Commensurating actions of birational groups and groups of pseudo-automorphisms

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Algebraic Geometry

J.S. Milne

Version 6.02 March 19, 2017

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algebraic varieties, and not just subvarieties of affine and projective space. This approach leads more naturally into scheme theory.

BibTeX information

@misc{milneAG,

author={Milne, James S.},

title={Algebraic Geometry (v6.02)}, year={2017},

note={Available at www.jmilne.org/math/}, pages={221}

}

v2.01 (August 24, 1996). First version on the web.

v3.01 (June 13, 1998).

v4.00 (October 30, 2003). Fixed errors; many minor revisions; added exercises; added two sections/chapters; 206 pages.

v5.00 (February 20, 2005). Heavily revised; most numbering changed; 227 pages.

v5.10 (March 19, 2008). Minor fixes; TEXstyle changed, so page numbers changed; 241 pages.

v5.20 (September 14, 2009). Minor corrections; revised Chapters 1, 11, 16; 245 pages.

v5.22 (January 13, 2012). Minor fixes; 260 pages.

v6.00 (August 24, 2014). Major revision; 223 pages.

v6.01 (August 23, 2015). Minor fixes; 226 pages.

v6.02 (March 19, 2017). Minor fixes; 221 pages.

Available at www.jmilne.org/math/

Please send comments and corrections to me at the address on my web page.

The curves are a tacnode, a ramphoid cusp, and an ordinary triple point.

Copyright c 1996–2017 J.S. Milne.

Single paper copies for noncommercial personal use may be made without explicit permission from the copyright holder.

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a. Definition of an algebraic set, 35 ; b. The Hilbert basis theorem, 36; c. The Zariski topology, 37; d.The Hilbert Nullstellensatz, 38; e.The correspondence between algebraic sets and radical ideals, 39 ; f. Finding the radical of an ideal, 43; g. Properties of the Zariski topology, 43; h.Decomposition of an algebraic set into irreducible algebraic sets, 44 ; i.Regular functions; the coordinate ring of an algebraic set, 47; j.Regular maps, 48; k.

Hypersurfaces; finite and quasi-finite maps, 48; l.Noether normalization theorem, 50 ; m.

Dimension, 52 ;Exercises, 56.

3 Affine Algebraic Varieties 57

a.Sheaves, 57 ; b.Ringed spaces, 58; c.The ringed space structure on an algebraic set, 59

; d.Morphisms of ringed spaces, 62 ; e.Affine algebraic varieties, 63; f.The category of affine algebraic varieties, 64; g.Explicit description of morphisms of affine varieties, 65 ; h. Subvarieties, 68; i. Properties of the regular map Spm.˛/, 69; j. Affine space without coordinates, 70; k.Birational equivalence, 71; l.Noether Normalization Theorem, 72; m.

Dimension, 73 ;Exercises, 77.

4 Local Study 79

a.Tangent spaces to plane curves, 79 ; b.Tangent cones to plane curves, 81 ; c.The local ring at a point on a curve, 83; d.Tangent spaces to algebraic subsets ofA^m, 84 ; e.The differential of a regular map, 86; f. Tangent spaces to affine algebraic varieties, 87 ; g.

Tangent cones, 91; h. Nonsingular points; the singular locus, 92 ; i. Nonsingularity and regularity, 94; j.Examples of tangent spaces, 95;Exercises, 96.

5 Algebraic Varieties 97

a.Algebraic prevarieties, 97; b.Regular maps, 98; c.Algebraic varieties, 99; d.Maps from varieties to affine varieties, 101; e.Subvarieties, 101 ; f.Prevarieties obtained by patching, 102; g. Products of varieties, 103 ; h. The separation axiom revisited, 108; i. Fibred products, 110 ; j. Dimension, 111; k.Dominant maps, 113; l. Rational maps; birational equivalence, 113; m. Local study, 114; n. Etale maps, 115 ; o.´ Etale neighbourhoods,´ 118 ; p. Smooth maps, 120 ; q.Algebraic varieties as a functors, 121 ; r. Rational and unirational varieties, 124 ;Exercises, 125.

6 Projective Varieties 127

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The homogeneous coordinate ring of a projective variety, 135; g.Regular functions on a projective variety, 136; h.Maps from projective varieties, 137; i.Some classical maps of projective varieties, 138 ; j.Maps to projective space, 143; k.Projective space without coordinates, 143; l.The functor defined by projective space, 144; m.Grassmann varieties, 144 ; n.Bezout’s theorem, 148; o.Hilbert polynomials (sketch), 149; p.Dimensions, 150;

q.Products, 152 ;Exercises, 153.

7 Complete Varieties 155

a.Definition and basic properties, 155 ; b.Proper maps, 157; c.Projective varieties are complete, 158 ; d. Elimination theory, 159 ; e. The rigidity theorem; abelian varieties, 163; f. Chow’s Lemma, 165 ; g. Analytic spaces; Chow’s theorem, 167; h. Nagata’s Embedding Theorem, 168 ;Exercises, 169.

8 Normal Varieties; (Quasi-)finite maps; Zariski’s Main Theorem 171 a. Normal varieties, 171 ; b. Regular functions on normal varieties, 174 ; c. Finite and quasi-finite maps, 176 ; d. The fibres of finite maps, 182; e. Zariski’s main theorem, 184 ; f.Stein factorization, 189; g.Blow-ups, 190 ; h.Resolution of singularities, 190 ; Exercises, 191.

9 Regular Maps and Their Fibres 193

a.The constructibility theorem, 193; b. The fibres of morphisms, 196; c.Flat maps and their fibres, 199 ; d. Lines on surfaces, 206; e. Bertini’s theorem, 211; f. Birational classification, 211;Exercises, 212.

Solutions to the exercises 213

Index 219

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Notations

We use the standard (Bourbaki) notations:ND f0; 1; 2; : : :g,ZDring of integers,RDfield of real numbers,CDfield of complex numbers,F^pDZ=pZDfield ofp elements,p a prime number. Given an equivalence relation,Œdenotes the equivalence class containing. A family of elements of a setAindexed by a second setI, denoted.ai/i2I, is a function i 7!aiWI !A. We sometimes writejSjfor the number of elements in a finite setS.

Throughout,kis an algebraically closed field. Unadorned tensor products are overk. For ak-algebraRandk-moduleM, we often writeMRforR˝M. The dual Homk-linear.E; k/

of a finite-dimensionalk-vector spaceEis denoted byE^_.

All rings will be commutative with1, and homomorphisms of rings are required to map 1to1.

