looijienga ag2013

8/11/2019 Looijienga AG2013

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Algebraic Geometry I

Fall 2013

Eduard Looijenga


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Rings are always supposed to possess a unit element 1 and a ring homomor-phism will always take unit to unit. We allow that 1 = 0, but in that case we getof course the zero ring {0}. We assume a ring to be commutative unless the con-trary is stated. If R is a ring, then we denote the multiplicative group of invertibleelements (units) of R by R. We say that R is a domain if R has no zero divisors,equivalently, if (0) is a prime ideal. An R -algebra is a ring A endowed with a ringhomomorphism : R A, but if is understood, then for every r R and aA,the product (r )a is often denoted by ra . We say that A is nitely generated asan R-algebra if we can nd a1 , . . . , a n in A such that every element of A can be written as a polynomial in these elements with coefcients in R ; in other words,the R-algebra homomorphism R[x1 , . . . , x n ] A which sends the variable x i to a iis onto. This is not to be confused with the notion of nite generation for an R -module M resp. which merely means the existence of a surjective homomorphismof R-modules Rn M . Similarly, a eld K is said to be nitely generated as aeld over a subeld k if we can nd elements b1 , . . . , bn in K such that every ele-ment of K can be written as a fraction of two polynomials in these elements withcoefcients in k.


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CHAPTER 1

Afne varieties

Throughout this chapter we x an algebraically closed eld k. Recall that thismeans that every polynomial f k[x] of positive degree has a root x1 k: f (x1) =0. This implies that we can split off the factor x x1 from f with quotient apolynomial of degree one less than f . Continuing in this manner we nd that f decomposes simply as f (x) = c(x x1) (x xd ) with c k {0}, d = deg( f )and x1 , . . . , x d

k. Since an algebraic extension of k is obtained by adjoining

to k roots of polynomials in k[x], this also shows that the property in question isequivalent to: every algebraic extension of k is equal to k.

A rst example you may think of is the eld of complex numbers C, but as weproceed you should be increasingly be aware of the fact that there are many others:it is shown in a standard algebra course that for any eld F an algebraic closure F is obtained by adjoining to F the roots of every polynomial f F [x] (this can notalways be done in one step, and might involve an innite process). So we couldtake for k an algebraic closure of the eld of rational numbers Q, of the nite eldFq , where q is a prime power 1 or even of the quotient eld of any domain such asC[x1 , . . . , x r ].

1. The Zariski topology

Any f k[x1 , . . . , x n ] determines in an evident manner a function kn

k.In such cases we prefer to think of kn not as vector spaceits origin and vectoraddition will be irrelevant to usbut as a set with a weaker structure. We shallmake this precise later, but it basically amounts to only remembering that elementsof k[x1 , . . . , x n ] can be understood as k-valued functions on it. For that reason itis convenient to denote this set differently, namely as An (or as Ank , if we feel that we should not forget about the eld k). We refer to An as the afne n -space overk. A k-valued function on An is then said to be regular if it is dened by somef k[x1 , . . . , x n ]. We denote the zero set of such a function by Z (f ) and thecomplement (the nonzero set) by U (f ) An . If f is not a constant polynomial(that is, f /k), then we call Z (f ) a hypersurface of An .

E XERC ISE 1. Prove that f k[x1 , . . . , x n ] is completely determined by the reg-ular function it denes. (Hint: do rst the case n = 1 .) So the ring k[x1 , . . . , x n ] canbe regarded as a ring of functions on An under pointwise addition and multiplica-tion. (This would fail to be so had we not assumed that k is algebraically closed:for instance the function on the nite eld Fq dened by xq x is identically zero.)

E XERC ISE 2. Prove that a hypersurface is nonempty.

1Since the elements of any algebraic extension of Fq of degree n 2 are roots of x ( qn ) x , we

only need to adjoin roots of such polynomials.

3


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4 1. AFFINE VARIETIES

It is perhaps somewhat surprising that in this rather algebraic context, the lan-guage of topology proves to be quite effective: algebraic subsets of An shall appearas the closed sets of a topology, albeit a rather peculiar one.

LEMMA -DEFINITION 1.1. The collection {U (f ) : f k[x1 , . . . , x n ]}is a basis of atopology on An , called the Zariski topology 2. A subset of An is closed for this topology if and only if it is an intersection of zero sets of regular functions.

PROOF. We recall that a collection {U } of subsets of a set X is a basis for atopology if and only if its union is all of X and any intersection U 1 U 2 is a unionof members of {U } . This is here certainly the case, for we have U (0) = X andU (f 1) U (f 2) = U (f 1f 2). Since an open subset of An is by denition a union of subsets of the form U (f ), a closed subset must be an intersection of subsets of theform Z (f ).

E XAM PLE 1.2. The Zariski topology on A1 is the conite topology: its closedsubsets = A1 are the nite subsets.

E XERC ISE 3. Show that the diagonal in A2 is closed for the Zariski topology,but not for the product topology (where each factor A1 is equipped with the Zariskitopology). So A2 does not have the product topology.

We will explore the mutual relationship between the following two basic maps:

{subsets of An } I

{ideals of k[x1 , . . . , x n ]}

{closed subsets of An } Z

{subsets of k[x1 , . . . , x n ]}. where for a subset X An , I (X ) is the ideal of f k[x1 , . . . , x n ] with f |X = 0 andfor a subset J k[x1 , . . . , x n ], Z (J ) is the closed subset of An dened by f J Z (f ).Observe that

I (X 1X 2) = I (X 1) I (X 2) and Z (J 1J 2) = Z (J 1) Z (J 2).In particular, both I and Z are inclusion reversing. Furthermore, I denes a sectionof Z : if Y An is closed, then Z (I (Y )) = Y . We also note that by Exercise 1I (An ) = (0) , and that any singleton { p = ( p1 , . . . , pn )} An is closed, as it is thecommon zero set of the degree one polynomials x1 p1 , . . . , x n pn .

E XERC ISE 4. Prove that I ({ p}) is equal to the ideal generated by these degreeone polynomials and that this ideal is maximal.E XERC ISE 5. Prove that the (Zariski) closure of a subset Y of An is equal to

Z (I (Y )) .

Given Y An , then f , g k[x1 , . . . , x n ] have the same restriction to Y if andonly if f g I (Y ). So the quotient ring k[x1 , . . . , x n ]/I (Y ) (a k-algebra) can beregarded as a ring of k-valued functions on Y . Notice that this k-algebra does notchange if we replace Y by its Zariski closure.

DEFINITION 1.3. Let Y An be closed. The k-algebra k[x1 , . . . , x n ]/I (Y ) iscalled the coordinate ring of Y and we denote it by A(Y ). A k-valued function onY is said to be regular if it lies in this ring.

2We shall later modify the denition of both An and the Zariski topology.


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1. THE ZARISKI TOPOLOGY 5

Given a closed subset Y An , then for every subset X An we have X Y if and only if I (X )I (Y ), and in that case I Y (X ) := I (X )/I (Y ) is an ideal of A(Y ):it is the ideal of regular functions on Y that vanish on X . Conversely, an ideal of A(Y ) is of the form J/I (Y ), with J an ideal of k[x1 , . . . , x n ] that contains I (Y ), andsuch an ideal denes a closed subset Z (J ) contained in Y . So the two basic mapsabove give rise to such a pair on Y :

{subsets of Y } I Y {ideals of A(Y )}

{closed subsets of Y }

Z Y {subsets of A(Y )}.We ask: which ideals of k[x1 , . . . , x n ] are of the form I (Y ) for some Y ? Clearly, if f k[x1 , . . . , x n ] is such that some positive power vanishes on Y , then f vanisheson Y . In other words: if f m I (Y ) for some m > 0, then f I (Y ). This suggests:

PROPOSITION -DEFINITION 1.4. Let R be a ring (as always commutative and with1 ) and let J R be an ideal. Then the set of aR with the property that am J for some m > 0 is an ideal of R , called the radical of J and denoted J .

We say that J is a radical ideal if J = J .We say that the ring R is reduced if the zero ideal (0) is a radical ideal (in other

words, R has no nonzero nilpotents: if aR is such that am = 0 , then a = 0 ).

PROOF. We show that J is an ideal. Let a, b J so that am , bn J forcertain positive integers m, n . Then for every r R, ra J , since (ra )m =r m am J . Similarly ab J , for (ab)m + n is a linear combination of monomialsthat are multiples of am or bn and hence lie in J .

E XERC ISE 6. Show that a prime ideal is a radical ideal.

Notice that J is a radical ideal if and only if R/J is reduced. The precedingshows that for every Y An , I (Y ) is a radical ideal, so that A(Y ) is reduced. Thedictionary between algebra and geometry begins in a more substantial manner with

THEOREM 1.5 (Hilberts Nullstellensatz). For every ideal J k[x1 , . . . , x n ] wehave I (Z (J )) = J .

We inclusion

is clear; the hard part is the opposite inclusion (which says thatif f k[x1 , . . . , x n ] vanishes on Z (J ), then f m J for some positive integer m ).We postpone its proof, but we discuss some of the consequences.

COROLLARY 1.6. Let Y An be closed. Then the maps I Y and Z Y restrict tobijections:{closed subsets of Y } {radical ideals of A(Y )}.These bijections are inclusion reversing and each others inverse.

PROOF. We rst prove this for Y = An . We already observed that for every closed subset X of An we have Z (I (X )) = X . The Nullstellensatz says that for aradical ideal J k[x1 , . . . , x n ], we have I (Z (J )) = J . An ideal of A(Y ) is of the form J/I (Y ). This is a radical ideal if and only if J is one. So the property also follows for an arbitrary closed Y .


