Complex Numbers form Subfield of Quaternions

Theorem
The field of complex numbers $\left({\C, +, \times}\right)$ is isomorphic to the subfields of the quaternions $\left({\mathbb H, +, \times}\right)$ whose underlying subsets are:


 * $(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: c = d = 0}\right\}$


 * $(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = d = 0}\right\}$


 * $(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = c = 0}\right\}$

That is:


 * $(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i \in \mathbb H}\right\}$


 * $(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + c \mathbf j \in \mathbb H}\right\}$


 * $(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + d \mathbf k \in \mathbb H}\right\}$

Proof
Let $\phi_i: \mathbb H_\mathbf i \to \C$ be defined as:
 * $\forall \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = a + b i$

where in this context $i$ is the imaginary unit.

Similarly we can define $\phi_j$ and $\phi_k$:


 * $\forall \mathbf x = \mathbf a \mathbf 1 + c \mathbf j \in \mathbb H_\mathbf j: \phi_j \left({\mathbf x}\right) = a + c i$


 * $\forall \mathbf x = \mathbf a \mathbf 1 + d \mathbf k \in \mathbb H_\mathbf k: \phi_k \left({\mathbf x}\right) = a + d i$

Proof of Bijectivity
First note that each of $\phi_i, \phi_j, \phi_k$ are bijections, as follows:


 * Injections:

Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$. Then:
 * $\phi_i \left({x_1}\right) = \phi_i \left({x_2}\right) \implies a_1 + b_1 i = a_2 + b_2 i \implies a_1 = a_2, b_1 = b_2 \implies \mathbf x_1 = \mathbf x_2$

and similarly with $\phi_j$ and $\phi_k$.


 * Surjections:

Let $x = a + b i \in \C$.

Then:
 * $\exists \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = x$

and similarly with $\phi_j$ and $\phi_k$.

Thus it is established that $\phi_i, \phi_j, \phi_k$ are bijections.

Proof of Morphism Property
Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$.

First we show that $+$ has the morphism property under $\phi_i$:

and similarly for $\phi_j$ and $\phi_k$.

Next we show that $\times$ has the morphism property under $\phi_i$:

Similarly for $\phi_j$ and $\phi_k$.

So we have shown that $\phi_i, \phi_j, \phi_k$ are ring homomorphisms.

So all of these mappings are bijective homomorphisms, that is, isomorphisms.