V (f) :={[x] 2 P n ( ) f(x) =0}. If (x) ( x) thenf( x) =


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1 20 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2. Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F vector subspaces of F n+. For example, P (R) is the set of lines through the origin in R 2. These are specified by slope 0 apple <. So, P (R) = R/ Z is a circle. However, there are many points missing. As an a ne scheme, the circle implicitly also contains all points in the punctured disk (the interior of the circle) and a generic point. As a projective variety, the puncture will be filled in. (These are the two circles at infinity identified to one point.) 2.. Lecture 7. Basic properties of projective space. In order to construct the scheme theoretic version of ndimensional projective space P n (k) we need to take P n ( ). This is defined to be the space of all one dimensional vector spaces in n+. This is also the quotient space of n+ \0 modulo the identification x ax for all a 6= 02. Note that the homogeneous coordinates X i are NOT functions P n ( )!. But homogeneous polynomials define subsets of P n ( ). First, a polynomial f[x] 2 k[x] ishomogeneous of degree d if for all 2 and x 2 n+, f( x) = d f(x). [Thanks to Mac for correcting an earlier version!] Given such a polynomial we can define If (x) ( x) thenf( x) = V (f) :={[x] 2 P n ( ) f(x) =0}. d f(x) = 0. So, V (f) iswelldefined. Definition 2... If S = {f } is a set of homogeneous polynomials f 2 k[x 0,,X n ] then let V (S) = \ V (f )={[x] 2 P n ( ) f (x) =08f 2 S}. f 2S The Zariski topology or ktopology on P n ( ) is defined by letting V (S) as above be the closed subsets. Let U i = {[x 0,,x n ] 2 P n ( ) : x i 6=0}. These are open subsets of P n ( ) since they are the complements of V (X i ). We claim that U i = n for all i, but we took i = 0 for simplicity. Then we have a bijection: ' 0 : n = U0 given by ' 0 (x,,x n )=[,x,,x n ]. The inverse mapping is given by x,, x n [x 0,,x n ] 2 U 0 x 0 x 0 So, ' 0 is a bijection. Proposition ' 0 : n = U 0 is a homeomorphism (using the Zariski topology on both sets). Proof. () As pointed out above, ' 0 is a bijection. (2) For any f(x) 2 k[x,,x n ], let f(x) 2 k[x 0,,X n ] be the unique homogeneous polynomial of the same degree as f so that f(,x,,x n )=f(x,,x n ). The formula for f is as follows. Suppose that f(x) = X =(a,,a n) c X where X = X a Xa 2 2 Xan n
2 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 2 Let d = max =degf where = P a i.then f(x) is given by f(x) = X Then f(x) is homogeneous of degree d. (3) For any f(x) 2 k[x,,x n ]wehave c X d 0 X ' 0 (V (f) n )=V ( f) \ U 0 This is easy to check: (x,,x n ) 2 V (f) i f(x,,x n ) = 0. But ' 0 (x,,x n )= [,x,,x n ]whichliesinv ( f) \ U 0 i f(,x,,x n ) = 0. But f(,x,,x n ) = f(x,,x n ). So, these are the same conditions and the two sets are equal as claimed. This proves that ' 0 is an open mapping (taking open sets to open sets) since it takes basic open sets in n to basic open sets in U 0. Next, suppose that g 2 k[x 0,,X n ] is homogeneous. Then V (g(,x,,x n )) = ' 0 (V (g) \ U 0 ) This again is easy to prove. So, ' 0 is continuous. Since ' 0 is continuous and open, it is a homeomorphism. This proposition ( n = U 0 ) is interpreted to mean that P n ( ) is given by adding points at to n. To understand this, let H i = {[x] 2 P n ( ) x i =0} Then P n ( ) = U i ` Hi and U i = n, H i = Pn ( ). The second isomorphisms is given by [x,,x n ] 2 P n ( ) $ [x,,x i, 0,x i+,,x n ] 2 H i The reason that H 0 is considered to be the set of points at is the following calculation which only makes sense when = C: apple lim [,tx,,tx n ]= lim t! t! t,x,,x n =[0,x,,x n ] since t! 0 as t!.thus [0,x,,x n ]= lim t! ' 0(tx) which corresponds to points in a straight line in n going to infinity. We will quickly verify that the theorems which hold in the a ne case also hold for projective space. The lemmas work as expected. But the theorem has a surprise that I need to explain. Definition A homogeneous ideal in k[x 0,,X n ] is an ideal generated by homogeneous polynomials. Every polynomial is a sum of homogeneous polynomials: f = P m d=0 f [d] where f [d] is called the homogeneous component of f of degree d. For example: f = {z} X 3 + XY {z + YZ } + {z} 5 f [3] f [2] f [0] Lemma An ideal a is homogeneous i it contains the homogeneous components f [d] of each of its elements f.
