an introduction to toric surfacesjsidman/pcmi08/book.pdf · 1.1. the ideal-variety correspondence....

AN INTRODUCTION TO TORIC SURFACES

JESSICA SIDMAN

1. An introduction to affine varieties

To motivate what is to come we revisit a familiar example from highschool algebra from a point of view that allows for powerful general-izations.

Example 1.1. In R2, the parabola is the set of all points satisfyingy = x2. We see that for any value of x, there is a unique value of y thatsatisfies the equation. In fact, we can parameterize the parabola witha map φ : R→ R2 where φ(t) = (t, t2).

For each value that the parameter t assumes we get a point in R2on the parabola. For example, φ(−5) = (−5, 25), φ(0) = (0, 0), φ(5) =(5, 25) are all points on the parabola. As t varies over all of the realline, φ traces out the entire parabola.

In R3 there is a natural generalization of the parabola called thetwisted cubic.

Example 1.2. This time, instead of starting off with an implicit equa-tion, we will begin by giving a parameterization of our curve in 3-space.Let φ : R→ R3 be defined by φ(t) = (t, t2, t3).

What does this curve look like? Suppose that you could fly wayabove the xy-plane and stare straight down in the z-direction. Youwouldn’t be able to distinguish between points at different heights. Ineffect, you would just be seeing a curve in the xy-plane given by theparameterization gotten by only looking at the first two coordinates ofφ. So, you would see the parabola with parameterization (t, t2).

However, if you were to look at this curve with the xy-plane at eye-level, you would see that when t is positive, the points twist up out ofthe xy-plane above points on the parabola, and when t is negative, thecurve twists down below the xy-plane in an analogous fashion. We see

The author is partially supported by NSF grant DMS 0600471 and the ClareBoothe Luce Program.

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2 JESSICA SIDMAN

the twisted cubic from two different points of view below.

10

2.5

5

25

30

0.0−5.0

15

20

−2.5

0

35

5.0

010

5.0

−20020

−100

2.5

0

100

0.0

200

−2.5

30

−5.0

This parameterized curve can also be defined as the solution setof a system of polynomial equations. We can see that every pointsatisfies the equation y = x2 because the square of the first coordinateof φ(t) = (t, t2, t3) is equal to its second coordinate.

Note also that the product of the first two coordinates of φ is equalto the third. Thus, if we let f(x, y, z) = xy − z we see that f(φ(t)) =f(t, t2, t3) = t · t2 − t3 = 0. In the picture below we see that the pointsof the image of φ, i.e., the points of R3 of the form (t, t2, t3), are exactlythe intersection of the surfaces y = x2 and xy − z = 0.

−5.0

−2.5

0.00

−200

2.510

20

−100

5.030

0

100

200

5.0

2.5x

0.0

−2.5

−5.0

−20

0y20−200

−100

0

100

200

5.0

−200

2.5

−100

0.0

x

−2.5

0

−5.0

100

20

0

y

200

−20

By computing t2− t · t = 0 and t · t2− t3 = 0, we can see that the imageof φ is contained in the intersection of the solution set of the equations.Morever, suppose that (u, v, w) is a solution of the system of equations.Since (u, v, w) is a solution of the first equation, we see that v = u2,so that (u, v, w) = (u, u2, w). But the second equation implies thatz = xy, and this implies that w = u · u2 = u3. Therefore, any solution

AN INTRODUCTION TO TORIC SURFACES 3

of the system of equations given by y−x2 = 0 and xy− z = 0 is of theform (u, u2, u3) for some u ∈ R. In other words, all solutions are pointsin the image of φ.

In the examples above, for any real number t, a corresponding pointis given explicitly by φ(t), and as t varies, the points φ(t) trace out acurve. There is another way of describing these curves. We may realizethem implicitly as the solution sets of systems of equations.

The basic object of study in algebraic geometry is a variety, or thesolution set of a finite system of polynomial equations. We will give aformal definition of a variety that will work in many different settings.To do this, we must introduce some notation.

Let k be a field. We will be interested in the geometry of sets in kn.Although kn is a vector space, we will not be performing any vectoroperations. To emphasize that we are thinking of kn just as a set ofpoints and not as a vector space, we call it affine n-space and writeAnk or An if the ground field k is understood. Given indeterminatesx1, . . . , xn we write k[x] = k[x1, . . . , xn] for the ring of polynomials inn variables with coefficients in the field k.

Definition 1.3. Let f1(x), . . . , fr(x) be polynomials in the ring k[x].The set

V (f1, . . . , fr) = {p ∈ An | f1(p) = · · · = fr(p) = 0}is the affine variety defined by the equations f1(x) = 0, . . . , fr(x) = 0.

Example 1.4. Let f(x, y) = x+y and g(x, y) = x−y. We can analyzethe variety contained in A3R using linear algebra. We know that thesolution set of the system of equations

x+ y = 0

x− y = 0is the same as the solution set of

x = 0

y = 0.

Thus V (x+ y, x− y) = V (x, y) = {(0, 0, c) | c ∈ R}.

We can see already with this linear example that different systems ofequations may define the same variety. This phenomenon also occurswith varieties defined by higher degree equations.

Example 1.5 (The twisted cubic II). We have seen that the equationsxy−z = 0 and y−x2 = 0 have common solution set equal to all pointsof the form (t, t2, t3). Looking at the parameterization, we can also see

4 JESSICA SIDMAN

that these points satisfy the equation y2 − xz = 0. In fact, the pair ofequations y2 − xz = 0 and xy − z = 0 also define the twisted cubic:Suppose that (r, s, t) satisfies both equations. Then (r, s, t) = (r, s, rs)by the second equation. Moreover, by the first, s2−r2s = s(s−r2) = 0.So, when s 6= 0, then s = r2, and we have (r, s, t) = (r, r2, r3). Whensecond coordinate is zero, xy − z = 0 says that the third coordinate isalso zero.

1.1. The ideal-variety correspondence. In Example 1.5 we sawthat the system of equations defining a variety may not be unique.In fact, it is never unique, even if a variety is defined by a single equa-tion f(x) = 0, we may multiply this equation by any nonzero scalarwithout changing the solution set.

As the equations defining a variety are not uniquely determined, itmakes sense then to consider the set of all equations that the points ofa variety satisfy.

Definition 1.6. Let X ⊆ kn. We define I(X) to be the set

I(X) = {f(x) ∈ k[x] | f(p) = 0 ∀ p ∈ X}.

Although this set is not finite, we will see that it has the structureof an ideal, which can be generated by a finite set of polynomials bythe Hilbert Basis Theorem.

Definition 1.7. Let R = k[x] be a polynomial ring. A nonemptysubset I contained in R is an ideal if

(1) The sum f(x) + g(x) ∈ I for all f(x), g(x) ∈ R.(2) For each f(x) ∈ I and r(x) ∈ R, r(x)f(x) ∈ I.

Theorem 1.8. Let X ⊆ kn. Then I(X) is an ideal.

Proof. The zero polynomial, z(x) = 0 is clearly in I(X) because itvanishes at every point of kn. Let f(x), g(x) ∈ I(X). If p ∈ X, thenf(p) = g(p) = 0. Therefore, when we evaluate (f + g)(x) at p wehave (f + g)(p) = f(p) + g(p) = 0 + 0 = 0. Hence, the polynomial(f + g)(x) ∈ I(X). Similarly, if r(x) ∈ R, then (rf)(x) evaluated atp is (rf)(p) = r(x)f(p) = r(p) · 0 = 0. We conclude that I(X) is anideal. �

We have now seen that we can associate an ideal I(V ) to any affinevariety V . It is natural to ask if the converse is true: can we associate avariety to every ideal in a polynomial ring? The answer to this questionis affirmative, and follows from the Hilbert Basis Theorem.


Theorem 1.9 (Hilbert Basis Theorem). Let I ⊂ k[x] be an ideal ina polynomial ring. Then there exist a finite set f1(x), . . . , fm(x) suchthat

I = {r1(x)f1(x) + · · ·+ rm(x)fm(x) | ri(x) ∈ k[x] ∀ i}.We say that the polynomials f1(x), . . . , fm(x) are a finite set of gener-ators for I. We write I = 〈f1(x), . . . , fm(x) rangle.

Example 1.10 (Parabola).

Example 1.11 (Twisted cubic III).

Definition 1.12. Suppose that I is an ideal in k[x]. The variety asso-ciated to I is the set

V (I) = {p ∈ kn | f(p) = 0 ∀f(x) ∈ I}.

Theorem 1.13. If I ⊆ k[x] is an ideal, then V (I) is an affine variety.

Proof. By the Hilbert Basis Theorem, I = 〈f1(x), . . . , fm(x) rangle.�

1.2. Singularities.

1.3. Dimension.

2. Projective space

2.1. Homogeneous coordinates.

2.2. An open affine covering of projective space.

2.3. Projective varieties and homogeneous ideals.

2.4. Limit points.

2.5. Degree.

3. Polytopes

We are familiar with two and three-dimensional polytopes. For ex-ample, triangles, rectangles, and pentagons are all two-dimensionalpolytopes, or polygons. In dimension three we have the tetrahedron,the cube, and many other three-dimensional polytopes.

The geometry and combinatorics of polytopes is an interesting sub-ject in its own right. For us, the point of interest is that polytopescarry a surprising amount of information about a certain class of vari-eties called projective toric varieties. We have seen that the algebraicdefinitions of some important properties of varieties are very difficult

6 JESSICA SIDMAN

to work with in practice. However, there is a dictionary relating prop-erties such as smoothness, dimension, and degree for toric varieties toproperties of polytopes that are easier to compute.

3.1. Polytopes as convex hulls. Although we will mainly be inter-ested in 2-dimensional polytopes, i.e., polygons, for which most of ushave a lot of intuition, we will give careful definitions for arbitrarydimensions as we will need them when we begin to decompose ourpolygons.

Perhaps the most basic issue to address is: How do we describe apolygon? Note that we need only know the vertices. In the planewe can construct a polygon from vertices by drawing edges betweenvertices in such a way that we enclose a region P that is convex, i.e., ifp, q ∈ P then so is the line segment joining p and q.

To think about how to make this process rigorous in general, let usrecall how to parameterize the line joining two points.

Example 3.1. Let p = (−2,−1) and q = (3, 1) in the plane. Ourparameter t will vary in the interval 0 ≤ t ≤ 1. The formula

`(t) = (1− t)p+ tq= (1− t)(−2,−1) + t(3, 1)= (−2 + 2t,−1 + t) + (3t, t))= (−2 + 5t,−1 + 2t)

parameterizes a line segment in R2 joining p and q as we see that`(0) = (−2,−1) = p and that `(1) = (3, 1) = q.

p

q

We want a generalization of this idea to an arbitrary finite set ofpoints. It may help to think of the construction in Example 3.1 inthe following way. Given two points p and q, `(t) is given by linearcombinations sp+ tq in which the values of r and q are constrained in


two ways. First, s is really a function of t where s = (1−t). This relationbetween the coefficients explains why `(t) is one-dimensional insteadof two-dimensional. We have also bounded the value of t between 0and 1. This explains why `(t) just describes a line segment instead ofa whole line.

