$\mathbb{C}$ is a one-dimensional complex vector space. What is its dimension when regarded as a vector space over $\mathbb{R}$? $\mathbb{C}$ is a one-dimensional complex vector space. What is its dimension when regarded as a vector space over $\mathbb{R}$?
I don't understand how $\mathbb{C}$ is one-dimensional. Please help me understand that.
Also, I'm pretty sure that when the field is reals we have $\dim(\mathbb{C})=2$. Since when $\alpha$ is real and $z=a+bi$ is complex we have $\alpha z=\alpha a+\alpha bi=\alpha a(1,0)+\alpha b(0,i)$.
How does this look? Any solutions or help is greatly appreciated.

$\mathbb C$ as an $\mathbb C$-vector space is one dimensional. Any field as a vector space over itself is one dimensional. It has a basis of $\{ 1 \}$. 
To see this, consider an element $f \in \mathbb F$ in a field $\mathbb F$. Write $\vec f$ when we are considering $f$ as a member of the vector space. Then as we can write $\vec f = f \vec 1$ for all $\vec f \in \mathbb F$, the multiplicative identity $\vec 1$ is in itself a perfectly good basis for $\mathbb F$.

Using that notation for $\mathbb C$ as an $\mathbb R$-vector space, we can write any vector $\vec z \in \mathbb C$ is
$$\vec z = a\vec 1 + b \vec i \quad \text{ with } a, b \in \mathbb R$$ 
Hence $\{ \vec 1, \vec i \}$ is a basis.

As for why the dimension of $\mathbb F^n$ is $n$, as you ask in the comments: that's because there is a basis $\{ (1,0,\cdots, 0), (0, 1, \cdots, 0), \cdots, (0,0 , \cdots, 0, 1) \}$ with $n$ elements. Hence by definition of dimension
$$\dim(\mathbb F^n) = n$$