3 22 KIYOSHI IGUSA BRANDEIS UNIVERSITY Proof. (() This implication is clear since then a is generated by the homogeneous components of its elements. ()) Let a be a homogeneous ideal. Then it is generated by homogeneous polynomials f with degree d. Any element h 2 a has the form h = X g f where g are arbitrary polynomials. Then, the homogeneous components of h are: h [d] = X g [d d ] f where the sum is over those so that d apple d. Since this is a linear combination of some of the f, h [d] 2 a for all d as claimed. Lemma The radical of any homogeneous ideal is homogeneous. Proof. Suppose that f 2 r(a). We need to show that each component f [d] of f lies in r(a). But f 2 r(a) impliesf n 2 a. Writef as a sum of components: Then f = f [d] + f [d ] + f n =(f [d] ) n + (lower terms) 2 a By the previous lemma, the top degree homogenous component lies in a: (f [d] ) n 2 a. But this implies that f [d] 2 r(a). But then f f [d] = f [d ] + (lower terms) 2 r(a) By induction on d, f [i] 2 r(a) for all i. So, r(a) is a homogeneous ideal by the previous lemma. Now we can follow all steps in the proof of Proposition 4 in [4]. Theorem There is an order reversing bijection: ( ) homogeneous radical ideal a k[x0,,x n ] ${closed subsets of P a 6= (X 0,,X n ) n ( )} a! V (a) I( ) where I( ) is the ideal generated by all homogenous polynomials f which vanish on. Proof. The reason that the homogeneous ideal (X 0,,X n ) is excluded in the bijection is because there are two ideals mapping to the same set (the empty set) in P n ( ), namely a = (X 0,,X n ) and a = k[x]. However, I(;) =k[x]. So, the required equation I(V (a)) = a does not hold for a =(X 0,,X n ). Also the proof does not work for this ideal. It is clear that = V (I( )) for all closed sets. So, we need to show that a = I(V (a)). But a I(V (a)) is clear. So, we just need to show that a I(V (a)). Let V (a) ={x 2 n+ f(x) =08f 2 a} Then we have a = r(a) =I(V (a)) So, we just need to show that I(V (a)) I(V (a)) when V (a) is nonempty. There are three cases:
4 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 23 () a = k[x], V (a) =;, V (a) =; (2) a =(X 0,,X n ), V (a) ={0}, V (a) =; (3) V (a) is nonempty and V (a) is the cone on V (a). I.e., (2.) V (a) ={ x [x] 2 V (a)}. For every [x] 2 V (a) and homogeneous f 2 I(V (a)) we have f(x) = 0. By (2.) this implies f( x) = 0 for all. So, f 2 I(V (a)). So, I(V (a)) I(V (a)) whenever (2.) holds which is Cases () and (3). So, the theorem holds in all cases but excluded Case (2). We glossed over the general discussion of homogeneous prime ideals (in the next subsection). I just said the following was clear using the obvious definitions: a closed subset of P n ( ) is irreducible if it is nonempty and not the union of two proper closed subsets, and, by a homogeneous prime ideal we mean an ideal having both of those properties. But we always need to exclude the case (X 0,,X n ). (We include it in the definition of homogeneous prime ideal but exclude it in all theorems.) Corollary In the correspondence above, irreducible closed subsets correspond to homogeneous prime ideals not equal to (X 0,,X n ). Proof. In Case 3 in the above proof, the homogeneous ideal s is prime i V (a) isirreducible i V (a) is irreducible. Cases,2 are excluded in the definition of irreducible set. Example Let f = XW k[x, Y, Z, W]. Let = V (XW it this way: YZ. This is a homogeneous polynomial of degree 2 in YZ) P 3 ( ). This may be more familiar when we write = apple x, y z,w :det=0 Since f cannot be factored as a product of polynomials of lower degree (necessarily linear), (XW YZ) is prime and is irreducible. Last week we discussed regular functions on open subsets of irreducible sets f : U!. By definition these are functions which are given locally by rational functions f(x) = g(x) h(x). Mumford emphasized that, in some cases, the same fraction cannot always be used throughout the set U. Here is an example. apple apple x, y x, y X = z,w : x 6= 0, Y = z,w : y 6= 0 Let U = X [ Y. Let f : U! be given by ( z f = x on X on Y on the intersection X \ Y w y we have that x, y are both nonzero and z x = zw xw = zw yz = w y so f is well defined. (We forgot to check the case when w = 0. But the equation xw = yz implies that z = 0 in that case and we get z x =0= w y.) This is an example of a regular function U! for which there is no single rational functional expression which works on all of U.