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Let m be a maximal ideal of k[x1 , . . . , x n ]. Such an ideal is certainly radicalas it is a prime ideal and so it is also maximal among the radical ideals that aredistinct from k[x

1, . . . , x

n]. Hence it is of the form I (Y ) for a closed subset Y . Since

the empty subset of An is dened by the radical ideal k[x1 , . . . , x n ], Corollary 1.6implies that Y will be nonempty and minimal for this property. In other words,Y is a singleton {y} (and if y = ( a1 , . . . , a n ), then m = ( x1 a1 , . . . , x n an ), by Exercise 4). Thus the above correspondence provides a bijection between the pointsof An and the maximal ideals of An . If we are given a closed subset Y An anda point y An , then y Y if and only if the maximal ideal dened by y containsI (Y ). But a maximal ideal containing I (Y ) is the preimage of a maximal ideal inA(Y ) = k[x1 , . . . , x n ]/I (Y ). We conclude that points of Y correspond to maximalideals of A(Y ). Via this (or a very similar) correspondence, algebraic geometry seeks to express geometric properties of Y in terms of algebraic properties of A(Y )and vice versa. In the end we want to forget about the ambient An completely.

2. Irreducibility and decomposition

We introduce a property which for most topological spaces is of little interest,but as we will see, is useful and natural for the Zariski topology.

DEFINITION 2.1. Let Y be a topological space. We say that Y is irreducible if it is nonempty and cannot be written as the union of two closed subsets = Y . Anirreducible component of Y is a maximal irreducible subset of Y .

E XERC ISE 7. Prove that an irreducible Hausdorff space must consist of a singlepoint. Prove also that an innite set with the conite topology is irreducible.

E XERC ISE 8. Let G1 , . . . , G s be closed subsets of a topological space Y . Provethat any irreducible subset of G1 Gs is contained in some Gi .

E XERC ISE 9. This exercise has been scrapped.

LEMMA 2.2. Let Y be a topological space.(i) If Y is irreducible, then every nonempty open subset of Y is dense in Y and

irreducible.(ii) Conversely, if C Y is an irreducible subspace, then C is also irreducible. In particular, an irreducible component of Y is always closed in Y .

PROOF. (i) Suppose Y is irreducible and let U Y be open and nonempty.Then Y is the union of the two closed subspaces Y U and U . Since Y is irreducible,and Y U = Y , we must have U = Y . So U is dense in Y . To see that U isirreducible, suppose that U is the union of two subsets that are both closed in U .These subsets will be of the form G i U with G i closed in Y . Then G1 G2 is aclosed subset of Y which contains U . Since U = Y , it follows that G1 G2 = Y .The irreducibility of Y implies that one of the G

i (say G

1) equals Y and then

U = G1 U .(ii) Let C Y be irreducible (and hence nonempty). If C is written as a unionof two closed subsets G1 , G2 of Y , then C is the union of the two subsets G1 C and G2 C that are both closed in C , and so one of these, say G1 C , equals C .This means that G1 C and hence G1 C . So C is irreducible. The following proposition tells us what irreducibility amounts to in the Zariski

topology.


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2. IRREDUCIBILITY AND DECOMPOSITION 7

PROPOSITION 2.3. A closed subset Y An is irreducible if and only if I (Y ) isa prime ideal (which, we recall, is equivalent to: A(Y ) = k[x1 , . . . , x n ]/I (Y ) is adomain).

PROOF. Suppose Y is irreducible and f, g k[x1 , . . . , x n ] are such that fg I (Y ). Then Y Z (fg ) = Z (f )Z (g). Since Y is irreducible, Y is contained inZ (f ) or in Z (g). So f I (Y ) or g I (Y ), proving that I (Y ) is a prime ideal.Suppose I (Y ) is a prime ideal, but that Y is the union of two closed subsets Y 1and Y 2 that are both = Y . Since Y i = Y , there exists a f i I (Y i ) I (Y ). Thenf 1f 2 vanishes on Y 1 Y 2 = Y , so that f 1f 2 I (Y ). The fact I (Y ) is a prime idealimplies that one of f 1 and f 2 is in I (Y ). Whence a contradiction.

One of our rst aims is to prove that the irreducible components of any closedsubset Y An are nite in number and have Y as their union. This may notsound very surprising, but we will see that this reects some nonobvious algebraicproperties. Let us rst consider the case of a hypersurface. Since we are goingto use the fact that k[x1 , . . . , x n ] is a unique factorization domain, we begin withrecalling that notion:

DEFINITION 2.4. We say that R is a unique factorization domain if R has nozero divisors and every principal ideal (a) := Ra in R which is neither the zeroideal nor all of R is in unique manner a product of principal prime ideals: (a) =( p1)( p2) ( ps ) (so the ideals ( p1), . . . , ( ps ) are unique up to order).

Observe that the principal ideal generated by p R is prime if and only if forany factoring of p, p = p p , one of the factors must be a unit and that the identity (a) = ( p1)( p2) ( ps ) amounts to the statement that a is a unit times p1 p2 ps .The factors are then unique up to order and multiplication by a unit.

For a eld (which has no proper principal ideals distinct from (0)) the imposedcondition is empty and hence a eld is automatically unique factorization domain. A more substantial example (that motivated this notion in the rst place) is Z: aprincipal prime ideal of Z is of the form ( p), with p a prime number. Every integern 2 has a unique prime decomposition and so Z is a unique factorization domain. A basic theorem in the theory of rings asserts that if R is a unique factorizationdomain, then so is its polynomial ring R[x]. This implies for instance (with induc-tion on n) that R[x1 , . . . , x n ] is one. Indeed, in the case when R is a eld (think of our k), then a nonzero principal ideal of this ring is prime precisely when it is gen-erated by an irreducible polynomial of positive degree and every f R[x1 , . . . , x n ]of positive degree then can be written as a product of irreducible polynomials:f = f 1f 2 f s , a factorization that is unique up to order and multiplication of eachf i by a nonzero element of R.

The following proposition connects two notions of irreducibility.

PROPOSITION 2.5. If f k[x1 , . . . , x n ] is irreducible, then the hypersurface Z (f )it denes is irreducible. If f k[x1 , . . . , x n ] is of positive degree and f = f 1f 2 f sis a factoring into irreducible polynomials, then Z (f 1), . . . , Z (f s ) are the irreduciblecomponents of Z (f ) and their union equals Z (f ) (but we are not claiming that theZ (f i )s are pairwise distinct). In particular, a hypersurface is the union of its ir-reducible components; these irreducible components are hypersurfaces and nite innumber.


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PROOF. If f k[x1 , . . . , x n ] is irreducible, then f generates a prime ideal andso Z (f ) is an irreducible hypersurface by Proposition 2.3 .If f

k[x

1, . . . , x

n] is of positive degree and f = f

1f

2 f

s is as in the propo-

sition, then it is clear that Z (f ) = Z (f 1) Z (f s ) with each Z (f i ) irreducible.To see that Z (f i ) is an irreducible component of Z (f ), suppose that C Z (f ) isirreducible. By Exercise 8, C is contained in some Z (f i ). Since C is a maximalirreducible subset of Z (f ), it follows that we then must have equality: C = Z (f i ).It remains to observe that if any inclusion relation Z (f i ) Z (f j ) is necessarily anidentity (prove this yourself) so that each Z (f i ) is already maximal and hence inirreducible component.

The discussion of the general case begins with the rather formal

LEMMA 2.6. For a partially ordered set (A, ) the following are equivalent:(i) (A, ) satises the ascending chain condition : every ascending chain a1 a2 a3 becomes stationary: an = an +1 = for n sufciently large.

(ii) Every nonempty subset B A has a maximal element, that is, an elementb0 B such that there is no bB with b > b0 .PROOF. (i)

(ii). Suppose (A, ) satises the ascending chain condition andlet B A be nonempty. Choose b1 B . If b1 is maximal, we are done. If not, thenthere exists a b2 B with b2 > b1 . We repeat the same argument for b2 . We cannotindenitely continue in this manner because of the ascending chain condition.

(ii)

(i). If (A, ) satises (ii), then the set of members of any ascending chainhas a maximal element, in other words, the chain becomes stationary. If we replace by , then we obtain the notion of the descending chain condi-tion and we nd that this property is equivalent to: every nonempty subset B Ahas a minimal element. These properties appear in the following pair of denitions.DEFINITIONS 2.7. We say that a ring R is noetherian if its collection of ideals

satises the ascending chain condition.We say that a topological space Y is noetherian if its collection of closed subsets

satises the descending chain condition.

E XERC ISE 10. Prove that a subspace of a noetherian space is noetherian. Provealso that a ring quotient of a noetherian ring is noetherian.

E XERC ISE 11. Prove that a noetherian space is compact: every covering of sucha space by open subsets contains a nite subcovering.

The interest of the noetherian property is that it is one which is possessed by almost all the rings we encounter and that it implies many niteness properties without which we are often unable to go very far. Let us give a nonexample rst:the ring R of holomorphic functions on C is not noetherian: if I k denotes the idealof f R vanishing on all the integers k, then I 1 I 2 is a strictly ascendingchain of ideals in R .

Obviously a eld is noetherian. The ring Z is noetherian: if I 1 I 2 is anascending chain of ideals in Z, then s =1 I s is an ideal of Z, hence of the form (n)for some n Z. But if s is such that n I s , then clearly the chain is stationary asof index s. (This argument only used the fact that any ideal in Z is generated by asingle element, i.e., that Z is a principal ideal domain.) That most rings we knoware noetherian is a consequence of


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THEOREM 2.8 (Hilberts basis theorem). If R is a noetherian ring, then so isR[x].

As with the Nullstellensatz, we postpone the proof and discuss some of itsconsequences rst.

COROLLARY 2.9. If R is a noetherian ring (for example, a eld) then so is every nitely generated R -algebra. Also, the space An (and hence any closed subset of An )is noetherian.

PROOF. The Hilbert basis theorem implies (with induction on n ) that the ringR[x1 , . . . , x n ] is noetherian. By Exercise 10 , every quotient ring R[x1 , . . . , x n ]/I isthen also noetherian. But a nitely generated R -algebra is (by denition actually)isomorphic to some such quotient and so the rst statement follows.

Suppose An Y 1 Y 2 is a descending chain of closed subsets. Then(0) I (Y 1) I (Y 2) is an ascending chain of ideals. As the latter becomesstationary, so will become the former. PROPOSITION 2.10. If Y is noetherian space, then its irreducible components are

nite in number and their union equals Y .

