calculus - the set of all complex numbers constitutes a one-dimensional complex vector space.

Show that the set of all real numbers, with the usual addition and
multiplication, constitutes a one-dimensional real vector space, and the
set of all complex numbers constitutes a one-dimensional complex
vector space.

I know that all properties to be vector space are fulfilled in real and complex but I have difficulty is in the dimension and the base of each vector space respectively. Scalars in the vector space of real numbers are real numbers and likewise with complexes? The basis for both spaces is $\{1\}$ or for the real ones it is $\{1\}$ and for the complexes it is $\{i\}$? How does one prove that these spaces have dimension $1$?
Thank you very much.

Answer

The set of real numbers is one dimensional over the reals (meaning our scalars are real) and the set of complex numbers is one dimensional over the complex numbers (meaning that our scalars are complex).

Why is $\{1\}$ a basis for each?

Well, consider any real number $x$. Then we simply choose the scalar $x$ and basis element $1$ to have $$x = x \cdot 1.$$

Similarly, for any complex number $z$, choose the complex scalar $z$ and basis element $1$ to get $$z = z \cdot 1.$$

Then it follows that $\{1\}$ spans $\mathbb{R}$ in the first case and it spans $\mathbb{C}$ in the second.

I leave it to you to see that a singleton set is always linearly independent (assuming the Singleton is non-zero).

Does this make sense?

Note: There are infinitely many bases for each vector space but the important thing is that every basis will have the same cardinality. In this case, any basis will consist of $1$ element. If we take the complex numbers with scalars from the complex plane then we could take $\{i\}$ as a basis since for any $z \in \mathbb{C}$, we have $$z = (-i z) \cdot i .$$ So, $\{i \}$ spans $\mathbb{C}$ in this case.