5 24 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2.2. Segre embedding. Mumford uses this example (in another book [5]) to introduce the following construction. (I changed the notation: elements of should be lower case letters. Capital letters are elements of the polynomial ring k[x 0,,X n ].) Definition The Segre embedding is given by : P n P m! P n+m+nm ([x i ], [y j ]) 7! [z ij = x i y j ]. In other words, each of the n + homogeneous variables X i is multiplied by each of the m + variables Y j to get (n + )(m + ) variables Z ij = X i Y j : For example, when n = m = we get: n+ m+! (n+)(m+) P ( ) P ( )! P 3 ( ) ([s, t], [u, v]) 7! [ {z} su, {z} sv, {z} tu, {z} tv ] x y z w Theorem The Segre embedding is a monomorphism with image (P n P m )=V (Z ij Z k` Z i`z kj ) Proof from [5]. (We did the case n = m = in class.) is injective: Suppose ([x], [y]) = ([x 0 ], [y 0 ]). Then there is a 6= 0 in so that x 0 i y0 j = x iy j for all i, j. Byrenumberingwe may assume x 0,y 0 are nonzero. Then x 0 0 y0 0 = x 0y 0 6= 0. So, we can rescale the coordinates of x, y, x 0,y 0 and assume that x 0 = y 0 = x 0 0 = y0 0 =. But then =. So, for every i, x 0 i y0 0 = x iy 0 making x 0 i = x i. Similarly yj 0 = y j for all j. So, ([x], [y]) = ([x 0 ], [y 0 ]). is surjective: Suppose [z] 2 V (Z ij Z k` Z i`z kj ) P n+m+nm ( ). Then, one of the coordinates is nonzero. Renumbering and rescaling we can assume that z 00 =. Then let x i = z i0 and y j = z 0j.Then z ij = z ij z 00 = z i0 z 0j = x i y j for all i, j and [z] = ([x], [y]). So, is a bijection. Lecture 8 Corollary The image of the Segre embedding is irreducible. The idea is to use the topology. The closed subsets of P n ( ) are V (S) ={[x] 2 P n ( ) : f(x) =08f 2 S [X] d } where [X] d denotes the set of degree d homogeneous polynomials in variables X with coe cients in. Since = we are in the realm of standard algebraic geometry. Since k, there are more closed set than kclosed sets. I.e., every kclosed set is closed. The following is clear. Lemma Every kclosed set which is irreducible is kirreducible. Lemma For every [c] 2 P m (k), themapping c([x]) = ([x], [c]) is continuous in the ktopology. c : P n ( )! P n+m+nm ( ) given by
6 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 25 Proof. Every kclosed set in P n+m+nm ( ) is an intersection of sets V (f) ={[z] 2 P n+m+nm ( ) : f(z) =0} for homogeneous f 2 k[z]. The inverse image of this in P n ( ) is c (V (f)) = {[x] 2 P n ( ) : f(x i c j )=0} Since f has coe cients in k and c j 2 k, thisisakclosed set in P n ( ). We need to go to the topology to get the statement that is continuous for all [c] 2 P m ( ). Lemma P n ( ) is irreducible in the topology. c : P n ( )! P n+m+nm ( ) Proof. P n ( ) = V (0) and 0 is a homogeneous prime ideal in [X]. I forgot one subtlety: We didn t prove the correspondence between prime ideals in [X] and irreducible subsets of P n ( ). But, we can go to V (0) = n+.thisis irreduciblesince isinfinite:ifn =0 this is clear since proper closed subsets of are finite. Suppose it is true for n. Then for any two proper closed subsets Y,Z, lety, z 2 n so that y 6 Y and z 6 Z. Then there are only finitely many points in y \ Y and z \ Z. Let w 2 so that (y, w) /2 Y,(z,w) /2 Z. ThenY \ n w and Z \ n w are proper closed subsets of n w whose union is the whole. But, by induction on n, n w is irreducible. Contradiction. We can now complete Mumford s proof [5] that the image of the Segre embedding : P n ( ) P m ( )! P n+m+nm ( ) is irreducible in the topology and