We use Gothic (fraktur) letters for ideals:

a b c m n p q A B C M N P Q

Finally

X ^defDY X is defined to beY, or equalsY by definition;

X Y X is a subset ofY (not necessarily proper, i.e.,X may equalY);

X Y X andY are isomorphic;

X 'Y X andY are canonically isomorphic (or there is a given or unique isomorphism).

A reference “Section 3m” is to Section m in Chapter 3; a reference “(3.45)” is to this item in chapter 3; a reference “(67)” is to (displayed) equation 67 (usually given with a page reference unless it is nearby).

Prerequisites

The reader is assumed to be familiar with the basic objects of algebra, namely, rings, modules, fields, and so on.

References

Atiyah and MacDonald 1969:Introduction to Commutative Algebra, Addison-Wesley.

CA:Milne, J.S., Commutative Algebra, v4.02, 2017.

FT:Milne, J.S., Fields and Galois Theory, v4.52, 2017.

Hartshorne 1977:Algebraic Geometry, Springer.

Shafarevich 1994: Basic Algebraic Geometry, Springer.

A reference monnnn (resp. sxnnnn) is to question nnnn on mathoverflow.net (resp.

math.stackexchange.com).

We sometimes refer to the computer algebra programs

CoCoA (Computations inCommutativeAlgebra)http://cocoa.dima.unige.it/.

Macaulay 2 (Grayson and Stillman)http://www.math.uiuc.edu/Macaulay2/.

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I thank the following for providing corrections and comments on earlier versions of these notes: Jorge Nicol´as Caro Montoya, Sandeep Chellapilla, Rankeya Datta, Umesh V. Dubey, Shalom Feigelstock, Tony Feng, B.J. Franklin, Sergei Gelfand, Daniel Gerig, Darij Grinberg, Lucio Guerberoff, Isac Hed´en, Guido Helmers, Florian Herzig, Christian Hirsch, Cheuk-Man Hwang, Jasper Loy Jiabao, Dan Karliner, Lars Kindler, John Miller, Joaquin Rodrigues, Sean Rostami, David Rufino, Hossein Sabzrou, Jyoti Prakash Saha, Tom Savage, Nguyen Quoc Thang, Bhupendra Nath Tiwari, Israel Vainsencher, Soli Vishkautsan, Dennis Bouke Westra, Felipe Zaldivar, Luochen Zhao, and others.

QUESTION: If we try to explain to a layman what algebraic geometry is, it seems to me that the title of the old book of Enriques is still adequate: Geometrical Theory of Equations . . . . GROTHENDIECK: Yes! but your “layman” should know what a system of algebraic equations is. This would cost years of study to Plato.

QUESTION: It should be nice to have a little faith that after two thousand years every good high school graduate can understand what an affine scheme is . . .

From the notes of a lecture series that Grothendieck gave at SUNY at Buffalo in the summer of 1973 (in 167 pages, Grothendieck manages to cover very little).

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Introduction

There is almost nothing left to discover in geometry.

Descartes, March 26, 1619

Just as the starting point of linear algebra is the study of the solutions of systems of linear equations,

n

X

jD1

aijXj Dbi; iD1; : : : ; m; (1) the starting point for algebraic geometry is the study of the solutions of systems of polynomial equations,

fi.X1; : : : ; Xn/D0; i D1; : : : ; m; fi 2kŒX1; : : : ; Xn:

One immediate difference between linear equations and polynomial equations is that theo- rems for linear equations don’t depend on which fieldkyou are working over,¹but those for polynomial equations depend on whether or notkis algebraically closed and (to a lesser extent) whetherkhas characteristic zero.

A better description of algebraic geometry is that it is the study of polynomial functions and the spaces on which they are defined (algebraic varieties), just as topology is the study of continuous functions and the spaces on which they are defined (topological spaces), differential topology the study of infinitely differentiable functions and the spaces on which they are defined (differentiable manifolds), and so on:

algebraic geometry regular (polynomial) functions algebraic varieties

topology continuous functions topological spaces

differential topology differentiable functions differentiable manifolds complex analysis analytic (power series) functions complex manifolds.

The approach adopted in this course makes plain the similarities between these different areas of mathematics. Of course, the polynomial functions form a much less rich class than the others, but by restricting our study to polynomials we are able to do calculus over any field: we simply define

d dX

XaiXⁱDX

i aiX^{i 1}:

Moreover, calculations with polynomials are easier than with more general functions.

1For example, suppose that the system (1) has coefficientsaij2kand thatKis a field containingk. Then (1) has a solution inkⁿif and only if it has a solution inKⁿ, and the dimension of the space of solutions is the same for both fields. (Exercise!)

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Consider a nonzero differentiable functionf .x; y; z/. In calculus, we learn that the equation

f .x; y; z/DC (2)

defines a surface S in R³, and that the tangent plane toS at a point P D.a; b; c/ has equation²

@f

@x

P

.x a/C @f

@y

P

.y b/C @f

@z

P

.z c/D0: (3)

The inverse function theorem says that a differentiable map˛WS!S⁰of surfaces is a local isomorphism at a pointP 2S if it maps the tangent plane atP isomorphically onto the tangent plane atP⁰D˛.P /.

Consider a nonzero polynomialf .x; y; z/with coefficients in a fieldk. In these notes, we shall learn that the equation (2) defines a surface ink³, and we shall use the equation (3) to define the tangent space at a pointP on the surface. However, and this is one of the essential differences between algebraic geometry and the other fields, the inverse function theorem doesn’t hold in algebraic geometry. One other essential difference is that1=Xis not the derivative of any rational function ofX, and nor isX^{np 1}in characteristicp¤0— these functions can not be integrated in the field of rational functionsk.X /.

These notes form a basic course on algebraic geometry. Throughout, we require the ground field to be algebraically closed in order to be able to concentrate on the geometry.

Additional chapters, treating more advanced topics, can be found on my website.

The approach to algebraic geometry taken in these notes

In differential geometry it is important to define differentiable manifolds abstractly, i.e., not as submanifolds of some Euclidean space. For example, it is difficult even to make sense of a statement such as “the Gauss curvature of a surface is intrinsic to the surface but the principal curvatures are not” without the abstract notion of a surface.

Until the mid 1940s, algebraic geometry was concerned only with algebraic subvarieties of affine or projective space over algebraically closed fields. Then, in order to give substance to his proof of the congruence Riemann hypothesis for curves and abelian varieties, Weil was forced to develop a theory of algebraic geometry for “abstract” algebraic varieties over arbitrary fields,³but his “foundations” are unsatisfactory in two major respects:

˘ Lacking a sheaf theory, his method of patching together affine varieties to form abstract varieties is clumsy.⁴

˘ His definition of a variety over a base fieldkis not intrinsic; specifically, he fixes some large “universal” algebraically closed field˝and defines an algebraic variety overk to be an algebraic variety over˝together with ak-structure.

