What is the "standard basis" for fields of complex numbers? What is the "standard basis" for fields of complex numbers?
For example, what is the standard basis for $\Bbb C^2$ (two-tuples of the form: $(a + bi, c + di)$)?  I know the standard for $\Bbb R^2$ is $((1, 0), (0, 1))$.  Is the standard basis exactly the same for complex numbers?
P.S. - I realize this question is very simplistic, but I couldn't find an authoritative answer online.
 A: The "most standard" basis is also $\left\lbrace(1,0),\, (0,1)\right\rbrace$. You just take complex combinations of these vectors. Simple :)
A: Just to be clear, by definition, a vector space always comes along with a field of scalars $F$.  It's common just to talk about a "vector space" and a "basis"; but if there is possible doubt about the field of scalars, it's better to talk about a "vector space over $F$" and a "basis over $F$" (or an "$F$-vector space" and an "$F$-basis").
Your example, $\mathbb{C}^2$, is a 2-dimensional vector space over $\mathbb{C}$, and the simplest choice of a $\mathbb{C}$-basis is $\{ (1,0), (0,1) \}$.
However, $\mathbb{C}^2$ is also a vector space over $\mathbb{R}$.  When we view $\mathbb{C}^2$ as an $\mathbb{R}$-vector space, it has dimension 4, and the simplest choice of an $\mathbb{R}$-basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.
Here's another intersting example, though I'm pretty sure it's not what you were asking about:
We can view $\mathbb{C}^2$ as a vector space over $\mathbb{Q}$.  (You can work through the definition of a vector space to prove this is true.)  As a $\mathbb{Q}$-vector space, $\mathbb{C}^2$ is infinite-dimensional, and you can't write down any nice basis.  (The existence of the $\mathbb{Q}$-basis depends on the axiom of choice.)
A: I'm not sure that this is what you want, but under the usual Argand-Gauss identification $\Bbb C=\Bbb R^2$ the standard basis of $\Bbb C$ would be $\{1,i\}$, the standard basis of $\Bbb C^2$ would be $\{(1,0),(i,0),(0,1),(0,i)\}$ and so on.