therefore irreducible in the ktopology. Proof of Corollary Suppose not. Then is the union of two proper closed nonempty subsets. The complements of these closed subsets are disjoint nonempty open sets U,U 2. We would like to say that the inverse images of these are open subsets of P n ( ) P m ( ). However, we do not (yet) know the topology on such a product space. (It is NOT the product topology!) But we can use the continuity in each variable. Since : P n P m! is onto by definition, we have disjoint nonempty subsets (U ), (U 2 ). Let ([a], [b]) 2 (U ) and ([c], [d]) 2 (U 2 ). This means that [a] 2 b (U ), [c] 2 d (U 2) So, we have two nonempty subsets b (U ), d (U 2) of P n ( ) which are open in the  topology. Since P n ( ) is irreducible, any two open sets meet. So, there is [e] intheintersection. So, ([e], [b]) 2 (U ) and ([e], [d]) 2 (U 2 ). This implies that e (U ), e (U 2 ) are nonempty open subsets of P m ( ). Since P m ( ) is irreducible, they meet at some point ([e], [f]). This means ([e], [f]) lies in U \ U 2 which is a contradiction. Another proof of the same thing can be given using graded rings and graded ideals Graded rings and graded ideals. Definition A graded ring is a ring R which is decomposed as an infinite direct sum: R = R 0 R R 2 so that R n R m R n+m. Elements in R d are said to be homogeneous of degree d. The condition R n R m R n+m means deg(fg)=degf +degg if f,g are homogeneous. The equation R = L R d means that every element of R is equal to a sum of homogeneous components which all lie in R.
7 26 KIYOSHI IGUSA BRANDEIS UNIVERSITY The basic example is R = k[x,,x n ]. A graded ring can be constructed from its pieces as follows. Let R 0 be a ring and let R n be R 0 modules. Let µ n,m : R n R m! R n+m be an R 0 bilinear mapping which is () commutative: µ n,m (a, b) =µ m,n (b, a) for all a 2 R n,b2 R m. (2) associative: µ(a, µ(b, c)) = µ(µ(a, b),c) for all a, b, c in R n,r m,r k. (3) µ : R 0 R n! R n gives the action of R 0 on R n. Then R = L R n is a graded ring. A graded ideal can be constructed using the same idea. Definition A graded ideal in a graded ring R = L R d is a sequence of R 0 submodules I n R n so that I n R m I n+m. Proposition Graded ideals are the same as homogeneous ideal in R = L R n. Proof. A graded ideal I = L I n is clearly an ideal. It is homogeneous since it is generated by elements of S I n. (Keep in mind that there is one element 0 which is in all I n and all R n but, otherwise, the sets R n are disjoint.) Conversely, any homogeneous ideal a in a graded ring R is a graded ideal since a contains I n = a \ R n and a = L I n. The grading makes it easier to talk about projective algebraic sets: V (I) ={[x] 2 P n ( ) : f(x) =08f 2 I d 8d} Definition Let R = L R n, S = L S m be graded kalgebras. Then a graded k homomorphism of degree d is a homomorphism of kalgebras ' : R! S so that '(R n ) S dn for all n. (Note that all rings are Zalgebras.) Proposition (a) The kernel of any graded homomorphism ' : R! S of graded rings is a graded ideal in R: ker ' = L (ker ' n : R n! S dn ) (b) Conversely, any graded ideal I R is the kernel of a degree graded homomorphism: ' : R S = L R n /I n given in degree n by the quotient map R n S n = R n I n. Going back to our example, one way to show that an ideal is prime is to show that it is the kernel of a graded homomorphism. In our case, I =(XW XY ) is the kernel of the (degree 2) graded homomorphism: ' : k[x, Y, Z, W]! k[s, T, U, V ] '(X) = SU '(Y ) = SV '(Z) = TU '(W ) = TV It is clear that XW YZ 2 ker '. To show that ker ' = I =(XW YZ), we should compare: ker ' d : R d! S 2d and I d =(XW YZ)R d 2
