This is a problem from Ideals, Varieties, and Algorithms by Cox et. al.

Let $V=V_1\cup \cdots \cup V_r$ be a decomposition of variety into its irreducible components.

Let $\Sigma$ be the singular locus of $V$ and let $\Sigma_i$ be the singular locus of $V_i$. If each $\Sigma_i$ is a proper subset of $V_i$, then show that $\Sigma$ contains no irreducible components of $V$.

In the previous part of this problem, I showed that if a point $p$ lies in a unique irreducible component, then it is nonsingular on $V$ if and only if it is nonsingular on $V_i$.

In another previous part of the problem, I showed $$\Sigma = (\bigcup_{i\ne j} (V_i\cap V_j))\cup (\bigcup_i \Sigma_i)$$

For the current part, here is my attempt:

Suppose $\Sigma$ contains a component $V_i$. Let $p$ be a point in $V_i$ that is not in $\Sigma_i$. Then $p$ is nonsingular in $V_i$ but singular in $\Sigma$. So it must lie on the intersection of two components $V_i\cap V_j$. Now I have no idea how to continue.

Thanks for any help!

  • 1
    $\begingroup$ Note that $p$ is any point in $V_i \setminus \Sigma_i$, and you concluded that $p \in V_i \cap V_j$ for some $j \neq i$. So what you've shown is that $V_i \setminus \Sigma_i \subset \bigcup_{j \neq i} V_j$. Is this possible? $\endgroup$
    – tracing
    Jan 11, 2016 at 12:55
  • $\begingroup$ @tracing: I see. Thank you for your answer! $\endgroup$
    – KittyL
    Jan 13, 2016 at 1:29


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