In the ensuing years, several attempts were made to resolve these difficulties. In 1955, Serre resolved the first by borrowing ideas from complex analysis and defining an algebraic variety over an algebraically closed field to be a topological space with a sheaf of functions that is locally affine.⁵ Then, in the late 1950s Grothendieck resolved all such difficulties by developing the theory of schemes.

2Think ofS as a level surface for the functionf, and note that the equation is that of a plane through .a; b; c/perpendicular to the gradient vector.Of /_P off atP.

3Weil, Andr´e. Foundations of algebraic geometry. American Mathematical Society, Providence, R.I. 1946.

4Nor did Weil use the Zariski topology in 1946.

5Serre, Jean-Pierre. Faisceaux alg´ebriques coh´erents. Ann. of Math. (2) 61, (1955). 197–278, commonly referred to as FAC.

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9 In these notes, we follow Grothendieck except that, by working only over a base field, we are able to simplify his language by considering only the closed points in the underlying topological spaces. In this way, we hope to provide a bridge between the intuition given by differential geometry and the abstractions of scheme theory.

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C

HAPTER

1 Preliminaries from commutative algebra

Algebraic geometry and commutative algebra are closely intertwined. For the most part, we develop the necessary commutative algebra in the context in which it is used. However, in this chapter, we review some basic definitions and results from commutative algebra.

a Rings and ideals

Basic definitions

LetAbe a ring. AsubringofAis a subset that contains1Aand is closed under addition, multiplication, and the formation of negatives. AnA-algebrais a ringBtogether with a homomorphismiBWA!B. Ahomomorphism ofA-algebrasB!C is a homomorphism of rings'WB!C such that'.iB.a//DiC.a/for alla2A.

Elementsx1; : : : ; xnof anA-algebraBare said togenerateit if every element ofBcan be expressed as a polynomial in thexi with coefficients iniB.A/, i.e., if the homomorphism ofA-algebrasAŒX1; : : : ; Xn!Bacting asiAonAand sendingXi toxi is surjective.

WhenAB andx₁; : : : ; x_n2B, we letAŒx₁; : : : ; x_ndenote theA-subalgebra ofB generated by thexi.

A ring homomorphismA!Bis said to be offinite-type, andBis afinitely generated A-algebra ifBis generated by a finite set of elements as anA-algebra.

A ring homomorphismA!B isfinite, and B is afinite¹ A-algebra, if B is finitely generated as anA-module.

Letkbe a field, and letAbe ak-algebra. When1_A¤0inA, the mapk!Ais injective, and we can identifykwith its image, i.e., we can regardkas a subring ofA. When1AD0 in a ringA, thenAis the zero ring, i.e.,AD f0g.

A ring is anintegral domainif it is not the zero ring and ifabD0implies thataD0or bD0; in other words, ifabDacanda¤0, thenbDc.

For a ringA,Ais the group of elements ofAwith inverses (the units in the ring).

Ideals

LetAbe a ring. AnidealainAis a subset such that

1The term “module-finite” is also used.

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(a) ais a subgroup ofAregarded as a group under addition;

(b) a2a,r2A)ra2a:

Theideal generated by a subsetSofAis the intersection of all idealsacontainingS — it is easy to see that this is in fact an ideal, and that it consists of all finite sums of the form Prisiwithri2A,si 2S. The ideal generated by the empty set is the zero idealf0g. When SD fs1; s2; : : :g, we write.s1; s2; : : :/for the ideal it generates.

Letaandbbe ideals inA. The setfaCbja2a; b2bgis an ideal, denoted byaCb.

The ideal generated byfabja2a; b2bgis denoted byab. Clearlyabconsists of all finite sums P

aibi withai 2a and bi 2b, and ifaD.a1; : : : ; am/and bD.b1; : : : ; bn/, then abD.a1b1; : : : ; aibj; : : : ; ambn/. Note that

aba\b: (4)

The kernel of a homomorphismA!Bis an ideal inA. Conversely, for any idealainA, the set of cosets ofainAforms a ringA=a, anda7!aCais a homomorphism'WA!A=a whose kernel isa. The mapb7!' ¹.b/is a one-to-one correspondence between the ideals ofA=aand the ideals ofAcontaininga.

An idealpisprimeifp¤Aandab2p)a2porb2p. Thuspis prime if and only if A=pis nonzero and has the property that

abD0 H) aD0orbD0;

i.e.,A=pis an integral domain. Note that ifpis prime anda1 an2p, then at least one of theai 2p.

An idealminAismaximalif it is maximal among the proper ideals ofA. Thusmis maximal if and only ifA=mis nonzero and has no proper nonzero ideals, and so is a field.

Note that

mmaximal H) mprime.

The ideals ofAB are all of the formabwithaandbideals inAandB. To see this, note that if c is an ideal in AB and .a; b/2c, then.a; 0/D.1; 0/.a; b/2c and .0; b/D.0; 1/.a; b/2c. Therefore,cDabwith

aD faj.a; 0/2cg; bD fbj.0; b/2cg:

IdealsaandbinAarecoprime(orrelatively prime) ifaCbDA. Assume thataandb are coprime, and leta2aandb2bbe such thataCbD1. Forx; y2A, letzDayCbx;

then

zbxxmoda zay ymodb;

and so the canonical map

A!A=aA=b (5)

is surjective. Clearly its kernel isa\b, which containsab. Letc2a\b; then cDc1DcaCcb2ab:

Hence, (5) is surjective with kernelab. This statement extends to finite collections of ideals.

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a. Rings and ideals 13 THEOREM1.1 (CHINESEREMAINDERTHEOREM). Leta1; : : : ;anbe ideals in a ringA.

Ifa_i is coprime toa_j wheneveri¤j, then the map

A!A=a1 A=an (6)

is surjective, with kernelQ

ai DT ai.

PROOF. We have proved the statement fornD2, and we use induction to extend it ton > 2.

Fori2, there exist elementsai 2a1andbi2ai such that aiCbi D1:

The productQ

i2.aiCbi/lies ina₁Ca₂ a_nand equals1, and so a₁Ca₂ a_nDA:

Therefore,

A=a1 a_n D A=a1.a2 a_n/

' A=a1A=a2 an by thenD2case

' A=a1A=a2 A=an by induction. ₂

We let spec.A/denote the set of prime ideals in a ringAand spm.A/the set of maximal ideals.

Noetherian rings

PROPOSITION1.2. The following three conditions on a ringAare equivalent:

(a) every ideal inAis finitely generated;

(b) every ascending chain of ideals a₁a₂ eventually becomes constant, i.e., a_mDa_m_C₁D for somem;

(c) every nonempty set of ideals inAhas a maximal element.