The line segment `(t) is the smallest convex region containing thepoints p and q. The following definition tells us how to construct thesmallest convex region containing an arbitrary finite set of points inRn.

Definition 3.2. Let V = {p1, . . . , pn} ⊂ Rm. The convex hull of V isthe set

convV = {r1p1 + · · ·+ rnpn | r1 + · · ·+ rn = 1, 0 ≤ ri ≤ 1}.

The set convV is a V−polytope.

Example 3.3 (The cube). The square is a 2-dimensional cube and isequal to conv{(0, 0), (1, 0), (0, 1), (1, 1)}. The n-dimensional cube maybe realized as the convex hull of the 2n 0-1 vectors in Rn.

Example 3.4 (The simplex). The 2-dimensional standard simplex isconv{(0, 0), (1, 0), (0, 1)}. The n-dimensional simplex is the convex hullof the standard basis vectors together with the origin.

Note that this definition ensures that the resulting set is indeed con-vex.

Proposition 3.5. Let P be the convex hull of points p1, . . . , pn. Theline segment joining any pair of points in P is contained in P.

Proof. Suppose that p and q are in P. This implies that p =∑ripi and

q =∑sipi where 0 ≤ ri, si,≤ 1 and

∑ri =

∑si = 1. A point on the

line joining p and q has the form

p+ (1− t)q = tr1p1 + · · · trnpn + (1− t)s1p1 + · · · ,+(1− t)snpn= (tr1 + (1− t)s1)p1 + · · ·+ (trn + (1− t)sn)pn.

8 JESSICA SIDMAN

As 0 ≤ t ≤ 1, we see that 0 ≤ tri, (1− t)si ≤ 1. Moreover,(tr1 + (1− t)s1) + · · ·+ (trn + (1− t)sn)= t(r1 + · · ·+ rn) + (1− t)(s1 + · · ·+ sn)

= t · 1 + (1− t) · 1= 1.

�

Theorem 3.6. Let V = {p1, . . . , pn} ⊂ Rm. The convex hull of V isthe smallest convex set containing V . In other words, if C ⊂ Rm is aconvex set containing V , then convV ⊆ C.

Proof. See the discussion on pages 3 and 4 of [?]. �

Up until now we have relied upon an intuitive notion of dimension,but we will need a rigorous definition. Since our definition of the convexhull of a finite set of points is based upon taking linear combinations itmakes sense that we may define dimension in terms of linear algebra.

Definition 3.7. Let P ⊆ Rm be the convex hull of points p1, . . . , pn.The affine span of P is

aff P = {r1p1 + · · · rnpn | r1 + · · ·+ rn = 1, ri ∈ R}.

To motivate the definition, let us consider the example of a linejoining two points again.

Example 3.8. Again, let p = (−2,−1) and q = (3, 1) in the plane. Ifwe allow our parameter t to range over all of R then `(t) = (1− t)p+tq = (−2 + 5t,−1 + 2t) parameterizes the line joining (−2,−1) and(3, 1) defined by y = 2

5x − 1

5. We can translate this line to the origin

by subtracting the point (−2,−1) from each point to get (5t, 2t) =t(5, 2), which is the span of a single nonzero vector and hence is one-dimensional

The affine span of a set of points is just a translation of a linearsubspace so that it does not necessarily contain the origin. If we thinkof the affine span in this way, then it makes sense to define dimensionby translating P to the origin and letting the dimension of the affinespan be the dimension of the linear subspace through the origin.

Observe that if, say, the point pn is at the origin, then

{r1p1 + · · · rnpn | r1 + · · ·+ rn = 1, ri ∈ R}.is just the span of the first n− 1 points as the value of rn is irrelevant,allowing r1, . . . , rn−1 to vary freely.


Definition 3.9. Let P = conv{p1, . . . , pn} ⊂ Rm. Then the dimensionof P is the dimension of the linear subspace aff(P − pn).3.2. Polytopes as intersections of halfspaces. We have a goodintuitive idea for the vertices and edges of a polygon in the plane.However, we will need a rigorous definition of these ideas that gener-alizes to higher dimensions. For this, an alternative way of describingpolytopes via intersections of halfspaces will be useful.

Example 3.10 (The simplex revisited). Consider the following in-equalities: x ≥ 0, y ≥ 0, x+ y ≤ 1.

Definition 3.11. Suppose we are given a set of vectors n1, . . . ,nk ⊂Rm and real numbers ai ∈ R. The vectors in Rm that satisfy the in-equalities x · ni ≥ ai for all i form an H- polyhedron. A boundedH-polyhedron is an H-polytope.Example 3.12 (Unbounded polyhedra). Consider the linear inequal-ities x ≥ 0, y ≥ 0, x + y ≥ 1. The resulting H-polyhedron depictedbelow is unbounded.

Theorem 3.13. EveryH-polytope is a V-polytope and every V−polytopeis an H-polytope. In other words, a polytope can be constructed from anintersection of halfspaces or as the convex hull of a finite set of points.

Proof. See section 1.1 of [?]. �

Example 3.14 (Equations for the standard simplex in Rn). Equationsfor the standard simplex are given by xi ≥ 0 for each i = 1, . . . , n and−x1 − · · · − xn ≥ 1.Example 3.15 (Equations for the n-dimensional cube). Equations aregiven by xi ≥ 0 and −xi ≥ 1 for each i = 1, . . . , n.Theorem 3.16.

10 JESSICA SIDMAN

3.3. The faces of a polytope. Now that we are able to describe poly-topes as the intersection of halfspaces, we can give a rigorous definitionof a face. It is instructive to think about the 2-dimensional case first.

Example 3.17 (Faces of the simplex in R2). . Consider the standardsimplex given by x ≥ 0, y ≥ 0, x + y ≤ 1. We want to define faces sothat we have three edges, or 1-dimensional faces, and three vertices, or0-dimensional faces.

What property distinguishes points on faces from points in the in-terior of the simplex? Suppose we consider an arbitrary halfspaceax + by ≥ c. The relationship between such a halfspace and our sim-plex falls into three different cases: the simplex does not lie in thehalfspace, the simplex lies entirely in the interior of the halfspace, thesimplex intersects the boundary of the halfspace and is contained inthe halfspace.

These situations are depicted below:

It is the third situation that interests us here. If the simplex lies ina halfspace and meets its boundary, then the intersection of the two iseither an edge or a vertex.

Definition 3.18. Let P ⊆ Rn be a polytope. We say that b · x = c isa supporting hyperplane of P if P is contained in the halfspace b ·x ≥ cand there exists a nonempty subset of points p ∈ P such that equationb · p = c. In other words, P intersects the boundary of the halfspace.If H is a supporting hyperplane of P , then H ∩ P is a face of P.

Example 3.19 (Supporting hyperplanes for the simplex). The simplex∆ in the plane has three edges and three vertices. It is easy to see thatfor each edge e there is exactly one supporting hyperplane He such thatHe ∩∆ = e.


It is more interesting to try to describe the supporting hyperplanesof a vertex. In the diagram below we depict several supporting hyper-planes which intersect the simplex in the same vertex.

Notice that each edge of a polygon is associated to a single sup-porting hyperplane, but that a vertex is associated to infinitely manysupporting hyperplanes. We next seek a way of organizing informationabout how supporting hyperplanes are related to one another.

To do this, we will think about multivariable calculus.

Example 3.20. Consider the function f(x, y) = −x− y. Its level setsare given by equations of the form x + y = c, where c is a constant.The value of c for which f(x, y) achieves its maximum on the simplexis exactly when c = 1 and −x− y = c is a supporting hyperplane.

−x−y=0−x−y=1/2

−x−y=1

If we want to talk about this family of hyperplanes, or lines, all weneed to do is give the normal vector (−1,−1).

Let us determine the set of vectors normal to the supporting hyper-plane of the origin and pointing into the simplex so that the supporting

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hyperplane maximizes a linear function on the simplex.

We see that if we translate the inward pointing normal vectors of thesupporting hyperplanes of the vertex to the origin we get a regionthat looks like a cone that lies between the rays spanned by (1, 0) and(−1,−1).3.4. The inner normal fan of a polytope. To discuss support-ing hyerplanes of vertices, and more generally, of faces of codimensiongreater than 1, we need a rigorous definition of a cone.

Definition 3.21. The cone generated by A = {p1, . . . ,pn} ⊂ Rm isthe set

coneA = {r1p1 + · · · rnpn | 0 ≤ ri}.A cone may also be described as the intersection of halfspaces. Thus

we can define supporting hyperplanes and faces of a cone in the sameway that we defined them for polytopes.

If P is a polytope and F is a face of P, then the inner normal vectorsof supporting hyplanes of P intersecting F form a cone. These conesfit together in a nice way.

Example 3.22. Below we depict the simplex in the plane togetherwith the cone of inner normals of supporting hyerplanes at each face.

Notice that the cones fit together in such a way that the intersectionof any pair of cones is another cone that is a face of each of them.

This brings us to the definition of a fan.

Definition 3.23. A polyhedral fan ∆ is a collection of cones with theproperty that if σ, τ ∈ ∆ then σ ∩ τ is a face of each and is also a conein ∆.


Theorem 3.24. The cones of inner normals of supporting hyperplanesof a polytope P fit together in a fan called the inner normal fan of P.

3.5. Exercises.

(1) Draw the convex hull of each set below.(a) {(0, 0), (1, 0), (0, 1)}(b) {(0, 0), (2, 0), (0, 2)}(c) {(0, 0), (1, 0), (1, 2)}(d) {(m, 0) | m ∈ Z>0} ∪ {(0,m) | m ∈ Z>0}

(2) Draw conv{(0, 0), (1, 0), (2, 0), (0, 1), (1, 1)}. Find a subset of thegiven points with the same convex hull.

(3) Draw conv{(1, 1, 0), (1,−1, 0), (−1, 1, 0), (−1,−1, 0), (0, 0, 1), (0, 0,−1)}.(4) Let p = (1, 0, 0) and q = (0, 1, 0).

(a) Draw the linear span of p and q.(b) Draw the affine span of p and q.(c) Can you explain the difference that you see?

(5) Consider the triangle P = conv(((0, 0), (2, 0), (0, 1), (1, 1)).(a) Draw contours of the function f(x, y) = x. At which points

of P does f attain its maximum? its minimum?(b) Repeat this for the function g(x, y) = y.(c) Now consider an arbitrary linear polynomial L(x, y) = ax+

by. Can you describe the points of P at which L attains itsminimum in terms of a and b?

(6) Consider the inequalities x ≥ 0, y ≥ 0,−y ≥ a, andx− y ≥ −b.(a) Draw the subset of R2 satisfying the inequalities if a = 1

and b = 2.(b) Draw the subset of R2 satisfying the inequalities if a = 3

and b = 2.(c) Draw the subset of R2 satisfying the inequalities if a = 1

and b = 3.(d) Do the three polyhedra above have the same inner normal

fans?

