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In many concrete cases I find it quite hard to show the irreducibility of a given variety.

Can I proceed in the following way:

Situation: We look at the closed subscheme $$X = V(f_1, \dots, f_n) \subset \mathbb{P}^m_k.$$ for a algebraically closed field $k$. We can show that the Jacobian has everywhere full rank and hence $X$ is smooth of codimension $n$.

Argument: Any regular local ring is integral. Hence $X$ is reduced. So $X$ is a complete intersection (as schemes) and $f_1, \dots, f_n$ generate the homogeneous ideal of $X$ by Hartshorne, Exercise II, 8.4. By Hartshorne, Exercise III, 5.5 the variety $X$ is connected. However, any connected noetherian scheme with integral local rings is integral.

Is this argument even correct? Can the proof simplified? What methods are there to show irreducibility in concrete situations?

Casaubon
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1 Answers1

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No: the transverse intersection of two conics in $\mathbb P^2_k$ is a complete intersection consisting of four smooth points but is certainly not irreducible.

Georges Elencwajg
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  • Thank you for the simple example! Do you mean set / scheme /ideal theoretical complete intersection? I don't see, where exactly Hartshorne, Ex II, 8.4 goes wrong, which tells you, that complete intersections are connected. – Casaubon Jul 21 '13 at 09:42
  • @Casaubon: in the case Georges mentions, the complete intersection is not of dimension $\geq 1$, which seems necessary for Hartshorne Ex.II.8.4(c) to apply. – Nils Matthes Jul 21 '13 at 10:14
  • What I wrote is true in any sense I'm aware of. In the exercise Hartshorne assumes that the complete intersection is of dimension $\geq 1$, whereas in my example it has dimension $0$. [I see that Nils gave the same answer while I was composing mine] – Georges Elencwajg Jul 21 '13 at 10:15
  • Ok, then my argument should work, if we add, that $X$ has Dimension $\ge 1$? – Casaubon Jul 21 '13 at 10:25
  • @Casaubon: If by "full rank" you mean rank $n$, then yes, your argument is correct when $m-n>1$. –  Jul 22 '13 at 21:41