PROOF. Suppose Y is a noetherian space. We rst show that every closed sub-set can be written as a nite union of closed irreducible subsets. First note thatthe empty set has this property, for a union with empty index set is empty. Let Bbe the collection of closed subspaces of Y for which this is not possible, i.e., thatcannot be written as a nite union of closed irreducible subsets. Suppose that Bis nonempty. According to 2.6 this collection has a minimal element, Z , say. ThisZ must be nonempty and cannot be irreducible. So Z is the union of two properclosed subsets Z and Z . The minimality of Z implies that neither Z nor Z isin B and so both Z and Z can be written as a nite union of closed irreduciblesubsets. But then so can Z and we get a contradiction.

In particular, there exist closed irreducible subsets Y 1 , . . . , Y k of Y such thatY = Y 1 Y k (if Y = , take k = 0 ). We may of course assume that noY i is contained in some Y j with j = i (otherwise, omit Y i ). It now remains toprove that the Y i s are the irreducible components of Y , that is, we must show thatevery irreducible subset C of Y is contained in some Y i . But this follows from anapplication of Exercise 8.

If we apply this to An , then we nd that every subset Y An has a nitenumber of irreducible components, the union of which is all of Y . If Y is closedin An , then so is every irreducible component of Y and according to Proposition2.3 any such irreducible component is dened by a prime ideal. This allows us torecover the irreducible components of a closed subset Y An from its coordinatering:

COROLLARY 2.11. Let Y An be a closed subset. If C is an irreducible componentof Y , then the image I Y (C ) of I (C ) in A(Y ) is a minimal prime ideal of A(Y ) andany minimal prime ideal of A(Y ) is so obtained: we thus get a bijective correspondencebetween the irreducible components of Y and the minimal prime ideals of A(Y ).

PROOF. Let C be a closed subset of Y and let I Y (C ) be the corresponding idealof A(Y ). Now C is irreducible if and only if I (C ) is a prime ideal of k[x1 , . . . , x n ],or what amounts to the same, if and only if I Y (C ) is a prime ideal of A(Y ). It is


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an irreducible component if C is maximal for this property, or what amounts to thesame, if I Y (C ) is minimal for the property of being a prime ideal of A(Y ).

E XAM PLE 2.12. First consider the set C := {(t, t 2 , t 3) A3 | t k}. This is aclosed subset of A3 : if we use (x,y,z ) instead of (x1 , x2 , x3), then C is the commonzero set of yx2 and z x3 . Now the inclusion k[x]k[x,y,z ] composed with thering quotient k[x,y,z ] k[x,y,z ]/ (y x2 , z x3) is easily seen to be an isomor-phism. Since k[x] has no zero divisors, (y x2 , z x3) must be a prime ideal. So C is irreducible and I (C ) = ( y x2 , z x3).We now turn to the closed subset Y A3 dened by xyz = 0 and y3 z2 = 0 .Let p = ( x,y,z ) Y . If y = 0 , then we put t := z /y ; from y3 = x2 , it follows thaty = t2 and z = t3 and xy = z implies that x = t. In other words, p C in thatcase. If y = 0 , then z = 0 , in other words p lies on the x-axis. Conversely, any pointon the x-axis lies in Y . So Y is the union of C and the x-axis and these are theirreducible components of Y .

We briey discuss the corresponding issue in commutative algebra. We begin with recalling the notion of localization and we do this in the generality that isneeded later.

2.13. L OCALIZATION. Let R be a ring and let S be a multiplicative subset of R : 1 S , 0 /S and S closed under multiplication. Then a ring S 1 R , together with a ring homomorphismR S 1 R is dened as follows: an element of S 1 R is written as a formal fraction r/s , withr R and s S , with the understanding that r/s = r /s if and only if s (s r sr ) = 0 forsome s S . This is a ring indeed: multiplication and subtraction is dened as for ordinary fractions: r/s.r /s = ( rr )/ (ss ) and r/s r /s = ( s r sr )/ (ss ); it has 0/ 1 as zero and1/ 1 as unit element. Since 0 / S , we have 0/ 1 = 1 / 1. The homomorphism R S 1 Ris simply r r/ 1. Notice that it maps any s S to an invertible element of S 1 R : theinverse of s/ 1 is 1/s . In a sense (made precise in part (b) of Exercise 12 below) the ringhomomorphism R S 1 R is universal for that property. This construction is called thelocalization away from S .3

It is clear that if S does not contain zero divisors, then r/s = r /s if and only if s r sr = 0 ; in particular, r/ 1 = r / 1 if and only if r = r , so that R S 1 R is theninjective. If we take S maximal for this property, namely take it to be the set of nonzerodivisors of R (which is indeed multiplicative), then S 1 R is called the fraction ring Frac( R)of R. In case R is a domain, S = R { 0} and so Frac( R) is a eld, the fraction eld of R . This gives the following corollary, which hints to the importance of prime ideals in thesubject.

COROLLARY 2.14. An ideal p of a ring R is a prime ideal if and only if it is the kernel of aring homomorphism from R to a eld

PROOF. It is clear that the kernel of a ring homomorphism from R to a eld is alwaysa prime ideal. Conversely, if p is a prime ideal, then it is the kernel of the composite R R/ p Frac( R/ p).

But the case of interest here is when we are given some a R which is not nilpotent.Then we can take S = {an | n 0} in which case we usually write R[1/a ] for S 1 R .

E XERC ISE 12. Let R be a ring and let S be a multiplicative subset of R .(a) What is the the kernel of R S 1R?(b) Prove that a ring homomorphism : R R with the property that (s)is invertible for every sS factors in a unique manner through S 1R .

3It is sometimes useful to allow 0S . We then stipulate that S 1 R is the zero ring.


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(c) Consider the polynomial ring R [xs : s S ] and the homomorphism of R-algebras R[xs : s S ] S 1R that sends x s to 1/s . Prove that thishomomorphism is surjective and that its kernel consists of the f

R[xs

:s S ] which after multiplication by an element of S lie in the idealgenerated the degree one polynomials sx s 1, sS .

E XERC ISE 13. Let R be a ring and let p be a prime ideal of R .(a) Prove that the complement R p is a multiplicative system. The result-ing localization (R p)1R is called the localization at p and is usually denoted R p .(b) Prove that pR p is a maximal ideal of R p and that it is the only maximal

ideal of Rp . (A ring with a unique maximal ideal is called a local ring .)(c) Prove that the localization map R R p drops to an isomorphism of eldsFrac( R/ p) Rp / pR p .(d) Work this out for R = Z and p = ( p), where p is a prime number.(e) Same for R = k[x, y ] and p = ( x).

LEMMA 2.15. Let R be a ring. Then the intersection of all the prime ideals of Ris the ideal of nilpotents (0) of R . Equivalently, for every nonnilpotent a R , thereexists a ring homomorphism from R to a eld that is nonzero on a .

PROOF. It is easy to see that a nilpotent element lies in every prime ideal. Nowfor nonnilpotent a R consider the homomorphism R R[1/a ]. The ring R[1/a ]has a maximal ideal 4 and hence admits a ring homomorphism to some eld F : : R[1/a ] F . Then the kernel of the composite R R[1/a ] F is a primeideal and a is not in this kernel (for its image is invertible with inverse (1/a )).

E XERC ISE 14. Let R be a ring. Prove that the intersection of all the maximalideals of a ring R consists of the a R for which 1 + aR R (i.e., 1 + ax isinvertible for every x

R). You may use the fact that every proper ideal of R is

contained in a maximal ideal.

We can do better if R is noetherian. The following proposition is the algebraiccounterpart of Proposition 2.10 . Note the similarity between the proofs.

PROPOSITION 2.16. Let R be a noetherian ring. Then any radical ideal of R isan intersection of nitely many prime ideals. Also, the minimal prime ideals of R are nite in number and their intersection is equal to the ideal of nilpotents (0) .

PROOF. Let B be the collection of the radical ideals I R that can not be writ-ten as an intersection of nitely many prime ideals and suppose that B is nonempty.Since R is noetherian, B contains a maximal member I 0 . We derive a contradictionas follows.

Since I 0 cannot be a prime ideal, there exist a1 , a2

R

I 0 with a1a2

I 0 .

Consider the radical ideal J i := I 0 + Ra i . We claim that J 1 J 2 = I 0 . Theinclusion

is obvious and

is seen as follows: if aJ 1J 2 , then for i = 1 , 2, thereexists an ni > 0 such that an i I 0 + Ra i . Hence an 1 + n 2 (I 0 + Ra 1)( I 0 + Ra 2) = I 0 ,so that aI 0 . Since J i strictly contains I 0 , it does not belong to B . In other words,4Every ring has a maximal ideal. For noetherian rings, which are our main concern, this is obvious,

but in general this follows with transnite induction, the adoption of which is equivalent to the adoptionof the axiom of choice.


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J i is an intersection of nitely many prime ideals. But then so is J 1 J 2 = I 0 and we get a contradiction.We thus nd that

(0) = p

1 p

s for certain prime ideals p

i. We may

of course assume that no pi contains some pj with j = i (otherwise, omit pi ). Itnow remains to prove that every prime ideal p of R contains some pi . If that isnot the case, then for i = 1 , . . . , s there exists a a i pi p. But then a1a2 a s p1 ps = (0) p and since p is a prime ideal, some factor a i lies in p. Thisis clearly a contradiction.

E XERC ISE 15. Let J be an ideal of the ring R . Show that J is the intersectionof all the prime ideals that contain J . Prove that in case R is noetherian, the primeideals that are minimal for this property are nite in number and that their commonintersection is still J . What do we get for R = Z and J = Zn?

3. Finiteness properties and the Hilbert theorems

The noetherian property in commutative algebra is best discussed in the con-text of modules, even if ones interest is only in rings. We x a ring R and rst recallthe notion of an R-module.

The notion of an R -module is the natural generalization of a K -vector space (where K is some eld). Let us observe that if M is an (additively written) abelian group, then theset End( M ) of group homomorphisms M M is a ring for which subtraction is pointwisedened and multiplication is composition (so if f, g End( M ), then f g : m M f (m) g(m) and f g : m f (g(m)) ); clearly the zero element is the zero homomorphismand the unit element is the identity. It only fails to obey our convention in the sense that thisring is usually noncommutative. We only introduced it in order to be able state succinctly:

DEFINITION 3.1. An R -module is an abelian group M , equipped with a ring homomor-phism R End( M ).

