# Generic point of closed subscheme meeting multiple irreducible components.

Suppose we have an affine, separated, integral $$k$$-scheme of finite type $$X = \text{Spec}(A)$$, and a regular sequence $$f_1, \dots, f_n \in \mathcal O_X(X)$$ and we consider the closed subscheme $$Z = V(f_1, \dots, f_n)$$. By an inductive argument $$Z$$ is of pure codimension $$n$$ in $$X$$. Take any generic point $$z$$ of $$Z$$.

If we write $$Y = V(f_1, \dots, f_{n-1})$$ (which is of pure codimension $$n-1$$ in $$X$$), I am interested in the points $$y \in Y$$ that 'lie above' $$z$$ in which $$z$$ has codimension $$1$$, i.e. such that $$z \in \overline{\{y\}}$$ and $$\text{codim}_{\overline{\{y\}}}(\overline{\{z\}}) = 1$$.

First of all, any such point $$y$$ must be a generic point of $$Y$$ (as its codimension is zero in some irreducible component, so $$\overline{\{y\}}$$ is already an irreducible component), hence we have $$\overline{\{z\}} = V(f_n) \cap \overline{\{y\}}$$. Now I am wondering if in general there might be multiple such irreducible components $$\overline{\{y\}}$$ containing $$z$$. My (very weak) intuition says yes, however if this is true: What additional assumptions would we need to make such that $$z$$ lies in only one irreducible component?

Thank you very much in advance!

EDIT: Closure of points $$\overline{\{x\}}$$ are considered with their integral subscheme structure.

## 1 Answer

If I'm understanding your question correctly, it seems to me that your question basically boils down to given $$Y$$, what conditions should you impose on $$f_n$$ such that the zero locus of $$f_n$$ is contained in a unique irred. component of $$Y$$. Note that if an irreducible component of $$Y$$ contains the generic point of $$z$$, then it contains $$Z$$ (with its reduced scheme structure).

As you said, this is definitely possible. For example, consider $$A = k[x,y]$$, $$f_1 = xy$$, $$f_2 = y-x$$. Then $$Z = V(xy,y-x)$$. This cuts out $$(0,0)$$ (with a non-reduced scheme structure), and $$(0,0)$$ (with its reduced scheme structure) is clearly contained in two irreducible components of $$V(xy)$$.

Let's call $$Y = \operatorname{Spec}(B)$$ and denote its irreducible components by $$Y_i = V(\mathfrak{p}_i)$$, where $$\mathfrak{p}_i$$ is a minimal prime ideal of $$Y$$. Then $$\overline{\{z\}} \subseteq Y_i = V(\mathfrak{p}_i)$$ if and only if $$\mathfrak{p}_i \subseteq \sqrt{(f_n)}$$. So for $$z$$ to lie in an unique irreducible component you want the radical of $$(f_n)$$ to contain exactly one minimal prime.

• Dear loch, thank you very much for your answer! To reformulate it a bit: $z$ lies exactly in one irreducible component if and only if the stalk $\mathcal O_{Y,z}$ is an integral domain, because otherwise the radical would contain more than one minimal prime, correct? Commented Jun 22, 2019 at 9:27