4. Affine toric varieties: from lattice points tomonomial mappings

In this chapter we introduce toric varieties embedded in affine space.We begin by giving embeddings and then show how to compute theideal of an affine toric variety from its parameterization. Throughoutsections 4.2 and 4.3, we follow Sturmfels’s book [3].

Notice that in the two examples above, the parameterizations hada particularly nice form; each coordinate was just a power of t. The

14 JESSICA SIDMAN

implicit equations defining these two varieties also had a special form;a monomial (product of indeterminates) minus another monomial.

In this section we will define affine toric varieties which may beviewed as higher dimensional analogues of these examples as they areparameterized by maps with monomial coordinates and may be cut outby binomial implicit equations.

4.1. Notation. We introduce the notation that we will use to describeour embeddings.

Definition 4.1. We call Zd ⊂ Rd the d-dimensional lattice and callelements of Zd lattice points.

Example 4.2. For example, (0, 0), (1, 0), (2, 1) are lattice points in R2.

Definition 4.3. Suppose that we have indeterminates x1, . . . , xd. Ifwe have a lattice point a = (a1, . . . ad) then we can associate to ita Laurent monomial in the xi’s: x

a11 · · ·xadn . (We call this a Laurent

monomial because we allow negative exponents.) We often denote thisLaurent monomial by xa using vector notation instead of writing outeach indeterminate and each exponent.

Example 4.4. Let C[x] = C[x1, x2, x3]. Then x“4 1 2

”= x41x

12x

23.

4.2. The variety XA.. Suppose we have a finite ordered set containingn elements, A = {a1, . . . , an} ⊂ Zd. We can use A to define a map toCn whose coordinates are the monomials ta. Since we allow monomialswith negative exponents, none of the elements in the domain can havea zero coordinate.

Definition 4.5. Let C∗ = C\{0}. Given A = {a1, . . . , an} ⊂ Zd, wedefine φA : (C∗)d → Cn by t = (t1, . . . , td) 7→ (ta1 , . . . , tan).

Example 4.6. Suppose that A = {1, 2} ⊂ Z. Then φA(t) = (t, t2).Restrict the domain to real numbers and think about the image of thismap in R2. It is a parameterization of the parabola in the plane minusthe point at the origin as in Example 1.1.

Example 4.7. Suppose thatA = {1, 2, 3} ⊂ Z. Then φA(t) = (t, t2, t3).Again, let us restrict the domain to real numbers and try to visualizewhat we get in R3. We see that we get the twisted cubic of Example1.2

Definition 4.8. The affine toric variety associated to A, denoted XA,is the Zariski closure of the image of φA. The image of φA is a denseopen subset of XA.


In our two previous examples, the variety XA is gotten by adding inthe origin. Here is a more interesting example in which we add morethan one point when we take the closure of the image of φA.

Example 4.9. Let A = {(1, 0), (0, 1), (1, 1)}. Then φA : (C∗)2 → C3 isgiven by φA(t1, t2) = (t1, t2, t1t2). Consider points of the form (0, y, 0),where y ∈ C∗. If we let t2 = y and consider limt1→0(t1, t2, t1t2) inthe complex numbers, we get (0, y, 0). As the Zariski closure of a set isclosed in the usual topology, these limit points must be inXA. Similarly,the line {(x, 0, 0) | x ∈ C} is contained in XA.

We can visualize the real points in the image of φA in R3 as a familiarsurface, the graph of f(t1, t2) = t1t2 minus two lines.

4.3. Computing the ideal of XA. A variety is not just a set of pointsin projective space. An algebraic variety is a set that arises as the set ofall points that are solutions to some finite set of polynomial equations.In this section we will learn how to compute the ideal of the varietyXA.

To do this, let’s think carefully about what it means for a polynomialon the ambient space to vanish at a point φA(t). If f(x1, . . . , xn) is apolynomial, then we need to compute f(φA(t)). This is done monomial-by-monomial. So, let’s focus on what it really means to substituteφA(t) into a monomial x

u. This is best done in an example. We willsee that if f = xu, the f(φA(t)) = t

Au where A is the d × n matrixwhose columns are the vectors a1, . . . , an.

Example 4.10. Let A = {(1, 0), (0, 1), (1, 1)} so that

A =

(1 0 10 1 1

).

Suppose that f(x1, x2, x3) = x3 − x1x2 = x(0,0,1) − x(1,1,0). Then

f(φA(t)) =f(t(1,0), t(0,1), t(1,1))

=t0·(1,0)t0·(0,1)t1·(1,1) − t1·(1,0)t1·(0,1)t0·(1,1)

=t

0@1 0 10 1 1

1A0BB@

001

1CCA− t

0@1 0 10 1 1

1A0BB@

110

1CCAWhat we observe is that substituting a monomial map into a mono-

mial xu is equivalent to applying a linear transformation to u usingthe matrix of the monomial map. From this it is clear that if we havea binomial f(x) = xu − xv, then f(φA(t)) is identically zero exactlywhen Au = Av. In fact, we have the following theorem:

16 JESSICA SIDMAN

Theorem 4.11 (Lemma 4.1 in [3]). Suppose A ⊂ Zd is a finite orderedset and that A is the associated d × n matrix. Let IXA be the ideal ofthe set XA. Then IXA is equal to the ideal

IA := 〈xu − xv | u− v ∈ kerA, u,v ∈ Zn≥0〉.

Proof. We will show that IA is contained in IXA with a computation.If f ∈ IA, then it is a linear combination (with polynomial coefficients)of elements of the form xu − xv where Au = Av. For each of thesebinomials

tAu − tAv = 0becuase Au = Av. Therefore, f(φA(t)) = 0 and f ∈ IXA .

Next, we will show that every element of IXA is in IA. Let U bethe set of all polynomials in IXA that are not in IA. For each f ∈ U,consider the lexicographic leading term of f , i.e., the term that wouldcome first in the dictionary. If U is not empty, then there is somenonzero f in U whose leading term is smallest. (The set of exponentvectors is bounded below by 0 and is discrete.)

Since f ∈ IA, f(φA(t)) = 0. Without loss of generality, assumethat the leading term of f is xu, so that the coefficient of the leadingterm is 1. If we expand f(φA(t)), we will see a term of the form t

Au.Since this term cancels when we simplify, there must be at least oneterm of the form γtAv with Au = Av in the expression. Now definef ′ = f−(xu−xv). Notice that the leading term of f ′ is lexicographicallysmaller than the leading term of f as we have gotten rid of our originalleading term xu and as xv appeared with nonzero coefficient in f, wehave not introduced any additional monomials. If f ′ is in U, then wehave contradicted the minimality of our choice of f. If f ′ is not in U,then neither is f which is again a contradiction. Therefore, U must beempty and the claim is proved. �

Corollary 4.12 (Corollary 4.3 in [3]). Assume the hypotheses of The-orem 4.11. Suppose that u ∈ kerA. Write u = u+ − u−, whereu+,u− ∈ Zn≥0. Then xu

+ − xu− ∈ IXA .

Example 4.13. Let A be as in 5.3. The vector u = (−1,−1, 1) is inthe kernel of A (and actually spans the kernel). We can write it as adifference of its positive and negative pieces: u = (0, 0, 1)− (1, 1, 0).

We will see in Exercise 1 that the binomials corresponding to a basisfor the kernel of A do not necessarily generate all of the ideal IA.However, if the kernel is one-dimensional, then the ideal IA is principal.

%beginclaim


4.4. Exercises.

(1) Consider the matrix

A =

(1 1 1 10 1 2 3

).

(a) Let v1 = (1,−2, 1, 0),v2 = (1,−1,−1, 1),v3 = (0, 1,−2, 1)Show that v1 is in the span of v2 and v3.

(b) Write down the binomials corresponding to v1,v2,v3.(c) Is v3 in the ideal generated by v1 and v2? Can you explain

your answer and how it relates to part (a)?(2) Consider the integer matrices

A1 =

1 1 10 1 00 0 1

, A2 =1 1 1 1 1 10 1 0 1 2 0

0 0 1 1 0 2

A3 =

1 1 1 1 10 1 0 1 20 0 1 1 0

, A4 =1 1 1 10 1 0 1

0 0 1 1

,(a) Let Ai be the set of columns of Ai. Write down φAi for each

i. (You might want to write each map using xa notation aswell as in the usual notation in which you write down eachvariable explicitly.)

(b) If IAi is nonzero, find at least three elements in IAi by hand.(3) Show that if B is an n×n integer matrix, then it has an inverse

(with integer entries) if and only if detB = ±1. We will denotethe set of all n× n invertible integer matrices by GLn(Z).

5. Projective toric varieties

We have defined an affine toric variety as the closure of the imageof a monomial map φA. Although we have a very nice description ofpoints in ImφA, it is not clear that the points that we add when wetake the closure of this set will have nice properties.

Let us revisit Example 1.1 again.

Example 5.1 (Parabola). Recall that a set that is closed in the Zariskitopology is also closed in the usual topology. Therefore, the Zariskiclosure of the image of the map φA(t) = (t, t

2) must contain its limitpoints in the usual sense. We can see that as t→ 0, φ(t)→ (0, 0). Wedo not get a limit point when we send to t to ∞.

However, if we think of our parabola in A2 as a parabola sitting insideof an affine open patch of P2, then we will be able to see the limits at

18 JESSICA SIDMAN

infinity by passing to other affine opens. For example, consider themap ψA : (C∗)2 → P2 given by ψA(t) = [1 : t : t2]. If t 6= 0 then[1 : t : t2] = [ 1

t2: 1t

: 1]. So, we can see that the limit as t → ∞ is[0 : 0 : 1].

The closure of the image of ψA is defined by the single equationxz = y2. When x is nonzero, we can multiply by a nonzero scalar sothat x = 1. In this affine open patch of P2, the equation reduces toz = y2 and the limit point in this patch is [1 : 0 : 0]. Similarly, whenz is nonzero, we get the equation x = y2 with limit point [0 : 0 : 1].When y is nonzero, then both x and z must also be nonzero if xz = y2.Therefore, we see that the closure of the image of ψA contains preciselytwo limit points not in ImψA.

More generally, giiven a finite set of lattice points P = {m0, . . . ,mn},define ψP : (C∗)2 → Pn by ψP(t) = [tm0 : · · · : tmn ]. Note that sincewe do not allow any ti = 0, this map is defined at every point of (C∗)2.Definition 5.2. The projective toric variety XP associated to a setof n + 1 lattice points P is the closure of the image of ψP in Pn. Theimage of a permutation of the coordinates of ψP is isomorphic to XPvia a linear change of coordinates on Pn. (See pg. 66 of [1], pg. 166 of[2], and pgs. 31 and 36 in [3].)

We would like to use Theorem 4.11 to compute the homogeneousideal of XP , but as Example 5.3 will show, some modification is neces-sary.

Example 5.3. Suppose we consider the points P = {(0, 0), (1, 0), (0, 1), (1, 1)}.Then ψP : (C∗)2 → P3 is given by ψP(t1, t2) = [1 : t1 : t2 : t1t2]. The ma-

trix

(0 1 0 10 0 1 1

)has a 1-dimensional kernel spanned by (0,−1,−1, 1).

