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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 $W_1\cap C$?

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    $\begingroup$ Why can't you just take $W_1$ to be any irreducible component of $A \cap B$ which contains $V$? $\endgroup$
    – Sasha
    Dec 14, 2020 at 19:35
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    $\begingroup$ Why would that necessarily work? There would then be an irreducible component of $W_1\cap C$ that contains $V$, but would there have to be an irreducible component of $W_1\cap C$ that equals $V$? $\endgroup$ Dec 14, 2020 at 19:54
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    $\begingroup$ Right, it is slightly more subtle. See my answer. $\endgroup$
    – Sasha
    Dec 14, 2020 at 20:11

3 Answers 3

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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".

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  • $\begingroup$ In this example, $C$ is reducible. $\endgroup$ Dec 14, 2020 at 21:43
  • $\begingroup$ Moreover, $V$ can only be the origin or the $y$-axis, so it seems this isn't a counterexample at all - not even if (as in D. Dona's answer) the condition that $C$ be irreducible is dropped. $\endgroup$ Dec 14, 2020 at 21:46
  • $\begingroup$ @HAHelfgott: I missed the requirement for irreducibility - now updated the answer. $\endgroup$
    – pinaki
    Dec 14, 2020 at 23:41
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    $\begingroup$ It just occurred to me that the same works even in $\mathbb{A}^{3}$, forgetting about the $w$ coordinate. In that case the counterexample is truly "minimal", since $\mathbb{A}^{2}$ doesn't have enough room to house a counterexample. $\endgroup$
    – D. Dona
    Dec 20, 2020 at 12:37
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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.

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  • $\begingroup$ Aah, I see you beat me to it! $\endgroup$
    – pinaki
    Dec 14, 2020 at 21:30
  • $\begingroup$ @auniket That's easy when you know the question before it's being posted on MO. But kudos to you for giving an all-irreducibles counterexample! $\endgroup$
    – D. Dona
    Dec 15, 2020 at 6:55
  • $\begingroup$ How should the weird sentence "a counterexample exists when $A,B,C$ are not all irreducible" be interpreted? that there's no counterexample with $A,B,C$ all irreducible? $\endgroup$
    – YCor
    Dec 15, 2020 at 10:05
  • $\begingroup$ @YCor My bad, I see how it can be misinterpreted. I mean "a counterexample exists if we drop the assumption that $A,B,C$ be all irreducible": the counterexample with $A,B,C$ all irreducible exists, see auniket's answer, although I didn't know that yet when I was writing the first time. I'll reformulate. $\endgroup$
    – D. Dona
    Dec 15, 2020 at 13:46
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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.

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    $\begingroup$ Is it clear that $V$ is the union of some $Z_{i,j}$? $\endgroup$ Dec 14, 2020 at 20:26

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