ALGEBRAIC NUMBER THEORY
J.S. MILNE
Abstract. These are the notes for a course taught at the University of Michigan in F92 as Math 676. They are available atwww.math.lsa.umich.edu/∼jmilne/.
Please send comments and corrections to me at jmilne@umich.edu.
v2.01 (August 14, 1996.) First version on the web.
v2.10 (August 31, 1998.) Fixed many minor errors; added exercises and index.
Contents
Introduction. . . .1 The ring of integers 1; Factorization 2; Units 4; Applications 5; A brief history of numbers 6; References. 7.
1. Preliminaries from Commutative Algebra. . . 10 Basic definitions 10; Noetherian rings 10; Local rings 12; Rings of fractions 12; The Chinese remainder theorem 14; Review of tensor products 15;
Extension of scalars 17; Tensor products of algebras 17; Tensor products of fields 17.
2. Rings of Integers. . . 19 Symmetric polynomials 19; Integral elements 20; Review of bases of A- modules 25; Review of norms and traces 25; Review of bilinear forms 26;
Discriminants 26; Rings of integers are finitely generated 28; Finding the ring of integers 30; Algorithms for finding the ring of integers 33.
3. Dedekind Domains; Factorization. . . 37 Discrete valuation rings 37; Dedekind domains 38; Unique factorization 39; The ideal class group 43; Discrete valuations 46; Integral closures of Dedekind domains 47; Modules over Dedekind domains (sketch). 48; Fac- torization in extensions 49; The primes that ramify 50; Finding factoriza- tions 53; Examples of factorizations 54; Eisenstein extensions 56.
4. The Finiteness of the Class Number. . . 58 Norms of ideals 58; Statement of the main theorem and its consequences 59; Lattices 62; Some calculus 67; Finiteness of the class number 69; Binary quadratic forms 71;
5. The Unit Theorem. . . 7 3 Statement of the theorem 73; Proof that UK is finitely generated 74; Com- putation of the rank 75; S-units 77; Finding fundamental units in real
c1996, 1998, J.S. Milne. You may make one copy of these notes for your own personal use.
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0 J.S. MILNE
quadratic fields 77; Units in cubic fields with negative discriminant 78;
Finding µ(K) 80; Finding a system of fundamental units 80; Regulators 80;
6. Cyclotomic Extensions; Fermat’s Last Theorem.. . . .82 The basic results 82; Class numbers of cyclotomic fields 87; Units in cyclo- tomic fields 87; Fermat’s last theorem 88;
7. Valuations; Local Fields. . . 91 Valuations 91; Nonarchimedean valuations 91; Equivalent valuations 93;
Properties of discrete valuations 95; Complete list of valuations for Q 95;
The primes of a number field 97; Notations 97; Completions 98; Com- pletions in the nonarchimedean case 99; Newton’s lemma 102; Extensions of nonarchimedean valuations 105; Newton’s polygon 107; Locally compact fields 108; Unramified extensions of a local field 109; Totally ramified exten- sions ofK 111; Ramification groups 112; Krasner’s lemma and applications 113; A Brief Introduction to PARI 115.
8. Global Fields. . . 116 Extending valuations 116; The product formula 118; Decomposition groups 119; The Frobenius element 121; Examples 122; Application: the quadratic reciprocity law 123; Computing Galois groups (the hard way) 123; Comput- ing Galois groups (the easy way) 124; Cubic polynomials 126; Chebotarev density theorem 126; Applications of the Chebotarev density theorem 128;
Topics not covered 130; More algorithms 130; The Hasse principle for qua- dratic forms 130; Algebraic function fields 130.
Exercises. . . .132.
It is standard to use Gothic (fraktur) letters for ideals:
a b c m n p q A B C M N P Q a b c m n p q A B C M N P Q I use the following notations:
X ≈Y X and Y are isomorphic;
X ∼=Y X and Y are canonically isomorphic or there is a given or unique isomorphism;
X =df Y X is defined to be Y, or equalsY by definition;
X ⊂Y X is a subset of Y (not necessarily proper).
