$C = \{(a,-b,b,a)\ | a,b \in \mathbb{R} \} \subset \mathbb{R}^4$ with
$(+): (a,-b,b,a) + (c,-d,d,c) = (a+c,-(b+d),b+d,a+c)$
$(\cdot): (a,-b,b,a) \cdot (c,-d,d,c) = (ac-bd, -(bc+ad), bc+ad, ac-bd)$
Prove $(C,+,\cdot)$ is an isomorphic field to $\mathbb{C}$
So to prove:
1) $(C,+)$ is a commutative group
2) Associative property for $(\cdot)$
3) Distributive property
4) Commutative property for $(\cdot)$
5) Neutral element for$(\cdot)$
6) Inverse element for $(\cdot)$
After 1)-6) we have proven that it is a field.
7) Isomorphism: $(C,+_1, \cdot_1)$ and $(\mathbb{C},+_2, \cdot_2)$, $f: C \rightarrow \mathbb{C}$
To show: $f((a,-b,b,a) +_1 (c,-d,d,c)) = f((a,-b,b,a)) +_2 f((c,-d,d,c))$ and $f((a,-b,b,a) \cdot_1 (c,-d,d,c)) = f((a,-b,b,a)) \cdot_2 f((c,-d,d,c))$
I've already proven 1,3,4,5 and 6. But I have a mistake somewhere in 2 and I'm not sure how to do 7.
2) $[(a,-b,b,a) \cdot (c,-d,d,c)] \cdot (f,-g,g,f) \\=(ac-bd, -(bc+ad), bc+ad, ac-bd) \cdot (f,-g,g,f)\\ =((ac-bd) \cdot f-(bc+ad)\cdot g, -((bc+ad)\cdot f + (ac-bd) \cdot g), \\ (bc+ad)\cdot f + (ac-bd) \cdot g, (ac-bd) \cdot f-(bc+ad)\cdot g) \\ =(acf-bdf-bcg-adg, -(bcf+adf+acg-bdg), \\ bcf+adf+acg-bdg, acf-bdf-bcg-adg) \\ = (a \cdot (cf-dg) -b \cdot (df+cg), -(b \cdot (cf-dg) + a \cdot (df+cg)), \\ b \cdot (cf-dg) + a \cdot (df+cg), a \cdot (cf-dg) -b \cdot (df+cg)) \\ = (a,-b,b,a) \cdot (cf-dg-(df+cg), -(cf-dg+df+cg), \\ cf-dg+df+cg, cf-dg-(df+cg)) \\ = (a,-b,b,a) \cdot (c \cdot (f-g)-d \cdot (g+f), -(c \cdot (f+g) + d \cdot (f-g)), \\ c \cdot (f+g) + d \cdot (f-g), c \cdot (f-g)-d \cdot (g+f)) \\ = (a,-b,b,a) \cdot [(c,-d,d,c) \cdot (f-g-g-f, -(f+g+f-g), \\ f+g+f-g, f-g-g-f)] \\ = (a,-b,b,a) \cdot [(c,-d,d,c) \cdot (-2g, -(2f), 2f, 2g)]$
I've checked it through but I can't find the mistake.
For 7):
$f((a,-b,b,a) +_1 (c,-d,d,c)) = f((a+_1c,-(b+_1d),b+_1d,a+_1c)) = ???$
How do I continue from here?
Best Answer
You can of course prove 1. - 6. by tedious computations. However, this is unnecessary. Simply define $$f: C \to \mathbb C, f(a,-b,b,a) = a + ib .$$ This is obviously a bijection and you can very easily verify that $f(\xi + \eta) = f(\xi) + f(\eta)$ and $f(\xi \cdot \eta) = f(\xi) \cdot f(\eta)$. This immediately implies that $(C,+,\cdot)$ is a field and $f$ is an isomorphism of fields.
Also note that you can identify $C$ with the set
$$C' = \left\lbrace\begin{pmatrix} a & -b\\b & a \end{pmatrix} \mid a,b \in \mathbb R \right\rbrace$$ of real $2 \times 2$-matrices via the association $(a,-b,b,a) \mapsto \begin{pmatrix} a & -b\\b & a \end{pmatrix}$. Under this association addition and mutiplication in $C$ correspond to matrix addition and matrix mutiplication. It is well-known that the set $C'$ is an alternative "model" of $\mathbb C$, and some authors have used it to introduce $\mathbb C$. To do so, simply show that $C'$ is closed wit respect to additon and multiplication of matrices which immediately shows that $C'$ is a ring (since the set $M_2(\mathbb R)$ of all real $2 \times 2$-matrices is one). The identity matrix is the multiplicative identity, and since each nonzero matrix in $ \mathcal A \in C'$ has determinant $\det(\mathcal A) = a^2 + b^2 > 0$, it has an inverse which is easily seen to be in $C'$. Finally you can show that multiplication is $C'$ is commutative. Therefore $C'$ is a field.