# Is the projective closure of a smooth variety still smooth?

Let $X$ be a closed subscheme of $\mathbb{A^n}$ (over a basefield) defined by an ideal $I$ and consider the immersion $\mathbb{A^n}\to \mathbb{P^n}$, $(x_1,\ldots, x_n)\mapsto [x_1,\ldots,x_n,1]$. One may consider the projective variety $\bar X$ in $\mathbb{P^n}$ given by the homogenized ideal $\bar I$. This ideal consists of the homogenized elements of $I$, so for example if $f=x_1^2+x_2+1$ is in $I$ then $\bar f=x_1^2+x_2x_{n+1}+x_{n+1}^2$ is in $\bar I$. Then $\bar X$ is the projective closure of the image of $X$ under $\mathbb{A^n}\to \mathbb{P^n}$, right? I wonder if smoothness is inherited.

If $X$ is a smooth affine scheme over the base field, is $\bar X$ smooth, too?

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Your addition to the question has rendered all the existing answers incomplete. You should ask it as a separate question. – Zhen Lin Jan 22 '12 at 1:23
Ok, sorry for the confusion. – Daniel Dreiberg Jan 22 '12 at 1:27

No, smoothness of $X$ is not inherited by $\bar X$.
Take two parallel lines in $\mathbb A^2_k$. Their union $X\subset \mathbb A^2_k$ is a smooth scheme but the lines meet at a singular point of $\bar X$ at infinity, so that $\bar X \subset \mathbb P^2_k$ is no longer smooth.
Renaissance artists, who invented perspective, have given us beautiful paintings of these schemes.

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In some sense the opposite is true. Namely, let $X \subset \mathbb{P}^n$ be an irreducible quasi-projective variety (over an algebraically closed field, let us say). Then there is a proper closed subvariety $S \subset X$ consisting of the singular points, so $Y = X \setminus S$ is a smooth quasi-projective variety.

In other words, if the answer to your question were "yes" then all varieties would be smooth!

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No. There is no reason for the curve to be smooth at the points at infinity. For simplicity I will give a counterexample for algebraically closed fields of characteristic not $2$ or $3$. Consider the curve $X$ in $\mathbb{A}^2$ defined by the equation $$x y^2 - 1 = 0$$ It is straightforward to verify that $$y^2 \, \mathrm{d}x + 2 x y \, \mathrm{d} y$$ is nowhere vanishing on $X$, so $X$ is non-singular.

The projective closure $\overline{X}$ in $\mathbb{P}^2$ is defined by the equation $$x y^2 - z^3 = 0$$ Now, let us consider the affine piece $\overline{X} \cap \{ x \ne 0 \}$. This is the affine curve defined by the equation $$y^2 - z^3 = 0$$ and it is readily checked that this curve has a cusp at $(0, 0)$. Thus the curve $\overline{X}$ is singular at the point $(1 : 0 : 0)$ (and, in fact, only there).

That said, at least for non-singular curves $X$, there is always a non-singular projective curve containing $X$ as a dense open subvariety. This is done in e.g. Hartshorne [Ch. I, §6]. The only thing is that this projective curve may live in a projective space of higher dimension than the affine space you started with!

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Lots of good answers here already, but here's another reason why your statement is false. Consider a singular projective variety $X \subseteq \mathbb{P}^n$ and suppose that the singular locus is contained in a hyperplane $H \subseteq \mathbb{P}^n$. (For example, any plane curve with a unique singular point fits this description.) Then $X' = X \setminus X \cap H$ is a smooth affine variety (sitting in $\mathbb{A}^n \cong \mathbb{P}^n \setminus H$) whose projective closure is the singular variety $X$.

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