Introduction 1
Introduction
An algebraic number field is a finite extension of Q; an algebraic number is an element of an algebraic number field. Algebraic number theory studies the arithmetic of algebraic number fields — the ring of integers in the number field, the ideals in the ring of integers, the units, the extent to which the ring of integers fails to be have unique factorization, and so on. One important tool for this is “localization”, in which we complete the number field relative to a metric attached to a prime ideal of the number field. The completed field is called alocal field — its arithmetic is much simpler than that of the number field, and sometimes we can answer questions by first solving them locally, that is, in the local fields.
Anabelian extension of a field is a Galois extension of the field with abelian Galois group. Global class field theory classifies the abelian extensions of a number field K in terms of the arithmetic ofK; local class field theory does the same for local fields.
This course is concerned with algebraic number theory. Its sequel is on class field theory (see my notes CFT).
I now give a quick sketch of what the course will cover. Thefundamental theorem of arithmetic says that integers can be uniquely factored into products of prime powers:
an m = 0 in Zcan be written in the form,
m =upr11· · ·prnn, u=±1, pi prime number,ri >0, and this factorization is essentially unique.
Consider more generally an integral domain A. An element a ∈A is said to be a unit if it has an inverse inA; I writeA× for the multiplicative group of units in A.
An elementpof A is said toprime if it is neither zero nor a unit, and if p|ab⇒p|a or p|b.
If A is a principal ideal domain, then every nonzero nonunit elementa of A can be written in the form,
a=pr11· · ·prnn, pi prime element, ri >0,
and the factorization is unique up to order and replacing each pi with an associate, i.e., with its product with a unit.
Our first task will be to discover to what extent unique factorization holds, or fails to hold, in number fields. Three problems present themselves. First, factorization in a field only makes sense with respect to a subring, and so we must define the “ring of integers” OK in our number fieldK. Secondly, since unique factorization will in general fail, we shall need to find a way of measuring by how much it fails. Finally, since factorization is only considered up to units, in order to fully understand the arithmetic ofK, we need to understand the structure of the group of unitsUK inOK. Resolving these three problems will occupy the first five sections of the course.
The ring of integers. Let K be an algebraic number field. Because K is of finite degree overQ, every elementα of K is a root of a monic polynomial
f(X) =Xn+a1Xn−1 +· · ·+a0, ai ∈Q.
2 Introduction
If α is a root of a monic polynomial with integer coefficients, then α is called an algebraic integer ofK. We shall see that the algebraic integers form a subring OK of K.
The criterion as stated is difficult to apply. We shall see that to prove that α is an algebraic integer, it suffices to check that its minimum polynomial (relative toQ) has integer coefficients.
Consider for example the field K = Q[
d], where d is a square-free integer. The minimum polynomial of α =a+b√
d,b = 0, a, b∈Q, is (X−(a+b√
d))(X−(a−b√
d)) =X2−2aX + (a2−b2d).
Thus α is an algebraic integer if and only if
2a∈Z, a2−b2d∈Z. From this it follows easily that
OK =Z[√
d] ={m+n√
d|m, n∈Z} if d≡2,3 mod 4, and
OK ={m+n1 +√ d
2 |m, n∈Z} if d≡1 mod 4, i.e., OK is the set of sums m +n√
d with m and n either both integers or both half-integers.
Letζdbe a primitivedth root of 1, for example,ζd = exp(2πi/d), and letK =Q[ζd].
Then we shall see that
OK =Z[ζd] = {
miζdi |mi ∈Z}. as one would hope.
Factorization. An element p of an integral domain A is said to be irreducible if it is neither zero nor a unit, and can’t be written as a product of two nonunits. For example, a prime element is (obviously) irreducible. A ringAis aunique factorization domain if every nonzero nonunit element of A can be expressed as a product of irreducible elements in essentially one way. Is OK a unique factorization domain?