So any r R denes a homomorphism M M ; we usually denote the image of m M simply by rm . If we write out the properties of an R -module structure in these terms, we get: r (m 1 m 2 ) = rm 1 rm 2 , (r 1 r 2 )m = r 1 m r 2 m , 1.m = m , r 1 (r 2 m) = ( r 1 r 2 )m .If R happens to be eld, then we see that an R -module is the same thing as an R-vectorspace.

The notion of an R -module is quite ubiquitous if you think about it. A simple exampleis any ideal I R . Any abelian group M is in a natural manner a Z-module. A R[x]-modulecan be understood as an real linear space V (an R-module) endowed with an endomorphism(the image of x in End( V )). A more involved example is the following: if X is a manifold,f is a C -function on X and a C differential p-form on X , then f is also a C

differential p-form on X . Thus the linear space of C -differential forms on X of a xeddegree p is naturally a module over the ring of C -functions on X .

Here are a few companion notions, followed by a brief discussion.

3.2. In what follows is M an R-module. A map f : M N from M to an R-module N is called a R-homomorphism if it is a group homomorphism with the property that f (rm ) =rf (m) for all r R and m M . If f is also bijective, then we call it an R -isomorphism ; inthat case its inverse is also a homomorphism of R -modules.

For instance, given a ring homomorphism f : R R , then R becomes an R -moduleby rr := f (r )r and this makes f a homomorphism of R -modules.

A subset N M is called an R -submodule of M if it is a subgroup and rn N for allr R and n N . Then the group quotient M/N is in a unique manner a R-module in sucha way that the quotient map M M/N is a R-homomorphism: we let r (m + N ) := rm + N


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3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 13

for r R and m M . Notice that a R -submodule of R (here we regard R as a R -module)is the same thing as an ideal of R .

Given a subset S M , then the set of elements m M that can be written as r 1 s1 + + r k sk with r i R and s i S is a R-submodule of M . We call it the R-submodule of M generated by S and we shall denote it by RS .

If there exists a nite set S M such that M = RS , then we say that M is nitely generated as an R -module.

DEFINITION 3.3. We say that an R -module M is noetherian if the collection of R-submodules of M satises the ascending chain condition: any ascending chainof R-submodules N 1 N 2 becomes stationary.

It is clear that then every quotient module of a noetherian module is also noe-therian. The noetherian property of R as a ring (as previously dened) coincides with this property of R as an R-module.

The following two propositions provide the passage from the noetherian prop-erty to nite generation:

PROPOSITION 3.4. An R -module M is noetherian if and only if every R-submoduleof M is nitely generated as an R-module.

PROOF. Suppose that M is a noetherian R-module and let N M be a R-submodule. The collection of nitely generated R -submodules of M contained inN is nonempty. Hence it has a maximal element N 0 . If N 0 = N , then N is nitely generated. If not, we run into a contradiction: just choose xN N 0 and considerN 0 + Rx . This is a R -submodule of N . It is nitely generated (for N 0 is), whichcontradicts the maximal character of N 0 .

Suppose now that every R-submodule of M is nitely generated. If N 1 N 2 is an ascending chain of R-modules, then the union N := i =1 N i is aR-submodule. Let {s1 , . . . , s k} be a nite set of generators of N . If s N i , and j := max{i1 , . . . , i k

}, then it is clear that N j = N . So the chain becomes stationary

as of index j .

PROPOSITION 3.5. Suppose that R is a noetherian ring. Then every nitely gen-erated R-module M is noetherian.

PROOF. By assumption M = RS for a nite set S M . We prove the propo-sition by induction on the number of elements of S . If S = , then M = {0} andthere is nothing to prove. Suppose now S = and choose sS , so that our induc-tion hypothesis applies to M := RS with S = S {s}: M is noetherian. But sois M/M , for it is a quotient of the noetherian ring R via the surjective R -modulehomomorphism R M/M , r rs + M .Let now N 1 N 2 be an ascending chain of R -submodules of M . ThenN 1 M N 2 M becomes stationary, say as of index j 1 . Hence we only need to be concerned with the stabilization of the submodules N

k/ (N

k M ) of

M/M . These stabilize indeed (say as of index j 2), since M/M is noetherian. Sothe original chain N 1 N 2 stabilizes as of index max{ j1 , j 2}.

We are now sufciently prepared for the proofs of the Hilbert theorems. They are jewels of elegance and efciency.

We will use the notion of initial coefcient of a polynomial, which we re-call. Given a ring R, then every nonzero f R[x] is uniquely written as rd xd +r d1xd1 + + r 0 with rd = 0 . We call rd R the initial coefcient of f and


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denote it by in(f ). For the zero polynomial, we simply dene this to be 0 R.Notice that in(f )in( g) equals in(fg ) when nonzero.PROOF OF THEOREM 2 .8 . The assumption is here that R is a noetherian ring.

In view of Proposition 3.4 we must show that every ideal I of R[x] is nitely gener-ated. Consider the subset in(I ) := {in( f ) : f I } of R . We rst show that this isan ideal of R . If r R , f I , then r in(f ) equals in( rf ) when nonzero and sincerf I , it follows that r in( f ) I . If f , g I with d := deg( f ) deg(g) 0, thenin( f ) in(g) = in( f td g). So in(I ) is an ideal as asserted.Since R is noetherian, in(I ) is nitely generated: there exist f 1 , . . . , f k I such that in(I ) = R in(f 1) + + R in(f k ). Let di be the degree of f i , d0 :=max{d1 , . . . , dk} and R[x]


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4. THE AFFINE CATEGORY 15

This has a consequence for eld extensions:

COROLLARY 3.7. A eld extension L/K is nite if and only if L is nitely gener-

ated as a K -algebra.PROOF. It is clear that if L is a nite dimensional K -vector space, then L is

nitely generated as a K -algebra.Suppose now b1 , . . . , bm L generate L as a K -algebra. We must show thatevery bi is algebraic over K . Suppose that this is not the case. After renumbering

we can and will assume that (for some 1 r m) b1 , . . . , br are algebraically inde-pendent over K and br +1 , . . . , bm are algebraic over the quotient eld K (b1 , . . . , br )of K [b1 , . . . , br ]. So L is a nite extension of K (b1 , . . . , br ). We apply Proposition3.6 to R := K , A := K (b1 , . . . , br ) and B := L and nd that K (b1 , . . . , br ) is asa K -algebra generated by a nite subset S K (b1 , . . . , br ). If g is a common de-nominator for the elements of S , then clearly K (b1 , . . . , br ) = K [b1 , . . . , br ][1/g ].Since K (b1 , . . . , br ) strictly contains K [b1 , . . . , br ], g must have positive degree. In

particular, 1 + g = 0 , so that 1/ (1 + g)K (b1 , . . . , br ) can be written as f /gN

, withf K [b1 , . . . , br ]. Here we may of course assume that f is not divisible by g inK [b1 , . . . , br ]. From the identity f (1+ g) = gN we see that N 1 (for g has positivedegree). But then f = g(f + gN 1) shows that f is divisible by g. We thus get acontradiction. E XERC ISE 16. Prove that a eld which is nite generated as a ring (that is,

isomorphic to a quotient of Z[x1 , . . . , x n ] for some n) is nite.

We deduce from the preceding corollary the Nullstellensatz.

PROOF OF THE NULLSTELLENSATZ 1 .5 . Let J k[x1 , . . . , x n ] be an ideal. Wemust show that I (Z (J )) J . This amounts to: for every f k[x1 , . . . , x n ] J there exists a pZ (J ) for which f ( p) = 0 . Consider k[x1 , . . . , x n ]/J and denote by f

k[x1 , . . . , x n ]/J the image of f . Since f is not nilpotent, we have dened

A := ( k[x1 , . . . , x n ]/J )[1/ f ].Notice that A is as a k-algebra generated by the images of x1 , . . . , x n and 1/ f (hence is nitely generated). Choose a maximal ideal m A. Then the eldA/ m is a nitely generated k-algebra and so by Corollary 3.7 a nite extensionof k. But then it must be equal to k, since k is algebraically closed. Denote by : k[x1 , . . . , x n ] A A/ m = k the corresponding surjection and put pi := (x i )and p := ( p1 , . . . , pn )An . So if we view xi as a function on An , then xi ( p) = pi =(x i ). The fact that is a homomorphism of k-algebras implies that it is given asevaluation in p: any g k[x1 , . . . , x n ] takes in p the value (g). Since the kernelof contains J , every g J will be zero in p, in other words, p Z (J ). On theother hand, f ( p) = (f ) is invertible, for it has the image of 1/ f in A/ m = k as itsinverse. So f ( p) = 0 .

4. The afne category

We begin with specifying the maps between closed subsets of afne spaces that we wish to consider.

DEFINITION 4.1. Let X Am and Y An be closed subsets. We say that amap f : X Y is regular if the components f 1 , . . . , f n of f are regular functionson X (i.e., are given by the restrictions of polynomial functions to X ).


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Composition of a regular function on Y with f yields a regular function on X (for if we substitute in a polynomial of n variables g(y1 , . . . , yn ) for every variabley

i a polynomial f

i(x

1, . . . , x

m) of m variables, we get a polynomial of m variables).

So f then induces a k-algebra homomorphism f : A(Y ) A(X ). There is also aconverse:PROPOSITION 4.2. Let be given closed subsets X Am and Y An and a k-algebra homomorphism : A(Y ) A(X ). Then there is a unique regular mapf : X Y such that f = .PROOF. Let f i := (yi |Y ) A(X ) ( i = 1 , . . . , n ) and dene f = ( f 1 , . . . , f n ) :X An , so that f yi = f i = (yi |Y ). If j : Y An denotes the inclusion, then wecan also write this as f yi = jyi . In other words, the k-algebra homomorphisms

f , j : k[y1 , . . . , yn ] A(X ) coincide on the generators yi . Hence they must beequal: f = j. It follows that f is zero on I (Y ), which means that f takes its values in Z (I (Y )) = Y , and that the resulting map A(Y ) A(X ) equals .

