# Algebraic Varieties. Chapter Algebraic Varieties

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1 Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) : f 1 (a 1,..., a n ) =... = f m (a 1,..., a n ) = 0} is called an affine algebraic variety defined over K (we will later actually add more conditions; in the literature, the objects above are often called prevarieties or algebraic sets). If we replace the affine by projective spaces (and the polynomials by homogeneous polynomials) we get the notion of projective algebraic prevarieties. Special cases include the space A n K (or P n K) itself, which is the zero set of an empty set of polynomials (or, if you don t like this, of the zero polynomial) and the empty set (which is the zero set of the (homogeneous) equation 1 = 0). Plane algebraic curves are zero sets in A 2 K of a single polynomial in two variables (or of a homogeneous polynomial in three variables in the case of the projective plane). An important class of affine algebraic prevarieties are hypersurfaces: these are the zero sets of a single polynomial. Another well studied class consists of quadrics, the zero sets of polynomials of degree 2; conics are examples of quadrics in the plane. Examples of higher dimensional quadrics are the sphere x 2 + y 2 + z 2 = 1 or elliptic curves (!) given as the intersection of two quadrics ax 2 + by 2 + cz 2 = 0, ey 2 + fz 2 + gw 2 = 0 in P 3 K under suitable conditions guaranteeing smoothness. As a first example, let us see how to find rational points on the sphere S : x 2 + y 2 + z 2 = 1 starting from the known point P = ( 1, 0, 0). The idea is clear: intersect lines through P with S. Lines through a point (u, v, w) are described by the equations x = u + at, y = v + bt, z = w + ct (the constants a, b, c describe the slopes of the lines; of course they are not allowed to vanish simultaneously). For our P we get x = 1 + at, y = bt, z = ct; plugging this 60

2 into the equation for S we find ( 1+at) 2 +b 2 t 2 +c 2 t 2 = 1. The value t = 0 gives the known point P, so we should factor out t: we get t(a 2 t + b 2 t + c 2 t 2a) = 0. 2a Thus the second point of intersection is given by t = a 2 +b 2 +c, and we find the 2 parametrization x = a2 b 2 c 2 a 2 + b 2 + c 2, y = 2ab a 2 + b 2 + c 2, z = 2ac a 2 + b 2 + c 2. Geometrically it is clear that every point P on S is parametrized by these equations, and P itself is parametrized for a = 0 and arbitrary values of b and c (not both zero). We will have to deal with singular points on algebraic varieties; postponing the general definition for now, let us simply remark that for hypersurfaces, singular points are those in which all the partial derivatives vanish. In order to find the singular points of the cubic surface X 2 + Y 3 Y 2 + Z 2 = 0, let us homogenize. We find F (X, Y, Z, W ) = X 2 W + Y 3 Y 2 W + Z 2 W, hence F X = 2XW, F Y = 3Y 2 2Y W, F Z = 2ZW, F W = X 2 Y 2 + Z 2. If w = 0, then y = 0, and the partials all vanish if and only if x 2 + z 2 = 0. Since x 0, we can rescale and assume that x = 1; then z = pmi, and we have exactly two singular points [1 : 0 : ±i : 0] at infinity. In the affine space, we have w = 1, and then we find x = y = z = 0, hence the origin [0 : 0 : 0 : 1] is the unique singular point in the affine part of the surface. This singularity can be used to parametrize the cubic surface: the lines through the origin are given by X = at, Y = bt, Z = ct. Intersection with the affine surface gives t 2 (a 2 + b 3 t b 2 + c 2 ) = 0, hence the second point of intersection satisfies t = b2 a 2 c 2 b. This provides us with the parametrization 3 x = a(b2 a 2 c 2 ) b 3, y = b2 a 2 c 2 b 2, z = c(b2 a 2 c 2 ) b 3. The point P is again parametrized infinitely often: just pick any rational numbers with a 2 + c 2 = b From Ideals to Varieties To any ideal I in K[X 1,..., X n ] we can attach a variety V(I) = {P A n K : f(p ) = 0 for all f I}. In fact, this is a variety by Hilbert s basis theorem (see below), according to which every ideal in K[X 1,..., X n ] is finitely generated. Thus V(I) really can be written as the zero set of a finite list f 1,..., f m of polynomials. 61

