A ne Algebraic Varieties Undergraduate Seminars: Toric Varieties

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1 A ne Algebraic Varieties Undergraduate Seminars: Toric Varieties Lena Ji February 3, 2016 Contents 1. Algebraic Sets 1 2. The Zariski Topology 3 3. Morphisms of A ne Algebraic Sets 5 4. Dimension 6 References 6 1. Algebraic Sets Let k be an algebraically closed field, for instance the complex numbers C. We define a ne n-space over k to be the set of n-tuples of elements in k. That is, A n = {(a 1,...,a n ) a i 2 k}. We denote by k[x 1,...,x n ] the polynomial ring in n variables with coe cients in k. Polynomials f 2 k[x 1,...,x n ] can be viewed as maps from A n to A by evaluating f at each point, and so we can consider the set of zeroes of a polynomial. More generally, for a collection of polynomials {f i } i2i,wedefinetheirzerosettobe V ({f i } i2i )={(a 1,...,a n ) 2 A n f i (a 1,...,a n )=0 8i 2 I}. Definition 1.1. A subset of A n of the form V ({f i } i2i ) is called an a set. ne algebraic Note that these are referred to as a ne algebraic varieties in Smith, et al. However, we will follow Fulton and adopt the following definition. Definition 1.2. An algebraic set V A n is irreducible if, for any expression V = V 1 [ V 2 where V i are algebraic sets in A n, V 1 = V or V 2 = V. Definition 1.3. An a ne algebraic variety is an irreducible a ne algebraic set. Example 1.4. (1) V (xy) C 2 is an a ne algebraic set, but it is not irreducible. (Figure 1) (2) V (y 2 x 2 x 3 ) C 2 is an a ne algebraic variety. (Figure 2) Here are many more examples of a pictures are over R. Example 1.5. A point (a 1,...,a n ) 2 A n is an a V (x 1 a 1,...,x n a n )={(a 1,...,a n )}. ne algebraic sets. Due to artistic limitations, 1 ne algebraic variety because

2 2 /(,,} f },,)\ Figure 1. V (xy) Figure 2. V (y 2 x 2 x 3 ) Figure 3. V (x 2 + y 2 z 2 ) Figure 4. V (y 2 x(x 2 1)) Example 1.6. A hypersurface in A n is the zero set of a single nonconstant polynomial, for example the quadratic cone V (x 2 + y 2 z 2 ) C 2. A hypersurface in A 2 is called an a ne plane curve. The a ne variety given in Figure 2 is an a ne plane curve, as is the elliptic curve V (y 2 x(x 2 1)). Example 1.7. This very nice heart is a hypersurface given by the solutions of the equation (x y2 + z 2 1) 3 x 2 z y2 z 3 = 0. Example 1.8. The Whitney umbrella is defined by the equation x 2 y 2 z = 0.

3 A ne Algebraic Varieties 3 Example 1.9. The torus with major radius R and minor radius r is defined by the equation (x 2 + y 2 + z 2 + R 2 r 2 ) 2 = R 2 (x 2 + y 2 ). Example The special linear group SL(n, C) ={A 2 M n (C) det(a) =1} is a hypersurface in M n (C) = C n2. This follows from the fact that the determinant is a polynomial in n 2 variables; for example when n = 3, then 0 a b c 1 d e fa = aei + bfg + cdh ceg bdi afh. g h i Example The unit sphere S n 1 C n is an a ne algebraic variety defined by the equation x x 2 n = 1. However, the unit open ball (in the Euclidean topology) and defined as the set {(a 1,...,a n ) 2 C n a a 2 n < 1} is not; if a polynomial vanishes on an open subset of C n in the Euclidean topology, then it is uniformly The Zariski Topology Recall that a collection T of subsets of a space X defines a topology on X if (1) X and ; are in T ; (2) the union of any subcollection of elements of T is contained in T ; (3) the intersection of any finite subcollection of elements of T is in T. We would like to use a ne algebraic sets to define the closed sets of a topology on A n, and so we must check that (1 F ) A n and ; are a ne algebraic sets; (2 F ) the arbitrary intersection of a ne algebraic sets is itself an a ne algebraic set; (3 F ) the finite union of a ne algebraic sets is itself an a ne algebraic set. Let s verify these! If a 2 A is nonzero, then the polynomial equation a = 0 has no solutions, and so V (a) =;. However, the equation 0 = 0 is satisfied by every point in A n, and so V (0) = A n. So condition (1 F ) is satisfied. For (2 F ), let {V } 2A be a collection of a ne algebraic sets, where each V = V ({f i } i 2I ). Then the intersection T 2A V is the common zero set of {f i } i 2I over all 2 A, i.e.! \ [ V = V {f i } i 2I. 2A 2A The twisted cubic curve in Figure 5 illustrates this, as it is given by the intersection of two surfaces: V (x 2 y) \ V (x 3 z)=v (x 2 y, x 3 z). So it remains to show (3 F ). By induction, it is enough to check the union of two a ne algebraic sets.

