$\mathbb{A}^1_\mathbb{C}$ and $\mathbb{A}^2_\mathbb{R}$ I wonder if there's a connection between $\mathbb{A}^1_\mathbb{C}$ and $\mathbb{A}^2_\mathbb{R}$ since one can also see $\mathbb{C}$ as $\mathbb{R}^2$. I'm just looking for general thoughts about this. Sorry if this question is not specific enough. 
Can we say that the topology on $\mathbb{A}^1_\mathbb{C}$ is coarser than the topology on $\mathbb{A}^2_\mathbb{R}$? My idea:
Let $z \in \mathbb{C}$ be in $V(f)$ for a polynomial $f = c_0 + ... + c_nX^n$ in $\mathbb{C}[X]$. Write $z = x + iy$ for $x,y \in \mathbb{R}$ and similarly $c_i = a_i + i b_i$. Then 
$$c_0 + ... + c_nz^n = 0 \Leftrightarrow a_o + i b_0 + ... + (a_n + i b_n)(x + iy)^n = 0 \Leftrightarrow f_r(x,y) + i f_i(x,y) = 0$$
for some real polynomials $f_r$ and $f_i$. So 
$$z \in V(f) \Leftrightarrow (x,y) \in V(f_r, f_i).$$ Showing that $V(f)$ is also closed in $\mathbb{A}^2_\mathbb{R}$. 
 A: We know that using the classical topology, $\mathbb{R}^2\cong \mathbb{C}$. Indeed, these spaces are isomorphic as real vector spaces. However, if we use the Zariski topology, $\mathbb{A}^2_{\mathbb{R}}$ and $\mathbb{A}^1_{\mathbb{C}}$ are quite different. Indeed, $\mathbb{A}^1_{\mathbb{C}}$ has the finite complement topology, whereas this is not the case for $\mathbb{A}^2_{\mathbb{R}}$. 
Of course, the differences between $\mathbb{A}^2_{\mathbb{R}}$ and $\mathbb{A}^1_{\mathbb{C}}$ are more substantial than just the non-homeomorphic topologies. They have distinct dimensions as topological spaces (in the algebraic sense) -  the former is dimension $2$ while the latter is dimension $1$. Lastly, at the level of rings of functions, we see that the affine coordinate ring of $\mathbb{A}^2_{\mathbb{R}}$ is $\mathbb{R}[x,y]$ while the affine coordinate ring of $\mathbb{A}^1_{\mathbb{C}}$ is $\mathbb{C}[x]$. The significant differences in these rings are reflected in the significant differences in the topologies of the spaces.
It's worth mentioning that $\mathbb{A}^1_{\mathbb{R}}$ is homeomorphic to $\mathbb{A}^1_{\mathbb{C}}$ since as sets they have the same cardinality and are equipped with the finite complement topology. Also, comparing rings of functions we see that $A(\mathbb{A}^1_{\mathbb{R}})=\mathbb{R}[x]$ while $A(\mathbb{A}^1_{\mathbb{C}})=\mathbb{C}[x]$. Since $\mathbb{R}[x]\otimes_{\mathbb{R}}\mathbb{C}\cong \mathbb{C}[x]$, we see that these spaces in some sense differ only by a "base change."
So, actually the real affine line is closer to the complex affine line in this context.
