Concept question about the local defining function for the irreducible hypersurface $Y$ Let $Y$ be a irreducible hypersurface in $\Bbb{C}^n$, therefore we can find the local defining function, which is holomorphic $f\in \mathcal{O}(U)$ such that $V(f) = Y\cap U$.
I was confused can we choose $f_y$ to be the irreducible for all $y \in U$? I know for the regular set $Y_{\text{reg}}$, we have the coordinate representation therefore $Y$ as hypersurface is defined by $f = z_1$, therefore it's irreducible for all $y\in U$, Can we find such defining $f$ which is irreducible in a neighborhood other than the regular point?
 A: No, this is not possible, because the hypersurface $Y$ need not be locally irreducible. Consider the nodal curve
$$C = \{(x,y) : y^2 = x^2(x+1)\} \subset \mathbb C^2,$$
which looks like this:

You see that locally around the origin, $C$ consists of two components which meet. The defining equation $f = y^2 - x^2(x+1)$ factorizes in $\mathcal O_0 = \mathbb C\{x,y\}$ as
$$f = (y-gx)(y+gx),$$
where
$$g = \sqrt{1+x} = \sum_{n=0}^\infty \binom{\frac 1 2}n x^n$$
is a converging power series.

Now, depending on how much algebraic geometry you already know, note that this is an analytic phenomenon (we had to use a power series to write down the decomposition!). If you think of $C$ as an algebraic curve, then
$$C = \operatorname{Spec} \mathbb C[x,y] / (f).$$
Since $f$ is an irreducible polynomial, $C$ is an irreducible variety, and so the stalks $\mathcal O_{C,y}$ will be integral domains. In particular, $f$ will not decompose in the algebraic stalks $\mathcal O_{\mathbb C^2, y}$, otherwise you would get zero divisors in $\mathcal O_{C,y}$.
