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Specifically, how to show that an affine variety over complex number is never compact in Euclidean topology unless it is a single point. I got a hint on this qiestion: Given an affine variety X, show that the image of X under the projection map onto the first coordinate is either a point or an open subset (in the Zariski topology).

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After proving what's asked for in the hint, do you see how to proceed? (Consider the images of X under all of the coordinate projections. What happens if X is compact?) –  Brad Feb 23 '11 at 18:57
    
In fact, I have no idea how to prove the image of X under the projection map onto the 1st coordinate is an open subset in Zariski topology. –  user7419 Feb 23 '11 at 19:11
    
Does someone help me? –  user7419 Feb 23 '11 at 19:56

3 Answers 3

The hint is a consequence of Chevalley's theorem on constructible sets: http://en.wikipedia.org/wiki/Constructible_set_%28topology%29

A non-empty Zariski-open susbset of the affine line is clearly not compact, so the image has to be a finite set. Projecting on each coordinate, you get that the variety is finite.

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As Plop states, the hint follows from Chevalley's theorem. However, in this context one shouldn't need to appeal to the full strength of that theorem.

In fact, Chevalley's theorem is a variation on Noether normalization (and both are variations on the Nullstellensatz --- see this MO answer), but Noether normalization is usually taught at an ealier stage than Chevalley's theorem, so you might consider using it instead. (Regard this as an alternative hint.)

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Thanks Matt E & Plop. Matt E, could you show me how to use the Noether normalization to solve my problem? –  user7445 Feb 24 '11 at 5:48
    
@charm: Dear Charm, Firstly, do you know the statement of Noether normalization? Regards, –  Matt E Feb 24 '11 at 5:50

Matt E, all I know about Noether normalization is the existence of an algebraically independent elements of a finitely generated commutative algebra.

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Dear Charm, First a practical comment: if you leave your comment as an answer, I don't get notified; if you leave it as a comment under my answer, as you did before, I get notified, which makes it easier to reply. –  Matt E Feb 24 '11 at 12:24
    
Okay, now for some mathematics! Noether normalization is a much stronger statement than what you wrote. The wikipedia entry en.wikipedia.org/wiki/Noether_normalization_lemmagives a good description which you might want to read; in particular, it gives the geometric interpretation, which is most relevant here. You could also look at this MO answer: mathoverflow.net/questions/42275/… which sketches a geometric proof of Noether normalization. ... –  Matt E Feb 24 '11 at 12:25
    
My suggestion is that you spend some time understanding Noether normalization and its geometric meaning first (and spend some time thinking about the general process of projection, perhaps by looking for and solving some exercises in Hartshorne Chapter I that are related to projections) before trying to solve your original question. This will have the added advantage that Chevalley's theorem discussed above (among other things) will make more sense, and the solution of your original question, will become reasonably obvious (following either the original hint or my Noether normalization hint). –  Matt E Feb 24 '11 at 12:35
    
Sorry, there was a typo in the above wikipedia link; it should be en.wikipedia.org/wiki/Noether_normalization_lemma –  Matt E Feb 24 '11 at 12:35
    
Matt E, could you give me a proof using the original hint (i.e. projection to the first coordinate is either open or a point)? –  user7445 Mar 1 '11 at 14:45

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