PROOF. (a) H) (b): Leta₁a₂ be an ascending chain of ideals. ThenS

a_i is an ideal, and hence has a finite setfa1; : : : ; angof generators. For somem, all theai belong to a_m, and then

a_mDa_m_C₁D DS a_i:

(b) H) (c): Let˙ be a nonempty set of ideals inA. If˙ has no maximal element, then the axiom of dependent choice²implies that there exists an infinite strictly ascending chain of ideals in˙, contradicting (b).

(c) H) (a): Letabe an ideal, and let˙ be the set of finitely generated ideals contained ina. Then˙ is nonempty because it contains the zero ideal, and so it contains a maximal elementcD.a1; : : : ; ar/. Ifc¤a, then there exists ana2aXc, and.a1; : : : ; ar; a/will be a finitely generated ideal inaproperly containingc. This contradicts the definition ofc, and

socDa. ₂

2This says the following: letRbe a binary relation on a nonempty setX, and suppose that, for eachainX, there exists absuch thataRb; then there exists a sequence.an/n2Nof elements ofXsuch thatanRanC1for alln. This axiom is strictly weaker than the axiom of choice (q.v. Wikipedia).

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A ringAisnoetherianif it satisfies the equivalent conditions of the proposition. On applying (c) to the set of all proper ideals containing a fixed proper ideal, we see that every proper ideal in a noetherian ring is contained in a maximal ideal. This is, in fact, true for every ring, but the proof for non-noetherian rings requires Zorn’s lemma (CA2.2).

A ringAis said to belocalif it has exactly one maximal idealm. Because every nonunit is contained in a maximal ideal, for a local ringADAXm.

PROPOSITION1.3 (NAKAYAMA’S LEMMA). LetAbe a local ring with maximal idealm, and letM be a finitely generatedA-module.

(a) IfM DmM, thenM D0:

(b) IfN is a submodule ofM such thatM DNCmM, thenM DN.

PROOF. (a) Suppose thatM ¤0. Choose a minimal set of generatorsfe1; : : : ; eng,n1, forM, and write

e1Da1e1C Canen; ai 2m.

Then

.1 a1/e1Da2e2C Canen

and, as.1 a1/is a unit,e2; : : : ; engenerateM, contradicting the minimality of the set.

(b) The hypothesis implies thatM=N Dm.M=N /, and soM=N D0. ₂ Now letAbe a local noetherian ring with maximal idealm. Thenmis anA-module, and the action ofAonm=m²factors throughkD^defA=m.

COROLLARY1.4. Elements a1; : : : ; an ofm generatem as an ideal if and only if their residues modulom²generatem=m²as a vector space overk. In particular, the minimum number of generators for the maximal ideal is equal to the dimension of the vector space m=m².

PROOF. Ifa1; : : : ; angeneratem, it is obvious that their residues generatem=m². Conversely, suppose that their residues generatem=m², so thatmD.a1; : : : ; an/Cm². BecauseAis noetherian,mis finitely generated, and Nakayama’s lemma shows thatmD.a1; : : : ; an/.₂ DEFINITION 1.5. LetAbe a noetherian ring.

(a) Theheightht.p/of a prime idealpinAis the greatest lengthd of a chain of distinct prime ideals

pDp_dp_{d 1} p₀: (7) (b) TheKrull dimensionofAis supfht.p/jpa prime ideal inAg.

Thus, the Krull dimension of a noetherian ringAis the supremum of the lengths of chains of prime ideals inA(the length of a chain is the number of gaps, so the length of (7) isd). For example, a field has Krull dimension0, and conversely an integral domain of Krull dimension0is a field. The height of every nonzero prime ideal in a principal ideal domain is1, and so such a ring has Krull dimension1(provided it is not a field).

The height of every prime ideal in a noetherian ring is finite, but the Krull dimension of the ring may be infinite because it may contain a sequence of prime idealsp₁;p₂;p₃; : : : such that ht.pi/tends to infinity (CA, p.13).

DEFINITION 1.6. A local noetherian ring of Krull dimensiond is said to beregularif its maximal ideal can be generated byd elements.

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b. Rings of fractions 15 It follows from Corollary1.4that a local noetherian ring is regular if and only if its Krull dimension is equal to the dimension of the vector spacem=m².

LEMMA1.7. In a noetherian ring, every set of generators for an ideal contains a finite generating subset.

PROOF. Letabe an ideal in a noetherian ringA, and letS be a set of generators fora. An ideal maximal among those generated by a finite subset ofSmust contain every element of

S (otherwise it wouldn’t be maximal), and so equalsa. ₂

In the proof of the next theorem, we use that a polynomial ring over a noetherian ring is noetherian (see Lemma2.8).

THEOREM1.8 (KRULLINTERSECTIONTHEOREM). LetAbe a noetherian local ring with maximal idealm; thenT

n1mⁿD f0g:

PROOF. Leta1; : : : ; argeneratem. Thenmⁿconsists of all finite sums X

i1CCirDn

ci1iraⁱ₁¹ a_rⁱ^r; ci1ir 2A:

In other words,mⁿconsists of the elements ofAof the formg.a1; : : : ; ar/for some homogeneous polynomialg.X1; : : : ; Xr/2AŒX1; : : : ; Xrof degreen. LetSmdenote the set of homogeneous polynomials f of degreemsuch thatf .a1; : : : ; ar/2T

n1mⁿ, and leta be the ideal inAŒX1; : : : ; Xrgenerated by the setS

mSm. According to the lemma, there exists a finite setff1; : : : ; fsgof elements ofS

mSmthat generatesa. Letdi Ddegfi, and let d Dmaxdi. Let b 2T

n1mⁿ; thenb2m^d^C¹, and so bDf .a1; : : : ; ar/ for some homogeneous polynomialf of degreedC1. By definition,f 2SdC1a, and so

f Dg1f1C Cgsfs

for somegi2AŒX1; : : : ; Xr. Asf and thefi are homogeneous, we can omit from eachgi

all terms not of degree degf degfi, since these terms cancel out. Thus, we may choose the gi to be homogeneous of degree degf degfi DdC1 di > 0. Thengi.a1; : : : ; ar/2m, and so

bDf .a₁; : : : ; a_r/DX

ig_i.a₁; : : : ; a_r/f_i.a₁; : : : ; a_r/2m\

n1mⁿ: Thus,T

mⁿDmT

mⁿ, and Nakayama’s lemma implies thatT

mⁿD0. ₂

ASIDE1.9. LetAbe the ring of germs of analytic functions at02R(see p.58for the notion of a germ of a function). Then Ais a noetherian local ring with maximal idealmD.x/, and mⁿ consists of the functionsf that vanish to ordernatxD0. The theorem says (correctly) that only the zero function vanishes to all orders at0. By contrast, the functione ^1=x²shows that the Krull intersection theorem fails for the ring of germs of infinitely differentiable functions at0(this ring is not noetherian).

