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Here is where I got lost .. I have a scheme Y over k (an algebraically closed field), in it I have an irreducible closed subscheme X of finite type (do I need finite type?). I also know that X is universally closed (over k) and separated (do I need separated?) .. then why is it that X should be either a point or Y itself?

flag	asked Feb 12 2010 at 12:45 quantum 41●2

To the downvoter: that's not really a nice welcome for quantum, so I upvoted the question again. – Wanderer Feb 12 2010 at 13:03

(Although I agree that this is not an ideal Mathoverflow question.) – Wanderer Feb 12 2010 at 13:05

Welcome to MO! It looks like your question comes from other sources than pure curiosity. Perhaps reproducing the exact question here would help us answer it better. – Hailong Dao Feb 12 2010 at 15:10

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hmm... I have the impression that you didn't read mathoverflow.net/howtoask in detail. Maybe you can get very helpful answers if you rephrase your question a little bit. – Konrad Voelkel Feb 12 2010 at 15:35

Are you mixing up the term "irreducible scheme" with the term "irreducible representation", where the meaning is very different? – Allen Knutson Feb 12 2010 at 19:35

Wanderer · Answer 1 · 2010-02-12 12:59:48Z UTC

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I don't get it. You are saying exactly that $X$ should be proper over $k$. Embed $X = \mathbb{P}^1(\mathbb{C})$ - which is proper over $\mathbb{C}$ - in $Y = \mathbb{P}^2(\mathbb{C})$ as a closed subscheme. That's a counterexample?

answered Feb 12 2010 at 12:59

Wanderer
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It also seems that taking $X$ disjoint union something else gives a counterexample, so I think we need more assumptions about these schemes. – Steven Sam Feb 13 2010 at 0:52

Schemy question..

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