I'm reading Rudin's book where he defines the complex field, initially as ordered pairs $(a,b)$. I'm trying to prove the assertion that $\mathbb{R}$ is a subfield of $\mathbb{C}$. I unfortunately haven't any background in field theory, but I believe that what I need to do is:
- embed $\mathbb{R}$ in $\mathbb{C}$
- make sure this embedding preserves the operations in $\mathbb{R}$
Is it correct that these are necessary and sufficient? With that said, here is my attempt.
We will write down an embedding of $\mathbb{R}$ in $\mathbb{C}$. Define $\varphi: \mathbb{R} \to \mathbb{C}$ by $x \mapsto (x,0)$. Notice that if $\varphi(x) = \varphi(y)$ for $x,y \in \mathbb{R}$, then $(x,0) = (y,0)$, hence $x = y$, so $\varphi$ is injective. Now, we show that $\varphi$ preserves the operations in $\mathbb{R}$. Let $x,y \in \mathbb{R}$. We have $$ \varphi(x+y) = (x+y,0) = (x,0) + (y,0) = \varphi(x) + \varphi(y), $$ and $$ \varphi(xy) = (xy,0) = (x,0)(y,0) = \varphi(x) \varphi(y), $$ as required.
Your idea is correct, but I'd suggest elaborating a bit on why the equality
$$(xy,0)=(x,0)\cdot (y,0)$$
should hold. Note that in the ordered pair definition of $\mathbb C$, we define multiplication by
$$(a,b)\cdot(c,d)=(ac-bd,bc+ad)\text{ for any }(a,b),(c,d)\in\mathbb R^2$$
and even though it is straightforward to deduce that $(xy,0)=(x,0)\cdot(y,0)$ from this definition, simply writing $(xy,0)=(x,0)\cdot(y,0)$ might fall short of some professors’/mathematicians’ standards for a rigorous proof because the form of the expression $(xy,0)$ doesn't immediately match that of $(xy-0\cdot0,0\cdot y+x\cdot 0)$. Fortunately, this is only one more line in your proof:
\begin{align*} (xy,0) &= (xy-0\cdot0,0\cdot y+x\cdot 0)\\ &=: (x,0)\cdot(y,0) \end{align*}