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.
 A: 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. 
