AN INTRODUCTION TO AFFINE TORIC VARIETIES: EMBEDDINGS AND IDEALS

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AN INTRODUCTION TO AFFINE TORIC VARIETIES: EMBEDDINGS AND IDEALS JESSICA SIDMAN. Affine toric varieties: from lattice points to monomial mappings In this chapter we introduce toric varieties embedded in affine space. We begin by giving embeddings and then show how to compute the ideal of an affine toric variety from its parameterization. We essentially follow Sturmfel s book [3]... Notation. We introduce the notation that we will use to describe our embeddings. Definition.. We call Z d R d the d-dimensional lattice and call elements of Z d lattice points. Example.2. For example, (, ), (, ), (, ), (2, 5) are some lattice points in Z 2 R 2. Definition.3. Suppose that we have indeterminates x,..., x d. If we have a lattice point a = (a,... a d ) then we can associate to it a Laurent monomial in the x is : x a x a d n. (We call this a Laurent monomial because we allow negative exponents.) We often denote this Laurent monomial by x a using vector notation instead of writing out each indeterminate and each exponent. Example.4. Let R = C[x] = C[x, x 2, x 3 ]. Then x 4 2 = x 4 x 2x 2 3..2. The variety X A. Suppose we have a finite ordered set A = {a,..., a n } Z d. We can use A to define a map to C n whose coordinates are the monomials x a. Since we allow monomials with negative exponents, none of the elements in the domain can have a zero coordinate. Definition.5. Let C = C\{}. Given A = {a,..., a n } Z d, we define φ A : (C ) d C n by t = (t,..., t d ) (t a,..., t an ). The author is partially supported by NSF grant DMS 647 and the Clare Boothe Luce Program.

2 JESSICA SIDMAN Example.6. Suppose that A = {, 2} Z. Then φ A (t) = (t, t 2 ). Restrict the domain to real numbers and think about the image of this map in R 2. It is a parameterization of the parabola in the plane minus the point at the origin Example.7. Suppose that A = {, 2, 3} Z. Then φ A (t) = (t, t 2, t 3 ). Again, let us restrict the domain to real numbers and try to visualize what we get in R 3. For t >, we trace out a curve that twists up out of the xy-plane above the parabola, and for t <, the curve lies below the parabola in the xy-plane. This gives the curve the name the twisted cubic. Definition.8. The affine toric variety associated to A, denoted X A, is the Zariski closure of the image of φ A. The image of φ A is a dense open subset of X A. In our two previous examples, the variety X A is gotten by adding in the origin. Here is a more interesting example in which we add more than one point when we take the closure of the image of φ A. Example.9. Let A = {(, ), (, ), (, )}. Then φ A : (C ) 2 C 3 is given by φ A (t, t 2 ) = (t, t 2, t t 2 ). Consider points of the form (, y, ), where y C. If we let t 2 = y and consider lim t (t, t 2, t t 2 ) in the complex numbers, we get (, y, ). As the Zariski closure of a set is closed in the usual topology, these limit points must be in X A. Similarly, the line {(x,, ) x C} is contained in X A. We can visualize the real points in the image of φ A in R 3 as a familiar surface, the graph of f(t, t 2 ) = t t 2 minus two lines..3. Computing the ideal of X A. A variety is not just a set of points in projective space. An algebraic variety is a set that arises as the set of all points that are solutions to some finite set of polynomial equations. In this section we will learn how to compute the ideal of the variety X A. To do this, let s think carefully about what it means for a polynomial on the ambient space to vanish at a point φ A (t). If f(x,..., x n ) is a polynomial, then we need to compute f(φ A (t)). This is done monomialby-monomial. So, let s focus on what it really means to substitute φ A (t) into a monomial x u. This is best done in an example. We will see that if f = x u, the f(φ A (t)) = t Au where A is the d n matrix whose columns are the vectors a,..., a n. Example.. Let A = {(, ), (, ), (, )} so that ( ) A =.

