  We construct a non-commutative scheme that behaves as if it is the space of
Penrose tilings of the plane.
  Let k be a field and B=k<x,y>(y^2). We consider B as the homogeneous
coordinate ring of a non-commutative projective scheme. The category of
"quasi-coherent sheaves" on it is, by fiat, the quotient category
QGr(B):=Gr(B)/Fdim(B) and the category of coherent sheaves on it is
qgr(B):=gr(B)/fdim(B), where gr(B) is the category of finitely presented graded
modules and fdim(B) is the full subcategory of finite dimensional graded
modules. We show that QGr B is equivalent to Mod S, the category of left
modules over the ring S that is the direct limit of the directed system of
finite dimensional semisimple algebras S_n=M_{f_n}(k) + M_{f_{n-1}}(k) where
f_{n-1} and f_n$ are adjacent Fibonacci numbers and the maps S_n \to S_{n+1}
are (a,b)--->(diag(a,b),a). When k is the complex numbers, the norm closure of
S is the C^*-algebra Connes uses to view the space of Penrose tilings as a
non-commutative space. Objects in QGr B have projective resolutions of length
at most one so the non-commutative scheme is, in a certain sense, a smooth
non-commutative curve. Penrose tilings of the plane are in bijection with
infinite sequences z=z_0z_1 ... of 0s and 1s with no consecutive 1s. We
associate to each such sequence a graded B-module, a "point module", that
becomes a simple object O_z in QGr B that we think of as a "skyscraper sheaf"
at a "point" on this non-commutative curve.
  Tilings T_z and T_{z'} determined by two such sequences are equivalent, i.e.,
the same up to a translation on R^2, if and only if O_z is isomorphic to
O_{z'}. A result of Herbera shows that Ext^1(O_z,O_{z'}) is non-zero for all z
and z'. This as an algebraic analogue of the fact that every equivalence class
of tilings is dense in the set of all Penrose tilings.
