How does $\dim \mathbb C$ work? In the Wikipedia page about Dimension (vector space), it says the dimension of complex numbers is 2 or 1 if it's complex or real vector space respectively. How does that work? How to I describe $\mathbb C$ in terms of one or two bases?
 A: If you consider $\mathbb C$ as an $\mathbb R$-vector space you need two base vectors for example $1$ and $i$. If you only had one of them you could not 'reach' all elements of $\mathbb C$ (= vector space) by multiplying this one vector by a real number (scalar of the field).
If you consider $\mathbb C$ as a $\mathbb C$-vector space, one vector is enough for spanning $\mathbb C$ (= vector space). For example if you take $1$ as base vector, you can get to every point in $\mathbb C$ (=vector space)by multiplying it by the corresponding scalar of $\mathbb C$ (= field).
A: You specify a point in $\mathbb C$ by specifying two real numbers or one complex number.
If "scalars" are real numbers, then you can take $\{1,i\}$ to be a basis, and you specify a complex number as a linear combination of those: $x+iy$, with two scalars $x$ and $y$, each of which is real.
If "sccalars" are complex numbers, then you can take $\{1\}$ to be a basis, and you specify a complex number as $z\cdot1$, with one scalar $z$, which is complex.
A: A vector space is described over a particular field $K$. It's finite dimensional if there is a finite set of symbols $b_i$ such that every element $v$ in the vector space is able to be written as $v = \displaystyle \sum_i k_i b_i$, where $k_i \in K$.
In particular, the coefficients $k_i$ are in $K$. So it matters very much what field we are thinking of the vector space as lying over. 
If we think of $\mathbb{C}$ over $\mathbb{C}$, then every element can be written as $z = z\cdot 1$, so that the single element $1$ is a basis. If we think of $\mathbb{C}$ over $\mathbb{R}$, then every element can be written as $a + bi$ where $a,b \in \mathbb{R}$. So it's fundamentally two-dimensional. If we think of $\mathbb{C}$ over $\mathbb{Q}$, then there is no finite basis -- this is an infinite dimensional vector space.
A: You have that backwards.
If we consider $\Bbb C$ as a vector space over the reals, then we have an isomorphism $\Bbb C\to\Bbb R^2$ given by $x+iy\mapsto\langle x,y\rangle.$ So, as a vector space over the reals, $\Bbb C$ acts just like $\Bbb R^2,$ meaning that $\dim_{\Bbb R}\Bbb C=2.$
On the other hand, we clearly have $\dim_{\Bbb C}\Bbb C=1.$ This is because for any fixed non-$0$ $v\in\Bbb C,$ we have for all $w\in\Bbb C$ that there is some unique $z\in\Bbb C$ such that $zv=w$--that is, $\{v\}$ is a basis for $\Bbb C$ over $\Bbb C$.
In fact, for any field $K\subseteq\Bbb C,$ we have that $\Bbb C$ is a vector space over $K.$ However, if $K\ne\Bbb R$ and $K\ne\Bbb C,$ then we will necessarily have that $\dim_K\Bbb C$ is infinite.
