I was taught in the previous thread "Is the projective closure of a smooth variety still smooth?", that the projective closure $\bar X\subseteq \mathbb{P}^n$ of a smooth closed subscheme $X\subseteq \mathbb{A}^n$ (over a basefield) needs not to be smooth again.

(-1-) I wonder, if one can test the smoothness of $\bar X$ on the closed ''complement'' $\mathbb{P}^{n-1}$ of $\mathbb{A}^n$ in $\mathbb{P^n}$: Is $\bar X$ smooth if $X$ and $\bar X\cap \mathbb{P}^{n-1}$ are smooth? If there is a counterexample, it would be nice to have an irreducible and reduced one, a good old variety.

In the thread ''Example of a non-smooth intersection'', I was taught that the converse is not true: If $\bar X$ is smooth, $\bar X\cap \mathbb{P}^{n-1}$ does not have to be, even if $\bar X\cap \mathbb{P}^{n-1}$ is reduced.

(-2-) Let $\bar X$ be smooth and suppose that $\bar X\cap \mathbb{P}^{n-1}$ is reduced. Is it true that in this case all the irreducible components of $\bar X\cap \mathbb{P}^{n-1}$ are smooth?


Yes, if $X$ is smooth and $\bar X\cap \mathbb P^{n-1}$ is smooth in the scheme theoretic sense, then $\bar X$ is indeed smooth.
In other words if $s\in \bar X$ is singular, so is $s\in \bar X\cap H $ for any hyperplane $s\in H\subset \mathbb P^{n}$ .

Beware however that you have to take scheme-theoretic intersections.

For example if $\bar X$ is the curve $z^2y=x^3$ in $\mathbb P^{2} $, then that curve is smooth in the affine plane $z\neq1$ and its intersection with the line at infinity $H$ given by $z=0$ is the single point $s=[0:1:0]$.
So, naïvely one might think that since a single point is a smooth variety one may conclude by the above that $\bar X$ is smooth.
In reality $\bar X$ has a singularity at $s$. The mistake was to not see that the intersection of $\bar X$ with $H$ is not the reduced point $s$ but $s$ with a nilpotent structure.

Edit: detailed calculation

Indeed the intersection of $\bar X$ with the line at infinity $H$ is best computed in the affine plane $\mathbb A^2_{x,z}=Spec (k[x,z])$ (=the points of $\mathbb P^{2}$ where we may choose $y=1$), whose coordinates are $(x,z)=[x:1:z]$.
The point $s$ then has coordinates $x=0,z=0$, $\bar X$ has equation $z^2=x^3$, $H$ has equation $z=0$ and the ideal in $k[x,z]$ of the intersection $\bar X \cap H$ is $(z^2-x^3,z)=(z,x^3)$.
So the intersection $\bar X \cap H$ is the affine subscheme $Spec (k[x,z]/(z,x^3)) \subset \mathbb A^2_{x,z}$, which is clearly isomorphic to $A=Spec(k[x]/(x^3)$, hence non reduced and thus singular.
Geometrically $\bar X$ is a cusp cut by every projective line in three points. The line at infinity however cuts it in what was classically known as a "triple point" before the introduction of schemes.

Second edit
Let me add a few words to the second sentence of my answer in order to address Daniel's comment.
The problem is local at $s$, so we may assume that $s\in X$ is a singularity and we must show that $X\cap H $ is singular at $s$ too.
For simplicity assume that $X$ is a hypersurface.
It thus has equation $f(x_1,...,x_n)=0$, with $f(x_1,...,x_n)=q_2(x_1,...,x_n)+q_3(x_1,...,x_n)+...$ where $q_i(x_1,...,x_n)$ is homogeneous of degree $i$.
The crucial point is that there is no linear term $q_1$: this is equivalent to $X$ being singular at $s$.
If the hyperplane $H$ has equation $x_n=l(x_1,...,x_{n-1})$ ( $l$ linear) the intersection $X\cap H$ is given by $q_2(x_1,...,l(x_1,...,x_{n-1})+q_3(x_1,...,l(x_1,...,x_{n-1}))+...=0$ in the affine coordinates $x_1,...,x_{n-1}$ and is thus also singular since it begins with a quadratic term.