Let f = z − xy. Note that f(1, t1, t2, t1t2) = 0, but this is not equiv-alent to the statement that f(ψP) = 0. The reason that these twostatements are not equivalent is that in projective space, the point[1 : t1 : t2 : t1t2] = λ[1 : t1 : t2 : t1t2] for any nonzero λ.

For example, suppose that ψP(1, 2) = [1 : 1 : 2 : 2] = [3 : 3 : 6 : 6].We see that f(1, 1, 2, 2) = 2−1 ·2 = 0, but that f(3, 3, 6, 6) = 6−3 ·6 =−12 6= 0. The problem, of course, is that f is not homogenous.Theorem 5.4. Let P = {m0, . . . ,mn} ⊂ Zm. Let ai be the d + 1-dimensional vector with first coordinate 1 and last d coordinates givenby mi. If A = {a0, . . . , an}, then IA is the ideal of XP .Proof. A homogeneous polynomial f is in the ideal of XP if and only iff(ψP(t) = 0 for every t ∈ (C∗)d. A homogeneous polynomial vanishesat ψP(t) if and only if it vanishes at λψP(t) for all nonzero λ.


From Theorem 4.11 we know that IA consists of all polynomialsf ∈ C[x0, . . . , xn] such that f(φA(t0, . . . , td)) = 0 for all (t0, . . . , td) ∈(C∗)d+1. As f(φA(t0, . . . , td)) = ft0(ψP(t1, . . . , td)), we are done. �

Remark 5.5. If A is as in Theorem 5.4, then the corresponding (d+1) × n matrix A has top row equal to the all 1’s vector. This implesthat any u in the kernel of A has the property that the sum of itscoordinates is zero. Equivalently, if we write u = u+u−, then the sumof the coordinates of u+ equals the sum of the coordinates of u−. Thisin turn implies that deg xu

+= deg xu

−and hence that xu

+ − xu− ishomogeneous.

Example 5.6. Let P = {0, 1, 2, 3} ⊂ Z. The corresponding matrix Ais (

1 1 1 10 1 2 3

).

In Exercise 1 you showed that v1 = (1,−2, 1, 0),v2 = (1,−1,−1, 1),v3 =(0, 1,−2, 1) are in the kernel of A. Note that the sum of the coor-dinates of each of these vectors is zero. Equivalently, the binomialswy − x2, wz − xy, xz − y2 are ll homogeneous of degree 2.

Typically, the sets P arise from lattice polytopes.

Notation 5.7. Suppose that P ⊂ Rd is a lattice polytope. Let P =P ∩ Zd. We will abuse notation and write ψP in place of ψP and XPfor the variety XP .

Example 5.8. The four lattice polygons below correspond to projec-tive toric varieties via the construction described above.

P1 = , P2 =

20 JESSICA SIDMAN

P3 = , P4 = ,

The corresponding maps are ψP1(t1, t2) = [1 : t1 : t2], ψP2(t1, t2) = [1 :t1 : t2 : t1t2 : t

21 : t

22], ψP3(t1, t2) = [1 : t1 : t2 : t1t2 : t

21], ψP4(t1, t2) = [1 :

t1 : t2 : t1t2]. We can compute their ideals using the following integermatrices.

A1 =

1 1 10 1 00 0 1

, A2 =1 1 1 1 1 10 1 0 1 2 0

0 0 1 1 0 2

A3 =

1 1 1 1 10 1 0 1 20 0 1 1 0

, A4 =1 1 1 10 1 0 1

0 0 1 1

.We will revisit these example frequently throughout the text to see howtheir toric realizations are related to more classical descriptions and toexplore how they are related to each other.

6. Limit points of projective toric varieties

In this chapter we will study the action of the torus (C∗)d on avariety XP and compute the orbits of the torus action. Restricting tothe case of surfaces for simplicity, we will see that this action helps usto decompose the limit points of XP into pieces that fit together in anice way.

6.1. The torus action. Recall the following definitions from grouptheory.

Definition 6.1. A group G acts on a set S if we have defined g · s ∈ Sfor every g ∈ G and s ∈ S so that

(1) e · s = s, where e ∈ G is the identity element.(2) (gh) · s = g · (h · s)).

The action of a group G on a set S defines an equivalence relation onS with s ≡ t if there exists some g ∈ G so that g · s = t. Given anelement s ∈ S, the orbit of s is

{t ∈ S | ∃g ∈ G with t = g · s}.

If s and t are in the same orbit, we write s ≡ t.


Definition 6.2. Let T = (C∗)d denote the d-dimensional algebraictorus. The set T is a group under coordinatewise multiplication.

Example 6.3. If T = (C∗)2, and (s1, s2), (t1, t2) ∈ T, then (s1, s2)(t1, t2) =(s1t1, s2t2). We will often use our monomial notation convention for el-ements of T, so that s denotes (s1, s2) and st = (s1t1, s2t2).The identityis (1, 1). The inverse of (s1, s2) is (

1s1, 1s2

), which is defined as neither s1nor s2 may be zero.

Definition 6.4. Given finite set of lattice points A ⊂ Rd, we get anaction of T = (C∗)d on the image of φA. If g ∈ T, define g · tAi bygai · tAi .

Note that g · φA(t) = φA(g1t1, . . . , gdtd). This shows that T leavesthe image of φA fixed.

Theorem 6.5. The action of T on ImφA is transitive.

Proof. Suppose that φA(s) and φA(t) are two points in the image ofφA. As T is a group, there is an element g such that gs = t. For thisg, we have g · φA(s) = φA(gs) = φA(t). �

Let’s look at some examples.

Example 6.6 (The triangle P1). The set P1∩Z2 = {(0, 0), (1, 0), (0, 1)}.These lattice vectors give the map φP1 : (C∗)2 → P2 given by t 7→ [1 :t1 : t2]. The torus action is g · [1 : t1 : t2] = [1 : g1t1 : g2t2].

Notice that the point [1 : 1 : 1] is in the image of ψP1 . We’ll determinethe orbit of the point [1 : 1 : 1] by computing g · [1 : 1 : 1] = [1 : g1 : g2].We see that {[1 : g1 : g2] | g ∈ T} is exactly the image of ψP1 . Therefore,the orbit of [1 : 1 : 1] is ImψP1 .

It follows from the definition that any group action is transitive oneach orbit. We can check that the action is transitive on the image ofψP1 directly. Indeed, suppose that [1 : t1 : t2] and [1 : s1 : s2] are twopoints with t, s ∈ (C∗)2. If we set gi = siti , then g · [1 : t1 : t2] = [1 : s1 :s2]. Observe that the orbit of [1 : 1 : 1] is exactly the set of points inP2 with no coordinate equal to zero since we are just multiplying thesecond two coordinates of P2 by nonzero elements of C∗. It follows thatthe orbits of this action on P2 must correspond to sets of points withfixed coordinates set to 0.

We can give a list of representatives for each orbit: {[1 : 1 : 1], [1 :1 : 0], [1 : 0 : 1], [0 : 1 : 1], [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1]}. The closureof the first orbit is all of P2, the closures of the next three orbits arethe three coordinate axes. The last three points are fixed by the torusaction, and each is the unique element in its orbit.

22 JESSICA SIDMAN

Example 6.7 (The square P4). As above, the orbit of [1 : 1 : 1 : 1] isall of ImψP4 . Each of the four points [1 : 0 : 0 : 0], [0 : 1 : 0 : 0], [0 :0 : 1 : 0], [0 : 0 : 0 : 1] is fixed because multiplying their coordinatesby nonzero coordinates leaves them unchanged and so each is its ownorbit.

There are four other orbits represented by [1 : 1 : 0 : 0], [1 : 0 :1 : 0], [0 : 1 : 0 : 1], and [0 : 0 : 1 : 1]. We can see that these orbitsare 1-dimensional. Indeed, g · [1 : 1 : 0 : 0] = [1 : g1 : 0 : 0], is anaffine line minus a point as are g · [1 : 0 : 1 : 0] = [1 : 0 : g2 : 0],g · [0 : 1 : 0 : 1] = [0 : g1 : 0 : g1g2] = [0 : 1 : 0 : g2] and g · [0 : 0 : 1 :1] = [0 : 0 : g2 : g1g2] = [0 : 0 : 1 : g1].

With the information we have seen so far it is not clear that we havefound all of the orbits of the action of T and XP4 . In fact, it is noteven clear that any of the orbit representatives other than [1 : 1 : 1 : 1]are in the toric variety. However, these points are limit points of theimage of ψP4 , and we will start to see how to compute them in the nextsection.

To generalize what we have seen in the examples above, notice thatthere are as many torus-fixed points as vertices in each polygon, andas many torus-invariant curves as edges. (Of course, for surfaces, thenumber or edges is always equal to the number of vertices, but asso-ciating points to vertices, etc is what is correct in higher dimensions.)The decomposition of a polygon into its 2-dimensional face, edges, andvertices corresponds to the orbit-closure decomposition of the corre-sponding toric variety.

6.2. Limits. In this section, we answer the question: what do we getwhen we take the closure of the image of our embeddings?

6.2.1. Affine varieties. We can use the torus action to find the limitpoints of XA. We will do this by parameterizing curves in (C∗)2 andtaking their limits as the parameter goes to zero. The limit will be apoint whose coordinates are all zero or 1. The orbit of such a pointwill either be the image of the torus, a dense subset of a torus-invariantcurve, or a torus-fixed point. The 1-dimensional torus-orbits togetherwith the torus-fixed points are the limit points that we add in whenwe take the closure of the image of φA.

The curves in (C∗)2 that we will use to find limit points have a specialform.

Definition 6.8 (pg. 37 in [1]). An integer vector v ∈ Z2 correspondsto a map λv : C∗ → (C∗)2 given by

λv(z) = (zv1 , zv2).


The image of λv is a subgroup of (C∗)2 and we call it a 1-parametersubgroup. If v 6= 0, then Imλv is a curve. (Note that λ0(z)(1, 1).)

Example 6.9. Suppose we start with A = {(1, 0), (0, 1)} so thatφA(t) = (t1, t2). We want to consider

limz→0

φA(λv(z))

for various lattice vectors v. If v = (v1, v2) then we get

limz→0

φA(λv(z)) = limz→0

(zv1 , zv2).

Let us see what is going on when v = (1, 2). Then λ(1,2)(z) = (z, z2)

is a parabola in the plane. We want to see what happens to thepoints φA(z, z

2) = (z, z2) on our surface as z → 0. In this case itis easy to see that the limit is (0, 0). If we had picked v = (−1, 0) thenlimz→0 φA(λv(z)) = limz→0(

1z, 1) which does not exist.

