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.