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I am trying to grapple with the basics of scheme theory. Is the scheme defined by Spec[C[x,y,z]/(xy,yx,zx)] a variety? What do the points look like?

I suspect it represents points satisfying xy = yz = zx = 0, so it should have three irreducible components {x = y = 0}, {y = z = 0} and {z = x = 0}. The motivation for this example comes from statistical mechanics and it has quite a bit more content:

Consider the space of 3x3 matrices (entries in C) with the following deformation of the matrix product: $P \circ Q = \sum_{i \leq j \leq k, cyc} P_{ij} P_{jk} $. Here we summing over j such that i, j, k appear in cyclic order mod 3. It appears in a set of slides on The Combinatorics of the Brauer Loop Scheme.

The paper then proceeds to define a scheme using equations in matrices. In the space of matrices with 0's along the diagonal, we consider the matrices with $M \circ M = 0$. In coordinates, the matrix product therefore looks like: $ \left( \begin{array}{ccc} 0 & b_{12} & b_{13} \\\\ b_{21} & 0 & b_{23} \\\\ b_{31} & b_{32} & 0 \end{array} \right) \circ \left( \begin{array}{ccc} 0 & b_{12} & b_{13} \\\\ b_{21} & 0 & b_{23} \\\\ b_{31} & b_{32} & 0 \end{array} \right) = \left( \begin{array}{lll} 0 & 0 & b_{12}b_{23} \\\\ b_{23}b_{31} & 0 & 0 \\\\ 0 & b_{31}b_{12} & 0 \end{array} \right)$ As all the entries on the right side vanish, this defines three equations in six unknowns. (Actually, only the "clockwise" matrix entries seem to be involved.)

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  • $\begingroup$ Varieties are usually defined to be irreducible, and your scheme is not irreducible. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 14, 2010 at 5:18

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As the coauthor of the relevant paper (and as Mariano already said), I consider irreducibility an important enough concept that it's worth reserving the word "variety" for only reduced, irreducible schemes.

Even if we only cared about reducedness, we don't know whether those equations define a radical ideal (though we conjecture so).

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  • $\begingroup$ Which equations? The first ideal (xy, yz, zx) John describes is radical. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2010 at 12:43
  • $\begingroup$ The equations "$P$ has zero diagonal, $P\circ P = 0$", from his second ideal, taken from our paper. $\endgroup$
    – Allen Knutson
    Commented Jan 14, 2010 at 15:40
  • $\begingroup$ Perhaps I should ask you these questions directly...? $\endgroup$
    – john mangual
    Commented Jan 15, 2010 at 23:02

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