8 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 27 Lemma If R = k[x 0,,X n ] then R 0 = k and R d a vector space over k of dimension with basis given by the monomials n+d n X = X a 0 0 Xan n where := P a i = d and all a i 0. n+d n Proof. There are monomials of degree d in n + variables. This is obvious from some examples: Take n =2,d= 7. The statement is that there are n+d n = 9 2 = 36 monomials in three letters of degree 7. To see this consider X 3 Y 2 Z 2 = XXX YY ZZ We put partitions between the letters. There are d letters and n partitions. The monomial is completely determined by the position of the partitions. For example: corresponds to Y 4 Z 3 (no X s): You are only allowed to put X i between the ith and i + st partitions. Since you choose n out of n + d placed to put your partitions, you get n+d n. We say that X has multidegree =(a 0,,a n ). In the case at hand we have: d +3 dim R d = = d + (d + 3)(d + 2) 3 6 d + dim I d =dimr d 2 = = d + d(d + ) 3 6 and the di erence is dim R d /I d = d + 6 (d2 +5d +6 d 2 d)=(d + ) 2. On the other hand, '(R d ) is spanned by monomials S a T b U c V e where a + b = d = c + e. There are (d + ) 2 such monomials. So, by linear algebra, I d =ker' d. So, I =ker' is a prime ideal. (Maybe not the most e cient way to prove this. But, when you get used to it, you can do this calculation very fast.) A similar calculation works for the general case k[z]! k[x, Y ] '(Z ij )=X i Y j Example Here is an example in [4]. The twisted cubic is V (I) P 3 ( ) where I is the kernel of the degree 3 graded homomorphism ' : k[x, Y, Z, W]! k[s, T ] '(X) = S 3 '(Y ) = S 2 T '(Z) = ST 2 '(W ) = T 3 Since ' is graded, the kernel I is a graded ideal in k[x, Y, Z, W]. The claim is: I =(XZ Y 2,YW Z 2,XW YZ) Certainly, these lie in ker '. It is also easy to see that ' d : R d! S 3d
9 28 KIYOSHI IGUSA BRANDEIS UNIVERSITY is surjective. Furthermore, any two monomials in the inverse image of S a T 3d a di er by a element of I d. So, I =ker'. (Proof: We can assume a apple 3d a. Then for any monomial X i Y j Z k W ` in ' (S a T 3d a ), we have k + ` 0. So, we can exchange all X terms using the relations XZ Y 2,XW YZ modulo I d.thenwehavei = 0. Next we can eliminate either Y or W in the expression using YW = Z 2. We are left with either Y Z or Z W which is unique.) The graded homomorphism ' : k[x, Y, Z, W]! k[s, T ] gives a morphism f ' : P ( )! P 3 ( ) defined by f ' ([s, t]) = [s 3,s 2 t, st 2,t 3 ] This is welldefined since, if we replace [s, t] with the equivalent [s 0,t 0 ]=[ s, t] then f ' ([s 0,t 0 ]) = [ 3 s 3, 3 s 2 t, 3 st 2, Claim: Let denote the image of this mapping. Then =V (I) where I =(XZ Y 2,YW Z 2,XW YZ). Proof: It is clear that V (I) since each generator of I is a homogeneous polynomial which is zero on the image of f '. Conversely, suppose that [x, y, z, w] 2 V (I). Then xz = y 2,yw = z 2,xw = yz. Also, either x 6= 0 or w 6= 0 (otherwise x = y = z = w =0 which is not allowed). Suppose w 6= 0. Then we may assume w =. Then, y = z 2 and x = yz = z 3. So, [x, y, z, ] = [z 3,z 2,z,] = f ' ([z,]) is in the image of f '. Lecture 8 ends here. We skipped the next subsection because it is routine Irreducible algebraic sets in P n. We need to verify that the arguments used in the a ne case carry over to the homogeneous case. Lemma A homogeneous ideal a in k[x] is prime i it satisfies the following. For any two homogenous polynomials