The same argument shows that if f : X Y and g : Y Z are regularmaps, then so is their composite gf : X Z . So we have a category (with objectsthe closed subsets and regular maps as dened above). In particular, we have anotion of isomorphism: a regular map f : X Y is an isomorphism if is hasa two-sided inverse g : Y X which is also a regular map. This implies thatf : A(Y ) A(X ) has a two-sided inverse g : A(X ) A(Y ) which is also anisomorphism of k-algebras, and hence is an isomorphism of k-algebras. Proposition4.2 implies that conversely, an isomorphism of k-algebras A(Y ) A(X ) comesfrom a unique isomorphism X Y .We complete the picture by showing that any nitely generated reduced k-algebra A is isomorphic to some A(Y ); the preceding then shows that Y is uniqueup to isomorphism. Since A is nitely generated as a k-algebra, there exists a surjective k-algebra homomorphism : k[x1 , . . . , x n ]

A. If we put I := Ker( ),

then induces an isomorphism k[x1 , . . . , x n ]/I = A. Put Y := Z (I ) An . SinceA is reduced, I is a radical ideal and hence equal to I (Y ) by the Nullstellensatz. Itfollows that factors through a k-algebra isomorphism A(Y )= A.

We may sum up this discussion in categorical language as follows.

PROPOSITION 4.3. The map which assigns to a closed subset of some An its coor-dinate ring denes an anti-equivalence between the category of closed subsets of afne spaces (and regular maps between them as dened above) and the category of reduced nitely generated k-algebras (and k-algebra homomorphisms).

E XAM PLE 4.4. Consider the regular map f : A1 A2 , f (t) = ( t2 , t 3). Theimage of this map is the hypersurface Z dened by x3 y2 = 0 . This is a home-omorphism on the image for it is a bijection of sets that have both the conitetopology. The inverse sends (0, 0) to 0 and is on Z

{(0, 0)

}given by (x, y )

y/x .

In order to determine whether the inverse is regular, we consider f . We haveA(Z ) = k[x, y ]/ (x3 y2), A(A1) = k[t] and f : k[x, y ]/ (x3 y2) k[t] is given by x t2 , y t3 . This homomorphism is not surjective for its image misses t k[t].So f is not an isomorphism.

E XAM PLE 4.5. An afne-linear transformation of kn is of the form x kn g(x) + a, where a kn and g GL(n, k ) is a linear transformation. Its inverse isy g1(ya) = g1(y)g1(a) and so of the same type. When we regard such an


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afne linear transformation as a map from An to itself, then it is regular: its coordi-nates (g1 , . . . , gn ) are polynomials of degree one. So an afne-linear transformationis also an isomorphism of An onto itself. But if n

2, not every isomorphism of

An onto itself need to be of this form. For example, (x, y ) (x, y + x2) denes anautomorphism of A2 with inverse (x, y) (x, y x2) (see also Exercise 18 ).E XERC ISE 17. Let C A2 be the circle, dened by x2 + y2 = 1 and let p0 :=(1, 0) C . For every p = ( x, y ) C { p0}, the line through p0 and p has slopef ( p) = y/ (x + 1) . Denote by 1k a root of the equation t2 + 1 = 0 .

(a) Prove that when char( k) = 2 , f denes an isomorphism 5 onto A1 { 1}.(b) Consider the map g : C A1 , g(x, y ) := x + 1y. Prove that whenchar( k) = 2 , g denes an isomorphism of C onto A1 {0}.(c) Prove that when char( k) = 2 , the dening polynomial x2 + y2 1 for C is the square of a degree one polynomial so that C is a line.

E XERC ISE 18. Let f 1 , . . . , f n k[x1 , . . . , x n ] be such that f 1 = x1 and f i x i k[x1 , . . . , x i1] for i = 2 , . . . , x n . Prove that f denes an isomorphism An An .4.6. Q UADRATIC HYPERSURFACES IN CASE char( k) = 2 . Let H An be a hyper-surface dened by a polynomial of degree two:

f (x1 , . . . , x n ) =1ij = n

a ij x i x j +n

i=1

a i x i + a0 .

By means of a linear transformation (this involves splitting off squares, hence re-quires the existence of 1/ 2 k), the quadratic form 1ij = n a ij x i x j can bebrought in diagonal form. This means that we can make all the coefcients aij with i = j vanish. Another diagonal transformation (which replaces x i by a ii x i when aii = 0 ) takes every nonzero coefcient aii to 1 and then renumberingthe coordinates (which is also a linear transformation) brings f into the formf (x1 , . . . , x n ) = ri=1 x2i + ni=1 a i x i + a0 for some r 1. Splitting off squaresonce more enables us to get rid of ri =1 a i x i so that we get

f (x1 , . . . , x n ) =r

i=1

x2i +n

i = r +1

a i x i + a0 .

We now have the following cases.If the nonsquare part is identically zero, then we end up with the equation

ri =1 x

2i = 0 for H .

If the linear part ni = r +1 a i x i is nonzero (so that we must have r < n ), then anafne-linear transformation which does not affect x1 , . . . , x r and takes

ni = r +1 a i x i +

a0 to xn yields the equation xn =

ri =1 x

2i . This is the graph of the function

r

i =1 x2i on An

1 and so

H is then isomorphic to An

1 .

If the linear part ni= r +1 a i x i is zero, but the constant term a0 is nonzero, then we can make another diagonal transformation which replaces x i by a0x i ) anddivide f by a0 : then H gets the equation ri =1 x2i = 1 .

5We have not really dened what is an isomorphism between two nonclosed subsets of an afnespace. Interpret this here as: f maps k[x, y ][1/ (x +1)] / (x2 + y2 1) (the algebra of regular functionson C { p0}) isomorphically onto k[t ][1/ (t2 +1)] (the algebra of regular functions on A1 { 1}).This will be justied by Proposition 4.10 .


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In particular, there are only a nite number of quadratic hypersurfaces up toisomorphism. This is also true in characteristic two, but the classication is moredelicate. (For instance, the hypersurfaces dened by x

1x

2 = 1 and x2

1 + x2

2 = 1 are

in this case not isomorphic.)

4.7. T HE MAXIMAL IDEAL SPECTRUM. The previous discussion (and in particularProposition 4.3 ) leads us to associate to an arbitrary ring R a space with a topology so that for R = A(Y ) we get a space homeomorphic to Y . Since the points of Y correspond to maximal ideals of A(Y ), we choose the underlying set of this space(which we shall denote by Specm( R)) to be the collection of maximal ideals of R .In order to avoid confusion about whether a maximal ideal is to be viewed as asubset of R or as a point of Specm(R) we agree that if m is a maximal ideal of R ,then the corresponding element of Specm(R) is denoted xm and if x Specm( R),then the corresponding maximal ideal of R is denoted mx . We write (x) for theresidue eld R/ mx and x : R (x) for the reduction modulo mx . We now havearranged that when R = A(Y ), x

Specm( A(Y )) is the same thing as a point of

Y . Since (x) is a nitely generated as a k-algebra, it is by Corollary 3.7 a niteextension of k and hence equal to k (k is algebraically closed).

Every r R denes a regular function f r on Specm( R) which takes in x Specm( R) the value x (r ) (x). So in general f r takes its values in a eld thatmay depend on x, but for R of the form A(Y ) we know this eld to be constantequal to k, so that f r : Specm(R) k. We denote by Z (r ) Specm(A) (or Z (f r ))the zero set of this function. So Z (r ) consists of the x Specm( R) with x (r ) = 0 ,or equivalently, r mx . We write U (r ) or (U (f r )) for its complement, the nonzeroset. We have U (rr ) = U (r ) U (r ) and so the collection of {U (r )}rR is the basisof a topology on Specm( R). Note that a subset Specm( R) is closed precisely if itis an intersection of subsets of the form Z (r ); this is equal to the common zero setof the set of functions dened by an ideal of R . The space Specm(R) is called the

maximal ideal spectrum6

of R (but our notation for it is less standard). Notice that if R = A(Y ), then the above discussion shows that Specm(R) can be identied withY as a topological space and that under this identication, A(Y ) becomes the ringof regular functions on Specm(R).

A ring homomorphism : S R will in general not give rise to a mapSpecm( ) : Specm( R) Specm(S ): if x Specm( R), then the composite homo-morphism S R (x) need not be onto and its kernel need not be a maximalideal (it will be a prime ideal, though). However, in case : S R is a homomor-phism of k-algebras, then k S maps onto (x) = k as the identity map so that1mx is a maximal ideal of S with residue eld k. We then do get a mapSpecm( ) : Specm( R) Specm( S ), xm x 1 m .

For sS , the preimage of U (s) is U ((s)) . This shows that Specm( ) is continuous.

We call the resulting pair (Specm( ), ) a morphism.E XERC ISE 19. Give an example of ring homomorphism : S R and a maxi-mal ideal m

R , such that 1m is not a maximal ideal of S . (Hint: take a look at

Exercise 13 .)

6I. Gelfand was presumably the rst to consider this in the context of functional analysis: hecharacterized the Banach algebras that appear as the algebras of continuous C-valued functions oncompact Hausdorff spaces and so it might be appropriate to call this the Gelfand spectrum.


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E XERC ISE 20. Prove that if R is a nitely generated k-algebra, then the map r R f r is a k-algebra homomorphism from R to the algebra of k-valued functionson

Specm(R) with kernel

(0). Show that for every subset

X Specm( R), the set

I (X ) of r R with f r |X = 0 is a radical ideal of R .E XERC ISE 21. Prove that Specm( R) is irreducible when R is a nitely generated

algebra over a eld and nonzero 7.

DEFINITION 4.8 (Preliminary denition). An afne k-variety 8 is a topologicalspace Y endowed with a nitely generated k-algebra A(Y ) of regular functions onY which identies Y with the maximal ideal spectrum of A(Y ).

