Singular irreducible affine plane curve is never a Riemann surface?

Question

Let $f\in \mathbb{C}[x,z]$ be a bivariate irreducible polynomial over the complex numbers. In case that $f$ is non-singular, one can endow the locus (zero set) of $f$ (considered with subspace topology) with the structure of a Riemann surface using the Implicit Function Theorem for holomorphic functions. What can we say in the case that $f$ is singular? In this case, is the locus (considered with subspace topology) never a Riemann surface?

Answer 1

A simple answer is that, at singular points the curve will not have structure of a Riemann surface, as is seen from Jacobian criterion (whose formulation is the same for algebraic vs. complex analytic settings). See e.g., Mumford's Red Book, III-4, Corollary 2.

But, from the way of your questioning, it might be better to go further: you fixed the (Euclidean) topology on the zero set $X$, but not the sheaf of holomorphic functions. What is hidden in the simple answer above is the choice of a structure sheaf $\mathcal{O}_X$. Usually $\mathcal{O}_X$ is naturally and implicitly assumed to be the quotient $\mathcal{O}_{\mathbb{C}^2}/\mathcal{I}_X$, where $\mathcal{I}_X$ is the subsheaf (ideal) of $\mathcal{O}_{\mathbb{C}^2}$ whose sections are functions vanishing along $X$. This is the canonical choice of the structure sheaf of $X$ and used in Jacobian criterion, and even in the definition of the associated analytic space of $X$. (In this sense, the simple answer should be considered as definitive.)

Based on the paragraph above, there may be a more technical answer: Sometimes we can choose a (non-canonical, not natural) structure sheaf of $X$, with which $X$ acquires a structure of a Riemann surface. I will exhibit two famous examples.

(I) Consider $X=\{y^2=x^3\}\subset \mathbb{C}^2$. The map $$\varphi \colon \mathbb{C}\rightarrow X,\ t\mapsto (t^2,t^3)$$ is a homeomorphism. (Its inverse is given by $(x,y)\mapsto y/x$ off origin, and is continuous at the origin by seeing $(y/x)^3=y\rightarrow 0$ as $(x,y)\rightarrow (0,0)$.) Therefore, we can put on $X$ the structure sheaf $\varphi_* \mathcal{O}_{\mathbb{C}}$ and the resulting Riemann surface structure on $X$ is trivially isomorphic to $\mathbb{C}$. Note again that this structure sheaf is different from what is called the associated structure of complex analytic space of $X$.

(II) Consider $X=\{y^2=x^2+x^3\}$. In this case, we cannot endow $X$ with a structure sheaf so that it becomes a Riemann surface, just because $X$ is not even topologically a manifold at the origin. To see it, let $D_{\varepsilon}$ be a small ball centered at $(0,0)$, namely $$D_{\varepsilon}=\{(x,y)\in \mathbb{C}^2\mid |x|^2+|y|^2 < \varepsilon^2\}.$$ The intersection $D_{\varepsilon}\cap X$ for small positive $\varepsilon$ can be written as $$ D_{\varepsilon}\cap X=\{(x, xz_+)\in D_{\varepsilon}\}\cup \{(x,xz_-)\in D_{\varepsilon}\}, $$ where $z_{\pm}$ denotes the two branches of the complex square root $z=\sqrt{1+x}$ near $x=0$, namely $$\begin{split} z_+&= 1+\frac{1}{2}x-\frac{1}{8}x^2+\cdots,\\ z_-&= -1-\frac{1}{2}x+\frac{1}{8}x^2-\cdots=-z_+. \end{split}$$ Since $xz_+=xz_-$ happens only if $x=0$, the punctured small neighborhood $D_{\varepsilon}\cap X-\{(0,0)\}$ at $(0,0)\in X$ decomposes into a disjoint union $\{(x,xz_+)\mid x\neq 0\}\sqcup \{(x,xz_-)\mid x\neq 0\}$ for all small positive $\varepsilon$. This shows that every small punctured neighborhood of $(0,0)\in X$ decomposes into two connected components, which cannot happen for a 2-dimensional topological manifold. Namely, $X$ is not a topological manifold at the origin.