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How to prove that an affine variety $X$is complete only if $\dim X=0$? It is clear that in this case $X$ must be a single point. But I don't known why its dimension should be zero. Could anyone help me? Thanks a lot!

Michael Hardy
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yang
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2 Answers2

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If an affine variety $X\subset \mathbb A^n_k$ has more than one point (which is certainly the case if $dim (X)\gt 0$) some coordinate function $x_i$ is not constant on $X$ and we thus get a non constant regular function $x_i\in \mathcal O(X)\setminus k$.
On the other hand, every regular function on a complete variety $Y$ is constant (in other words $\mathcal O(Y)= k$) and thus our positive-dimensional affine variety $X$ cannot be complete.

Edit
In the above answer, set in an elementary context not assuming schemes, I Assumed $X$ irreducible and $k$ algebraically closed.
At a more advanced level one can only say that $\operatorname {dim} _k \Gamma(X, \mathcal O_X) \lt \infty $ for a scheme $X$ proper over a ( maybe not algebraically closed) arbitrary field $k$ . Thanks to @Alex for his comment which caused this edit.

Georges Elencwajg
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  • Original post was about complete varieties, but it is also true that the only regular functions on complete varieties are constant, so your argument answer the original post as well. – Callus - Reinstate Monica Dec 30 '13 at 03:44
  • Dear @Callus, you are absolutely right and I corrected my answer. Since the question was probably written by a beginner in algebraic geometry (welcome to this enchanted world, yang!), I guess that I subconsciously substituted "projective" in place of "complete". Anyway, thanks a lot for your attentive reading. – Georges Elencwajg Dec 30 '13 at 03:57
  • Thank you for your answer! But I still can't understand why every regular function on a complete variety is constant. I only know that "complete" means that the corresponding projection morphism is closed. So could you give me some more details? Thank you! – yang Dec 30 '13 at 06:02
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    @yang Think about a regular function as a morphism to $\mathbb{A}^1$. If your variety is irreducible (which you certainly can assume for this problem) then the image is closed (since your variety is complete), and irreducible and so must either be a point or all of $\mathbb{A}^1$. But, $\mathbb{A}^1$ is certainly not complete (and the image of complete varieties are complete) since there are tons of non-closed projections. – Alex Youcis Dec 30 '13 at 06:11
  • @GeorgesElencwajg I feel a little silly asking this, but could you clarify something? When someone says "regular function", what precisely do they mean? If one defines a variety over $k$ to be a reduced finite type separated scheme over $\text{Spec}(k)$, and by regular function they mean "global section", then I don't understand the statement that all regular functions on a complete variety are constant (despite my previous comment :]). For example, $\text{Spec}(\mathbb{C})\to\text{Spec}(\mathbb{R})$ is proper, but the global sections of $\text{Spec}(\mathbb{C}))$ isn't $\mathbb{R}$. Thanks! – Alex Youcis Dec 30 '13 at 06:16
  • Dear @Alex, I assumed that the question was from a relative beginner in algebraic geometry and I tried to answer it at that level. In other words I interpreted "affine variety" as "irreducible closed subset of affine space over an algebraically closed field" in the style of, say, Fulton's book on algebraic curves. At a more advanced level you can only say that for a scheme $X$ proper over a field $k$, the $k$-vector space $\Gamma(X, \mathcal O_X)$ is finite-dimensional, as is $\Gamma(X, \mathcal F_X)$ for any coherent sheaf $\mathcal F$. Your example beautifully illustrates this. – Georges Elencwajg Dec 30 '13 at 09:23
  • @GeorgesElencwajg Of course, that makes sense. I just wanted to make sure I was making the terminology connection. Thank! – Alex Youcis Dec 30 '13 at 18:33
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If I understand properly you are asking why an affine variety which is a single point has dimension zero?

It is very clear from the definition of dimension: A variety X has dimension n is there exists an chain of irreducible closed subsets of X $Z_0 \subsetneq Z_1\subsetneq...\subsetneq Z_n \subsetneq X$.

If X has to have dimension 1 then u need at least two points because you need two irreducible closed subsets of X one containing the other properly.

Babai
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