In fact, it is easy to see that the limit exists if and only if v1, v2 ≥ 0,i.e., that v is in the first quadrant (including the axes). Moreover, wecan see that there are four different possible limits, according to thecases v1 = v2 = 0, v1 > 0, v2 = 0, v1 = 0, v2 > 0, and v1, v2 > 0. Pickinga representative for each possibility we can summarize the results withthe table below.

v φA(λv(z)) limz→0 φA(λv(z))

(0, 0) (1, 1) (1, 1)(1, 0) (z, 1) (0, 1)(0, 1) (1, z) (1, 0)(1, 1) (z, z) (0, 0)

Let us look at another example.

Example 6.10. Suppose that A = {(−1, 0), (−1, 1)}. For which vdoes limz→0 φA(λv(z)) exist? Notice that φA(λv(z)) = φA(z

v1 , zv2) =(z−v1 , z−v1+v2). Therefore, the limit as z → 0 exists if and only if−v1,−v1 +v2 ≥ 0. Geometrically, this means that (v1, v2) lies in the in-tersection of the halfplanes v1 ≤ 0 and v1 ≤ v2 depicted below. Again,there are four possibilities: v1 = v2 = 0, v1 < 0, v2 = v1, v1 = 0, v2 > 0,

24 JESSICA SIDMAN

and v1 < 0, v2 > v1. Choosing representatives of each case we have

v λv(z) φA(λv(z)) limz→0 φA(λv(z))

(0, 0) (1, 1) (1, 1) (1, 1)(−1,−1) (1

z, 1z) (z, 1) (0, 1)

(0, 1) (1, z) (1, z) (1, 0)(−1, 1) (1

z, z) (z, z) (0, 0)

In general, the lattice points inA generate a cone. The limit limz→0 φA(λv(z))exists if and only if v is in the cone dual to the cone generated by A.This is just the set of vectors whos dot produce with every vector inconeA is nonnegative. There will be a special limit point for each faceof the dual of coneA. If coneA is 2-dimensional then it has precisely 4faces: one 2-dimensional face, two edges, and the cone point which isa zero-dimensional face. The limit points form a set of representativesfor the orbits of the action of T on XA. In the 2-dimensional case, thereis one dense orbit, one fixed point, and two orbits that are curves.

6.2.2. Projective varieties.

Fact 6.11. If P is a lattice polygon, then XP is the union of the imageof ψP together with a projective torus-invariant curve for each edge ofP. Two edges meet in a vertex in P if and only if the correspondingcurves in XP meet in a point that is fixed by the torus action.

Example 6.12 (The triangle P1). We have (t1, t2) ∈ (C∗)2 mappingto [1 : t1 : t2].

v λv(z) φP1(λv(z)) limz→0 φP1(λv(z))

(1, 1) (z, z) [1 : z : z] [1 : 0 : 0](1,−1) (z, 1

z) [1 : z : 1

z] [0 : 0 : 1]

(−1, 1) (1z, z) [1 : 1

z: z] [0 : 1 : 0]

(1, 0) (z, 1) [1 : z : 1] [1 : 0 : 1](0, 1) (1, z) [1 : 1 : z] [1 : 1 : 0]

(−1,−1) (1z, 1z) [1 : 1

z: 1z] [0 : 1 : 1]

Notice that the first three limit points are fixed by the torus action.The last three points have orbits equal to C∗. We obtain the closures ofthese orbits by adding in the appropriate fixed points to get projectivelines.

Example 6.13 (The big triangle P2). The lattice points in P2 give themap φP2 : (C∗)2 → P5 sending

(t1, t2) 7→ [1 : t1 : t2 : t1t2 : t21 : t22].


Again, we make a table

v λv(z) φP2(λv(z)) limz→0 φP2(λv(z))

(1, 1) (z, z) [1 : z : z : z2 : z2 : z2] [1 : 0 : 0 : 0 : 0 : 0](1,−1) (z, 1

z) [1 : z : 1

z: 1 : z2 : 1

z2] [0 : 0 : 0 : 0 : 0 : 1]

(−1, 1) (1z, z) [1 : 1

z: z : 1 : 1

z2: z2] [0 : 0 : 0 : 0 : 1 : 0]

(1, 0) (z, 1) [1 : z : 1 : z : z2 : 1] [1 : 0 : 1 : 0 : 0 : 1](0, 1) (1, z) [1 : 1 : z : z : 1 : z2] [1 : 1 : 0 : 0 : 1 : 0]

(−1,−1) (1z, 1z) [1 : 1

z: 1z

: 1z2

: 1z2

: 1z2

] [0 : 0 : 0 : 1 : 1 : 1]

The first three points correspond to fixed points. What happens to thelast three under the torus action? We have

g · [1 : 0 : 1 : 0 : 0 : 1] = [1 : 0 : g2 : 0 : 0 : g22],

g · [1 : 1 : 0 : 0 : 1 : 0] = [1 : g1 : 0 : 0 : g21 : 0],

g · [0 : 0 : 0 : 1 : 1 : 1] = [0 : 0 : 0 : g1g2 : g21 : g22].

What we see then, is that the closure of the image of φP2 containsthree plane conics. In fact, what we are seeing is an embedding of P2in which the three coordinate lines that we found earlier are all mappedto conics.

Note that the three edges in P1 all had length one measured alongthe lattice, and in the limit we got three lines. The edges of P2 all havelattice length 2, and in the limit we got three degree 2 curves. We seethat P3 has 3 edges of length 1 and one edge of length 2, so we expectto see limit points falling along three lines and one plane conic.

Example 6.14 (The trapezoid P3). For the trapezoid, we get the map(t1, t2) 7→ [1 : t1 : t2 : t1t2 : t21]. We find the torus invariant curves:

v λv(z) φP3(λv(z)) limz→0 φP3(λv(z)) = p g · p

(1, 0) (z, 1) [1 : z : 1 : z : z2] [1 : 0 : 1 : 0 : 0] [1 : 0 : g2 : 0 : 0](0, 1) (1, z) [1 : 1 : z : z : 1] [1 : 1 : 0 : 0 : 1] [1 : g1 : 0 : 0 : g

21]

(−1,−1) (1z, 1z) [1 : 1

z: 1z

: 1z2

: 1z2

] [0 : 0 : 0 : 1 : 1] [0 : 0 : 0 : g2 : g1](0,−1) (1, 1

z) [1 : 1 : 1

z: 1z

: 1] [0 : 0 : 1 : 1 : 0] [0 : 0 : 1 : g1 : 0]

We see that the orbit closures will be three projective lines and oneplane conic.

26 JESSICA SIDMAN

Example 6.15 (The square P4). We have the map φP4 : (C∗)2 → P3sending t 7→ [1 : t1 : t2 : t1t2].

v λv(z) φP4(λv(z)) limz→0 φP4(λv(z)) g · p

(1, 0) (z, 1) [1 : z : 1 : z] [1 : 0 : 1 : 0] [1 : 0 : g2 : 0](0, 1) (1, z) [1 : 1 : z : z] [1 : 1 : 0 : 0] [1 : g1 : 0 : 0]

(−1, 0) (1z, 1) [1 : 1

z: 1 : 1

z] [0 : 1 : 0 : 1] [0 : 1 : 0 : g2]

(0,−1) (1, 1z) [1 : 1 : 1

z: 1z] [0 : 0 : 1 : 1] [0 : 0 : 1 : g1]

We see that the orbit closures are 4 lines.

Right now the choices of v are a bit mysterious, but we will see laterhow to make sense of them.

6.3. Exercises.

(1) Suppose that a group G acts on a set S and that s, t ∈ S. Showthat if there exists a g ∈ G such that g · s = t, then there is anh ∈ G such that h · t = s.

(2) Prove that T = (C∗)d is a group.(3) Let P be as in Definition 6.4. Prove that the action defined

there extends to projective space: in other words, that g · [x0 :· · · : xn] = [gm0x0 : · · · : gmnxn] is well-defined.

(4) Show that if v ∈ Z2, then the set {(zv1 , zv2)} is a subgroup ofT.

(5) Find the torus fixed points of XPi , i = 3, 4.(6) Substitute t1 =

x1x0

and t2 =x2x0

into the map φP2 . Homogenizethe map to show that this allows us to identify XP2 with theimage of a map ν2 : P2 → XP2 . The map ν2 is the quadraticVeronese emebdding of P2 into P5. Show that ν2 is injective.

More generally, we have d-uple Veronese embeddings νd :Pn → PN whose coordinates may be given by a basis for themonomials in the homogeneous coordinates on Pn of degree d.What is N? Can you show that all of these maps are injective?

(7) We can define maps ψi from P2 to XPi where i = 3, 4 by deletingthe appropriate coordinates from ν2. These maps are not definedon all of P2. Determine the points where each map is undefined.

(8) Consider the closure of the maps ψi above. Can you identifythe points that we add when we take the closure? Are the mapsinjective? If not, can you determine which points map to thesame image?

(9) Notice that the map φP3 can be decomposed as

[1 : t1 : t2 : t1t2 : t21] = [1 : t1 : 0 : 0 : t

21] + t2[0 : 0 : 1 : t1 : 0].


Substitute t2 =y1y0

into each summand. Convince yourself that

XP3 has the following description: Map P1 into P4 simultane-ously as a plane conic and a line where the line and plane donot meet. Then construct a surface by taking the union of alllines connecting the image of p ∈ P1 on the conic to its imageon the line.

(10) Substitute t1 =x1x0

and t2 =y1y0

into the map φP4 and deho-

mogenize the map. Show that this gives an injective map fromP1 × P1 = {([x0 : x1], [y0 : y1]) | [x0 : x1] ∈ P1, [y0 : y1]) ∈ P1}to XP4 . This is the Segre embedding of P1 × P1 → P3. Can yougive a natural description of the 1-dimensional orbit closures inXP4 in terms of P1 × P1?

In general, we have the Segre embeddings Pm × Pn → PNwhose coordinates are all products of homgeneous coordinatesfor Pm with homogeneous coordinates for Pn. What is N? Canyou show that the Segre embeddings are injective?

7. Coordinate rings of affine toric varieties

In this section we will explore the connection between the set A andthe monomials that span the coordinate ring of XA.

We can view the coordinate ring of affine space as a model for thesetup for toric varieties more generally.

Example 7.1. The set A = {1, 2} gives rise to the familiar parame-terization of the parabola in the plane φA(t) = (t, t

2). We can think

of this map as giving a ring homomorphism φ#A : C[x, y]→ C[t] whereφ#A(x) = t and φ

#A(x) = t

2. Theorem 4.11 shows that IA = 〈y − x2〉 isthe kernel of this homomorphism. What is its image? As t is in theimage of φ#A, so is any polynomial in t. So we see that Imφ

#A = C[t].

By the First Isomorphism Theorem for rings C[x, y]/〈y − x2〉 ∼= C[t].