b Rings of fractions

Amultiplicative subsetof a ringAis a subsetS with the property:

12S; a; b2S H) ab2S:

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Define an equivalence relation onAS by

.a; s/.b; t / ” u.at bs/D0for someu2S:

Write ^a_s ora=sfor the equivalence class containing.a; s/, and define addition and multiplication of equivalence classes in the way suggested by the notation:

a

sC^bt Dât^Cst^bs; â_s^b_t Dâbst:

It is easy to check that these do not depend on the choices of representatives for the equivalence classes, and that we obtain in this way a ring

S ¹ADna

s ja2A; s2So and a ring homomorphisma7! ^a1WA!S ¹A, whose kernel is

fa2AjsaD0for somes2Sg:

For example, ifAis an integral domain an0…S, thena7!^a₁ is injective, but if02S, then S ¹Ais the zero ring.

Writei for the homomorphisma7!^a₁WA!S ¹A.

PROPOSITION1.10. The pair.S ¹A; i /has the following universal property: every ele- ments2S maps to a unit inS ¹A, and any other homomorphismA!Bwith this property factors uniquely throughi:

A S ¹A

B

i

9Š

PROOF. Ifˇexists,

sâ_s Da H) ˇ.s/ˇ.â_s/Dˇ.a/ H) ˇ.â_s/D˛.a/˛.s/ ¹; and soˇis unique. Define

ˇ.^a_s/D˛.a/˛.s/ ¹: Then

a

c D_d^b H) s.ad bc/D0somes2S H) ˛.a/˛.d / ˛.b/˛.c/D0

because˛.s/is a unit inB, and soˇis well-defined. It is obviously a homomorphism. ₂ As usual, this universal property determines the pair.S ¹A; i /uniquely up to a unique isomorphism.

WhenAis an integral domain andSDAX f0g, F DS ¹Ais the field of fractions ofA, which we denoteF .A/. In this case, for any other multiplicative subsetT ofAnot containing0, the ringT ¹Acan be identified with the subringf^a_t 2F ja2A,t2SgofF.

We shall be especially interested in the following examples.

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b. Rings of fractions 17 EXAMPLE1.11. Leth2A. ThenS_hD f1; h; h²; : : :gis a multiplicative subset ofA, and we letAhDS_h¹A. Thus every element ofAhcan be written in the forma= h^m,a2A, and

a

h^m D _h^bⁿ ” h^N.ahⁿ bh^m/D0; someN:

Ifhis nilpotent, thenA_hD0, and ifAis an integral domain with field of fractionsF and h¤0, thenA_his the subring ofF of elements of the forma= h^m,a2A,m2N:

EXAMPLE1.12. Letpbe a prime ideal inA. ThenS_pDAXpis a multiplicative subset of A, and we letA_pDS_p¹A. Thus each element ofA_pcan be written in the form ^a_c,c…p, and

a

c D_d^b ” s.ad bc/D0, somes…p:

The subsetmD f^a_s ja2p; s…pgis a maximal ideal inA_p, and it is the only maximal ideal, i.e.,A_pis a local ring.³ WhenAis an integral domain with field of fractionsF,A_pis the subring ofF consisting of elements expressible in the form^a_s,a2A,s…p.

LEMMA1.13. For every ringAandh2A, the mapP

aiXⁱ 7!Pa_i

hⁱ defines an isomorphism

AŒX =.1 hX / ^'!A_h:

PROOF. IfhD0, both rings are zero, and so we may assumeh¤0. In the ringAŒxD AŒX =.1 hX /,1Dhx, and sohis a unit. Let˛WA!B be a homomorphism of rings such that˛.h/is a unit inB. The homomorphismP

aiXⁱ7!P

˛.ai/˛.h/ ⁱWAŒX !B factors throughAŒxbecause1 hX 7!1 ˛.h/˛.h/ ¹D0, and, because˛.h/is a unit in B, this is the unique extension of˛ toAŒx. ThereforeAŒxhas the same universal property asA_h, and so the two are (uniquely) isomorphic by an isomorphism that fixes elements ofA

and makesh ¹correspond tox. ₂

LetSbe a multiplicative subset of a ringA, and letS ¹Abe the corresponding ring of fractions. Any idealainA, generates an idealS ¹ainS ¹A. Ifacontains an element ofS, thenS ¹acontains a unit, and so is the whole ring. Thus some of the ideal structure ofAis lost in the passage toS ¹A, but, as the next proposition shows, much is retained.

PROPOSITION1.14. LetSbe a multiplicative subset of the ringA. The map p7!S ¹pD.S ¹A/p

is a bijection from the set of prime ideals ofAdisjoint fromS to the set of prime ideals of S ¹Awith inverseq7!(inverse image of qinA).

PROOF. For an idealbofS ¹A, letb^c be the inverse image ofbinA, and for an idealaof A, leta^eD.S ¹A/abe the ideal inS ¹Agenerated by the image ofa.

For an idealbofS ¹A, certainly,bb^ce. Conversely, if ^a_s 2b,a2A,s2S, then

a

12b, and soa2b^c. Thus^a_s 2b^ce, and sobDb^ce.

For an idealaofA, certainlyaaêc. Conversely, ifa2aêc, then â₁2aê, and soâ₁ Dâs⁰

for somea⁰2a,s2S. Thus,t .as a⁰/D0for somet2S, and soast2a. Ifais a prime ideal disjoint fromS, this implies thata2a: for such an ideal,aDa^ec.

3First checkmis an ideal. Next, ifmDAp, then12m; but if1D ^a_s for somea2pands…p, then u.s a/D0someu…p, and souaDus…p, which contradictsa2p. Finally,mis maximal because every element ofApnot inmis a unit.

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Ifbis prime, then certainlyb^c is prime. For any idealaofA,S ¹A=a^e' NS ¹.A=a/

whereSN is the image ofSinA=a. Ifais a prime ideal disjoint fromS, thenSN ¹.A=a/is a subring of the field of fractions ofA=a, and is therefore an integral domain. Thus,a^eis prime.

We have shown thatp7!p^eandq7!q^care inverse bijections between the prime ideals

ofAdisjoint fromS and the prime ideals ofS ¹A. ₂

LEMMA1.15. Letmbe a maximal ideal of a ringA, and letnDmA_m. For alln, the map aCmⁿ7!^a₁CnⁿWA=mⁿ!A_m=nⁿ (8) is an isomorphism. Moreover, it induces isomorphisms

m^r=mⁿ!n^r=nⁿ for allr < n.

