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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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    Ok, thanks. I'll follow your advice.2012-01-22