Example 7.2. Let A = {(1, 0), (0, 1)}. Then φA : (C∗)2 → C2 is givenby φA = (t1, t2). The map φA gives rise to a ring homomorphism φ

#A :

C[x1, x2] → C[t1, t2] as follows. Define φ#A(xi) = ti on the generatorsof C[x1, x2] and define φ#A(α) = α for each α ∈ C. If φ

#A is a ring

homomorphism, this information completely determines the map. Foreach monomial xa11 x

a22 we have

φ#A(xa11 x

a22 ) = φ

#A(x1)

a1φ#A(x2)a2 = ta11 t

a22

as φ#A preserves multiplication. Therefore, for a polynomial f(x) =∑αax

a we have φ#A(f(x)) =∑αat

a, and we see that the map is anisomorphism. Its image is spanned by the set of all monomials in t1

28 JESSICA SIDMAN

and t2. Note that the exponent vectors of these monomials form theset {(a, b) | a, b ∈ N}. This is the intersection of the cone generated byA with the lattice Z2.

More generally, suppose we have A = {a1, . . . , an} ⊂ Zd. The pa-rameterization φA gives rise to a ring homorphism φ

#A : C[x] → C[t]

where φ#A(xi) = tai . Theorem 4.11 implies that the kernel of φ#A is ex-

actly IA. Thus, the First Isomorphism Theorem tells us that C[x]/IA,which is the affine coordinate ring of XA, isomorphic to Imφ

#A.

From the description of φ#A, we can see that Imφ#A is spanned by

monomials. We want to describe the exponent vectors of these mono-mials.

Definition 7.3. Given A = {a1, . . . , an} ⊂ Zd we define the semigroupgenerated by A to be SA = {r1a1 + · · ·+ rnan | ri ∈ N}.

Theorem 7.4. The set of exponent vectors of monomials in Imφ#A isSA.

Proof. Since φ#A is a homomorphism, φ#A(x

r11 · · ·xrnn ) = tr1a1 · · · trnan =

tr1a1+···+rnan . Therefore, we can see that SA has the form claimed above.�

Suppose that σ is the cone generated by A. Let Sσ = σ ∩ Zd. ThenSA ⊆ Sσ. In general, SA may not be equal to Sσ.

Example 7.5. Let A = {(2, 0), (0, 2)}. Then Sσ = N2, but SA ={(a, b) | 2|a, d|b}.

In the following chapters we will restrict our attention to affine toricvarieties such that SA = Sσ. This means that the semigroup generatedby A is saturated, or that the affine coordinate ring of XA is normal.

Example 7.6. Suppose we consider the set

A′ = {(1, 0), (0, 1), (1, 1), (2, 0), (0, 2)}.Then SA′ = SA where A was defined in Example 7.2. Here, φA′ :(C∗)2 → C5, and the kernel of φ#A is the ideal IA′ = 〈x22 − x5, x1x2 −x3, x

21−x4〉. Note that in the quotient ring C[x]/IA′ , the coset x3 + IA′

is equal to x1x2 + IA′ as x1x2 − x3 ∈ IA′ . In fact, each indeterminatex3, x4, x5 can each be rewritten in terms of x1 and x2 modulo IA′ . Thus,the quotient ring has a basis of monomials in x1 and x2. Indeed, by theFirst Isomorphism Theorem for rings, C[x]/IA′ ∼= C[t] again.

In fact, if B ⊂ Z2 and SB = N2, then the ring C[x]/IB is isomorphicto a polynomial ring in two variables, which is the coordinate ring ofaffine 2-space. What we are seeing then, is that if we embedd A2 into


another An, then the ring C[x1, . . . , xn] modulo the ideal of the imageof A2 is isomorphic to the polynomial ring in two variables. Indeedeach embedding of A2 into affine n-space gives us another realizationof C[t1, t2] as the quotient ring of a polynomial ring.

Example 7.7. Suppose we let B = {(−1, 0), (−1, 1)}. Again, we havea ring homomorphism φ#B : C[x, y] → C[t1, t2] that is determinedby φ#B (x) =

1t1

and φ#B (y) =t2t1. The image of φ#B is the subring of

C[t1.t2, 1t1 ,1t2

] spanned by 1t1

and t2t1. The exponent vectors of monomi-

als in Imφ#B lie in the cone depicted below.

We can see that Imφ#B is isomorphic to C[x, y] as the kernel of B =(−1 −10 1

)is 0, which implies that IB is also zero.

This isomorphism of rings also gives us an isomorphism of semi-groups. The matrix B sends the standard basis vectors to (−1, 0) and(−1, 1). Since B is invertible, this map has an inverse.

Let σ ⊂ R2 be a rational convex polyhedral cone and Sσ = σ ∩ Z2be the semigroup of lattice points in the cone. If XA is an arbitraryaffine toric surface, the subring of C[t1, t2] spanned by the monomialswith exponents in SA can be thought of as the coordinate ring of XAwithout reference to any embedding. A specific choice of generators ofthe semigroup SA corresponds to a specific embedding

Example 7.8. Suppose now that we let A = {(1, 0), (1, 1), (1, 2)}. Letσ = coneA. The matrix

A =

(1 1 10 1 2

)

30 JESSICA SIDMAN

has kernel with basis (1,−2, 1).

7.1. Exercises.

(1) Let A = {a1, . . . , an} ⊂ Zd. We wish to define a ring homomor-phism φ#A : C[x]→ C[t]. Let φ

#A(xi) = t

a and φ#A be the identity

on C. If φ#A is a ring homomorphism, what is φ#A(∑αmt

m)?(2) Consider the quotient ring C[x]/IA′ as in Example 7.6. Find a

representative of the coset of x3x4 − x25 using only x1 and x2.

8. Affine open covers and inner normal fans

Example 8.1. Let Q0 = conv{(0, 0), (1, 0), (0, 1)} be the standardsimplex in the plane. We get the map ψQ0(t) = [1 : t1 : t2] whichmaps to U0. To see the affine variety XQ0 ∩ U0, as above let A0 ={(1, 0), (0, 1)}. Then φA0(t) = (t1, t2), and we see that XA0 = C2. Inthis case C[t1, t2] ∼= C[x, y] is just a polynomial ring. Moreover, notethat the exponent vectors of monomials in C[t1, t2] are precisely thelattice vectors in the first quadrant.

Now consider Q1 = conv{(−1, 0), (0, 0), (−1, 1)}, which is just atranslation of Q0. This gives rise to the map ψQ1(t) = [

1t1

: 1 : t2t1

].

Of course, [ 1t1

: 1 : t2t1

] = [1 : t1 : t2] if t1 6= 0, so then XQ0 = XQ1 eventhough ψQ1 maps (C∗)2 to U1. The affine variety that we see is again C2,


although in this case A1 = {(−1, 0), (−1, 1)}. and C[ 1t1 ,t2t1

] ∼= C[x, y]has exponent vectors contained in the cone below.

The affine toric variety gotten from translating the top vertex of thetriangle to the origin can be analyzed in a smilar fashion.

We introduce language that will allow us to discuss the exampleabove more precisely.

Definition 8.2. A finite set of vectors A = {a1, . . . , an} ⊂ Rd gener-ates a convex polyhedral cone

coneA = {λ1a1 + · · ·+ λnan | λi ∈ R≥0}.The dimension of a cone is the dimension of its linear span. A coneis rational if it can be generated by lattice vectors. We will simplywrite “cone” in these notes but will always mean a rational convexpolyhedral cone. We say that the cone is strongly convex if it doesn’tcontain v and −v for any v ∈ Rd. (See Proposition 3 on pg. 14 of [1].)

Definition 8.3. If σ ⊂ Rd is a cone, we can define the ring C[ta | a ∈Zd ∩ σ]. In other words, we have defined a ring that is generated bymonomials whose exponent vectors lie in the cone σ.

Assumption 8.4. Assume that A generates the set (coneA) ∩ Zd asa semigroup.

Fact 8.5. Let P ⊂ Rd be a lattice polygon with P ∩ Zd = {a0 =0, . . . , an}. If σ = cone{a1, . . . , an} andA = (a1 · · · an), then C[ta1 , . . . , tan ] ∼=C[x1, . . . , xn]/IA.

8.0.1. Keeping track of cones: fans and duals.

Example 8.6 (XP3). Suppose we try to draw the cone at each vertexof P3. Notice that we can’t draw all of them in the same picture withoutoverlaps.

32 JESSICA SIDMAN

There is a dual picture that involves all of the same data, but givesus better intuition about how the different affine patches interact witheach other.

Definition 8.7. If σ ⊂ Rd is a cone, then its dual is defined to be

σ̂ = {u ∈ Rd | u · v ≥ 0,∀v ∈ σ}.

Example 8.8 (Dual cones for P1). On the right we have depicted thecones that we get when we translate each vertex of P1 to the origin.Their duals are on the left. Notice that the dual cones fit togethernicely. Note that we denote the cones on the left by σi and the coneson the right by σ̂i, which might seem odd from our point of view sofar. However, in the literature one typically starts with the picture onthe left and then dualizes to produce the picture on the right. Sincethe dual of the dual of a cone is the original cone, the pictures conveyequivalent data, and we follow the standard conventions to ease thetransition to other readings.

γ

σ

γ

τ

σ

τ

Observe that the rays of the dual cones are exactly the normal vectorsto the edges of the polygon. The definitions below give us preciselanguage for describing how these cones fit together.

Definition 8.9. If σ ⊂ Rd is a cone, we define a face of σ to be a setof the form

{v ∈ σ | u · v = 0}

for some u ∈ σ̂.

Example 8.10 (Faces of some cone). We have depicted the cone gen-erated by (1, 0), (0, 1) together with each of its faces. Below each face


we give its dual.

Definition 8.11. A fan ∆ ⊂ Rd is a collection of cones satisfying(1) If σ ∈ ∆ and τ is a face of σ, then τ ∈ ∆.(2) If σ, τ ∈ ∆, then σ∩ τ is simultanenously a face of σ and a face

of τ.

Definition 8.12. If P ⊂ Rd is a polytope, its inner normal fan is thefan whose maximal cones are the duals of the cones at each vertex ofP.

If P ⊂ R2 is a lattice polygon, then we can get its inner normal fanas follows. For each edge e of P, translate the normal vector pointinginto P to the origin and let ρe be the ray spanned by this vector. Theserays are the rays of the inner normal fan of P.

8.1. Exercises.

(1) Draw the inner normal fans for each of the polygons that wehave seen so far.

(2) Pick a fan above. Draw the dual of each cone in the fan.(3) Blowup warmup: Let B ⊂ C2×P1 given by the equation x0y1 =

x1y0 be the blowup of C2 at 0 with projection map π : B →C2. When we blowup we replace 0 with a copy of P1 whichwe call the exceptional divisor. Moreover, the lines through 0get separated in the blowup and instead of intersecting eachother at 0, each now intersects the new copy of P1, at a pointcorresponding to its slope. We can see this explicitly as follows:(a) Let x1 −mx0 = 0 be the equation of a line with nonzero

slope. Recall that U0 is the coordinate patch where y0 = 1.Use the equation of the blowup to rewrite the equation ofthe line in the coordinates for U0.

(b) The new equation should factor in U0. What does this tellus geometrically about the zero set of this equation in U0?

(c) Use the coordinates of U0 to write down the points of π−1

of the line in C2 defined by x1 −mx0 = 0. You should seethat the inverse image “remembers” the slope.