In more elementary terms, there should exist a homeomorphism h of Y onto aclosed subset of some An such that A(Y ) = hk[x1 , . . . , x n ]. We shall often denotean afne variety simply by the underlying topological space, e.g., by Y , where it isthen understood that A(Y ) is part of the data. Likewise a morphism will simply bedenoted by f : X

Y ; it is then understood that composition with f takes A(Y )

to A(X ) and denes f : A(Y ) A(X ).With this terminology Proposition 4.3 takes now a rather trivial form:4.9. The functor which assigns to an afne k-variety its k-algebra of regular

functions identies the category of afne k-varieties with the dual of the category of reduced nitely generated k-algebras, the inverse being given by the functor Specm.

From our previous discussion it is clear that if Y is an afne variety, then every closed subset Y Y determines an afne variety Y with A(Y ) := A(Y )/I (Y ).Here I (Y ) is of the course the ideal of regular functions on Y that vanish on Y .

Our next aim is to give any basic open subset U of Y the structure of an afne variety.

PROPOSITION 4.10. Let Y be an afne variety. Then every basic open subsetU Y is in a natural manner an afne variety so that an inclusion U U of any two such is a morphism and the algebra of regular functions on U = U (g) is A(Y )[1/g ](assuming U = , so that g is not nilpotent).

PROOF. We observe that A(Y )[1/g ] is dened as a k-algebra and as such it isnitely generated (just add to a generating set for A(Y ) the generator 1/g ) andis without zero divisors. So we have an afne variety Specm(A(Y )[1/g ]). Any maximal ideal of A(Y )[1/g ] intersects A(Y ) in a maximal ideal of A(Y ) which doesnot contain g. Conversely, if m is a maximal ideal of A(Y ), then mA(Y )[1/g ] isa maximal ideal of A(Y )[1/g ] unless g m. So the injection A(Y ) A(Y )[1/g ]induces an injection of Specm(A(Y )[1/g ]) in Y with image U (g) = U . This allowsus to regard U as an afne variety whose coordinate ring is A(Y )[1/g ].

Assume that we have an inclusion U (g )

U (g). Then Z (g )

Z (g) and so

by the Nullstellensatz, g (g). This implies that (g )n (g)

for some positiveinteger n. So we have a natural homomorphism A(Y )[1/g ] A(Y )[1/ (g )n ] =A(Y )[1/g ]. This also shows that A(Y )[1/g ] only depends on U , for if U (g ) = U (g), we also a homomorphism in the opposite direction which serves as inverse so thatA(Y )[1/g ] = A(Y )[1/g ].

7This corrects an earlier formulation, with thanks to my students.8Since we xed k, we will often drop k and speak of an afne variety.


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Let f : X Y be a morphism of afne varieties. Since f is continuous, a berf 1(y), or more generally, the preimage f 1Y of a closed subset Y Y , will beclosed in X . It is the zero set of the ideal in A(X ) generated by f

I (Y ). But thisideal need not be a radical ideal. Here is a simple example:

E XAM PLE 4.11. Let f : X = A1 A1 = Y be dened by f (x) = x2 . Thenf : k[y] k[x] is given by f y = x2 . If we assume k not to be of characteristic 2,and we take a Y {0}, then the ber f 1(a) consists of two distinct points andis dene by the ideal generated by f (y a) = x2 a . This a radical ideal and theber has coordinate ring k[x]/ (x2 a). However, the ber over 0 Y = A1 is thesingleton {0}X = A1 and the ideal generated by f y = x2 is not a radical ideal.This example indicates that there might good reason to accept nilpotent elementsin the coordinate ring of f 1(0) by endowing f 1(0) with the ring of functionsA(f 1(0)) := k[x]/ (x2). Since this is a k-vector space of dimension 2, we thusretain the information that two points have come together and the ber should bethought of as a point with multiplicity 2.

Example 4.4 shows that a homeomorphism of afne varieties need not be anisomorphism. Here is another class of examples.

E XAM PLE 4.12 (T HE FROBENIUS MORPHISM ). Assume that k has positive char-acteristic p and consider the morphism F p : A1 A1 , x x p. If we rememberthat A1 can be identied with k, then we observe that under this identication, F pis a eld automorphism: F p(a b) = ( a b) p = a p b p = F p(a) F p(b) and of course F p(ab) = ( ab) p = F p(a)F p(b) and F p is surjective (every element of k has a pth root since k is algebraically closed) and injective. But the endomorphism F p of k[x] induced by F p sends x to x p and has therefore image k[x p]. Clearly, F p is notsurjective.

The xed point set of F p (so the set of a

A1 with a p = a) is via the iden-

tication of A1 with k just the prime subeld F p k and we therefore denote itby A1(F p) A1 . Likewise, the xed point set A1(F pr ) of F r p is the subeld of k with pr elements. Since the algebraic closure F p of F p in k is the union of the nitesubelds of k, the afne line over F p equals r 1A1(F pr ). This generalizes in astraightforward manner to higher dimensions: by letting F p act coordinatewise onAn , we get a morphism An An (which we still denote by F p) which is also abijection. The xed point of F r p is An (F pr ) and An (F p) = r 1An (F pr ).

E XERC ISE 22. Assume that k has positive characteristic p. Let q = pr be a powerof p with r > 0 and denote by Fq k the subeld of a k satisfying aq = a. We write F for F r p : An An . Suppose Y An is the common zero set of polynomials with coefcients in Fq k.

(a) Prove that f k[x1 , . . . , x n ] has its coefcients in Fq if and only if F f =f q .(b) Prove that an afne-linear transformation of An with coefcients in Fq

commutes with F .(c) Prove that F restricts to a bijection F Y : Y Y and that the xed pointset of F mY is Y (Fqm ) := Y An (Fqm ).(d) Suppose that k is an algebraic closure of F p. Prove that every closed

subset of An is dened over a nite subeld of k.


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5. THE PRODUCT 21

R EMARK 4.13. After this exercise we cannot resist to mention the Weil zetafunction. This function and its relativesamong them the Riemann zeta functioncodify arithmetic properties of algebro-geometric objects in a very intricate manner.In the situation of Exercise 22 , we can use the numbers |Y (Fqm )| (= the number of xed points of F m in Y ) to dene a generating series m 1 |Y (Fqm )|tm . It appearsto be more convenient to work with the Weil zeta function :

Z Y (t) := exp

m =1|Y (Fqm )|

tm

m,

which has the property that t ddt log Z Y yields the generating series above. Thisseries has remarkable properties. For instance, a deep theorem due to BernardDwork (1960) asserts that it represents a rational function of t. Another deeptheorem, due to Pierre Deligne (1974), states that the roots of the numerator anddenominator have for absolute value a nonpositive half-integral power of q and thatmoreover, these powers have an interpretation in terms of an algebraic topology for algebraic geometry. All of this was predicted by Andr e Weil in 1949. (This canbe put in a broader context by making the change of variable t = q s . Indeed,now numerator and denominator have their zeroes when the real part of s is anonnegative half-integer and this makes Delignes result reminiscent of the famousconjectured property of the Riemann zeta function.)

E XERC ISE 23. Compute the Weil zeta function of afne n -space relative to theeld of q elements.

5. The product

Let m and n be nonnegative integers. If f k[x1 , . . . , x m ] and g k[y1 , . . . , yn ],then we have a regular function f g on Am + n dened by f g(x1 , . . . , x m , y1 , . . . , yn ) := f (x1 , . . . , x m )g(y1 , . . . , yn ).It is clear that U (f g) = U (f ) U (g), which shows that the Zariski topology onAm + n renes the product topology on Am An . Equivalently, if X Am andY An are closed, then X Y is closed in Am + n . We give X Y the topology it inherits from Am + n (which is usually ner than the product topology). For the

coordinate rings we have dened a map:

A(X ) A(Y ) A(X Y ), (f, g ) f g which is evidently k-bilinear (i.e., k-linear in either variable). We want to provethat the ideal I (X Y ) dening X Y in Am + n is generated by I (X ) and I (Y )(viewed as subsets of k[x1 , . . . , x m , y1 , . . . yn ]) and that X Y is irreducible whenX and Y are. This requires that we translate the formation of the product into al-gebra. This centers around the notion of the tensor product, the denition of which we recall. (Although we here only need tensor products over k, we shall dene thisnotion for modules over a ring, as this is its natural habitat. This is the setting thatis needed later anyhow.)

If R is a ring and M and N are R -modules, then we can form their tensor product overR , M R N : as an abelian group M R N is generated by the expressions a R b, a M ,bN and subject to the conditions (ra )R b = a R (rb), (a + a )R b = a R b+ a R band a R (b + b ) = a R b + a R b . So a general element of M R N can be written


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like this: N i =1 a i R bi , with a i M and bi N . We make M R N an R -module if westipulate that r (a R b) := ( ra )R b (which is then also equal to r R (rb)). Notice that themap

R : M N M R N, (a, b) a R b,is R -bilinear (if we x one of the variables, then it becomes an R -linear map in the other variable).

In case R = k we shall often omit the sufx k in k .

E XERC IS E 24. Prove that R is universal for this property in the sense that every R-bilinear map M N P of R-modules is the composite of R and a unique R -homomorphismM R N P . In other words, the map

Hom R (M R N, P ) BilR (M N, P )) , f f Ris an isomorphism of R -modules.

E XERC IS E 25. Let m and n be nonnegative integers. Prove that Z/ (n ) Z Z/ (m) can beidentied with Z/ (m, n ).

If A is an R -algebra and N is an R -module, then A R N acquires the structure of anA-module which is characterized by

a. (a R b) := ( aa ) R b.

For instance, if N is an R-vector space, then CR N is a complex vector space, the complexi- cation of N . If A and B are R-algebras, then AR B acquires the structure of an R-algebracharacterized by

(a R b).(a R b ) := ( aa ) R (bb ).Notice that A A R B , a a R 1 and B A R B , b 1R b are R -algebra homo-morphisms. For example, AR R[x] = A[x] as A-algebras (and hence AR R[x1 , . . . , x n ] =A[x1 , . . . , x n ] with induction).

E XERC IS E 26. Prove that CR C is as a C-algebra isomorphic to CC with componen-twise multiplication.

