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Say $X\subset \mathbb{C}^{n}$ is an algebraic set. Is it true the irreducible components of $X$ in the algebraic sense coincide with irreducible components of $X$ in the analytic sense?

A way to rephrase this question is to ask if the irreducible components of $X$ in the algebraic sense coincide with the closure of the connected components (In the euclidian topology) of $X\backslash Sing(X)$?

Over $\mathbb{R}$ this is incorrect, for instance $y^2=x^3$ is algebraically irreducible but has 2 analytic components. Over $\mathbb{C}$ I suspect it is correct.

Thank you!

Edit: As the answer bellow shows, it is true. However I actually was interested in this question when $\mathbb{C}^{n}$ is switched by some domain like the unit disk. In this case it is false, since one algebraic component may intersect a domain along many analytic components.

Espace' etale
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1 Answers1

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Yes, this is true. The set of smooth points of an irreducible algebraic set over $\Bbb C$ are connected, and thus the connected components of $X\setminus X^{sing}$ correspond exactly to the irreducible components of $X$. This characterization works in both the algebraic and analytic worlds.

By the way, your example over $\Bbb R$ does not work - the cuspidal cubic is the image of $\Bbb R$ under the map $t\mapsto (t^2,t^3)$, and the continuous image of an irreducible set is again irreducible. $\Bbb R$ is irreducible, as the analytic closed sets are discrete - if we had an analytic function who's zero set had an accumulation point, it would define an analytic function in a small complex neighborhood of $\Bbb R\subset \Bbb C$, and this function would be identically zero by the properties of analytic functions over $\Bbb C$.

KReiser
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  • Thank you for your answer.
    1. If so, then over $\mathbb{R}$ analytic components are not the same as connected componenst of $reg(X)$.
    2. Is there a simple example for this to be wrong over $\mathbb{R}$?
     – Espace' etale Dec 12 '19 at 11:10
  • Unless I'm missing something, your example already shows this: $y^2=x^3$ has two connected components of the regular set, corresponding to $y>0$ and $y<0$, but is analytically irreducible. – KReiser Dec 12 '19 at 11:18
  • Yes, this is what I meant, that if $y^2=x^3$ is analytically irreducible then the charachterization of components as topological components of the regular points is wrong over the reals.

    Anyway, is there an example of an analytically reducible but algebraically irreducible algebraic set over $\mathbb{R}$?

     – Espace' etale Dec 12 '19 at 11:23