# cohen-macaulay property for reducible schemes

I have the following question about certain schemes being Cohen-Macaulay.

1. Let $X$ be the union of all coordinate $k$-planes in ${\mathbb A}^N$. Is it CM?

2. Let $R$ be a collection of $k$-element subsets of $\{1,\ldots,N\}$ such that for each $J\in R$ there exists $J'\in R$ which differs from $J$ at just ONE element. Such $R$ will be called an admissible collection. Let us denote the union of corresponding k-planes by $A(R)$. Is it $CM$?

3. The following statement appears in Eisenbud's book: if $X$ and $Y$ are equidimensional CM subschemes, and $X\cap Y$ has codimension 1, and is CM, then $X\cup Y$ is CM. In case X has a few irreducible components, and $Y$ intersects some in higher codimension (but such intersections are embedded into other intersections of codimension 1, so that the total intersection in a sense does have codimension 1), does the statement still hold? How to prove this statement?

4. Is the intersection of an admissible $A(R)$ with a k-plane $A_J$ corresponding to a subset $J$ reduced?

If all the answers are positive, one can deduce 2 by induction from 3 and 4.

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For 1, the answer is yes. It is the locus of all points where at least n-k coordinates are equal to 0 (if I understand you). The reduced ideal is generated by all squarefree monomials of degree k+1. Such monomial ideals are Cohen-Macaulay in any characteristic by Stanley-Reisner theory. See for example Example 3.2(b) of this paper http://arxiv.org/abs/math/0409097

(A general reference being Miller-Sturmfels, Combinatorial Commutative Algebra, specifically Theorem 5.53. This might help answer question 2, but I was unable to check.)

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Thank you very much! – Alexander Braverman Aug 30 '11 at 20:24

Counter-example to (2): Take $N=6$, $k=3$ and $R$ to be $156$, $126$, $246$, $234$, $345$, $135$. The point is that, if you draw the simplicial complex with vertices $1$, ... $6$ and these faces, it is two dimensional and has nontrivial cohomology in $H^1$. (Specifically, if I didn't make any typos, it is a triangulation of the annulus.) By Reisner's criterion, if this were to be CM, it would only have top dimensional cohomology.

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Thanks a lot! Your link to Reisner's criterion doesn't work - could you maybe give me a reference? – Alexander Braverman Aug 30 '11 at 20:24
The standard reference for all this material is Richard Stanley, Combinatorics and commutative algebra. Progress in Mathematics, 41. Birkhäuser (1996). But the Wikipedia page is pretty good. If I send you to en.wikipedia.org/wiki/Stanley%E2%80%93Reisner_ring , does that link work for you? – David Speyer Aug 30 '11 at 21:10
Yes, now it works; thanks a lot! – Alexander Braverman Aug 30 '11 at 21:30

(3) is nonsensical. For schemes of dimension $>0$, CM implies S1 which means no embedded components, like you were suggesting $X\cap Y$ to have.

(4) is true. A subscheme of affine space is a union of coordinate planes iff its ideal is generated by squarefree monomials. Concatenate the generators to intersect the schemes, and it's still generated by squarefree monomials, so againa union of coordinate planes.

(Sillier proof, over ${\mathbb F}_p$, and then by tricks any field: a subscheme is a union of coordinate planes iff it's compatibly split w.r.t. the standard Frobenius splitting on affine space. And intersections of compatibly split subschemes are again compatibly split.)

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No, I didn't mean that $X\cap Y$ has embedded components. I meant that $X\cap Y$ is reduced and has codimension 1 in both, but this is not necessarily true for the corresponding intersections of their irreducible components. – Alexander Braverman Aug 31 '11 at 3:20
So $X = X_1 \cup\ldots\cup X_k$, and some $Y\cap X_i$ are high codimension? Then, I don't understand what you're asking. Does Eisenbud assume $X$ or $Y$ is irreducible? I wouldn't expect so. – Allen Knutson Sep 1 '11 at 2:57
Probably you are right that there is no question - I was just surprised by that statement... – Alexander Braverman Sep 1 '11 at 3:37