f,g, iffg 2 a then either f 2 a or g 2 a. Remark This can be rephased (and generalized) in two equivalent ways: () A graded ring R = L R n is a domain i there are no nonzero elements f 2 R n,g 2 R m so that fg = 0. (2) A graded ideal I = L I d in a graded ring R = L R d is prime i for all f 2 R n \I n, g 2 R m \I m, fg 2 R n+m \I n+m. Proof. Prime ideals certainly have this property. Conversely, suppose that a is homogeneous and has this property. Let f,g be arbitrary polynomials so that fg 2 a. Let f [n],g [m] be the highest degree terms in f,g. Then f [n] g [m] 2 a since it is the highest degree term in fg. So, either f [n] or g [m] lies in a, sayf [n] 2 a. Then(f f [n] )g 2 a. By induction, either f f [n] 2 a (which implies f 2 a) or g 2 a. So, a is prime. By a similar argument we have the following. Lemma A homogeneous ideal a in k[x] is primary i it satisfies the following. For any two homogenous polynomials f,g, iffg 2 a then either f 2 r(a) or g 2 a. Write this in terms of graded ideals in any graded ring. 3 t 3 ]
10 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 29 Proof. We may assume that none of the homogeneous components of g lie in a. Then f [n] g [m] 2 a implies that f [n] 2 r(a). Let k be minimal so that (f [n] ) k g 2 a. Then (f f [n] )(f [n] ) k g =(f [n] ) k fg (f [n] ) k g 2 a and (f [n] ) k g/2 a. So, f f [n] 2 r(a) (whichimpliesf 2 r(a) by induction on the number of homogeneous components of f). Using these lemmas, the proof in the a ne case carries over to show the following. Proposition Every homogeneous radical ideal in any graded Noetherian ring is the intersection of finitely many homogeneous prime ideals. Theorem Irreducible closed subsets of P n ( ) correspond to homogeneous prime ideal in k[x] not equal to (X 0,,X n ). Furthermore, every closed subset of P n ( ) can be expressed uniquely as a finite union of irreducible subsets. Note: the uniqueness is a topological property. The proof of Proposition.4.2 works in any topological space. Lecture Projective schemes. The purpose of this section is to construct the projective analogue of Spec(R) and prove the analogue of Theorem.9.3. We continue using the color code: red for schemes, blue for varieties. Definition Let R = L R d be a graded ring. Then Proj(R) is defined to be the following topological space (together with a sheaf which will be constructed later). As a set Proj(R) is the set of homogeneous prime ideals in R which are not equal to the ideal L d>0 R d. In the special case R = k[x 0,,X n ], this is the ideal (X 0,X,,X n )which was excluded before. The topology on Proj(R) is given as follows. For any f 2 R, let V (f) ={P 2 Proj(R) :f 2 P } These sets are defined to be closed. General closed sets are intersections of these sets: V (S) = \ V (f) ={P 2 Proj(R) :S P } f2s In the case R = k[x] we have a bijection: irreducible closed subsets of P n ( ) $ P 2 Proj(k[X]) Theorem There is a continuous epimorphism : P n ( ) Proj(k[X]) given by ([x]) = P where P 2 Proj(k[X]) corresponds to the irreducible set [x]. Proof. We first show is surjective. Let P be a homogeneous prime ideal in R = k[x 0,,X n ], P 6= (X 0,,X n ). Then P 6= (X 0,,X n ) prime ) V (P ) 6= {0} n+ irreducible We know that V (P ) has a generic point y 2 n+. And y 6= 0 since, otherwise, V (P )= 0={0}, the excluded case. So, y gives a point [y] 2 P n ( ). Let = V (P ). Claim: [y] =.(Thus ([y]) = P.) Pf: [y] = 0 = V (I) wherei is some graded ideal. But y 2 V (I) closed V (P )implies V (I) =V (P )impliesi = P. So, 0 =.