PROOF. The second statement follows from the first, because of the exact commutative diagram.r < n/:

0 m^r=mⁿ A=mⁿ A=m^r 0

0 n^r=nⁿ A_m=nⁿ A_m=n^r 0:

' '

LetSDAXm. ThenA_mDS ¹AandnⁿDmⁿA_mD f^b_s 2A_mjb2mⁿ; s2Sg. In order to show that the map (8) is injective, it suffices to show that

a

1 D^b_s witha2A; b2mⁿ; s2S H) a2mⁿ.

But if ^a₁ D ^bs, thent asDt b2mⁿ for some t 2S, and sot as D0in A=mⁿ. The only maximal ideal inAcontainingm^mism(becausem⁰m^m H) m⁰m/, and so the only maximal ideal inA=mⁿism=mⁿ. Asst is not inm=mⁿ, it must be a unit inA=mⁿ, and as st aD0inA=mⁿ,amust be0inA=mⁿ, i.e.,a2mⁿ:

We now prove that the map (8) is surjective. Let â_s 2A_m,a2A,s2S. Because the only maximal ideal ofAcontainingmⁿism, no maximal ideal contains bothsandmⁿ. It follows that.s/CmⁿDA. Therefore, there existb2Aandq2mⁿsuch thatsbCqD1 inA. It follows thatsis invertible inA_m=nⁿ, and so â_s is theuniqueelement of this ring such thatsâ_s Da. Ass.ba/CqaDa, the image ofbainA_m=nⁿalso has this property and

therefore equals ^a_s inA_m=nⁿ. ₂

PROPOSITION1.16. In every noetherian ring, only 0 lies in all powers of all maximal ideals.

PROOF. Letabe an element of a noetherian ringA. Ifa¤0, thenfbjbaD0gis a proper ideal, and so is contained in some maximal ideal m. Then ^a₁ is nonzero inA_m, and so

a

1….mA_m/ⁿfor somen(by the Krull intersection theorem1.8), which implies thata…mⁿ.₂ NOTES. For more on rings of fractions, see CA5.

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b. Rings of fractions 19

Modules of fractions

Let S be a multiplicative subset of the ring A, and let M be an A-module. Define an equivalence relation onMS by

.m; s/.n; t / ” u.t m sn/D0for someu2S:

Write ^m_s for the equivalence class containing.m; s/, and define addition and scalar multiplication by the rules:

m

s Cⁿ_t D^mt_st^C^ns; ^a_s^m_t D^am_st ; m; n2M; s; t2S; a2A:

It is easily checked that these do not depend on the choices of representatives for the equivalence classes, and that we obtain in this way anS ¹A-module

S ¹M D f^ms jm2M; s2Sg

and a homomorphismm7! ^m₁WM ⁱ!^S S ¹M ofA-modules whose kernel is fa2M jsaD0for somes2Sg:

1.17. The elements ofSact invertibly onS ¹M, and every homomorphism fromM to anA-module N with this property factors uniquely throughiS.

M S ¹M

N:

i_S

9Š

PROPOSITION1.18. The functorM S ¹M is exact. In other words, if the sequence of A-modules

M⁰ ^˛!M ^ˇ!M⁰⁰ is exact, then so also is the sequence ofS ¹A-modules

S ¹M⁰ ^S

1˛

!S ¹M ^S

1ˇ

!S ¹M⁰⁰:

PROOF. Becauseˇı˛D0, we have0DS ¹.ˇı˛/DS ¹ˇıS ¹˛. Therefore Im.S ¹˛/

Ker.S ¹ˇ/. For the reverse inclusion, let ^m_s 2Ker.S ¹ˇ/wherem2M ands2S. Then

ˇ .m/

s D0and so, for somet2S, we havetˇ.m/D0. Thenˇ.t m/D0, and sot mD˛.m⁰/ for somem⁰2M⁰. Now

m

s D^{t m}t s D^˛.mt s⁰^/ 2Im.S ¹˛/: ₂ PROPOSITION1.19. LetAbe a ring, and letM be anA-module. The canonical map

M !Y

fM_mjma maximal ideal inAg is injective.

PROOF. Letm2M map to zero in allM_m. The annihilatoraD fa2AjamD0gofm is an ideal inA. Becausemmaps to zeroM_m, there exists ans2AXmsuch thatsmD0.

Thereforeais not contained inm. Since this is true for all maximal idealsm,aDA, and so

it contains1. NowmD1mD0. ₂

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COROLLARY1.20. AnA-moduleM D0ifM_mD0for all maximal idealsminA.

PROOF. Immediate consequence of the lemma. ₂

PROPOSITION1.21. LetAbe a ring. A sequence ofA-modules

M⁰ ^˛!M ^ˇ!M⁰⁰ (*)

is exact if and only if

M_m⁰ ^˛!^m M_m ^˛!^m M_m⁰⁰ (**)

is exact for all maximal idealsm.

PROOF. The necessity is a special case of (1.18). For the sufficiency, letNDKer.ˇ/=Im.˛/.

Because the functorM M_mis exact,

N_mDKer.ˇ_m/=Im.˛_m/:

If (**) is exact for allm, thenN_mD0for allm, and soN D0(by1.20). But this means

that (*) is exact. ₂

COROLLARY1.22. A homomorphismM !N ofA-modules is injective (resp. surjective) if and only ifM_m!N_mis injective (resp. surjective) for all maximal idealsm:

PROOF. Apply the proposition to0!M !N (resp.M !N !0). ₂

Direct limits

Adirected setis a pair.I;/consisting of a setI and a partial orderingonI such that for alli; j 2I, there exists ak2I withi; j k.

Let.I;/be a directed set, and letAbe a ring. Adirect systemofA-modules indexed by.I;/is a family.Mi/i2I ofA-modules together with a family.˛_jⁱWMi !Mj/ij of A-linear maps such that˛ⁱ_i DidMi and ˛_k^jı˛_jⁱ D˛_kⁱ alli j k.⁴ An A-moduleM together withA-linear maps˛ⁱWMi !M such that˛ⁱ D˛^jı˛_jⁱ for alli j is thedirect limitof the system.Mi; ˛_i^j/if

(a) M DS

i2I˛ⁱ.Mi/, and

(b) mi 2Mi maps to zero inM if and only if it maps to zero inMj for somej i. Direct limits ofA-algebras are defined similarly.

PROPOSITION1.23. For every multiplicative subsetSofA,S ¹A'lim

!A_h, wherehruns over the elements ofS (partially ordered by division).