3 Proposition The map V from ideals in R = K[X 1,..., X n ] to affine varieties in A n K has the following properties: 1. V(0) = A n K; V(R) = ; 2. I J implies V(I) V(J); 3. V(I J) = V(I) V(J); 4. V( I ν ) = V(I ν ). Here I, J, I ν are ideals in R, and I ν denotes the ideal consisting of finite R-linear combinations of elements in the I ν. Proof. 1. These properties are clear. 2. If I J, then V(J) V(I). 3. We have V(I) V(I J) since I J I. Similarly V(J) V(I J), hence V(I) V(J) V(I J). Now assume that P V(I J). If P is not in V(I), then there is some f I with f(p ) 0. Similarly, if P is not in V(J), then there is some g J with g(p ) 0. Then fg I J and fg(p ) 0: contradiction. 4. Let P V(I ν ); then f ν (P ) = 0 for all f ν I ν, hence f(p ) = 0 for all finite linear combinations f I ν. This shows that V(I ν ) V( I ν ). On the other hand, I µ I ν for every µ, hence V( I ν ) V(I µ ) for every µ, and therefore V( I ν ) V(I µ ). Note that this proposition (in particular (1), (3) and (4)) implies that the algebraic varieties in A n K form the closed sets of a topology on A n K; this topology is called the Zariski topology. Lemma The Zariski-closed sets in A 1 K are the finite subsets (finite sets of points) and A 1 K itself From Varieties to Ideals: The Vanishing Ideal Assume that V is an affine algebraic variety, defined as the zero set of polynomials f 1,..., f m K[X 1,..., X n ]. For any P = (a 1,..., a n ) V we have f j (P ) = 0. Thus f(p ) = 0 for any K-linear combination f of the f i. This shows that f(a) = 0 for all f (f 1,..., f n ). Thus we are led to define the vanishing ideal I = {f K[X 1,..., X n ] : f(p ) = 0 for all P V } of the variety V. Note that I( ) = K[X 1,..., X n ]. If K has infinitely many elements, then I(A n K) = (0). 62

5 12.4 Hilbert s Basis Theorem We start with the following result: Proposition Let R be a ring. The following assertions are equivalent: 1. Every ideal I in R is finitely generated, i.e., can be written in the form I = (a 1,..., a m ) for a i R. 2. R satisfies the ascending chain condition, i.e., every chain I 1 I 2 I 3... becomes stationary: there is some index N such that I N = I N+1 =.... A ring R is called Noetherian if the conditions in Prop are satisfied. Proof. (1) = (2). Let I 1 I 2... be an ascending chain of ideals and put I = I ν. Then I is an ideal, hence finitely generated: I = (a 1,..., a m ). We have a i I ν(i) for some ν(i); let N = max ν(i). Then a 1,..., a m I ν for every ν N, hence I I ν I, and we are done. Conversely let I be a nonzero ideal in R, and let a 1 I \ (0). If I = (a 1 ), we are done, if not then pick a 2 I \ (a 1 ) and continue. In this way we get a strictly increasing ascending chain of ideals (a 1 ) (a 1, a 2 ) (a 1, a 2, a 3 ).... This shows that the process must terminate. Clearly, every field K is Noetherian since K has only two ideals (0) and (1). Every principal ideal ring (such as Z) is also Noetherian because in such rings every ideal is generated by a single element. Proposition For every ideal I in some Noetherian ring R, the quotient R/I is also Noetherian. Hilbert s Basis Theorem is a method for constructing lots of Noetherian rings: Theorem If R is Noetherian, then so is R[X]. In particular, the rings K[X 1,..., X n ] are Noetherian. In fact, K[X] is Noetherian because K is, and then K[X, Y ] is Noetherian because K[X] is etc. Proof. Let I be an ideal in R[X]. We claim that I is finitely generated. Choose f 1 I \ (0) with minimal degree; if I (f 1 ), choose f 2 I \ (f 1 ) with minimal degree. If this process terminates, we are done. If not, let a i be the leading coefficient of f i, and form the ideal J = (a 1, a 2, a 3,...) in R. Since R is Noetherian, we have J = (a 1,..., a N ) for some N. We now claim that I = I N := (f 1,..., f N ). If not, then f N+1 I \ I N. Moreover, a N+1 = N i=1 r ia i since a N+1 J. Now set g = N r i f i x deg f N+1 deg f i. i=0 64

