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Let $X^n \subset \mathbb{P}^N$ to be a toric projective variety. Is $X$ a local complete intersection? Is being a local complete intersection an intrinsic property, independent of embedding?

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Any point of a toric variety has an affine neighborhood which has coordinate ring of the form $\mathbb{C} [x_1,\cdots,x_k]/I$ where $I$ is an ideal generated by $k-n$ monomials. So this gives an embedding of this affine neighborhood in $\mathbb{C}^k$ as a local complete intersection. But I am not sure if it implies $X$ is a local complete intersection in $\mathbb{P}^N$ ! – Mohammad F. Tehrani Apr 12 '11 at 17:50
up vote 6 down vote accepted

No, there are projective toric varieties that are not even Gorenstein (and hence not l.c.i). In fact the Gorenstein property translates into a geometric property of the defining polytope, namely that it is reflexive under taking the dual polytope.

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In fact I am interested in Gorenstein, toric varieties. What can we say then? – Mohammad F. Tehrani Apr 12 '11 at 18:08
A variety is Gorenstein if it is Cohen-Macaulay and it's canonical divisor is a line bundle. These properties does not depend on the embedding. – J.C. Ottem Apr 12 '11 at 18:12
I am not expert but this seems to contradict the Rmk. on page 185 of Hartshorne that being an lci in a smooth variety is a property independent of the embedding, and that one proves it using the cotangent complex of Lichtenbaum and Schlessinger. – roy smith Apr 12 '11 at 18:56
@J.C. Ottem, I don't think it does depend on the embedding, see for example Chapter 21 of Matsumura's commutative ring theory. There they basically define a local ring to be an LCI if its completion is an LCI (in the usual sense). However, this is done because not all rings are quotients of regular rings. But by Theorem 21.1(iii), for an LCI which is a quotient of ANY regular ring, it can be cut out by the right number of generators. – Karl Schwede Apr 12 '11 at 20:10
@Mohammed, (Gorenstein Toric => LCI?) I suspect the answer is no. See exercise 21.3 in Matsumura's book. That ring is not normal, but it's really close to being toric (it is a binomial ring). You could try googling things like "affine semi-group ring" (commutative algebraists version of toric) and "complete intersection", I suspect there are examples in the literature which explain when this happens. – Karl Schwede Apr 12 '11 at 20:18

In "All toric l.c.i.-singularities admit projective crepant resolutions", the authors say (second page)

Families of Gorenstein non-l.c.i. toric singularities that have such special full resolutions seem to exist only rarely.

I would like to think they would not say this if there were no examples of Gorenstein non-l.c.i. toric singularities.

Just to emphasize what everyone else is saying, being l.c.i. is an intrinsic and local condition on a ring. So the natural question is to ask about an affine toric variety, since the question for a projective toric varitey just comes down to checking in every coordinate chart. And l.c.i. implies Gorenstein, which not all toric varieties are, so you at least want to include the condition that your toric variety be Gorenstein.

I think I have an example of a Gorenstein toric non-l.c.i. Let $D$ be the lattice $\{(i,j,k,\ell) \in \mathbb{Z}^4 : i+j+k+\ell \equiv 0 \mod 2 \}$. Let $C$ be the cone $\{ i,\ j,\ k,\ \ell \geq 0 \}$ in $D \otimes \mathbb{R}$. Our toric singularity will be $S:=k[C \cap D]$. Note that we can think of this as the subring of $k[w,x,y,z]$ consisting of even degree polynomials. So $S$ is generated by $w^2$, $wx$, ..., $z^2$ -- ten monomials in all.

Normal semi-group rings are always Cohen-Macaulay, and the dualizing module is $\mathrm{Span}_k (C^{\circ} \cap D)$ where $C^{\circ}$ is the interior of $C$. In other words, the dualizing module consists of the $k$-span of the even degree monomials where each variable appears with even degree. This module is clearly generated by $wxyz$, so the dualizing module is free of rank $1$, and we see that $S$ is Gorenstein.

Now, let's check that $S$ is not l.c.i. We see $S$ as a quotient of the polynomial ring in $10$ variables, coming from the generators $w^2$, $wx$ etc. Let $R$ be the polynomial ring in $10$ variables. I'll write $[w^2]$ and so forth for the generators of $R$. Let $I$ be the kernel of $S \to R$; we want to show that $I$ is NOT generated by any $6$ elements. Let $\mathcal{M}$ be the maximal ideal of $R$. Since $I$ is graded, it is enough to check that $I/\mathcal{M}I$ is NOT $6$ dimensional.

Well, there are already $6$ degree $2$ elements of $I$ which are the $S_4$ symmetries of $[w^2] [x^2] = [wx]^2$. In addition, we have two more relations $[wx] [yz] = [wy][xz] = [wz][xy]$. These $8$ elements are linearly independent in $I/\mathcal{M} I$, so we see that $S$ is not an l.c.i.

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So considering all this to be true, The discussion for toric varieties reduces to just one cone $\sigma$ in the fan of toric variety. Then the question is when $Spec \mathbb{C}[\check{\sigma}]$ is l.c.i ?! which is a bit wierd? – Mohammad F. Tehrani Apr 12 '11 at 21:29
Right. I'm not sure why you think it is weird. The word "local" is in "l.c.i." for a reason. – David Speyer Apr 12 '11 at 21:32
So I guess may be my comment right after question is wrong and the ideal $I$ is not necessarily always generated by $k-n$ (codimension) elements. I am confused :) – Mohammad F. Tehrani Apr 12 '11 at 21:56
So, why did you think that the ideal should be generated by codimension many elements? – David Speyer Apr 12 '11 at 22:48
I just considered a basis for the space of relations. Let me think more on this problem, I'll come back if I found any thing. – Mohammad F. Tehrani Apr 13 '11 at 2:00

Nakajima classifies L.C.I toric varieties in:

But it s hard to read.

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