# Looking for elementary proof that irreducible/smooth curve in $\mathbb C^2$ is connected in Euclidean topology of $\mathbb C^2$

Let $f(X,Y)\in \mathbb C[X,Y]$ be an irreducible polynomial. I know that the zero set of $f$ , $V(f):=\{(a,b)\in \mathbb C^2 : f(a,b)=0\}$ is connected in the usual Euclidean topology of $\mathbb C^2$ . Is there a proof of this result without using too much algebraic geometry ? What are the most elementary proof of this result ?

If we assume that $V(f)$ is smooth i.e. $$\forall p \in \mathbb C^2\colon\quad(\partial f/\partial X , \partial f/\partial Y)|_p\ne (0,0),$$ can we make a proof substantially more elementary ?

• Will you share a reference for an advanced proof of this result? – Matt F. Apr 2 '18 at 1:36
• @MattF.: You may take a look here math.stackexchange.com/questions/242535/… ... zero set of irreducible polynomial over algebraically closed field is an irreducible algebraic set in Zariski topology, hence connected in Zariski topology. – user Apr 2 '18 at 15:31
• Coincidentally Terry Tao recently posted notes on this at the end of terrytao.wordpress.com/2018/03/28/… – Dap Apr 4 '18 at 4:56

How about this? By Noether normalization, after a generic change of coordinates, our curve $C$ is of the form $$y^d + f_1(x) y^{d-1} + \cdots + f_d(x)=0$$ for various polynomials $f_j$.

Let $\Delta(x)$ be the discriminant of this polynomial as a polynomial in $y$. Since $f$ is squarefree as a polynomial in $\mathbb{C}[x,y]$, we know that $\Delta(x)$ is not identically zero. So it has finitely many zeroes $z_1$, $z_2$, ..., $z_N$. Let $U = \mathbb{C} \setminus \{ z_1, z_2, \ldots, z_N \}$. For $x \in U$, there are $d$-distinct roots $y$ of the above polynomial.

Let $\pi$ be projection onto the $x$-coordinate. Then the implicit function theorem shows that $\pi^{-1}(U) \to U$ is a $d$-fold covering map. We want to show $\pi^{-1}(U)$ is connected. Suppose not. Then it breaks into components $Y_1$, $Y_2$, ..., $Y_r$ which cover $U$ with degrees $k_1$, $k_2$, ..., $k_r$ where $\sum k_i = d$.

Choose a component $Y$ of $\pi^{-1}(U)$, with $Y \to U$ of degree $k$. For $1 \leq j \leq k$ and $x \in U$, let $e_j(x)$ be the $j$-th elementary symmetric function in the $y$-coordinates of the $k$ points of $Y \cap \pi^{-1}(x)$.

The function $e_j$ is holomorphic on $U$. At the punctures $z_1$, ..., $z_N$, the function $e_j$ is bounded, since $C \to \mathbb{C}$ is proper. So, by the Riemann extension theorem, $e_j$ extends to a holomorphic function on all of $\mathbb{C}$. Also, $e_j$ is a polynomial in algebraic (multivalued) functions, so its growth rate as $|x| \to \infty$ is bounded by $|x|^N$ for some $N$. We see that $e_j(x)$ is an entire function bounded by a polynomial, so it is a polynomial. Thus, $y^k - e_1(x) y^{k-1} + \cdots \pm e_k(x)$ is a polynomial of degree $k$ vanishing on $Y$.

We can repeat this for each component $Y_i$ and get a polynomial $g_i(x,y)$ with $y$-degree $k_i$ vanishing on the corresponding points of the cover $\pi^{-1}(U) \to U$. Then $\prod g_i(x,y)$ vanishes on all of $\pi^{-1}(U)$, and hence is divisible by $f$. Since $f$ and $\prod g_i(x,y)$ have the same $y$-degree, they are proportional. This gives a nontrivial factorization of $f$, giving our contradiction.

• Very nice answer! I am trying to understand your proof: $U$ is connected. The connected components of $\pi^{-1}(U)$ are covers of $U$. Each produces a polynomial and their product is the original one. It must be only one, so $\pi^{-1}(U)$ is connected. $C$ is the closure of $\pi^{-1}(U)$ so connected. – Orest Bucicovschi Apr 2 '18 at 18:45
• @orangeskid That's right. I'm thinking I introduced a red herring by bringing in the symmetric group explicitly; I might edit to make it more geometric later if I get a chance. – David E Speyer Apr 2 '18 at 22:30
• Removed references to covering groups and orbits in place of more geometric language. – David E Speyer Apr 3 '18 at 2:58