6 Since g is a linear combination of the f i, we have g I N. Next deg g = deg f N+1, and both polynomials have the same leading term. Thus deg(f N+1 g) < deg f N+1, hence f N+1 g I \ I N ; but this contradicts the minimality of deg f N Hilbert s Nullstellensatz Recall that an ideal p in a ring is prime if ab p implies a p or b p; it is then easily seen that p is prime if and only if R/p is a domain. In fact, assume that R/p is a domain, and let ab p. Then 0 = ab + p = (a + p)(b + p), and since R/p is a domain, we get a p or b p: thus p is prime. Conversely, if p is prime and (a + p)(b + p) = 0, then ab p, hence a p or b p, and this shows that R/p is a domain. Moreover, an ideal m is maximal if m n R for some ideal n implies that m = n or n = R; again it is easy to see that m is maximal if and only if R/m is a field. For the rest of this chapter, let K be an algebraically closed field. Theorem (Weak Nullstellensatz). Let K be an algebraically closed field. Then every maximal ideal in the polynomial ring R = K[X 1,..., X n ] has the form M = (X 1 a 1,..., X n a n ) for some point P = (a 1,..., a n ) A n K. Note that the weak Nullstellensatz gives us a bijection between maximal ideals in K[X 1,..., X n ] and points in A n K. Also observe that the theorem does not hold for arbitrary fields: in R[X], the ideal M = (X 2 + 1) is maximal because R[X]/(X 2 + 1) C is a field, yet M is not of the form (X a) for some a R. As a consequence we have Corollary (Nullstellensatz). Let K be an algebraically closed field and I (1) be an ideal in K[X 1,..., X n ]. Then V(I). Proof. If I R = K[X 1,..., X n ], then there is some maximal ideal M with I M (this follows from the ascending chain condition: if I is maximal, we are done; if not then there is some I 1 sandwiched between I and R; now repeat and observe that this process must end eventually). Now M = (X 1 a 1,..., X n a n ) by the weak Nullstellensatz, hence f(p ) = 0 for P = (a 1,..., a n ) and any f I. In particular, P V(I). Another consequence is Corollary For any ideal I in R = K[X 1,..., X n ] we have I(V(I)) = rad I. In particular, if I is a radical ideal, then I(V(I)) = I. Thus I and V set up a one-to-one correspondence between radical ideals in K[X 1,..., X n ] and affine algebraic varieties in A n K. 65

