Isomorphism between C and another ring

Question

Let the operations of addition and multiplication on the set $K = {at+bu : a,b ∈ R}$, where $t$ and $u$ are formal symbols, be defined as follows:

$(at+bu)+(ct+du) = (a+c)t+(b+d)u$,

$(at+bu)·(ct+du) = (ac+ad+bc−bd)t+(−ac+ad+bc+bd)u$

Specify a bijection $f : C → K$ such that $f(α+β) = f(α)+ f(β)$ and $f(αβ) = f(α)f(β)$ for all complex numbers $α$ and $β$. [Such a bijection is called an isomorphism of rings.]

Is there a quick way-trick-golden rule to find such a bijection easily? I've tried several linear functions (since they will always satisfy the additive property and are also easy to prove bijective), but multiplication goes haywire whatever I try. Mind you that this is an exam question, so trial and error is not an efficient way to solve it, especially when the calculations take so much time out of the 2 hours.

Answer 1

Hint: You need to find complex numbers (for the preimages of) $t, u$ such that $$tu=t+u\\ t^2=t-u\\ u^2=-t+u$$

Answer 2

As a first step, you can search for an element $at+bu\in K$ such that$$(\forall ct+du\in K):(at+bu)(ct+du)=(ct+du)(at+bu)ct+du.$$After a few computations (Linear Algebra stuff), you will get $a=b=\frac12$. So, your ring isomorphism should map $1$ into $\frac12u+\frac12v$.

The second step would consists in finding an element $ct+du\in K$ whose square is $-\left(\frac12t+\frac12u\right)$. Afer a few more computations, you will be able to prove that $-\frac12t+\frac12u$ has that property.

So, define$$\begin{array}{rccc}\varphi\colon&\mathbb C&\longrightarrow&K\\&a+bi&\mapsto&a\left(\frac12t+\frac12u\right)+b\left(-\frac12t+\frac12u\right)\end{array}$$and prove that it is indeed an isomorphism.