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?


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.

  • $\begingroup$ 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 $\mathbb{P}^n$? $\endgroup$ – Daniel Dreiberg Jan 22 '12 at 19:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Jan 22 '12 at 19:43
  • $\begingroup$ 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? $\endgroup$ – Daniel Dreiberg Jan 22 '12 at 20:01
  • 1
    $\begingroup$ 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 . $\endgroup$ – Georges Elencwajg Jan 22 '12 at 20:56
  • $\begingroup$ Ok, thanks. I'll follow your advice. $\endgroup$ – Daniel Dreiberg Jan 22 '12 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.