7 Proof. Let I = (f 1,..., f m ). Then g I(V(I)) if g(p ) = 0 for any P V(I). We have to prove that some power of g lies in I. The following proof is called the trick of Rabinowitsch. Adjoin a new variable X n+1 to R, i.e., let S = R[X n+1 ]. Then the polynomials f 1,..., f m and X n+1 g 1 have no common zeros in A n+1 K, so by the weak Nullstellensatz we must have 1 = p 1 f p m f m + p m+1 (X n+1 g 1), where p j S. Now plug in X n+1 = 1 g K(X 1,..., X n ). This gives (the result will be an element in 1 = p 1 (x 1,..., x n, 1/g)f p m (x 1,..., x n, 1/g)f m. Now multiply through by a power of g to clear denominators; the resulting equation then shows that this power of g is in I. For the proof of the Nullstellensatz we need the following lemma (whose proof is due to D. Allcock): Lemma Let k be a field and K/k an extension which is finitely generated as a k-algebra. Then K/k is algebraic. We will give the proof later; for now let me explain what this means. Recall that an algebra is, cum grano salis, a vector space in which vectors can be multiplied. For example, the polynomial ring A = K[X 1,..., X n ] is also a K- vector space and therefore a K-algebra: elements in A can be added (as vectors), multiplied by scalars from K, and can be multiplied as polynomials. If K/k is a field extension, then we can interpret K as a k-vector space: we can add and subtract elements of K and multiply them by the scalars from k. Since the vectors in K can also be multiplied as elements of K, we see that K actually is a k-algebra. A k-algebra K is now said to be finitely generated if there are elements α 1,..., α m such that every α K can be written as a k-linear combination of products of the α i ; in other words: if the k-algebra generated by the α i is equal to K. As an example, consider the Q-algebra Q(X) of rational functions in one variable X. The Q-algebra generated by f = 1 X 1 X and g = X+1 consists of all Q-linear combinations of f, g, f 2, fg, g 2,.... Note that all such elements have poles at most at x = 0 and x = 1. Now we claim that Q(X) is not finitely generated as a Q-algebra. If it were, then there would be rational functions f 1,..., f n such that every rational function f Q(X) can be written as a linear combination of products of the f j. But this is not possible: the f i have finitely many poles, and by forming products and linear combinations no new poles are created. Thus if a Q is not one of the finitely many poles, then 1 X a is not an element of the Q-algebra generated by the f i. 66

8 Proof of the Weak Nullstellensatz. Let m be a maximal ideal in R. Then R/m is a field. Moreover, it is obviously finitely generated as a k-algebra, namely by the elements X 1 + m,..., X n + m: every element in R is a k-linear combination of products of the X i, hence every element in R/m is a k-linear combination of products of the X i + m. Observe that R/m contains k as a subfield: the map φ : k R/m; a a + m is a ring homomorphism, and we have ker φ = {a k : a m} = k m. If m contains a nonzero element a k, then it contains 1 and is therefore equal to R. Since m is maximal, this cannot happen, hence k m = (0), and φ is injective. Thus R/m is a field extension of k, and by Lemma we find that it is algebraic over k. But k is algebraically closed, hence R/m = k. Now consider the natural projection π : R R/m = k sending f R to the coset f +m. Since π(x j ) = a j for a j k, the elements x j a j are in the kernel of π, which is by definition just m. Thus m contains (X 1 a 1,..., X n a n ); this last ideal is maximal, and since m is also maximal, we must have equality: m = (X 1 a 1,..., X n a n ). In order to get an idea for the proof of Lemma , let us first consider the special case where k is infinite and where K = k(x) is a simple transcendental extension. Our claim is that if f 1,..., f m K, then the k-algebra A they generate is strictly smaller than K. In fact, choose a k such that a is not a pole of any of the rational functions f j ; then no element of A can have a pole 1 at a, hence X a K \ A. Now we complete the proof of the Nullstellensatz by proving the central lemma: Proof of Lemma In order to simplify the presentation, we will assume that k is algebraically closed, although the proof for general fields k is only slightly more complicated. We first give the proof for fields K with transcendence degree 1 over k and then do the general case later. Recall that K/k has transcendence degree 1 if there is some X K such that K/k(X) is finite. As before, let f 1,..., f m K, and let A be the k-algebra these rational functions generate. Since K/k(X) is finite, there exist elements e 1,..., e n K that form a basis for the vector space K over k(x). The product e i e j of two basis elements can be expressed as a k(x)-linear combination of the e h, hence there exist polynomials a ijh, b ijh k[x] with e i e j = h a ijh (X) b ijh (X) e h. We also can express our f i in terms of this basis: f i = j c ij (X) d ij (X) e j for polynomials c ij, d ij k[x]. 67

9 Now observe that the poles of the elements in A must come from the finitely many poles of the e j, the roots of the finitely many polynomials d ij, or the roots of the finitely many polynomials b ijh. Thus there is some a k (here we use that k is algebraically closed) that is not among these poles, and then 1 X a K \ A. Now let us do the general case of fields K with arbitrary transcendence degree 1. Let F be a subextension of K such that K/F has transcendence degree 1. From what we have proved we know that K is not a finitely generated F -algebra; this immediately shows that it is not a finitely generated k-algebra. 68

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