$\mathbb{C}$ 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.

To answer your question of why the dimension of $\Bbb C$ as a vector space over itself is $1$:
Do you agree that the dimension of a vector space is the same as the number of basis elements?  So to find the dimension of a vector space over a scalar field, you just have to find a basis for the vector space and then the dimension will be the number of elements in the basis.
Well, if $S$ is a vector space and $F$ is our scalar field, then a basis of $S$ (over $F$) is a set of elements $\{s_{1}, s_{2}, \dots, s_{n} \} \subseteq S$ such that the set is linearly independent, and for each $s \in S$, we can find scalars $f_{1}, f_{2}, \dots, f_{n} \in F$ such that $s = \sum \limits_{i = 1}^{n} f_{i}s_{i}$.
Now that we have that out of the way, replace the above stuff with $S = \Bbb C$ and $F = \Bbb C$. The dimension of $S$ over $F$ is a collection of elements in $S$ such that for each $s \in S$, we can write it as a linear combination of the collection of elements, and the collection of elements is linearly independent.
Here is my candidate for a basis of $\Bbb C$ over $\Bbb C$.  It is $\{ 1 \}$.  Don't believe me that this is a basis?  Check the two conditions:
1) Is it a linearly independent set?  Well, for which scalars $c \in \Bbb C$ do we have $c \cdot 1 = 0$?  Only $c = 0$, so yes, it is a linearly independent set.
2)  Does it span $\Bbb C$ (over the field of scalars $\Bbb C$)?  Well, let $\alpha \in C$.  Can we find a scalar $c \in \Bbb C$ such that $\alpha = c \cdot 1$?  Yes, just let $c = \alpha$.  Obviously, $\alpha \cdot 1 = \alpha$.  So $\{1 \}$ spans $\Bbb C$ as a vector space over itself.
Since $\{ 1 \}$ satisfies the criteria to be a basis of $\Bbb C$ over itself, and since the set only has one element, then $\Bbb C$ has dimension $1$ as a vector space over itself.