PROPOSITION 5.1. For closed subsets X Am and Y An the bilinear mapA(X )A(Y ) A(X Y ) , (f, g ) f g induces an isomorphism : A(X )A(Y ) A(X Y ) of k-algebras (so that in particular A(X )A(Y ) is reduced). If X and Y are irreducible, then so is X Y , or equivalently, if A(X ) and A(Y )are domains, then so is A(X )A(Y ).PROOF. Since the obvious map

k[x1 , . . . , x m ]k[y1 , . . . , yn ] k[x1 , . . . , x m , y1 , . . . , yn ]is an isomorphism, it follows that is onto. In order to prove that is injective, letlet rst observe that every u A(X )A(Y ) can be written u =

N i =1 f igi withg1 , . . . , gN are k-linearly independent. Given p X , then the restriction of (u) =N

i =1 f i

gi to

{ p

} Y = Y is the regular function u p :=

N i =1 f i ( p)gi

A(Y ).

Since the gi s are linearly independent, we have u p = 0 if and only if f i ( p) = 0 forall i. In particular, the subset X (u) X of p X for which u p = 0 , is equal toN i =1 Z (f i ) and hence closed.If (u) = 0 , then u p = 0 for all pX and hence f i = 0 for all i. So u = 0 . Thisproves that is injective.

Suppose now X and Y irreducible and suppose that u is zero divisor: uv = 0for some v A(X Y ) with u, v = 0 . Since the restriction of uv = 0 to { p}Y = Y is u pv p and A(Y ) is a domain, it follows that u p = 0 or v p = 0 . So X = X (u)X (v).


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6. FUNCTION FIELDS AND RATIONAL MAPS 23

Since X is irreducible we have X = X (u) or X (v). This means that u = 0 or v = 0 , which contradicts our assumption.

E XERC ISE 27. Let X and Y be closed subsets of afne spaces. Prove that eachirreducible component of X Y is the product of an irreducible component of X and one of Y .

It is clear that the projections X : X Y X and Y : X Y Y areregular. We have observed that X Y has not a topological product in general.Still it is the right product in the sense of category theory: it has the followinguniversal property, almost seems too obvious to mention: if Z is a closed subset of some afne space, then any pair of regular maps f : Z X , g : Z Y denes aregular map Z X Y characterized by the property that its composite with Xresp. Y yields f resp. g (this is of course (f, g )).

6. Function elds and rational maps

In this section we interpret the formation of the fraction ring of an algebra of regular functions.

Let Y be an afne variety. Recall that for f, g A(Y ), a fraction f /g Frac( A(Y )) is dened if g not a zerodivisor. We need the following lemma.LEMMA 6.1. Let Y be an afne variety and let g A(Y ). Then g is not a zerodivisor if and only if g is nonzero on every irreducible component of Y . This is also

equivalent to U (g) being dense in Y .

PROOF. Suppose g is zero on some irreducible component C of Y and supposeC = Y . If Y denotes the union of the irreducible components of Y distinct form C ,then either Y = (and so g = 0 in A(Y )) or Y = and so there exists a nonzerog I Y (Y ). Then gg vanishes on C Y = Y and so gg = 0 . It follows that g isthen a zero divisor.

Conversely, assume that g is a zerodivisor of A(Y ). So there exists a nonzerog A(Y ) with gg = 0 . Let C be an irreducible component of Y on which g isnonzero. Then we must have g|C = 0 .The proof of the last clause is left to you.

This lemma tells us that this means that the afne variety U (g) is open-densein Y . Clearly, f /g denes a regular function U (g) k. Another such fraction f /g yields a regular function on an open-dense U (g ) and their product f/g.f /g =ff /gg resp. difference f /g f /g = ( fg f g)/gg in Frac( A(Y )) denes aregular function on U (gg ) which is there just the product resp. difference of theassociated regular functions. In particular, if f /g = f /g in A(Y ), then the twoassociated regular functions coincide on U (gg ). There is a priori no best way to represent a given element of Frac( A(Y )) as a fraction (as there is in a UFD),

and so we must be content with the observation that an element of Frac( A(Y ))denes a regular function on some (unspecied) open dense subset of Y , with theunderstanding that two such functions are regarded as equal if they coincide on anopen-dense subset contained in a common domain of denition. When an elementof Frac( A(Y )) is thus understood, then we call it a rational function on Y . Thisis the algebro-geometric analogue of a meromorphic function in complex functiontheory. When Y is irreducible, then Frac( A(Y )) is a eld, called the function eldof Y , and will be denoted k(Y ).


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We shall now give a geometric interpretation of nitely generated eld exten-sions of k and the k-linear eld homomorphisms between them. We begin with anotion that generalizes that of a rational function.

DEFINITION 6.2. Let X and Y be afne varieties. A rational map from X to Y isgiven by a pair (U, F ), where U is an afne open-dense subset of X and F : U Y is a morphism, with the understanding that a pair (V, G) denes the same rationalmap if F and G coincide on U V . We denote a rational map like this f : X Y .We say that the rational map is dominant if for a representative pair (U, F ),F (U ) is dense in Y . (This is then also so for any other representative pair. Why?)

PROPOSITION 6.3. Any nitely generated eld extension of k is k-isomorphic tothe function eld of an irreducible afne variety. If X and Y are irreducible and afne,then a dominant rational map f : X Y determines a k-linear eld embeddingf : k(Y ) k(X ) and every k-linear eld embedding k(Y ) k(X ) is of this form(for a unique f ).

PROOF. Let K/k be a nitely generated eld extension of k: there exist ele-ments a1 , . . . , a n K such that every element of K can be written as a fractionof polynomials in a1 , . . . , a n . So if R denotes the k-subalgebra R of K generatedby a1 , . . . , a n , (a domain since K is a eld), then K is the eld of fractions of R .Since R is the coordinate ring of a closed irreducible subset X An (dened by the kernel of the obvious ring homomorphism k[x1 , . . . , x n ] R), it follows thatK can be identied with k(X ).

Suppose we are given an open-dense afne subvariety U X and a morphismF : U Y with F (U ) dense in Y . Now F : A(Y ) A(U ) will be injective, forif g A(Y ) is in the kernel: F (g) = 0 , then F (U ) Z (g ). Since F (U ) is densein Y , this implies that Z (g) = Y , in other words, that g = 0 . Hence the compositemap A(Y ) A(U ) k(U ) = k(X ) is an injective homomorphism from a domainto a eld and therefore extends to a eld embedding k(Y ) k(X ).It remains to show that every k-linear eld homomorphism : k(Y ) k(X ) isso obtained. For this, choose generators b1 , . . . , bm of A(Y ). Then (b1), . . . , (bm )are rational functions on X and so are regular on an open-dense afne subsetU X . Since b1 , . . . , bm generate A(Y ) as a k-algebra, it follows that maps A(Y )to A(U ) k(X ). This is a k-algebra homomorphism and so denes a morphismF : U Y such that F = |A(Y ). The image of F will be dense by the argumentabove: if the image of F is contained in a closed subset of Y distinct from Y , thenits maps to some Z (g) with g A(Y ){0}. But this implies that (g) = F (g) = 0 , which cannot happen, since is injective. It is clear that is the extension of F to the function elds.

As to the uniqueness: if (V, G) is another solution, then choose a nonempty afne open-dense subset W

U

V so that F and G both dene morphisms

W Y . The maps A(Y ) A(W ) they dene are equal, for are given by . HenceF and G coincide on W . Since W is dense in U V , they also coincide on U V . E XERC ISE 28. Let f k[x1 , . . . , x n +1 ] be irreducible of positive degree. Its zeroset X An +1 is then closed and irreducible. Denote by d the degree of f in xn +1 .

(a) Prove that the projection : X An induces an injective k-algebrahomomorphism : k[x1 , . . . , x n ] A(X ) = k[x1 , . . . , x n ]/ (f ) if andonly if d > 0.


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(b) Prove that for d > 0, is dominant and that the resulting eld homomor-phism k(x1 , . . . , x n ) k(X ) is a nite extension of degree d.

COROLLARY 6.4. There is a category with objects the irreducible afne varietiesand morphisms the rational dominant maps. Assigning to an irreducible afne va-riety its function eld makes this category anti-equivalent to the category of nitely generated eld extensions of the base eld k.

PROPOSITION -DEFINITION 6.5. A rational map f : X Y is an isomorphismin the above category (that is, induces a k-linear isomorphism of function elds) if andonly if there exists a representative pair (U, F ) of f such that F maps U isomorphically onto an open subset of Y . If these two equivalent conditions are satised, then f iscalled a birational map . If a birational map X Y merely exists (in other words,if there exists a k-linear eld isomorphism between k(X ) and k(Y ) ), then we say thatX and Y are birationally equivalent .

PROOF. If f identies a nonempty open subset of X with one of Y , then f :k(Y ) k(X ) is clearly a k-algebra isomorphism.Suppose now we have a k-linear isomorphism k(Y ) = k(X ). Represent thisisomorphism and its inverse by (U, F ) and (V, G) respectively. Then F 1V is afneand open-dense in U and GF : F 1V X is dened. Since GF induces theidentity on k(X ), GF is the identity on a nonempty open subset of F 1V andhence on all of F 1V . This implies that F maps F 1V injectively to G1U . Forthe same reason, G maps G1U injectively to F 1V . So F denes an isomorphismbetween the open subsets F 1V X and G1U Y .

E XERC ISE 29. We here assume k not of characteristic 2. Let for X A2 bedened by x21 + x22 = 1 . Prove that X is birationally equivalent to the afne line A1 .(Hint: take a look at Exercise 17 .) More generally, prove that the quadric in An +1dened by x21 + x22 +

x2n +1 = 1 is birationally equivalent to An . What about the

quadric cone in An +1 dened by x21 + x22 + x2n +1 = 0 (n 2)?In case k(X )/k (Y ) is a nite extension, one may wonder what the geometric

meaning is of the degree d of that extension, perhaps hoping that this is just thenumber of elements of a general ber of the associated rational map X Y . We will see that this often true (namely when the characteristic of k is zero, or moregenerally, when this characteristic does not divide d), but not always, witness thefollowing example.

