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ALGEBRAIC GEOMETRY NOTES 1. Conventions and Notation Fix A ALGEBRAIC GEOMETRY NOTES E. FRIEDLANDER J. WARNER 1. Conventions and Notation Fix a field k. At times we will require k to be algebraically closed, have a certain charac- teristic or cardinality, or some combination of these. AN and PN are affine and projective spaces in N variables over k. That is, AN is the set of N-tuples of elements of k, and PN N+1 is the set of equivalence classes of A − 0 under the relation (a0; : : : ; aN ) ∼ (b0; : : : ; bN ) if and only if there exists c 2 k with cai = bi for i = 0;:::;N. 2. Monday, January 14th Theorem 2.1 (Bezout). Let X and Y be curves in P2 of degrees d and e respectively. Let fP1;:::;Png be the points of intersection of X and Y . Then we have: n X (1) i(X; Y ; Pi) = de i=1 3. Wednesday, January 16th - Projective Geometry, Algebraic Varieties, and Regular Functions We first introduce the notion of projective space required to make Bezout's Theorem work. Notice in A2 we can have two lines that don't intersect, violating (1). The idea is that these two lines intersect at a "point at infinity". Projective space includes these points. 3.1. The Projective Plane. Define the projective plane, P2, as a set, as equivalence classes 3 of A under the relation (a0; a1; a2) ∼ (b0; b1; b2) if and only if there exists c 2 k with cai = bi for i = 0; 1; 2. It is standard to write [a0 : a1 : a2] for the equivalence class represented by 3 (a0; a1; a2) 2 A . As a set, the projective plane can be written in many ways as the union of three copies of 2 the affine plane, A . Here is the standard way of doing so. Let Ui = f[a0 : a1 : a2] j ai 6= 0g for i = 0; 1; 2, and notice the following bijections of sets 2 2 2 A ! U0 A ! U1 A ! U2 (a; b) 7! [1 : a : b](a; b) 7! [a : 1 : b](a; b) 7! [a : b : 1] 3 The above maps are bijections because each element of Ui has a unique representative in A whose ith coordinate is 1. Also, they cover P2 because every element of P2 has some nonzero 2 coordinate. The points of P − U0 are all of the form [0 : a : b], and we consider these to be the points at infinity for the affine plane U0. Notice that these points are in one-to-one correspondence with P1, the projective line. P1 in turn can be covered by two copies of A1, Date: Spring 2013. 1 the affine line, each of whose complement in P1 is just a single point at infinity P0 = A0. Hence, we can decompose the projective plane as follows: 2 2 1 2 1 0 2 1 0 P = A t P = A t A t P = A t A t A 3.2. Algebraic Varieties. Let f1; : : : ; fm 2 k[x1; : : : ; xN ], and denote by Z(f1; : : : ; fm) the N N set of common zeroes of the fi in A . That is, Z(f1; : : : ; fm) = fa 2 A j fi(a) = 0 8ig: Such a set is called an affine algebraic variety. Notice that if all fi vanish at a, then so does any linear combination of the fi so that any element in the ideal generated by the fi also vanishes at a. The Hilbert Basis Theorem says that k[x1; : : : ; xN ] is Noetherian, so all ideals are finitely generated. Thus, there is no loss of generality in starting with finitely many polynomial equations. That is, any collection of polynomial equations generates an ideal that is also generated by finitely many polynomial equations, and the common zeroes of each will be the same. Example 3.1. Consider the polynomial f(x; y) = x − y2 2 C[x; y]. Z(f) = f(a; b) 2 C2 j a = b2g. The real points of the variety can be drawn as a sideways parabola. To each affine algebraic variety we can assign a ring, called the coordinate ring. To do so, let X be an affine algebraic variety, and define the ideal I(X) = ff 2 k[x1; : : : ; xN ] j f(P ) = 0 8P 2 Xg. Then the coordinate ring is the quotient k[x1; : : : ; xN ]=I(X). We would like to define a similar concept for projective space. A problem arises due to the fact that a point in projective space is represented by many points in affine space, that is, the zeroes of a polynomial aren't well defined in projective space. For example, consider the polynomial 1 + x + y 2 C[x; y], and notice that it's values at the points (1; −2) and (2; −4) are 0 and −1 respectively. But [1 : −2] = [2 : −4] in P1 so that the polynomial