No, not in general!
In fact, we shall see that each element of OK can be written as a product of irreducible elements (this is true for all Noetherian rings) — it is the uniqueness that fails.
For example, in Z[√
−5] we have
6 = 2·3 = (1 +√
−5)(1−√
−5).
To see that 2, 3, 1 +√
−5, 1−√
−5 are irreducible, and no two are associates, we use the norm map
Nm :Q[√
−5]→Q, a+b√
−5→a2+ 5b2. Forα ∈ OK, we have
Nm(α) = 1 ⇐⇒ αα¯ = 1 ⇐⇒ α is a unit. (*)
Introduction 3
If 1 +√
−5 =αβ, then Nm(αβ) = Nm(1 +√
−5) = 6. Thus Nm(α) = 1,2,3, or 6. In the first case, α is a unit, the second and third cases don’t occur, and in the fourth case β is a unit. A similar argument shows that 2,3, and 1−√
−5 are irreducible.
Next note that (*) implies that associates have the same norm, and so it remains to show that 1 +√
−5 and 1−√
−5 are not associates, but 1 +√
−5 = (a+b√
−5)(1−√
−5) has no solution with a, b∈Z.
Why does unique factorization fail inOK? The problem is that irreducible elements inOK need not be prime. In the above example, 1 +√
−5 divides 2·3 but it divides neither 2 nor 3. In fact, in an integral domain in which factorizations exist (e.g. a Noetherian ring), factorization is unique if all irreducible elements are prime.
What can we recover? Consider
210 = 6·35 = 10·21.
If we were naive, we might say this shows factorization is not unique inZ; instead, we recognize that there is a unique factorization underlying these two decompositions, namely,
210 = (2·3)(5·7 ) = (2·5)(3·7).
The idea of Kummer and Dedekind was to enlarge the set of “prime numbers” so that, for example, in Z[√
−5] there is a unique factorization, 6 = (p1·p2)(p3·p4) = (p1·p3)(p2·p4),
underlying the above factorization; here thepi are “ideal prime factors”.
How do we define “ideal factors”? Clearly, an ideal factor should be character- ized by the algebraic integers it divides. Moreover divisibility by a should have the following properties:
a|0; a|a,a|b⇒a|a±b; a|a⇒a|ab for all b ∈ OK. If in addition division by ahas the property that
a|ab⇒a|a or a|b,
then we call aa “prime ideal factor”. Since all we know about an ideal factor is the set of elements it divides, we may as well identify it with this set. Thus an ideal factor is a set of elementsa⊂ OK such that
0∈a; a, b∈a⇒a±b ∈a; a∈a⇒ab∈a for all b∈ OK; it is prime if an addition,
ab∈a⇒a ∈aor b∈a.
Many of you will recognize that an ideal factor is what we now call an ideal, and a prime ideal factor is a prime ideal.
There is an obvious notion of the product of two ideals:
ab|c ⇐⇒ c=
aibi, a|ai, b|bi. In other words,
ab={
aibi |ai ∈a, bi ∈b}.
4 Introduction
One see easily that this is again an ideal, and that if
a= (a1, ..., am) andb= (b1, ..., bn) then
a·b = (a1b1, a1b2, ..., aibj, ..., ambn).
With these definitions, one recovers unique factorization: ifa= 0, then there is an essentially unique factorization:
(a) =pr11· · ·prnn with eachpi a prime ideal.
In the above example, (6) = (2,1 +√
−5)(2,1−√
−5)(3,1 +√
−5)(3,1−√
−5).
In fact, I claim
(2,1 +√
−5)(2,1−√
−5) = (2) (3,1 +√
−5)(3,1−√
−5) = (3) (2,1 +√
−5)(3,1 +√
−5) = (1 +√
−5) (2,1−√
−5)(3,1−√
−5) = (1−√
−5).