AN INTRODUCTION TO AFFINE TORIC VARIETIES: EMBEDDINGS AND IDEALS3 Suppose that f(x, x 2, x 3 ) = x 3 x x 2 = x (,,) x (,,). Then f(φ A (t)) =f(t (,), t (,), t (,) ) =t (,) t (,) t (,) t (,) t (,) t (,) @ =t AB C @ A @ t AB C @ A What we observe is that substituting a monomial map into a monomial x u is equivalent to applying a linear transformation to u using the matrix of the monomial map. From this it is clear that if we have a binomial f(x) = x u x v, then f(φ A (t)) is identically zero exactly when Au = Av. In fact, we have the following theorem: Theorem. (Lemma 4. in [3]). Suppose A Z d is a finite ordered set and that A is the associated d n matrix. Let I XA be the ideal of the set X A. Then I XA is equal to the ideal I A := x u x v u v ker A, u, v Z n. Proof. We will show that I A is contained in I XA with a computation. If f I A, then it is a linear combination (with polynomial coefficients) of elements of the form x u x v where Au = Av. For each of these binomials t Au t Av = becuase Au = Av. Therefore, f(φ A (t)) = and f I XA. Next, we will show that every element of I XA is in I A. Let U be the set of all polynomials in I XA that are not in I A. For each f U, consider the lexicographic leading term of f, i.e., the term that would come first in the dictionary. If U is not empty, then there is some nonzero f in U whose leading term is smallest. (The set of exponent vectors is bounded below by and is discrete.) Since f I A, f(φ A (t)) =. Without loss of generality, assume that the leading term of f is x u, so that the coefficient of the leading term is. If we expand f(φ A (t)), we will see a term of the form t Au. Since this term cancels when we simplify, there must be at least one term of the form γt Av with Au = Av in the expression. Now define f = f (x u x v ). Notice that the leading term of f is lexicographically smaller than the leading term of f as we have gotten rid of our original leading term x u and as x v appeared with nonzero coefficient in f, we have not introduced any additional monomials. If f is in U, then we have contradicted the minimality of our choice of f. If f is not in U,

4 JESSICA SIDMAN then neither is f which is again a contradiction. Therefore, U must be empty and the claim is proved. Corollary.2 (Corollary 4.3 in [3]). Assume the hypotheses of Theorem.. Suppose that u ker A. Write u = u + u, where u +, u Z n. Then x u+ x u I XA. Example.3. Let A be as in.. The vector u = (,, ) is in the kernel of A (and actually spans the kernel). We can write it as a difference of its positive and negative pieces: u = (,, ) (,, )..4. Exercises. () Consider the matrix A = ( ). 2 3 (a) Let v = (, 2,, ), v 2 = (,,, ), v 3 = (,, 2, ) Show that v is in the span of v 2 and v 3. (b) Write down the binomials corresponding to v, v 2, v 3. (c) Is v 3 in the ideal generated by v and v 2? Can you explain your answer and how it relates to part (a)? (2) Consider the integer matrices A =, A 2 = 2 2 A 3 = 2, A 4 =, (a) Let A i be the set of columns of A i. Write down φ Ai for each i. (You might want to write each map using x a notation as well as in the usual notation in which you write down each variable explicitly.) (b) If I Ai is nonzero, find at least three elements in I Ai 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 det B = ±. We will denote the set of all n n invertible integer matrices by GL n (Z). References [] William Fulton, Introduction to Toric Varieties, Annals of Math. studies no. 3, Princeton Univ. Press, Princeton, 993.

AN INTRODUCTION TO AFFINE TORIC VARIETIES: EMBEDDINGS AND IDEALS5 [2] Israel M. Gelfand, Mikhail M. Kapranov, and Andrei V. Zelvinsky, Discriminants, Resultants and Multidimensional Determinants, Birkhäuser, Boston, 994. [3] Bernd Sturmfels, Gröbner Bases and Convex Polytopes, Univ. Lecture Series, v. 8, Amer. Math. Soc., Providence, 996. E-mail address: jsidman@mtholyoke.edu Department of Mathematics and Statistics, Mount Holyoke College, South Hadley, MA 75