34 JESSICA SIDMAN

(d) The closure of the inverse image of the nonzero points onyour line is its strict transform. Find the coordinates ofthe intersection of the strict transform of your line withthe exceptional divisor.

9. Next

Suppose that P ⊂ Rd is a lattice polytope with vertex a0 = 0. Letσ be the dual of the cone generated by P. The lattice points in thecone generated by the lattice points in P are the exponent vectors ofLaurent monomials in Aσ ∼= C[ta1 , . . . , tan ] ∼= C[x1, . . . , xn]/IA.

Remark 9.1. Since we are working in the dimension two today, andwe are not going to be looking at the equations of our toric varieties,we will let our semigroup algebras be subsets of S[x, y, x−1, y−1].

Observation 9.2. The ring Aσ corresponds to an “abstract” affinetoric variety Uσ. (See section 1.3 of [1] for a treatment of toric varietieswhich begins with this point of view.) In the setup above, XA ⊂ Cn isan embedding of Uσ into affine space.

Goals:

(1) Understand how a fan ∆ tells us how varieties Uσ fit togetherto give an abstract toric variety X∆.

(2) Understand how subdivisions of cones give maps of toric vari-eties.

(3) Use (1) and (2) to resolve the singularities of a toric surface!

9.1. Fans and duals II. In this section we will see that the dualpicture is useful not just because the cones fit together nicely. Thepoint is that faces of the dual cones correspond algebraically to thecoordinate rings of open subsets in a nice way. We follow the notationof [1] as closely as possible to help students who wish to transtion fromthese notes to the other text.

Assumption 9.3. In what follows today we will assume that d = 2throughout. This is the situation that we will depict pictorially. Thereare natural generalizations for arbitrary d, with the added complicationthat if d ≥ 3 specifying a collection of rays does not automaticallyspecify a complete fan.

Since we will be moving between cones and their duals frequently, itwill be useful to introduce notation that will help us to remember thata cone and its dual ought to live in different spaces.


Definition 9.4. As in [1], let MR denote the copy of R2 that containsP (and the cones at the vertices of P ). We let NR denote the copy ofR2 containing the inner normal fan of P. We denote the lattice in MRby M and write N for the lattice in NR.

Although Aσ can be realized as C[x1, . . . , xn]/IA (in many differentways, in fact), it is a ring in its own right independent of its realizationas a quotient of a polynomial ring. In fact, Aσ corresponds to an“abstract” affine toric variety Uσ ∼= XA via its spectrum of prime ideals.For our purposes, we will think only of the set of all maximal idealsof Aσ, for these ideals correspond to points of Uσ in the usual sense(closed points in the language of algebraic geometry). We will denotethis abstract affine toric variety by SpecmAσ, following pg. 14 of [1].

Our first goal today is to understand how to think about these ab-stract affine toric varieties Uσ and how a fan tells us how they gluetogether to give an arbitrary toric variety.

Here are some useful facts.

Lemma 9.5. (1) If τ ⊂ σ ⊂ N, then σ̂ ⊂ τ̂ and Aσ ↪→ Aτ .(2) If τ is a face of σ, then Aτ is gotten from Aσ by adjoining the

inverse of a monomial in Aσ.(3) The maximal ideals of Aτ correspond precisely to the maximal

ideals of Aσ which do not contain the monomial that is inverted.

Proof. We leave part (1) as an exercise. for part (2), see Proposition2 on pg. 13 of [1]. Part (3) requires a bit of commutative algebra.Let f ∈ Aσ be the monomial that we invert to get Aτ . If m ⊂ Aτis maximal, let m be its intersection with Aσ. If m is not maximal,then it is contained in some maximal ideal n. If n does not containf , then its expansion to Aτ is a nontrivial ideal strictly containing m,which contradicts the maximality of m. Therefore, every maximal idealcontaining m must contain f. But, Aσ is a Jacobson ring, which meansthat any prime ideal is the intersection of all of the maximal idealscontaining it. As m must be prime, this would imply f ∈ m, which isa contradiction because then m would contain 1. �

Example 9.6 (Cone for C2). Let σ = cone{e1, e2} ⊂ NR and note thatit is its own dual. The vectors in σ̂ ⊂MR correspond to the monomialsin C[x, y]. This is the coordinate ring of C2.

What do the faces of σ correspond to? The face cone{e1} has dualequal to the right half-plane. The corresponding algebra is C[x, y, 1

y].

This is the ring of polynomial functions on C2\{(x, 0) | x ∈ C}. Sim-ilarly, the face cone{e2} corresponds to C[x, y, 1x ] which is the ring ofpolynomial functions on C2 minus the y-axis.

36 JESSICA SIDMAN

9.2. Limit points and the fan. Suppose that P is a lattice polygon.How do we find the limit points corresponding to the faces of P?

Let ∆ be the inner normal fan of P. Each face of P corresponds toa cone in ∆ in a natural way. The maximal face of P corresponds tothe origin. If e is an edge of P , it corresponds to the 1-dimensionalcone spanned by an inner normal vector to e. If v is a vertex of P itcorresponds to the 2-dimensional cone that is dual to the 2-dimensionalcone generated by P when we translate v to the origin.

Theorem 9.7 (Claim 1 on pg. 38 of [1]). Let P ⊂ R2 be a latticepolygon with inner normal fan ∆. For a face Q of P, let σQ ∈ ∆ be thecone corresponding Q. If v is in the relative interior of σQ, then

limz→0

ψP (λv(z))

is the distinguished orbit representative of the orbit corresponding to Q.

Proof. See [1]. �

9.3. Fans and gluing varieties. Suppose that we have a fan ∆ inN. For each cone σ ∈ ∆ we have a ring Aσ that is the algebra onmonomials with exponent vectors in σ̂.

If cones σ1 and σ2 intersect in a face τ , then Aσi ↪→ Aτ for each i.These ring maps correspond to Uτ ⊂ Uσi as an open subset. So, we seethat the cones in N correspond to affine varieties which glue togetherover open sets corresponding to faces of cones in N.

Example 9.8 (P2). The three cones σ0, σ1, σ2 correspond to affinevarieties Uσi which correspond to the standard affine open cover of P2by coordinate patches Ui. Let’s see this and see how the ring maps giverise to the usual patching maps.

We see that Aσ0 = C[x, y], Aσ1 = C[ 1x ,yx], and Aσ2 = C[xy ,

1y]. Each

of these rings is the affine coordinate ring of a copy of C2 as they eachrequire two generators that do not satisfy any relations.

The intersection of σ0 and σ1 is the face τ equal to the nonnegative y-axis. Therefore, the dual of τ is the upper halfplane andAτ = C[x, 1x , y].The maximal ideals of Aτ have the form 〈x−a, y− b〉. Notice however,that such an ideal contains 1− a

x. Therefore, if 〈x−a, y−b〉 is a maximal

ideal, a 6= 0. (Otherwise, it contains 1 so is equal to the whole ring.)i.e., Uτ ∼= C∗ × C.

We know that Aσ0 ↪→ Aτ . In the map of varieties from Uτ → Uσ0 thepoint (a, b) ∈ Uτ goes to the point (a, b) ∈ Uσ0 since the inverse imageof 〈x− a, y − b〉 ⊂ Aτ is just the ideal 〈x− a, y − b〉 ⊂ Aσ0


We also have Aσ1 ↪→ Aτ . To see how to pull back the ideal 〈x−a, y−b〉we do a little computation.

〈x− a, y − b〉 = 〈1− ax,y

x− bx〉 = 〈1

a− 1x,y

x− ba〉

This shows us that (a, b) ∈ Uτ is the image of ( 1a ,ba) ∈ Uσ1 . (Or, that

(c, d) ∈ Uσ1 maps to (1c ,dc) ∈ Uτ .) We see that the points (a, b) in Uσ0

with a 6= 0 glue to the points ( 1a, ba) ∈ Uσ1 .

We’ll check the other transitions in the exercises.

9.4. Refining a fan.

Example 9.9 (P2 vs XP3). Consider the inner normal fans of P1 andP3. Notice that the only difference is that one of the cones has beensplit in two.

Definition 9.10. We say that a fan ∆′ is a refinement of ∆ if everycone in ∆ is the union of cones in ∆′.

The important point is that if σ′ ⊂ σ, then Aσ ↪→ Aσ′ . Therefore,we get a map of varieties going the other direction. More precisely,

Proposition 9.11. If ∆′ is a refinement of ∆, then there is birationalmorphism X∆′ → X∆.

This is discussed in [1]. See the exercise and hint on pg. 18 and moreon pgs. 22-23.

Example 9.12 (The blowup of C2 at 0). Consider the fan ∆ consistingof cone{(1, 0), (0, 1)} and all of its faces. Let ∆′ be the fan gotten bysubdividing the maximal cone of ∆ by the nonnegative span of (1, 1).So, ∆′ is a refinement of ∆.

Let σ be the maximal cone of ∆ so that Aσ = C[x, y]. Let σ0 =cone{(0, 1), (1, 1)} and σ1 = cone{(1, 1), (1, 0)}. We see that Aσ0 =C[x, y

x] and Aσ1 = C[xy , y]. Each of these rings is the coordinate ring of

a copy of C2. Moreover, since Aσ ↪→ Aσi we get maps of affine varietiesUσi → Uσ.

Let’s see what these maps do. The point (a, b) ∈ Uσ0 corresponds tomaximal ideal 〈x− a, y

x− b〉. Since

〈x− a, yx− b〉 = 〈x− a, y − bx〉 = 〈x− a, y − ba〉

we see that (a, b) 7→ (a, ab) and that the points (0, b) all map to theorigin.

Similarly, a point (c, d) ∈ Uσ1 corresponds to the ideal 〈xy − a, y − b〉which pulls back to the ideal 〈x− ab, y − b〉 in Aσ. Therefore, (c, d) 7→(cd, d) and all points of the form (c, 0) go to the origin.

38 JESSICA SIDMAN

In fact, the Uσi are exactly the standard affine patches covering theblowup of C2 at the origin. To see this, let’s glue them together. Thecones intersect in the ray τ spanned by (1, 1) so Aτ = C[x, xy ,

yx].

Let’s see which points (a, b) ∈ Uσ0 gets glued to points of Uσ1 . Theideal 〈x−a, y

x− b〉 ⊂ Aσ0 is the inverse image of an ideal with the same

generators in Aτ . The ideal in Aτ can be rewritten so that we can tellwhat its inverse image in Aσ1 is as follows. We want to see an ideal ofthe form 〈x

y− c, y − d〉 ⊂ Aσ1 .

Since

−1b

x

y(y

x− b) = x

y− 1b

andy

x(x− a) + a(y

x− b) = y − ab

we see that (a, b) ∈ Uσ0 corresponds to (1b , ab) ∈ Uσ1 and that for thiscorrespondence to hold, b 6= 0.

Here is a chart that summarizes the information of which pointsfrom Uσ0 and Uσ1 get glued together and where they go in Uσ underthe “blow down” map.