E XAM PLE 6.6. Suppose k has characteristic p > 0. We take X = A1 = Y and letf be the Frobenius map: f (a) = a p. Then f is homeomorphism, but f : A(Y ) A(X ) is given by y x p and so induces the eld extension k(y) = k(x p) k(x), which is of degree p. From the perspective of Y , we have enlarged its algebra of regular functions by introducing a formal pth root of its coordinate (which yieldsanother copy of A1 , namely X ). From the perspective of X , A(Y ) is just f A(X ).

This is in fact the basic example of a purely inseparable eld extension, i.e., aeld extension L/K with the property that every element of L has a minimal poly-nomial in K [T ] that has at most one root in L. Such a polynomial must be of theform T p

r

c, with c K not a p-th power of an element of K when r > 0, where p is the characteristic of K (for p = 0 we necessarily have L = K ). So if L = K ,then the Frobenius map F p : a K a p K is not surjective. Purely inseparable


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extensions are not detected by Galois theory, for they have trivial Galois group: af-ter all, there is only one root to move around.

For any algebraic extension L/K , the elements of L that are separable over K makeup an intermediate extension L sep /K that is (of course) separable and is such that L/L sep

is purely inseparable. When L is an algebraic closure of K , then L sep is called a separablealgebraic closure of K : it is a separable algebraic extension of K which is maximal for thatproperty. Then L is as an extension of Lsep obtained by successive adjunction of p-powerroots of elements of Lsep : it is an extension minimal for the property of making the Frobeniusmap a L ap L surjective. In case L/K is a nite normal extension (i.e., a niteextension with the property that every f K [x] that is the minimal polynomial of someelement of L has all roots in L), then the xed point set of the Galois group of L/K is asubextension Linsep /L that is purely inseparable, whereas L/L insep separable. In that casethe natural map Lsep K L insep L is an isomorphism of K -algebras.

E XERC ISE 30. Let f : X

Y be a dominant morphism of irreducible afne

varieties which induces a purely inseparable eld extension k(Y )/k (X ). Prove thatthere is an open-dense subset U Y such that f restricts to a homeomorphismf 1U U . (Hint: show rst that it sufces to treat the case when k(X ) is obtainedfrom k(Y ) by adjoining the pth root of an element f k(Y ). Then observe thatif f is regular on the afne open-dense U Y , then Y contains as an open densesubset a copy of the locus of (x, t )U A1 satisfying t p = f .)

If K/k is a nitely generated eld extension and generated by b1 , . . . , bm , say,then we may after renumbering assume that for some r m, b1 , . . . , br are alge-braically independent (so that the k-linear eld homomorphism k(x1 , . . . , x r ) K which sends x i to bi is injective) and br +1 , . . . , bm are algebraic over k(b1 , . . . , br ).This number r is invariant of K/k , called its transcendence degree . So if K = k(X ),then this denes a dominant rational map X Ar . We now recall the theorem of the primitive element:

THEOREM 6.7. Every separable eld extension L/K of nite degree is generatedby a single element (which we then call a primitive element for L/K ).

It follows that if br +1 , . . . , bm are separable over k(b1 , . . . , br ), then we cannd a b K that is algebraic over k(b1 , . . . , br ) such that K is obtained fromk(b1 , . . . , br ) by adjoining b. This b is a root of an irreducible polynomial F k(b1 , . . . , br )[T ]. If we clear denominators, we may assume that the coefcients of F lie in k[b1 , . . . , br ][T ] = k[b1 , . . . , br , T ]. Then we can take for X the hypersurfacein Ar +1 dened by F , when viewed as an element of k[x1 , . . . , x r , x r +1 ].

In particular, every irreducible afne variety Y is birationally equivalent to ahypersurface in Ar +1 , where r is the transcendence degree of k(Y ). This suggeststhat this transcendence degree might be a good way to dene the dimension of Y .We shall return to this.

Much of the algebraic geometry in the 19th century and early 20th century was of a birational nature: birationally equivalent varieties were regarded as notreally different. This sounds rather drastic, but it turns out that many interestingproperties of varieties are an invariant of their birational equivalence class.


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Here is an observation which not only illustrates how afne varieties over al-gebraically nonclosed elds can arise when dealing with afne k-varieties, but onethat is also a tell-tale sign that we ought to enlarge the maximal ideal spectrum.Let f : X Y be a dominant morphism of irreducible afne varieties. This impliesthat f : A(Y ) A(X ) is injective and that f (X ) contains an open-dense subset of Y . Then we may ask whether there exists something like a general ber: is therean open-dense subset V Y such that the bers f 1(y), y V all look the same?The question is too vague for a clear answer and for most naive ways of makingthis precise, the answer will be no. But there is a cheap way out by simply refusingto specify one such V and allowing it to be arbitrarily small: we would like to takea limit over all open-dense subsets of Y . This we cannot do (yet) topologically, but we can do this algebraically by making all the nonzero elements of A(Y ) in A(X )invertible: we take

(A(Y ) {0})1A(X ) = k(Y )A(Y ) A(X )(this equality follows from the fact that A(X ) = A(Y ) A (Y ) A(X )). This isa reduced nitely generated k(Y )-algebra and we may regard its maximal idealspectrum Specm( k(Y ) A(Y ) A(X )) as an afne variety over the (algebraically nonclosed) eld k(Y ): now every regular function on X which comes from Y istreated as a scalar (and will be invertible when nonzero). If we decide to add toY = Specm( k(Y )) a point (Y ) with residue eld k(Y ), then we can simply say that the generic ber of f is the ber X (Y ) over (Y )9. But once we decide todo this for Y , we should of course do this for every irreducible afne variety. Inparticular, we must then add for every irreducible subset Z of Y (or equivalently,for every prime ideal p A(Y )) a point to Specm(k(Y )) with residue eld k(Z )(= Frac( A(Y )/ p)). We do this below in the same kind of generality as the maximalideal spectrum, namely for an arbitrary ring.

6.8. T HE PRIME IDEAL SPECTRUM. Let R be a ring. Its spectrum Spec(R) isa topological space whose underlying set is the set of its prime ideals, where weemploy the following notation: if p is a prime ideal of R, then we denote thecorresponding element of Spec(R) by xp and if x Spec(R), we denote by px Rthe corresponding prime ideal. If r R, then we write U (r ) for the set of x Spec(R) with r / px . We have U (r ) U (r ) = U (rr ) and so this collection of subsets is a basis for a to topology on Spec(R), called (from now on) the Zariskitopology . It is also characterized by the property that the closed sets are thosedened by an ideal of R: if we put Z (r ) := Spec( R) U (r ), then any closed subsetof Spec(R) is of the form Z (S ) = rS Z (r ) for some subset S R (and vice versa).Notice that this just the set of xSpec(R) with px S and hence only depends onthe ideal I (S ) generated by S : Z (S ) = Z (I (S )) .

Since the zero ring (for which we have 0 = 1 ) has no prime ideals, its spectrumis the empty set.

As the following exercise shows, points of Spec(R) need not be closed:

9It is often preferable to work over an algebraically closed eld. We can remedy this simply by choosing an algebraic closure L of k(Y ) . The maximal ideal spectrum of L A ( Y ) A (X ) is then anafne L-variety, and yields a notion of a general ber that is even closer to our geometric intuition.


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E XERC ISE 31. Let x, y Spec(R). Prove that x lies in the closure of y if andonly if px py . Conclude that x is closed in Spec(R) if and only if px is a maximalideal of R .We can think (at least for now) of r R a dening a regular function onSpec(R), but its value eld varies with the point: to x Spec(R) we associate theeld (x) := Frac( R/ px ), called the residue eld of x . Then r R determines aregular function on Spec(R): it assigns to x the image of r in (x) (denoted r (x)).

(We will come up with something better when we discuss sheaves.)One immediate advantage of the prime ideal spectrum over the maximal ideal

spectrum is that any ring homomorphism : R R determines a mapSpec() : Spec(R) Spec(R ), xp xp ,

where p := 1p = ker( R R R/ p). This is a prime ideal of R indeed, for now induces an embedding of R / p in R/ p and so R / p has no zero divisors(but if p were a maximal ideal, then we would not be able to conclude that

1

pis maximal). The embedding R / p R/ p of domains of course extends to anembedding of their elds of fractions, (xp ) (xp ). For r R , the regularfunction it denes on Spec(R ) as above precomposed with yields the regularfunction on Spec(R) dened by (r ) and so Spec()1(U (r )) = U ((r )) . HenceSpec() is continuous.

A generic point of Spec(R) is a point that does not lie in the closure of anotherpoint. In other words, it is associated to a minimal prime ideal of R, or equivalently,to an irreducible component Spec(R). If R is a domain, there is only one such point,namely x(0) with residue eld Frac( R).

Notice that for as above, the ber of Spec() over x = xp Spec(R ) is(as a subspace of Spec(R)) the set of x Spec(R) with for which p = 1px , orequivalently, for which p (R ) = (p ). In Exercise 33 you will show that suchprime ideals correspond bijectively to prime ideals of the (x )-algebra (x )

R R.Hence it makes sense to regard Spec((x )R R) as the ber over x . (Here withthe given of the (x )-algebra (x ) R R); we will later employ denitions for which this is automatic.) If R is a domain, then the ber over the generic point,Spec(Frac( R )R R), is called the generic ber of Spec().

E XAM PLE 6.9. The spectrum of Z has a point x p for every prime number p(with residue eld (x p) = F p), and one for the zero ideal, yielding the genericpoint x0 (with residue led (x0) = Q). For every ring R we have a natural ringhomomorphism Z R. This denes a map Spec(R) Spec(Z). In case R isnot the zero ring (so that Spec(R) is nonempty), the ber over x p is Spec(F p R) = Spec( R/pR ) (which is empty if p is invertible in R ) and the ber over x0 isSpec(QR).

E XERC ISE 32. Determine the points and their residue elds of Spec(k[X ]) andSpec(R[X ]). Also try your hand at Spec(Z[X ]).

E XERC ISE 33. Let : R R be a ring homomorphism and p R a primeideal. Prove that for every prime ideal p R with 1p = p , the image of Frac( R / p )R p in Frac( R / p )R R is a prime ideal of latter. Prove that thisdenes a bijection between the prime ideals p R with 1p = p and the primeideals of Frac( R / p )R R .

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