doesn't well-define a function to C: We can solve this problem by only considering homogeneous polynomials, that is, polyno- mials all of whose monomials have the same degree. Homogeneous polynomials of degree 0 are just constants. Consider the following proposition. Proposition 3.2. In projective space, the zero locus of a polynomial is well-defined if and only if the polynomial is homogeneous. Proof. Let f be a polynomial. What we mean by a well-defined zero locus in projective space is if any representative of a point is zero in f, then all such representatives are zero. Suppose f is homogeneous of degree d. Then the equation f(ca) = cdf(a) for all c 6= 0, a 2 AN+1, shows that the zero locus is well-defined. Next, suppose f is not homogeneous. Write f as the sum of its homogeneous components, P f = fi. Choose some a with f(a) = 0 but fi(a) 6= 0 for some i (if no such a existed, then Z(f) ⊂ Z(fi) for all i. Applying the Nullstellenzats shows then that f must be homogeneous). Then notice that f(λa) is a non-constant polynomial in λ, and thus is not zero for all λ. Hence, the zero locus of f is not well-defined if f is not homogeneous. The above proposition shows that the zero set of a homogeneous polynomial is well-defined N in P . Now let F1;:::;Fm 2 k[x0; : : : ; xN ] be a collection of homogeneous polynomials, not necessarily of the same degree, and denote by Z(F1;:::;Fm) the set of common zeroes of N N the Fi in P . That is, Z(F1;:::;Fm) = fa 2 P j Fi(a) = 0 8ig: Such a set is called a 2 projective algebraic variety. Again by Hilbert's Basis theorem, we only require finitely many polynomial equations to define varieties. Similar to the case of affine varieties, each projective variety has a homogeneous coordinate ring. Let Y be a projective variety, and define the ideal I(Y ) generated by all homogeneous polynomials vanishing at all points of Y . Since I(Y ) is generated by homogeneous ele- ments, it is a homogeneous ideal. The desired homogeneous coordinate ring is the quotient k[x0; : : : ; xN ]=I(Y ). 3.3. Regular Functions. The concept of a regular functions on a variety differs slightly depending on if the variety is projective or affine. We describe both here. Definition 3.3. Let X ⊂ AN be an affine variety, and let P be a point on X. A function f : X ! k is said to be regular at P if there is a Zariski open neighborhood U ⊂ X containing P , and polynomials g; h 2 k[x1; : : : ; xn] with h nowhere zero on U such that f = g=h on U. If f is regular at all points of some open subset U, we say f is a regular function on U. The regular functions on any open set U form a ring under pointwise multiplication and addition. Denote this ring by OX (U). Example 3.4. Consider the variety X = Z(f) from example 3.1. Any polynomial g 2 C[x; y] determines a regular function on X, and another polynomial g0 determines the same regular function if and only if g0 − g 2 I(X). This shows that the coordinate ring of X embeds into the ring of regular functions. In fact, in the affine case, the map is surjective and the coordinate ring of X is isomorphic to the ring of regular functions. That is, any function that is locally a rational function everywhere on X is actually a polynomial. Notice we have yet to use anything specific about our variety X. Consider the rational function f(x) = 1=(x − 1). This defines a regular function at all points P on X except when x = 1. This corresponds to the two real points (1; 1) and (1; −1) on X. To see this, notice that the Zariski open set U = X T Z(x − 1)C contains all such points, and the function f is a ratio of two polynomials, of which the denominator doesn't vanish on X. Above we constructed a regular function on an open set U by taking a rational function and discarding the points at which the denominator vanishes. In this way, rational functions yield regular functions on open sets U ⊂ X, but sometimes different rational functions yield the same regular function, as the following example illustrates. 4 Example 3.5. Consider the hypersurface X = Z(x1x2 − x3x4) inside A . Let U be the Zariski open set of X given by the complement of the equation x2x3 = 0.
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