For example, (2,1 +√
−5)(2,1−√
−5) = (4,2 + 2√
−5,2−2√
−5,6). Since every generator is divisible by 2, (2,1 +√
−5)(2,1−√
−5)⊂(2). Conversely, 2 = 6−4∈(4,2 + 2√
−5,2−2√
−5,6) and so (2,1 +√
−5)(2,1−√
−5) = (2). Moreover, the four ideals (2,1 + √
−5), (2,1−√
−5), (3,1 +√
−5), and (3,1−√
−5) are all prime. For example Z[√
−5]/(3,1−√
−5) =Z/(3), which is an integral domain.
How far is this from what we want, namely, unique factorization of elements? In other words, how many “ideal” elements have we had to add to our “real” elements to get unique factorization. In a certain sense, only a finite number: we shall see that there is a finite set of idealsa1, ...,ah such that every ideal is of the form ai·(a) for some i and some a ∈ OK. Better, we shall construct a group I of “fractional”
ideals in which the principal fractional ideals (a),a∈K×, form a subgroupP of finite index. The index is called the class number hK of K. We shall see that
hK = 1 ⇐⇒ OK is a principal ideal domain ⇐⇒ OK is a unique factorization domain.
Units. Unlike Z, OK can have an infinite number of units. For example, (1 +√ 2) is a unit of infinite order in Z[√
2] : (1 +√
2)(−1 +√
2) = 1; (1 +√
2)m = 1 for m≥1.
In fact Z[√
2]× ={±(1 +√
2)m |m∈Z}, and so Z[√
2]×≈ {±1} × {free abelian group of rank 1}.
Introduction 5
In general, we shall show (unit theorem) that the roots of 1 in K form a finite group µ(K), and that
OK× ≈µ(K)×Zr (as an abelian group);
moreover, we shall find r.
Applications. I hope to give some applications. One motivation for the development of algebraic number theory was the attempt to prove Fermat’s last “theorem”, i.e., that there are no integer solutions to the equation
Xm+Ym =Zm when m≥3, except for the obvious solutions.
When m = 3, this can proved by the method of “infinite descent”, i.e., from one solution, you show that you can construct a smaller solution, which leads to a contradiction1. The proof makes use of the factorization
Y3 =Z3−X3 = (Z−X)(Z2+XZ+X2),
and it was recognized that a stumbling block to proving the theorem for larger m is that no such factorization exists into polynomials with integer coefficients. This led people to look at more general factorizations.
In a very famous incident, the French mathematician Lam´e gave a talk at the Paris Academy in 1847in which he claimed to prove Fermat’s last theorem using the following ideas. Let p >2 be a prime, and suppose x,y, z are nonzero integers such that
xp+yp =zp. Write
xp =zp−yp =
(z−ζiy), 0≤i≤p−1, ζ =e2πi/p.
He then showed how to obtain a smaller solution to the equation, and hence a contra- diction. Liouville immediately questioned a step in Lam´e’s proof in which he assumed that, in order to show that each factor (z −ζiy) is a pth power, it suffices to show that the factors are relatively prime in pairs and their product is a pth power. In fact, Lam´e couldn’t justify his step (Z[ζ] is not always a principal ideal domain), and Fermat’s last theorem remains unproven to the present day2. However, shortly after Lam´e’s embarrassing lecture, Kummer used his results on the arithmetic of the fields Q[ζ] to prove Fermat’s last theorem for all “regular primes”.
Another application is to finding Galois groups. The splitting field of a polynomial f(X) ∈ Q[X] is a Galois extension of Q. In the basic graduate algebra course (see FT), we learn how to compute the Galois group only when the degree is very small (e.g., ≤ 3). By using algebraic number theory one can write down an algorithm to do it for any degree.
1The simplest proof by infinite descent is that showing that √
2 is irrational.
2Written in 1992.
