• There exists a number $i \in \mathbb{F}$ such that $i^2 = -1$. (This is reasonable: The whole point is that we want to have closure, to have roots even for negative numbers.)

From this, can we deduce that $i = (0,1)$? And that $\times$ is the usual complex multiplication? In short, is $\mathbb{F} = \mathbb{C}$?

Preferably, the deduction should come from elementary observations of the various algebraic properties of $\mathbb{F}$, instead of citing results like "$\mathbb{C}$ is the unique field extension of $\mathbb{R}$ that is 2-dimensional over $\mathbb{R}$..." etc., unless necessary.

My attempt:

By the distributive and associative laws, $z(x + y) = zx + zy$ and $z(cx) = c(zx)$ for all complex $z, x, y$ and scalars $c$. This means that each complex number $z = (a,b)$ can be viewed as a linear transformation, mapping vectors of the plane to other vectors of the plane. Thus, we can think of $z$ as a $2 \times 2$ matrix. From the identity $ze_1 = z$, the first column of the matrix is $(a,b)$. But here is where I am stuck. Aren’t there infinitely many possibilities for the second column? That is, why can’t we allow $ze_2$ to take on a value other than $(-b,a)$?

Answer 1

You can take any $\mathbb{R}$-linear isomorphism $A: \mathbb{R}^2 \to \mathbb{C}$, where $\mathbb{C}$ is the standard complex structure on $\mathbb{R}^2$. Then define $(x,y)\cdot (x',y') := A^{-1}(A(x,y)\cdot A(x',y'))$. Then $A^{-1}1 = 1_{\mathbb{R}^2}$ and $A^{-1}i = i_{\mathbb{R}^2}$. For example $A= \begin{pmatrix} 1 & -1 \\ 1 & 0\end{pmatrix}$. Then $1_{\mathbb{R}^2} = (0,-1)$ and $i_{\mathbb{R}^2} = (1,1)$.

Answer 2

No, you can check that if we set $(0,1)=a+b*i$ with $b \neq 0,$ than all the properties you require are satisfied.

In fact this field is isomorphic to $\mathbb{C}$ and it can be shown that up to isomorphism this is the only field structure on $\mathbb{R}^2$ that extends the field structure of $\mathbb{R}$ and respects the vector field structure.

This is because this field is a finite dimensional $\mathbb{R}-$ vector space and hence algebraic over $\mathbb{R}.$ And so over $\mathbb{C}$ as well. But $\mathbb{C}$ is algebraically closed by the fundamental theorem of algebra, so the only possibilities are $\mathbb{R}$ or $\mathbb{C}.$ However it can't be $\mathbb{R}$ because $\mathbb{R}$ has no subfields isomorphic to itself.

The choice to visualize $i$ as $(0,1)$ is purely for convenience as it gives a nice geometric interpretation to multiplication as rotation plus scaling.