Real points of a complex curve
Solution 1:
You say that you Feel free to assume all these varieties are defined over $\mathbb{R}$ if it helps. If you don't assume that $a$ and $b$ are real, then there's no reason that $C$ should have real dimension one. For example, if $a = 0$ and $b$ is not real, then, equating imaginary parts, the only point on $C$ is $\infty$.
Even if $S$ is defined over $\mathbb{R}$, then your question is still not quite well defined, because classes in $H^1(S,\mathbb{Z})$ come with a natural orientation, but $C$ is simply a sub-manifold. If $C$ is connected, then this is just an issue of signs, but if $C$ is not connected then there is no a priori way to sum up the components with well defined signs.
One can still understand the topology of $C$, however. Suppose that $E$ is an elliptic curve over $\mathbb{R}$, and suppose that $S = E(\mathbb{C})$ and $C = E(\mathbb{R})$. The group law on $E$ makes $S$ and $C$ into topological groups, and $C$ is a closed subgroup of $S$. The complex points on an elliptic curve are, by Weierstrass, given by a complex torus $\mathbb{C}/\Lambda$. In particular, as a real topological group, $$S = S^1 \times S^1.$$ The real points form a closed subgroup. The closed $1$-dimensional subgroups of $S$ are all of the form $S^1 \times \mathbb{Z}/N \mathbb{Z}$ for some positive integer $N$. Using the Weil paring, one can show that not all the $N$-torsion points of $S$ are real for $N > 2$. Hence $C$ will be determined by $C[2] \subset S[2]$. Using the standard description of the two torsion of an elliptic curve, one has:
If $x^3 + a x + b$ has three real roots, then $C = S^1 \times \mathbb{Z}/2 \mathbb{Z}$.
If $x^3 + a x + b$ has exactly one real root, then $C = S^1$.
Here's another characterization. Write $S = \mathbb{C}/(\mathbb{Z} + \tau \mathbb{Z})$. Then the complex conjugate of $S$ replaces $\tau$ by $\overline{\tau}$, and $S$ is defined over $\mathbb{R}$ if and only if
$$\overline{\tau} = m + n \tau$$ For integers $m$ and $n$. Looking at imaginary parts forces $n = -1$. The only solutions to this, up to translation by integers, are given by $\tau \in i \mathbb{R}$ and $\tau \in \frac{1}{2} + i \mathbb{R}$.
If $\tau \in i \mathbb{R}$, then the real points in $S = \mathbb{C}/\Lambda$ are the images of the points in $\mathbb{R}$ and $\mathbb{R} + \tau/2$, and $C$ has two real components.
If $\tau \in \frac{1}{2} + i \mathbb{R}$, then the only real points in $S$ are given by the image of $\mathbb{R}$, and hence there is only one real component.
Yet one more characterization. In case $1$, it is clear that $\tau$ is equivalent to an element of the fundamental domain $\Omega$ of $\mathrm{PSL}_2(\mathbb{Z})$ to an element $\tau$ which is also of the form $i \mathbb{R}$ (by either doing nothing or by replacing $\tau$ by $-1/\tau$.) It follows that $q = e^{2 \pi i \tau} \ge e^{-2 \pi}$, and $j(\tau) \ge j(i) = 1728$. Hence we also have:
If $j(E) \ge 1728$, then $C$ has two components.
If $j(E) < 1728$, then $C$ has only one component.
(If you wonder why this will not be a contradiction by "writing down" a family of smooth elliptic curves and then using continuity, note that the difficulty of realizing $X(1)$ as a fine moduli space rather than a stack is exactly due to automorphisms arising at the points $j = 1728$ and $j = 0$.)
Returning to homology, note that, in the disconnected case, both $S^1$ classes are just translates of eachother, and so represent the same "class" in homology. So, writing $H_1(S,\mathbb{Z}) = \mathbb{Z}^2$, one could write $[C] = \pm (1,0)$ in the connected case, depending on the sign, and $[C] = \pm (1,0) \pm (1,0)$ in the disconnected case.
More generally, if you are interested in the topology of real curves, you should look at the following paper ("Real algebraic curves", Gross-Harris, 1981):
http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1981_4_14_2/ASENS_1981_4_14_2_157_0/ASENS_1981_4_14_2_157_0.pdf
In particular, take a look at section 3.