PROOF. When hjh⁰, say, h⁰Dhg, there is a canonical homomorphism â_h 7! âg_h0WA_h! A_h0, and so the rings A_h form a direct system indexed by the setS. Whenh2S, the homomorphismA!S ¹Aextends uniquely to a homomorphism â_h 7! â_hWA_h!S ¹A (1.10), and these homomorphisms are compatible with the maps in the direct system. Now it is easy to see thatS ¹Asatisfies the conditions to be the direct limit of theA_h. ₂

4RegardI as a category with Hom.a; b/empty unlessab, in which case it contains a single element.

Then a direct system is a functor fromI to the category ofA-modules.

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c. Unique factorization 21

c Unique factorization

LetAbe an integral domain. An elementaofAisirreducibleif it is not zero, not a unit, and admits only trivial factorizations, i.e.,

aDbc H) borcis a unit.

An elementais said to beprimeif.a/is a prime ideal, i.e., ajbc H) ajborajc.

An integral domainAis called aunique factorization domain(or afactorial domain) if every nonzero nonunit inAcan be written as a finite product of irreducible elements in exactly one way up to units and the order of the factors:Principal ideal domains, for example, ZandkŒX , are unique factorization domains,

PROPOSITION1.24. Let Abe an integral domain, and let abe an element ofA that is neither zero nor a unit. Ifais prime, thenais irreducible, and the converse holds whenAis a unique factorization domain.

PROOF. Assume that ais prime. If aDbc, thenadivides bc and so a dividesb orc.

Suppose the first, and writebDaq. NowaDbcDaqc, which implies thatqcD1because Ais an integral domain, and soc is a unit. Thereforeais irreducible.

For the converse, assume thatais irreducible and thatAis a unique factorization domain.

Ifajbc, then

bcDaq, someq2A:

On writing each ofb,c, andqas a product of irreducible elements, and using the uniqueness of factorizations, we see thatadiffers from one of the irreducible factors ofborc by a unit.

Thereforeadividesborc. ₂

COROLLARY1.25. LetAbe an integral domain. IfAis a unique factorization domain, then every prime ideal of height1is principal.

PROOF. Letpbe a prime ideal of height1. Thenpcontains a nonzero element, and hence an irreducible elementa. We havep.a/.0/. As.a/is prime andphas height1, we

must havepD.a/. ₂

PROPOSITION1.26. LetAbe an integral domain in which every nonzero nonunit element is a finite product of irreducible elements. If every irreducible element ofAis prime, thenA is a unique factorization domain.

PROOF. Suppose that

a1 amDb1 bn (9)

with theai andbi irreducible elements inA. Asa1is prime, it divides one of thebi, which we may suppose to beb1. Asb1is irreducible,b1Dua1for some unitu. On cancellinga1

from both sides of (9), we obtain the equality

a2 amD.ub2/b3 bn:

Continuing in this fashion, we find that the two factorizations are the same up to units and

the order of the factors. ₂

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PROPOSITION1.27 (GAUSS’S LEMMA). LetAbe a unique factorization domain with field of fractionsF. Iff .X /2AŒX factors into the product of two nonconstant polynomials inF ŒX , then it factors into the product of two nonconstant polynomials inAŒX .

PROOF. Letf DghinF ŒX . For suitablec; d2A, the polynomialsg1Dcgandh1Ddh have coefficients inA, and so we have a factorization

cdf Dg1h1inAŒX .

If an irreducible elementpofAdividescd, then, looking modulo.p/, we see that 0Dg1h1in .A=.p// ŒX .

According to Proposition1.24, .p/ is prime, and so.A=.p// ŒX is an integral domain.

Therefore,pdivides all the coefficients of at least one of the polynomialsg1; h1, sayg1, so thatg1Dpg2for someg22AŒX . Thus, we have a factorization

.cd=p/f Dg2h1inAŒX .

Continuing in this fashion, we can remove all the irreducible factors ofcd, and so obtain a

factorization off inAŒX . ₂

LetAbe a unique factorization domain. A nonzero polynomial f Da0Ca1XC CamX^m

inAŒX is said to beprimitiveif the coefficientsai have no common factor (other than units).

Every polynomialf inF ŒX can be writtenf Dc.f /f1withc.f /2F andf1primitive.

The elementc.f /, which is well-defined up to multiplication by a unit, is called thecontent off. Note thatf 2AŒX if and only ifc.f /2A.

LEMMA1.28. The product of two primitive polynomials is primitive.

PROOF. Let

f Da0Ca1XC CamX^m gDb0Cb1XC CbnXⁿ;

be primitive polynomials, and letpbe an irreducible element ofA. Letai0,i0m, be the first coefficient off not divisible byp, and letbj₀, j0n, the first coefficient ofg not divisible byp. Then all the terms in the sumP

iCjDi0Cj0aibj are divisible byp, except ai0bj0, which is not divisible byp. Therefore,pdoesn’t divide the.i0Cj0/th-coefficient offg. We have shown that no irreducible element ofAdivides all the coefficients offg,

which must therefore be primitive. ₂

PROPOSITION1.29. LetAbe a unique factorization domain with field of fractionsF. For polynomialsf; g2F ŒX ,

c.fg/Dc.f /c.g/I

hence every factor inAŒX of a primitive polynomial is primitive.

PROOF. Letf Dc.f /f1andgDc.g/g1withf1andg1primitive. Then fgDc.f /c.g/f1g1

withf₁g₁primitive, and soc.fg/Dc.f /c.g/. ₂

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c. Unique factorization 23 COROLLARY1.30. The irreducible elements inAŒX are the irreducible elementsaofA and the nonconstant primitive polynomialsf such thatf is irreducible inF ŒX .

PROOF. Obvious from (1.27) and (1.29). ₂

PROPOSITION1.31. IfAis a unique factorization domain, then so also isAŒX .

PROOF. We shall check thatAsatisfies the conditions of (1.26).

Letf 2AŒX , and writef Dc.f /f1. Thenc.f /is a product of irreducible elements in A, andf1is a product of irreducible primitive polynomials. This shows thatf is a product of irreducible elements inAŒX .

Letabe an irreducible element ofA. Ifadividesfg, then it dividesc.fg/Dc.f /c.g/.

Asais prime (1.24), it dividesc.f /orc.g/, and hence alsof org.

Letf be an irreducible primitive polynomial inAŒX . Thenf is irreducible inF ŒX , and so iff divides the productghofg; h2AŒX , then it dividesgorhinF ŒX . Suppose the first, and writef qDgwithq2F ŒX . Thenc.q/Dc.f /c.q/Dc.f q/Dc.g/2A, and soq2AŒX . Thereforef dividesginAŒX .