11 30 KIYOSHI IGUSA BRANDEIS UNIVERSITY Next we show is continuous. Take a closed subset V (f) Proj(R). Then (V (f)) ={[y] 2 P n ( ) : f 2 I([y])} = V (f). But V (f) P n ( ) is a closed subset by definition. So, is continuous. Corollary The continuous epimorphism : P n ( ) Proj(R) is also an open mapping. Therefore, Proj(R) has the quotient topology. Proof. Every open subset of Y = P n is a union of basic open sets Y f where f 2 k[x] is homogeneous. But Y f is the complement of V (f). And (V (f)) =V (f) implies that (Y f ) is the complement of V (f) in Proj(R). So, is an open mapping as claimed. Example Let k = R and consider P ( )! Proj(R[X, Y ]). Then P (R) =R [ is a circle and P ( ) = [. The elements of P ( ) are [,x] for x 2 and =[0,x]= [0,y] 8x, y. P ( ) has three kinds of points: () P (R) is the set of closed point x = {x}, x 2 R and. (2) P (C) is the set of x =[y] so that x is finite (consisting of [y] and [y], the complex conjugate of [y]. (3) P ( )\P (C) is the set of generic points. This comes from dimension theory. The only homogeneous ideal in R[X, Y ] which is not maximal is 0. Consider Proj(R[X, Y ]) as the quotient space of P ( ) under the identification [x] [y] if ([x]) = ([y]). This gives the following. P (C) =C [ is a sphere. The equator is P (R). The identification is z z. This identifies the top hemisphere with the bottom hemisphere. All other points are generic and therefore are identified to one point in Proj. So, Proj(R[X, Y ]) is a 2dimensional disk with one generic point which we view as being smeared over the entire disk (like peanut butter on a cracker). Spec(R) = [ (generic point) We did another example. The key point was: Given ' : R! S a graded homomorphism of degree d, there is an induced mapping ' : Proj(S)! Proj(R) sending P to Q = ' (P ). This mapping is continuous since since f 2 ' (P ), '(f) 2 P. (' ) (V (f)) = {P : f 2 ' (P )} = V ('(f)) 2.6. Projective space as union of a ne spaces. Recall that P n ( ) = S U i where U i = {x i 6=0}. Then n = Ui (x,,x n ) 7! [x,,x i,,x i+,,x n ] Projective space (together with its structure sheaf) is the basic example of a variety which is not a ne but rather made up of a ne varieties pasted together. We need to study this so
12 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 3 that we can generalize the construction later. The a ne pieces U i are pasted together along their intersection. This pasting is given by isomorphisms of open subsets of irreducible sets. We want to see how U i,u j are pasted together. By symmetry we can assume i =0,j =. We look at U 0 \ U and see where it goes to under the equivalence n = U i. x 0 6=0 x 6=0 U 0 U = U 0 \ U x 0,x 6=0 = V 0 V x y n x 6=0 x 6=0 n The image of U 0 \U under the equivalences U i = n are shaded. The image under both maps is the set V 0 = V = {x 6=0}. The mapping : V 0! V is the pasting map: :(x,,x n ) 7! [,x,,x n ]/x 7!, x 2,, x n x x x This is a regular function with 2 = id. In the drawing, (x) =y. Since V 0,V are given by X 6= 0, the ring of regular functions on V 0 and V is V 0 = V = k[x 0,X,X 2,,X n ]/(X 0 X ) = R Then ' : V = V0 is given by ' (X 0 )=X,' (X )=X 0 and ' (X i )=X 0 X i for i 2. This is an automorphism of R as kalgebra with (' ) 2 = id. In my notes I verified that the ring isomorphism ' induces an isomorphism (' ) : Spec( V 0 )! Spec( V ) which is compatible with the homeomorphism in the sense that (' ) sends the prime ideal (x) to (y) ify = (x). But I don t think I got to that in class. We will do this again later (verify that the pasting together of a ne varieties is compatible with the pasting together of corresponding a ne schemes).
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