6 Introduction
A brief history of numbers. Prehistory(??-1600). Basic arithmetic was devel- oped in many parts of the world thousands of years ago. For example, 3,500 years ago the Babylonians apparently knew how to construct the solutions to
X2+Y2 =Z2. At least they knew that
(4961)2 + (6480)2 = (8161)2
which could scarcely be found by trial and error. The Chinese remainder theorem was known in China, thousands of years ago. The Greeks knew the fundamental theorem of arithmetic, and, of course, Euclid’s algorithm.
Fermat (1601–1665). Apart from his famous last “theorem”, he invented the method of infinite descent. He also posed the problem of finding integer solutions to the equation,
X2−AY2 = 1, A ∈Z, (*) which is essentially the problem3 of finding the units in Z[√
A]. The English math- ematicians found an algorithm for solving the problem, but neglected to show that the algorithm always works.
Euler(1707–1783). Among many other works, he discovered the quadratic reci- procity law.
Lagrange(1736–1813). He proved that the algorithm for solving (*) always leads to a solution.
Legendre (1752–1833). He proved the “Hasse principle” for quadratic forms in three variables over Q: the quadratic form Q(X, Y, Z) has a nontrivial zero in Q if and only if it has one inRand the congruenceQ≡0 modpn has a nontrivial solution for allp and n.
Gauss(1777–1855). He found many proofs of the quadratic reciprocity law:
p q
q p
= (−1)(p−1)(q−1)/4, p, q odd primes.
He studied the Gaussian integers Z[i] in order to find a quartic reciprocity law. He studied the classification of binary quadratic forms over Z which, as we shall see, is closely related to the problem of finding the class numbers of quadratic fields.
Dirichlet (1805–1859). He proved the following “unit theorem”: letα be a root of a monic irreducible polynomial f(X) with integer coefficients; suppose that f(X) has r real roots and 2s complex roots; then Z[α]× is a finitely generated group of rank r+s−1. He proved a famous analytic formula for the class number.
Kummer(1810–1893). He made a deep study of the arithmetic of cyclotomic fields, motivated by a search for higher reciprocity laws. His general result on Fermat’s last theorem is the most important to date.
Hermite (1822–1901).
Eisenstein (1823–1852).
3The Indian mathematician Bhaskara (12th century) knew general rules for finding solutions to the equation.
Introduction 7
Kronecker (1823–1891). He developed an alternative to Dedekind’s ideals. He also had one of the most beautiful ideas in mathematics, the Kronecker liebster Ju- gendtraum, for generating abelian extensions of number fields.
Riemann (1826–1866). Made the Riemann hypothesis.
Dedekind (1831–1916). He was the first mathematician to formally define fields
— many of the basic theorems on fields in basic graduate algebra courses were proved by him. He also found the correct general definition of the ring of integers in a number field, and he proved that ideals factor uniquely into products of prime ideals.
Moreover, he improved the Dirichlet unit theorem.
Weber (1842–1913). Made important progress in class field theory and the Kro- necker Jugendtraum.
Hensel(1861–1941). He introduced the notion of thep-adic completion of a field.
Hilbert(1862–1943). He wrote a very influential book on algebraic number theory in 1897, which gave the first systematic account of the theory. Some of his famous problems were on number theory, and have also been influential.
Takagi(1875–1960). He made very important advances in class field theory.
Hecke(1887–1947). Introduced Hecke L-series.
Artin (1898–1962). He found the “Artin reciprocity law”, which is the main theorem of class field theory.
Hasse(1898–1979). Proved the Hasse principle for all quadratic forms over number fields.
Weil (1906–1998). Defined the Weil group, which enabled him to give a common generalization of Artin L-series and HeckeL-series.
Chevalley (1909–??). The main statements of class field theory are purely al- gebraic, but all the earlier proofs used analysis. Chevalley gave a purely algebraic proof.
Iwasawa(1917– ). He introduced an important new approach into the study of algebraic number theory which was suggested by the theory of curves over finite fields.
Tate (1925– ). With Artin, he gave a complete cohomological treatment of class field theory. With Lubin he introduced a concrete way of generating abelian exten- sions of local fields.