4 4 /(,,} f },,)\ Figure 5. An intersection Figure 6. A union Proposition 2.1 ([3] Exercise 1.2.1). The union of two a is an a ne algebraic set. ne algebraic sets in A n Proof. Let V ({f i } i2i ) and V ({g j } j2j ) be a ne algebraic sets. We claim that V ({f i } i2i ) [ V ({g j } j2j )=V ({f i g j } (i,j)2i J ). Certainly holds, since if p =(a 1,...,a n ) 2 V ({f i } i2i )[V ({g j } j2j ), then f i (p) = 0 for all i or g j (p) = 0 for all j. In either case, then f i g j (p) = 0 for all i and all j, and so p 2 V ({f i g j } (i,j)2i J ). Now let q 2 V ({f i g j } (i,j)2i J ) and suppose that q 62 V ({f i } i2i ) [ V ({g j } j2j ). Then there exist i and j with f i (q) 6= 0 and g j (q) 6= 0. But this implies f i g j (q) 6= 0, a contradiction, so q must be in V ({f i } i2i )[V ({g j } j2j ) and we have shown. An example of this is the union of the x-axis and yz-plane in Figure 6: V (x) [ V (y, z) =V (xy, xz). Definition 2.2. The topology on A n where the closed sets are of the form V ({f i } i2i ) is called the Zariski topology. The Zariski topology on A n is very di erent from the Euclidean topology. Open subsets in this topology are very big; in fact they are dense and quasi-compact. Additionally, two non-empty open sets will always intersect, and so the Zariski topology is not Hausdor on A n for n>0. Example 2.3. Let k = C. Then the Zariski topology on A 1 is the cofinite topology on C closed sets are ;, C, and finite sets since polynomials in one variable have finitely many roots. Example 2.4 ([3] Exercise 1.2.2). The Zariski topology on A 2 is not the product topology on A 1 A 1. Recall that the product topology on X X is generated by open sets of the form U 1 U 2,whereU 1,U 2 are open subsets of X. SoifA 1 A 1 is equipped with the product topology, where each A 1 has the Zariski topology, open sets are of the form A 2 {finitely many horizontal lines [ vertical lines [ points} The diagonal of A 1 A 1,defined A 1 A 1 = {(a 1,a 2 ) 2 A 1 A 1 a 1 = a 2 }, is not closed in the product topology, where each A 1 is endowed with the Zariski topology, since A 1 is not Hausdor. However, it is the zero set of the polynomial x y 2 A[x, y], so A 1 A 1 = V (x y) A2 is closed in the Zariski topology on A 2.

5 A ne Algebraic Varieties 5 If V A n is an a ne algebraic set, then we can endow V with the subspace topology induced by the Zariski topology on A n. Then closed subsets of V are of the form V \ W,whereW A n is an a ne algebraic set. 3. Morphisms of Affine Algebraic Sets Definition 3.1. Let V A n and W A m be a ne algebraic varieties. A morphism of algebraic varieties is a map F : V! W given by the restriction of a polynomial map A n! A m (meaning that each of the m components is given by a polynomial in k[x 1,...,x n ]). So when we compose F with the inclusion i : W,! A m, the resulting map will be of the form i F =(F 1,...,F m ) where each F i is the restriction to V of a (non-unique) polynomial in k[x 1,...,x n ]. Definition 3.2. A morphism F : V! W is an isomorphism if it has an inverse morphism. In this case we say that V and W are isomorphic. Example 3.3. Let C be the plane parabola given by the equation y x 2 = 0. Then C is isomorphic to A 1 via the maps ' C, where ' : A 2! A 1 A 1! C (x, y) 7! x t 7! (t, t 2 ). Example 3.4 ([3] Exercise 1.3.2). The twisted cubic V = V (x 2 y, x 3 z)in Figure 5 is isomorphic to A 1.Since V = {(t, t 2,t 3 ) 2 A 3 t 2 A}, we can define a morphism A 1! V by t 7! (t, t 2,t 3 ). The restriction to V of the projection A 3! A 1 onto the first factor, defined (x, y, z) 7! x, gives an inverse morphism. Proposition 3.5 ([3] Exercise 1.3.1). If F : V! W is a morphism of a ne algebraic sets, then F is continuous in the Zariski topology. Proof. Any closed subset of W is of the form V ({f i } i2i )\W where f i 2 k[x 1,...,x m ]. F 1 (V ({f i } i2i ) \ W )=F 1 (V ({f i } i2i )) \ F 1 (W )=V ({f i F }i2i ) \ V is a closed subset of V,where F : A n to F,soF is continuous.! A m is any polynomial map that restricts

6 6 /(,,} f },,)\ 4. Dimension Definition 4.1. The dimension of an a ne algebraic set V is the length of the longest chain of distinct nonempty a ne closed subvarieties of V sup{d V d ) V d 1 ) ) V 0 }. Hence the dimension of an a ne algebraic set is equal to the maximum of the dimensions of its irreducible components (maximal irreducible subsets). Example 4.2. The quadratic cone V (x 2 + y 2 z 2 ) has dimension 2. Example 4.3. The a ne algebraic set V (xy, xz) has dimension 2 = max{1, 2}. Definition 4.4. An a ne algebraic set is equidimensional if all of its irreducible components have the same dimension. So in the earlier examples, V (xy, xz) is not equdimensional but V (x 2 + y 2 z 2 ) and V (xy) (Figure 1) are. Definition 4.5. The codimension of an a codim V = n dim V. ne algebraic sets of codi- Example 4.6. Hypersurfaces in A n are precisely the a mension 1. References ne algebraic set V A n is defined [1] William Fulton. Algebraic Curves, [2] Herwig Hauser. Algebraic Surfaces Gallery. bildergalerie/gallery.html. [3] Karen Smith, Lauri Kahanpää, Pekka Kekäläinen, and William Traves. An Invitation to Algebraic Geometry. Springer,2004.