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In the answers to this question I was taught that the projective closure $\bar X$ of an affine scheme $X$ (over a field) need not to be smooth since for example two distinct parallel lines in $\mathbb{A^2}$ meet in a singular point when considered in $\mathbb{P^2}$.

I have two questions on determining when the projective closure $\bar X$ of an affine scheme $X$ is smooth. Thank you in advance for any support!

(-1-) Suppose we embed $\mathbb{A}^2$ into $\mathbb{P}^2$ by $(x,y)\mapsto [x:y:1]$ and another copy $\mathbb{A}^2_'$ by $(y,z)\mapsto [1:y:z]$. Consider a projective closure $\bar X\subset\mathbb{P}^2$ of a smooth $X\subset \mathbb{A}^2$. Is $\bar X$ smooth if the preimage in $\mathbb{A}^2_'$ is smooth? Is this a valid criterion for all $n$ (here $n=2$)?

(-2-) Drawing a picture, I have the impression that smoothness of some kind of ''bounded'' smooth schemes in $\mathbb{A}^n$ like for example the circle $Spec~k[x,y]/(x^2+y^2-1)$ is preserved. But also for some ''non-bounded'' smooth schemes like $Spec~k[x,y]/(xy-1)$, the projective closure seem to be smooth (here I used the criterion of (-1-), which may be false). A classification of affine smooth schemes with this property is probably a little too ambitious but are there some sufficient properties? And can the term ''bounded'' be made precise?

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I am sorry I have to disappoint you, dear Daniel:

(-1-) is not correct. For example the curve $yx^2-z^3=0$ has smooth pull-backs (= restrictions) to your two copies of $\mathbb A^2$, but is singular at $[x:y:z]=[0:1:0]$ as you can check in the third embedding $\mathbb A^2 \hookrightarrow \mathbb A^2: (x,z)\mapsto [x:1:z]$ .

(-2-) The real picture is treacherously misleading !
For example, over $\mathbb C$ your affine curve $x^2+y^2-1=0$ is not bounded: the point $(1000i,(1000001)^{1/2})$ is at distance $(2000001)^{1/2}$ from the origin if you consider the usual euclidean distance on $\mathbb C^2=\mathbb R^4$.
More generally every affine subvariety of $\mathbb C^n $ of positive dimension is unbounded.

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Thank you very much for the answer. Can I rescue the first statement by checking smoothness at all $n+1$ embeddings of $\mathbb{A}^n$ into $\mathbb{P}^n$? –  Daniel Dreiberg Jan 22 '12 at 19:13
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Dear @Daniel, good news at last: yes! You can check smoothness by restricting to the open sets of any open covering. You see, smoothness is what is called a local property. Other interesting attributes of varieties cannot be checked on the opens of a covering: for example being affine or projective. –  Georges Elencwajg Jan 22 '12 at 19:43
    
Thank you. So I can deduce that the projective closure of the circle given by $x^2+y^2-1$ is smooth, at least. But even if the real picture is misleading, (-2-) seems to ''work'' for this example, so (-2-) my be still correct or is there a ''bounded'' (in the real picture, something like a closed curve) smooth affine scheme such that it's projective closure is singular? –  Daniel Dreiberg Jan 22 '12 at 20:01
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Dear @Daniel, the affine curve $x^4+x^2+y^2+1=0$ is bounded and smooth in $\mathbb R^2$ because it is empty! However it is very singular at infinity in the complex projective plane. Please, give up this circle of ideas, it is a dead end and will only mislead you . –  Georges Elencwajg Jan 22 '12 at 20:56
    
Ok, thanks. I'll follow your advice. –  Daniel Dreiberg Jan 22 '12 at 21:14

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