Langlands(1936– ). “Langlands’s philosophy” is a vast series of conjectures that, among other things, contains a nonabelian class field theory.
References. Books on algebraic number theory.
Artin, E., Theory of Algebraic Numbers, G¨ottingen notes, 1959. Elegant; good exam- ples; but he adopts a valuation approach rather than the ideal-theoretic approach we use in this course.
Artin, E.,Algebraic Numbers and Algebraic Functions, Nelson, 1968. Covers both the number field and function field case.
Borevich, Z. I., and Shafarevich, I. R., Number Theory, Academic Press, 1966.
In addition to basic algebraic number theory, it contains material on diophantine equations.
8 Introduction
Cassels, J.W.S., and Fr¨ohlich, A., Eds., Algebraic Number Theory, Academic Press, 1967. The proceedings of an instructional conference. Many of the articles are excel- lent, for example, those of Serre and Tate on class field theory.
Cassels, J.W.S.,Local fields, London Math. Soc., 1986. Concentrates on local fields, but does also deal with number fields, and it gives some interesting applications.
Cohn, P.M., Algebraic Numbers and Algebraic Functions, Chapman and Hall, 1991.
The valuation approach.
Dedekind, R.,Theory of Algebraic Integers, Cambridge Univ. Press, 1996 (translation of the 1877 French original). Develops the basic theory through the finiteness of the class number in a way that is surprising close to modern approach in, for example, these notes.
Edwards, H., Fermat’s Last Theorem: A Genetic Introduction to Algebraic Number Theory, Springer, 1977. A history of algebraic number theory, concentrating on the efforts to prove Fermat’s last theorem. Edwards is one of the most reliable writers on the history of number theory.
Fr¨ohlich, A., and Taylor, M.J., Algebraic Number Theory, Cambridge Univ. Press, 1991. Lots of good problems.
Goldstein, L.J., Analytic Number Theory, Prentice-Hall, 1971. Similar approach to Lang 1970, but the writing is a bit careless. Sometimes includes more details than Lang, and so it is probably easier to read.
Janusz, G. Algebraic Number Fields, Second Edn, Amer. Math. Soc., 1996. It covers both algebraic number theory and class field theory, which it treats from a lowbrow analytic/algebraic approach. In the past, I sometimes used the first edition as a text for this course and its sequel.
Lang, S. Algebraic Numbers Theory, Addison-Wesley, 1970. Difficult to read unless you already know it, but it does contain an enormous amount of material. Covers alge- braic number theory, and it does class field theory from a highbrow analytic/algebraic approach.
Marcus, D. Number Fields, Springer, 1977. This is a rather pleasant down-to-earth introduction to algebraic number theory.
Narkiewicz, W. Algebraic Numbers, Springer, 1990. Encyclopedic coverage of alge- braic number theory.
Samuel, P.,Algebraic Theory of Numbers, Houghton Mifflin, 1970. A very easy treat- ment, with lots of good examples, but doesn’t go very far.
Serre, J.-P.Corps Locaux, Hermann, 1962 (Translated as Local Fields). A classic. An excellent account of local fields, cohomology of groups, and local class field theory.
The local class field theory is bit dated (Lubin-Tate groups weren’t known when the book was written) but this is the best book for the other two topics.
Weil, A.,Basic Number Theory, Springer, 1967. Very heavy going, but you will learn a lot if you manage to read it (covers algebraic number theory and class field theory).
Weiss, R.,Algebraic Number Theory, McGraw-Hill, 1963. Very detailed; in fact a bit too fussy and pedantic.
9
Weyl, H.,Algebraic Theory of Numbers, Princeton Univ. Press, 1940. One of the first books in English; by one of the great mathematicians of the twentieth century. Id- iosyncratic — Weyl prefers Kronecker to Dedekind, e.g., see the section “Our disbelief in ideals”.
Computational Number Theory.
Cohen, H.,A Course in Computational Number Theory, Springer, 1993.