We have shown that every element ofAŒX is a product of irreducible elements and that every irreducible element ofAŒX is prime, and soAŒX is a unique factorization domain

(1.26). ₂

Polynomial rings

Letkbe a field. The elements of the polynomial ringkŒX1; : : : ; Xnare finite sums Xca1anX₁^a¹ X_n^aⁿ; ca1an2k; aj 2N;

with the obvious notions of equality, addition, and multiplication. In particular, the monomials form a basis forkŒX1; : : : ; Xnas ak-vector space.

Thedegree, deg.f /, of a nonzero polynomialf is the largest total degree of a monomial occurring inf with nonzero coefficient. Since deg.fg/Ddeg.f /Cdeg.g/,kŒX1; : : : ; Xn is an integral domain and kŒX1; : : : ; XnDk. An element f of kŒX1; : : : ; Xnis irreducible if it is nonconstant andf Dgh H) gorhis constant.

THEOREM1.32. The ringkŒX1; : : : ; Xnis a unique factorization domain.

PROOF. Note that

kŒX1; : : : ; Xn 1ŒXnDkŒX1; : : : ; Xn:

This simply says that every polynomialf innsymbolsX1; : : : ; Xncan be expressed uniquely as a polynomial inX_nwith coefficients inkŒX₁; : : : ; X_{n 1},

f .X1; : : : ; Xn/Da0.X1; : : : ; Xn 1/X_n^rC Car.X1; : : : ; Xn 1/:

Since, as we noted,kŒX is a unique factorization domain, the theorem follows by induction

from Proposition1.31. ₂

COROLLARY1.33. A nonzero proper principal ideal.f /inkŒX1; : : : ; Xnis prime if and only iff is irreducible.

PROOF. Special case of (1.24). ₂

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d Integral dependence

LetAbe a subring of a ringB. An element˛ofB is said to be⁵integraloverAif it is a root of a monic⁶polynomial with coefficients inA, i.e., if it satisfies an equation

˛ⁿCa1˛^{n 1}C CanD0; ai 2A:

If every element ofBis integral overA, thenBis said to beintegraloverA.

In the next proof, we shall need to apply a variant of Cramer’s rule: ifx1; : : : ; xmis a solution to the system of linear equations

m

X

jD1

cijxj D0; i D1; : : : ; m;

with coefficients in a ringA, then

det.C /xj D0; j D1; : : : ; m; (10) whereC is the matrix of coefficients. To prove this, expand out the left hand side of

det 0 B

@

c11 : : : c1 j 1 P

ic1ixi c1 jC1 : : : c1m

::: ::: ::: ::: :::

cm1 : : : cm j 1 P

icmixi cm jC1 : : : cmm

1 C AD0 using standard properties of determinants.

AnA-moduleM isfaithfulifaM D0,a2A, implies thataD0.

PROPOSITION1.34. LetAbe a subring of a ringB. An element˛ofBis integral overA if and only if there exists a faithfulAŒ˛-submoduleM ofBthat is finitely generated as an A-module.

PROOF. )WSuppose that

˛ⁿCa1˛^{n 1}C CanD0; ai 2A:

Then theA-submoduleM ofBgenerated by1,˛, ...,˛^{n 1}has the property that˛M M, and it is faithful because it contains1.

(W Let M be a faithful AŒ˛-submodule of B admitting a finite set fe1; : : : ; eng of generators as anA-module. Then, for eachi,

˛ei DP

aijej, someaij 2A:

We can rewrite this system of equations as

.˛ a11/e1 a12e2 a13e3 D0 a21e1C.˛ a22/e2 a23e3 D0 D0:

5More generally, iffWA!Bis anA-algebra, an element˛ofBisintegraloverAif it satisfies an equation

˛ⁿCf .a1/˛^{n 1}C Cf .an/D0; ai2A:

Thus,˛is integral overAif and only if it is integral over the subringf .A/ofB.

6A polynomial ismonicif its leading coefficient is1, i.e.,f .X /DXⁿCterms of degree less thann.

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d. Integral dependence 25 LetC be the matrix of coefficients on the left-hand side. Then Cramer’s formula tells us that det.C /eiD0for alli. AsM is faithful and theei generateM, this implies that det.C /D0.

On expanding out the determinant, we obtain an equation

˛ⁿCc1˛^{n 1}Cc2˛^{n 2}C CcnD0; ci 2A: ₂ PROPOSITION1.35. AnA-algebraBis finite if it is generated as anA-algebra by a finite set of elements each of which is integral overA.

PROOF. Suppose thatBDAŒ˛1; : : : ; ˛mand that

˛_iⁿⁱCai1˛_iⁿⁱ ¹C Cai n_i D0; aij 2A; iD1; : : : ; m.

Any monomial in the˛idivisible by some˛ⁿ_iⁱ is equal (inB) to a linear combination of monomials of lower degree. Therefore,Bis generated as anA-module by the finite set of

monomials˛^r₁¹ ˛_m^r^m,1ri < ni. ₂

COROLLARY1.36. AnA-algebraBis finite if and only if it is finitely generated and integral overA.

PROOF. (: Immediate consequence of (1.35).

): We may replaceAwith its image inB. ThenBis a faithfulAŒ˛-module for all

˛2B(because1B2B), and so (1.34) shows that every element ofBis integral overA. As B is finitely generated as anA-module, it is certainly finitely generated as anA-algebra. ₂ PROPOSITION1.37. Consider ringsABC. IfBis integral overAandC is integral overB, thenC is integral overA.

PROOF. Let2C. Then

ⁿCb1^{n 1}C CbnD0

for somebi2B. NowAŒb1; : : : ; bnis finite overA(see1.35), andAŒb1; : : : ; bnŒ is finite overAŒb1; : : : ; bn, and so it is finite overA. Therefore is integral overAby (1.34). ₂ THEOREM1.38. LetAbe a subring of a ringB. The elements ofBintegral overAform anA-subalgebra ofB.

PROOF. Let˛andˇbe two elements ofBintegral overA. ThenAŒ˛; ˇis finitely generated as anA-module (1.35). It is stable under multiplication by˛˙ˇand˛ˇand it is faithful as anAŒ˛˙ˇ-module and as anAŒ˛ˇ-module (because it contains1A). Therefore (1.34)

shows that˛˙ˇand˛ˇare integral overA. ₂

DEFINITION 1.39. LetAbe a subring of the ringB. Theintegral closureofAinBis the subring ofBconsisting of the elements integral overA.

PROPOSITION1.40. LetAbe an integral domain with field of fractionsF, and let˛be an element of some field containingF. If˛is algebraic overF, then there exists ad 2Asuch thatd˛is integral overA.