Irreducible components: associativity for intersections?

Question

Let $A$, $B$, $C$ be closed irreducible subvarieties of $\mathbb{A}^n$. Let $V_1$ be an irreducible component of $B\cap C$, and $V$ an irreducible component of $A\cap V_1$. Must there necessarily be an irreducible component $W_1$ of $A\cap B$ such that $V$ equals some irreducible component of $W_1\cap C$?

Answer 1

No.

This is an example with irreducible (and nonsingular) $A,B,C$: Consider $\mathbb{A}^4$ with coordinates $(w,x, y, z)$. Let $B$ be the $(x,y)$-plane (i.e. the set $w = z = 0$), $C$ be the hypersurface $z = xy$, and $A$ be the $(y,z)$-plane. Then $B \cap C$ is the union of the "$x$-axis" and "$y$-axis". Let $V_1$ be "$x$-axis" and $V$ be the origin, which is the only irreducible component of $A \cap V_1$. Since $A \cap B$ is the "$y$-axis", which is irreducible, the only possibility for $W_1$ is $y$-axis. But $W_1 \cap C = W_1 \neq V$.

Note that each of $A, B, C$ is isomorphic to $\mathbb{A}^k$ (for an appropriate $k$).

Original Example: Consider $\mathbb{A}^4$ with coordinates $(w,x, y, z)$. Let $B$ be the $(x,y)$-plane (i.e. the set $w = z = 0$), $C$ be the union of the $(w,x)$-plane and the $(w,y)$-plane and $A$ be the $(y,z)$-plane. Then $B \cap C$ is the union of the "$x$-axis" and "$y$-axis", whereas $A \cap B = A \cap C = A \cap B \cap C$ is the "$y$-axis", so that the property fails with $V_1$ equal to the "$x$-axis".

Answer 2

To give more context for possible answers: at the very least, a counterexample exists if we drop the assumption that $A,B,C$ be all irreducible (arisen in conversations with Harald). In $\mathbb{A}^{5}$, take \begin{align*} A & =\{x_{4}=0,x_{3}^{2}-x_{5}-1=0\}, \\ B & =\{x_{5}=0\}, \\ C & =\{x_{1}^{2}-x_{2}^{2}x_{3}^{2}=0,x_{1}-x_{2}x_{3}+x_{4}=0\}, \\ V_{1} & =\{x_{1}+x_{2}x_{3}=0,-2x_{2}x_{3}+x_{4}=0,x_{5}=0\}, \\ V & =\{x_{1}=x_{2}=x_{4}=x_{5}=0,x_{3}=1\}. \end{align*} Then the two candidates for $W_{1}$ are $W_{1}=\{x_{4}=x_{5}=0,x_{3}=1\}$ and $W_{1}=\{x_{4}=x_{5}=0,x_{3}=-1\}$: the second one does not contain $V$, while the first one gives \begin{equation*} W_{1}\cap C=\{x_{1}-x_{2}=0,x_{3}=1,x_{4}=x_{5}=0\}, \end{equation*} which is irreducible already. In this example, $A,B$ are irreducible but $C$ is not.

Answer 3

Let $A \cap B = \cup W_i$, where $W_i$ are the irreducible components. Then $$ A \cap B \cap C = (\cup W_i) \cap C = \cup (W_i \cap C). $$ Assume that $W_i \cap C = \cup Z_{i,j}$, where $Z_{i,j}$ are the irreducible components. Then $$ A \cap B \cap C = \cup Z_{i,j}. $$ In particular, the irreducible component $V$ is the union of some $Z_{i,j}$. But every $Z_{i,j}$ is closed in $A \cap B \cap C$, hence, irreducibility of $V$ imples that $V = Z_{i,j}$ for some $i$ and $j$. Then the component $W_i$ is the one you need.