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  • $\begingroup$ That's really good news, thanks Georges! Isn't this a surprise somehow? Normally you have to check smoothness on an open cover, but here the $\bar X\cap\mathbb{P}^{n-1}$ has a closed immersion into $\bar X$! There is no ''overlap'' with $X$. Could you do me a big favour and tell me, where I can find this in the literature? Thank you! $\endgroup$ – Daniel Dreiberg Aug 8 '12 at 21:02
  • $\begingroup$ Did you mean ''...in the affine plane $z=1$''? $\endgroup$ – Daniel Dreiberg Aug 8 '12 at 21:10
  • $\begingroup$ Sorry to bother you again, but I don't understand the example: Is the intersection of $\bar X$ with $z=0$ not just given by setting $z=0$ in the equation, i.e. $0^2*y=x^3$ and hence $A=k[x,y]/(x^3)$? $\endgroup$ – Daniel Dreiberg Aug 8 '12 at 21:21
  • $\begingroup$ Dear Daniel, I have tried to make the calculations more explicit in an Edit and hope you will find them clearer now.[Your calculation for $A$ is incorrect because your $A$ is the ring of a triple line instead of a triple point: do not make substitutions in equations but come back to ideals. Also, I didn't mean $z=1$ anywhere]. Unfortunately I don't know a reference to the literature for what is indeed a rather surprising fact. $\endgroup$ – Georges Elencwajg Aug 8 '12 at 22:42
  • $\begingroup$ Dear Georges, I tried to understand why (-1-) is true, as you said, but failed. Could you give me a little insight in your reasoning? $\endgroup$ – Daniel Dreiberg Aug 13 '12 at 14:53

Your second question put slightly more canonically is:

Let $\bar X\subset \mathbb P^n$ be smooth and suppose that for some hyperplane $H\subset \mathbb P^n$ the intersection $\bar X\cap H$ is reduced.
Is it true that in this case all the irreducible components of $\bar X\cap H$ are smooth?

The answer is no: take for $\bar X$ a smooth hypersurface and for $H$ its tangent space $H=T_P(H)$ at some point $P\in \bar X$.
The intersection $\bar X \cap H$ will in general be a reduced singular irreducible variety.

A completely explicit example
Let $\bar X\subset \mathbb P^3$ be the the smooth cubic surface $x^3+y^3+z^3-w^3=0$ and $P=[-1:b:0:1]$ where $b=\sqrt[3]2$.
We will do our calculations in the affine space $\mathbb A^3_{xyz}$ given by $w=1$ and with coordinates $(x,y,z)=[x:y:z:1]$.
There $X=\bar X\cap \mathbb A^3$ has equation $$x^3+y^3+z^3-1=0\quad (1)$$
The tangent plane $H=T_P(X)$ at $P=(-1,b,0)$ is given by $$x+b^2y-1=0\quad (2)$$ and the intersection $X\cap H$ has an equation in the plane $\mathbb A^2_{yz}$ obtained by eliminating $x$ from $(1)$ thanks to (2) [and don't forget that $b^3=2$ and that $b^4=2b $ !].
One gets: $$(1-b^2y)^3+y^3+z^3-1=z^3-3y^3+6by^2-3b^2y =z^3-3y(y-b)^2=0 , $$
the equation of a reduced irreducible singular cubic (a cissoid or cusp in traditional terminology.)

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  • $\begingroup$ Thanks for this instructive answer! $\endgroup$ – Daniel Dreiberg Aug 12 '12 at 20:32
  • $\begingroup$ You are welcome, Daniel! $\endgroup$ – Georges Elencwajg Aug 12 '12 at 20:56

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