Uσ0 Uσ1 Uσ conditions for gluing(a, b) (1

b, ab) (a, ab) b 6= 0

(a, 0) − (a, 0) −(0, b) (1

b, 0) (0, 0) b 6= 0

(cd, 1d) (c, d) (cd, d) d 6= 0

− (c, 0) (0, 0) −(0, 1

d) (0, d) (0, d) d 6= 0

10. Smoothness I

Fact 10.1. If P is a lattice polygon, then the singular points of XPare isolated points. This implies that if p is a singular point its orbitmust be zero-dimensional and hence that p is a fixed point.

Suppose that P is a lattice polygon and v is a vertex which we mayassume is at the origin in R2.

Proposition 10.2. If we can find an element of GL2(Z) that bringsthe edges incident at v to the positive coordinate axes of R2, then XPis smooth at the fixed point corresponding to v.

Proof. As P lies in the first quadrant of R2, the fixed point correspond-ing to v is the image of (0, 0) in the natural extension of ψP to C2.Moreover, t1 and t2 appear as coordinates of ψP . Therefore,

∂ψP∂t1|(0,0)


and ∂ψP∂t2|(0,0) are linearly independent. These two vector span a 2-plane

that is tangent to XP at ψP (0, 0). Thus, the point is smooth. �

In the next lecture we will work on developing the local vocabularyto discuss smoothness algebraically and we will eventually see that theproposition above is an if and only if statement.

11. Smoothness II

Suppose we want to investigate whether the point at the origin in anaffine variety XA is singular. In C[x1, . . . , xn]/IA, the ideal correspond-ing to the origin is m = 〈x1, . . . , xn〉. (Of course, although x1, . . . , xnform a set of minimal generators for the ideal of the origin in Cn, theymay not be minimial generators of the ideal they generate in the quo-tient ring.)

In the theorem below we translate the computation of dimCm/m2

into terminology using only the lattice.

Theorem 11.1 (pgs. 28-29 in [1]). Let A be a d×n integer matrix andσ be the cone generated by the columns of A. Assume that the columnsof A generate the set σ∩Zd as a semigroup, i.e., that every element ofσ ∩ Zd is a finite sum of some subset of the columns of A.

The point 0 ∈ XA is smooth if and only if the dimension of XA isequal to the size of the set

{v ∈ (σ ∩ Zd) | v 6= 0, v 6= u + w, u,w ∈ (σ ∩ Zd)\{0}}.

Proof. Letm0 denote the ideal of the origin in C[ta1 , . . . , tan ] ∼= C[x1, . . . , xn]/IA.Note that m0 is spanned by monomials with exponents in (σ∩Zd)\{0}.The ideal m20 is spanned by products of monomials in m0. Hence, theexponent vectors of elements of m20 are precisely the set of elements inin (σ∩Zd)\{0} that are the nontrivial sum of two other such elements.

Therefore, to compute the vector space dimension of m0/m20, we just

need to count the vectors in (σ ∩ Zd)\{0} that cannot be written asthe nontrivial sum of two other elements in (σ∩Zd)\{0}. By definitionthe variety XA is smooth at 0 if and only if dimXA = dimCm0/m

20,

so we are done. �

Fact 11.2. If σ ⊂ Rd is a d-dimensional cone, then σ ∩ Zd generatesZd as a group. Essentially this is because there are no “holes” in thecone σ. For a converse to this see the Exercise on pg. 19 of [1].

Corollary 11.3 (Proposition on pg. 29 of [1]). The variety XA issmooth at 0 if and only if the first lattice vectors along the rays of σare a Z-basis for Zd.

40 JESSICA SIDMAN

Proof. The vector space m0/m20 has a basis consisting of monomials,

and this basis must contain the first lattice vectors along the rays of σ. Ifσ is d-dimensional, it must have at least d generators. This vector spacehas dimension d = dimXA if and only if 0 is a smooth point. Therefore,0 is a smooth point if and only if m0/m

20 has a basis consisting of the

first lattice vectors along the rays of σ. These vectors clearly generateσ ∩Zd if all of the elements of m20 are the sum of at least two of them.Since σ ∩ Zd generates Zd as a group, these lattice vectors must alsogenerate Zd as a group. �

Here is an example of a singular point on a toric surface.

Example 11.4 (Cone over a conic). Let P be the convex hull of{(0, 0), (1, 0), (1, 2)} depicted below.

We see the region containing the monomials in m0 and the regioncontaining the monomials in m20. The three black dots in the cone onthe right represent the monomials in m0 that are not in m

20. We see

that dimCm0/m20 = 3 6= dimXP = 2. Therefore we see that the fixed

point corresponding to 0 is a singular point.

12. Toric resolution of singularities

We have seen that a toric surface is nonsingular at a fixed pointcorresponding to vertex v if and only if the first lattice vectors alongthe cone edges meetig at v formed a Z-basis for Z2.

Question 12.1. Can we resolve these singularities staying in the toricworld? i.e., can we find a smooth toric surface that maps birationallyto a given singular toric variety?

The answer is – yes! Given any fan, we can subdivide cones untileach cone corresponds to a smooth affine toric patch. We get a toricresolution of singularities. We’ll work out examples of toric resolutionof singularities of surfaces tomorrow.


Remark 12.2. Recall our basic setup. We have a lattice polytopeP ⊂MR ∼= Rd with P ∩M = {a0 = 0, . . . , an}. Then if A = (a1 · · · an),the affine variety XA = XP ∩ U0 has coordinate ring C[ta1 , . . . , tan ] ∼=C[x1, . . . , xn].

Note: Let σ be the dual of the cone generated by P. For the ringC[ta1 , . . . , tan ] to be the algebra spanned by monomials whose exponentvectors are the elements of σ̂ ∩ M, we must be able to write everyelement of σ̂ ∩M as a sum of elements from the set {a1, . . . , an}. Thisis the same as saying that the lattice points in P must generate σ̂ ∩Mas a semigroup.

We need the condition above to hold for P at every vertex in order tosay that XP is isomorphic to the abstract toric variety that we associateto inner normal fan ∆ of P. In more advanced language, we need Pto correspond to a very ample divisor. This is always true if P is apolygon or if P has an inner normal fan corresponding to a smoothvariety. However, in dimensions greater than 2, P may correspond toa divisor that is merely ample on the abstract variety associated to itsinner normal fan. (See pgs. 70-72 of [1] for a discussion.)

The upshot of this remark is that everything we’ve done works inthe 2-dimensional case, and in higher dimensions more care is needed.

In today’s lecture we are following sections 2.2, 2.5, an 2.6 of [1].

Lemma 12.3. Let σ be a 2-dimensional cone and v1,v2 be the firstlattice vectors along its 1-dimensional faces (listed counter clockwise).Then there exists an invertible 2 × 2 integer matrix B so that Bv1 =(m,−k) where 0 ≤ k < m are integers and Bv2 = (0, l).

Proof. This is an exercise. �

Example 12.4. Consider the cone below where v1 = (2,−1) and v2 =

(0, 1). Since det

(2 0−1 1

)= 2, the vectors are not a Z-basis for Z2.

Therefore, the origin is a singular point of the corresponding affinevariety.

Subdivide this cone by adding in the ray spanned by (1, 0). Tak-

ing since det

(2 1−1 0

)= 1 and det

(1 00 1

)= 1, the affine varieties

associated to the new 2-dimensional cones are both smooth!From yesterday, we know that adding in this new ray exactly corre-

sponds to blowing up the origin in our original affine variety. In thiscase, we get a smooth variety after just one blowup.

42 JESSICA SIDMAN

In some cases, we may need to repeat this subdivision proceduremultiple times, but we can give an algorithm (a set of steps whichterminates) to resolve any singularity .

Algorithm 12.5. INPUT: A cone σ generated by (m0,−k0), with0 ≤ −k0 < m0 and (0, 1).

OUTPUT: A fan ∆ corresponding to a smooth toric variety X∆ thatis gotten by starting with Uσ and then blowing up finitely many times.

Let ∆ be the fan consising of σ and all of its faces.

(1) Set ∆ equal to the subdivision of the current fan ∆ gotten byadding in the ray spanned by (1, 0).

(2) Let ∆ =

(0 −11 0

)∆. (Rotate ∆ by 90 degrees.)

(3) Let ∆ =

(1 0−a 1

)∆ for some value of a that takes (ki,mi)

to (mi+1,−ki+1) where mi+1 = ki and 0 ≤ ki+1 < mi+1. (Thisshears the plane.)

(4) If ki+1 = 0, then stop. Otherwise, let i = i+ 1 and repeat.

Example 12.6. Let σ be the cone generated by (3,−2) and (0, 1).After we add the ray generated by (1, 0), the cone in the first quadrantis smooth, but the one below the first quadrant is not. After the

rotation in step (2) of the algorithm, we shear by

(1 0−2 1

).

The nonsmooth cone is now generated by (2,−1) and (0, 1). Applyingthe algorithm again, we insert the ray spanned by (1, 0). We now checkthat each cone is generated by a Z-basis for Z2.

Theorem 12.7. If X∆ is a smooth projective toric surface, then itcan be constructed by blowing up finitely many times starting with ei-ther P2 or a Hirzebruch surface corresponding to the fan with rays(1, 0), (0, 1), (−1, t), (0,−1).

Proof. A proof is outlined in the exercises in section 2.5 of [1]. �

References

[1] William Fulton, Introduction to Toric Varieties, Annals of Math. studiesno. 131, Princeton Univ. Press, Princeton, 1993.

[2] Israel M. Gelfand, Mikhail M. Kapranov, and Andrei V. Zelvinsky, Discrim-inants, Resultants and Multidimensional Determinants, Birkhäuser, Boston,1994.

[3] Bernd Sturmfels, Gröbner Bases and Convex Polytopes, Univ. Lecture Series,v. 8, Amer. Math. Soc. , Providence, 1996.


E-mail address: [email protected]

Department of Mathematics and Statistics, Mount Holyoke Col-lege, South Hadley, MA 01075

1. An introduction to affine varieties1.1. The ideal-variety correspondence1.2. Singularities1.3. Dimension

2. Projective space2.1. Homogeneous coordinates2.2. An open affine covering of projective space2.3. Projective varieties and homogeneous ideals2.4. Limit points2.5. Degree

3. Polytopes3.1. Polytopes as convex hulls3.2. Polytopes as intersections of halfspaces3.3. The faces of a polytope3.4. The inner normal fan of a polytope3.5. Exercises

4. Affine toric varieties: from lattice points to monomial mappings4.1. Notation4.2. The variety XA.4.3. Computing the ideal of XA4.4. Exercises

5. Projective toric varieties6. Limit points of projective toric varieties6.1. The torus action6.2. Limits6.3. Exercises

7. Coordinate rings of affine toric varieties7.1. Exercises

8. Affine open covers and inner normal fans8.1. Exercises

9. Next9.1. Fans and duals II9.2. Limit points and the fan9.3. Fans and gluing varieties9.4. Refining a fan

10. Smoothness I11. Smoothness II12. Toric resolution of singularitiesReferences

an introduction to toric surfacesjsidman/pcmi08/book.pdf · 1.1. the ideal-variety correspondence....

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