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Lets say $X$ is a conical affine algebraic variety (conical meaning $X$ is the zero set of homogeneous polynomials of positive degree, equivalenty $X \subsetneq k^n$ and $x \in X \Rightarrow ax \in X$ for all $a \in k$) and $\mathbb P(X)$ is the projectivization of $X$ (the projective variety defined by the same homogeneous equations as $X$).

Clearly the origin $0 \in X$ is a singular point. If $a \neq 0$ is a nonsingular point of $X$ then is the line through $a$ a nonsingular point of $\mathbb P(X)$?

More generally, if $A$ is a graded Noetherian ring generated in degree $1$, and $\mathfrak p \in \mathrm{Proj} \ A$ is a homogeneous prime ideal such that the point $\mathfrak p \in \mathrm{Spec} \ A$ is regular, then is $\mathfrak p$ regular in $\mathrm{Proj} \ A$?

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up vote 2 down vote accepted

The answer is yes (and is if and only if) in the general case. Indeed, let $f\in A$ be a homogeneous element of degree $1$. Then we have a canonical isomorphism $$ A_{(f)}[T, 1/T] \simeq A_f$$ (see proof below). Now choose such an $f$ with $\mathfrak p\in D_+(f)$. Denote by $X=\mathrm{Spec}(A)$, $Y=\mathrm{Proj}(A)$ and $x\in X$ (resp. $y\in Y$) the point corresponding to $\mathfrak p$. Then $O_{X,x}$ is a localization of $O_{Y,y}[T,1/T]$, hence flat over $O_{Y,y}[T,1/T]$. Thus $O_{Y,y}\to O_{X,x}$ is flat, and the regularity of $O_{X,x}$ implies that of $O_{Y,y}$ (see Matsumura, Commutative Algebra, 21.D, Theorem 51).

More globally, let $m=\oplus_{d\ge 1} A_d$ be the irrelevent ideal. Note that the above isomorphism for various $f$ implies that the canonical morphism $$\mathrm{Spec}(A) \setminus V(m) \to \mathrm{Proj} A$$ is smooth (even a $\mathbb G_m$-bundle). If $y$ is regular in $Y$, then any point of $X$ lying over $y$ is regular.


If $X$ is the affine cone over a field, then the above isomorphisms says that $$X\setminus \{ 0\} \simeq \mathbb P(X)\times \mathbb G_m$$ Zariski locally on $\mathbb P(X)$. If the field is algebraically closed, the regularity at closed points can be checked with Jacobian criterion, and the if and only if statement is clear (without invoking flatness).

End of Edit.

Now let us prove the above isomorphism. The homogeneous localization $A_{(f)}$ is a subring of $A_f$. Define $\phi : A_{(f)}[T]\to A_f$ by taking $T$ to $f$. It induces canonically a ring homomorphism $\phi: A_{(f)}[T,1/T]\to A_f$ because $f$ is invertible in $A_f$. As $f$ has degree $1$, it is easy to see $\phi$ is surjective. Suppose $\phi(\sum_{k} b_k T^k)=0$ with $b_k\in A_{(f)}$ (the sum is a finite sum in $k\in\mathbb Z$). There exists $N\ge 1$ such that $b_k=a_k/f^N$ with $a_k\in A$ homogeneous of degree $N$ for all $k$. Then $\sum_k a_kf^{k}=0$ in $A_f$. This means for some $m\ge 1$, we have $$\sum_k a_k f^{k+m}=0 \in A.$$ As $\deg(a_kf^{k+m})=N+k+m$, this implies that $a_kf^{k+m}=0$ for all $k$, hence $b_k=a_k/f^N=0$ in $A_{(f)}$ and $\phi$ is injective.

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Very impressive:+1. The theorem quoted can also be found in the easier to find other book by Matsumura Commutative ring theory, as Theorem 23.7, page 182. – Georges Elencwajg Jan 11 '13 at 10:12
A little unimportant remark: it seems to me that $\phi$ is an isomorphism also if $f$ is homogeneous of arbitrary degree $d$. Your proof goes through by just replacing $N$ by $dN$ in a few places. Do you agree ? Anyway I think your answer is a quite important result that I only knew in the classical case: I'll try to remember your general version in terms of graded rings . – Georges Elencwajg Jan 11 '13 at 11:23
@GeorgesElencwajg: Thansk ! The problem with $\deg f>1$ is the surjectivity of $\phi$. If say $\deg f=2$, then in the image of $\phi$ we only have fractions of even degrees (and sums of such fractions). But when $A$ is generated by elements of degree 1 as in the question, then we don't really need to care about element of degree $>1$. But if $A$ is not generated in degree 1, the result on the morphism from the affine cone to the projective scheme is probably false (consider $A=\mathbb C[T_0, T_1, T_2]$ with the variables of respective degrees $2, 3, 5$. Then the projective scheme is singular – user18119 Jan 11 '13 at 12:39
continued: but the affine scheme is regular. – user18119 Jan 11 '13 at 12:39
Perfect, thanks so much! – Jim Jan 11 '13 at 19:56

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