Need help: Isomorphism vector spaces

Question

For $a+bi\in\mathbb{R}(i)$, we have that $a+bi=a\times1+b\times i$. Then, if $a+bi=0$, then $a=0=b$. Thus $\{1, i\}$ is a basis of $\mathbb{R}(i)$.

Then the isomorphism $a+bi\to(a, b)$ is what needs to be proven... I am working on making this more detailed because I think I am still being too vague. Could anyone help me out with finishing this out?

Answer 1

What you did is fine. And there is really not much left. The map$$\begin{array}{ccc}\mathbb R(i)&\longrightarrow&\mathbb R^2\\a+bi&\mapsto&(a,b)\end{array}$$is an isomorphism precisely because $a+bi=0\iff a=0\wedge b=0$.

Answer 2

What you did is to make a big assumption: every element of $\mathbb{R}(i)$ can be written as $a+bi$. This is part of the question -- that $\{1,i\}$ actually spans $\mathbb{R}(i)$ as an $\mathbb{R}$-vector space.

Compared with a similar situtation $\mathbb{Q}(\sqrt[3]2)$. Here $\{1,\sqrt[3]2\}$ is not a basis of $\mathbb{Q}(\sqrt[3]2)$ as a $\mathbb{Q}$-vector space. So there is something to prove here.