Deriving the addition and product on $\mathbb C$ I am studying Stillwell's Elements of Algebra. In Chapter 3, Section 3.6, he writes

There is a unique extension of $+$, $-$, $\times$, $\div$ to $\mathbb C$ satisfying the field properties since if these properties hold we necessarily have
\begin{align*}
(\alpha_1 + i\beta_1) + (\alpha_2+ i\beta_2) & = (\alpha_1 + \alpha_2) + i(\beta_1 + \beta_2)\text{,}\\
(\alpha_1 + i\beta_1)(\alpha_2 + i\beta2) & = (\alpha_1\alpha_2 - \beta_1\beta_2) + i(\alpha_1\beta_2 + \beta_1\alpha_2)
\end{align*}
(using the fact that $i^2 = -1$).

Now what I understood from this is the following.
If we try to make $\mathbb R\times\mathbb R$ a field such that for each $\alpha_1, \alpha_2\in\mathbb R$, we have

*

*"extension" of addition and multiplication from $\mathbb R$ so that

*

*$(\alpha_1, 0) + (\alpha_2, 0) = (\alpha_1 + \alpha_2, 0)$, and

*$(\alpha_1, 0)(\alpha_2, 0) = (\alpha_1\alpha_2, 0)$,



*"$i^2 = -1$", that is,

*

*$(0, 1)(0, 1) = (-1, 0)$,



then we must have the familiar definitions of the addition and product rules for $\mathbb C$. (Note that I have denoted the addition and multiplication in $\mathbb R$ and $\mathbb{R\times R}$ by same notations.)
I have realized that I just need to show that for any $(\alpha, \beta)\in\mathbb{R\times R}$, we have $(\alpha, \beta) = (\alpha, 0) + (0, 1)(\beta, 0)$.
Question: Can someone please help showing this? Thanks for your time!
 A: The claim you are trying to prove is certainly not true.  Indeed, the assumptions you have made impose no restrictions whatsoever on what the operations are for elements that are not of the form $(a,0)$ or $(0,1)$ (besides just the field axioms).  So, for instance, you could take any bijection $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}\times\mathbb{R}$ which satisfies $f(a,0)=(a,0)$ for all $a$ and $f(0,1)=(0,1)$ (but could wildly permute the ordered pairs that are not of this form) and transport the usual complex number operations along this bijection $f$ to get a different field structure on $\mathbb{R}\times\mathbb{R}$ which satisfies your requirements.  Very concretely, for instance, your field operations could be the usual ones except that the elements $(1,1)$ and $(1,2)$ are swapped (so the operations are computed as though $(1,1)$ was $1+2i$ and $(1,2)$ was $1+i$, instead of the other way around). You could in fact get such a field structure that is not even isomorphic to the field structure of $\mathbb{C}$, since you could instead choose a bijection between $\mathbb{R}\times\mathbb{R}$ and some other field like $\mathbb{C}(x)$ of the same cardinality which preserves the copy of $\mathbb{R}$ and maps $(0,1)$ to a square root of $-1$.
Instead, Stillwell's claim should be interpreted to be assuming that $(\alpha, \beta) = (\alpha, 0) + (0, 1)(\beta, 0)$ is also required to be true.  The idea here is that the notation $\alpha+i\beta$ implicitly assumes that this formal object (which you represent as an ordered pair $(\alpha,\beta)$) should actually be equal to the sum of $\alpha$ and the product $i\cdot \beta$ with respect to the field operations.  Or, in your ordered pair notation, it should be equal to $(\alpha,0)+(0,1)(\beta,0)$.
A: They simply write $\alpha+i\beta$ for the pair $(\alpha,\beta)$. This seems to already presume the statement you want to prove, nevertheless it gives the right motivation for abstractly defining the operations.
With your notation, the rule for product translates as
$$(\alpha_1,\beta_1)\cdot(\alpha_2,\beta_2)=(\alpha_1\alpha_2-\beta_1\beta_2,\ \alpha_1\beta_2+\alpha_2\beta_1)$$
So, in particular it implies
$$i\cdot\beta=(0,1)\cdot(\beta,0)=(0,\beta)$$
and thus, $\alpha+i\cdot\beta=(\alpha,0)+(0,\beta)=(\alpha,\beta)$ indeed.
