Given a quasi-smooth toric variety $X$ in the sense of Gelfand-Kapranov-Zelevinsky, i.e. a (not necessarily normal) toric variety $X$ whose normalization is a $\mathbb{Q}$-factorial toric variety and such that the normalization morphism is bijective. It is known that such non-normal toric varieties share nice properties with smooth toric varieties. What kind of singularities can we expect from $X$? Are they Cohen-Macaulay?
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asked Jun 16, 2016 at 0:11
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1$\begingroup$ Cohen-Macaulay schemes are S2, hence satisfy the S2 extension property. So, beginning with $\mathbb{P}^n$, $n\geq 2$, "pinch" it infinitesimally at one point. The resulting scheme is not Cohen-Macaulay, even though the normalization morphism is universally bijective on points. $\endgroup$– Jason StarrCommented Jun 16, 2016 at 7:46
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$\begingroup$ That's a good point. More generally, Serre's criterion says normal=R1+S2. Since CM implies S2 this means that $$\text{quasi-smooth toric}+\text{CM}+\text{smooth in codimension one}\Rightarrow\text{normal}$$ In other words, quasi-smooth non-normal toric varieties are almost never CM. $\endgroup$– Friedrich KnopCommented Jun 17, 2016 at 7:51
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1 Answer
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Counterexample to CM: Let $M$ be the monoid $\{x^iy^j\mid i,j\ge0\}\setminus\{x,y\}$ and $X=\text{Spec}\ \mathbb C[M]$ the corresponding toric variety. Then $x^2,y^2$ is a system of parameters of $X$. But $\mathbb C[M]$ is not a free $\mathbb C[x^2,y^2]$-module since the fiber $\mathbb C[M]/(x^2,y^2)=\langle 1,xy,x^3,x^2y,xy^2,y^3\rangle$ is of dimension $6$ instead of $4$.
answered Jun 16, 2016 at 7:48