Question regarding irreducible components of affine variety $(\cup W_j) \cap Z$ Let $W = \cup W_j$ be an affine variety in $\mathbb{A}_{\mathbb{C}}^n$
where the $W_j$'s are the irreducible components. Let $Z$ be another affine variety, and let $X$ be an irreducible component of $Z \cap W$. Is it then the case that $X$ must be an irreducible component of $W_j \cap Z$ for one of the $j$'s?
I would appreciate any comments or explanations!
Thank you very much!
PS Here I don't assume irreducible when I say variety.  
 A: $\newcommand{\ideal}[1]{\mathfrak{#1}}$
Let $A=\mathbb{C}[x_1,\ldots,x_n]$ and $W_j = V(\ideal{p}_j)$ with prime ideals $\ideal{p}_j \subseteq A$. Furthermore $Z = V(\ideal{a})$ so that $W = V(\bigcap_j \ideal{p}_j)$ and $Z \cap W = V(\ideal{a} + \bigcap_j \ideal{p}_j)$. If $X=V(\ideal{q})$ with $\ideal{q} \subseteq A$, prime, is an irreducible component of $Z \cap W$, we have
$$\ideal{q} \supseteq \ideal{a} + \bigcap_j\ideal{p}_j$$
and $\ideal{q}$ is minimal with respect to this property.
So a fortiori $\ideal{q} \supseteq \bigcap_j \ideal{p}_j$ and by a standard theorem on prime ideals $\ideal{q} \supseteq \ideal{p}_{j'}$ for a certain $j'$. (Proof of the standard theorem: If $a_j \in \ideal{p}_j \notin\ideal{q}$ for all $j$, then $\prod_j a_j \notin \ideal{q}$ and $\prod_j a_j \in \bigcap_j\ideal{p}_j$, contradiction).
So $\ideal{q} \supseteq \ideal{a} + \ideal{p}_{j'}$ and as $V(\ideal{a} + \ideal{p}_{j'}) = Z \cap W_{j'}$, we have $X \subseteq Z \cap W_{j'}$.
Now $\ideal{q}$ is minimal with respect to $\ideal{q} \supseteq \ideal{a} + \ideal{p}_{j'}$, because if $\ideal{q} \supsetneq \ideal{q}' \supseteq \ideal{a} + \ideal{p}_{j'}$ then $\ideal{q}' \supseteq \ideal{a} + \bigcap_j \ideal{p}_j$ although $\ideal{q}$ is minimal among these prime-ideals.
So $X$ is an irreducible component of $Z \cap W_{j'}$.