Lenstra, H., Algorithms in Algebraic Number Theory, Bull. Amer. Math. Soc., 26, 1992, 211–244.
Pohst and Zassenhaus, Algorithmic Algebraic Number Theory, Cambridge Univ.
Press, 1989.
The two books provide algorithms for most of the constructions we make in this course. The first assumes the reader knows number theory, whereas the second de- velops the whole subject algorithmically. Cohen’s book is the more useful as a sup- plement to this course, but wasn’t available when these notes were first written, and so the references are to Pohst and Zassenhaus. While the books are concerned with more-or-less practical algorithms for fields of small degree and small discriminant, Lenstra’s article concentrates on finding “good” general algorithms.
Additional references
Atiyah, M.F., and MacDonald, I.G.,Introduction to Commutative Algebra, Addison- Wesley, 1969. I use this as a reference on commutative algebra.
Washington, L., Introduction to Cyclotomic Fields, 1982. This is the best book on cyclotomic fields.
I will sometimes refer to my other course notes:
GT: Group Theory (594)
FT: Fields and Galois Theory (594) EC: Elliptic Curves (679).
CFT: Class Field Theory (776).
10
1. Preliminaries from Commutative Algebra
Many results that were first proved for rings of integers in number fields are true for more general commutative rings, and it is more natural to prove them in that context.
Basic definitions. All rings will be commutative, and have an identity element (i.e., an element 1 such that 1a =a for all a∈A), and a homomorphism of rings will map the identity element to the identity element.
A ring B together with a homomorphism of rings A→B will be referred to as an A-algebra. We use this terminology mainly whenAis a subring ofB. In this case, for elementsβ1, ..., βmof B,A[β1, ..., βm] denotes the smallest subring ofB containingA and the βi. It consists of all polynomials in the βi with coefficients inA, i.e., elements of the form
ai1...imβ1i1...βmim, ai1...im ∈A.
We also refer toA[β1, ..., βm] as theA-subalgebra ofB generated by theβi, and when B =A[β1, ..., βm] we say that the βi generate B as an A-algebra.
For elementsa1, a2, . . . of A, (a1, a2, . . .) denotes the smallest ideal containing the ai. It consists of finite sums
ciai, ci ∈ A, and it is called the ideal generated by a1, a2, . . .. When a and b are ideals inA, we define
a+b={a+b|a ∈a, b ∈b}.
It is again an ideal inA — in fact, it is the smallest ideal containing both aand b. If a= (a1, ..., am) and b= (b1, ..., bn), thena+b= (a1, ..., am, b1, ..., bn).
Given an ideal a in A, we can form the quotient ring A/a. Let f: A → A/a be the homomorphism a→a+a; thenb →f−1(b) defines a one-to-one correspondence between the ideals of A/a and the ideals of A containinga, and
A/f−1(b)→≈ (A/a)/b.
A proper ideal a of A is prime if ab ∈a ⇒a or b ∈ a. An ideal a is prime if and only if the quotient ring A/a is an integral domain. An elementp of A is said to be prime if (p) is a prime ideal; equivalently, if p|ab⇒p|a or p|b.
A proper ideala inA is maximal if there does not exist an ideal b,ab A.An idealais maximal if and only ifA/a is a field. Every proper idealaof Ais contained in a maximal ideal — if A is Noetherian (see below) this is obvious; otherwise the proof requires Zorn’s lemma. In particular, every nonunit in A is contained in a maximal ideal.
There are the implications: A is a Euclidean domain ⇒ A is a principal ideal domain⇒Ais a unique factorization domain (see any good graduate algebra course).
Noetherian rings.
Lemma 1.1. The following conditions on a ring A are equivalent:
(a) Every ideal in A is finitely generated.
1. Preliminaries from Commutative Algebra 11
(b) Every ascending chain of ideals
a1 ⊂a2 ⊂ · · · ⊂an⊂ · · ·
becomes stationary, i.e., after a certain point an =an+1 =· · ·.
(c) every nonempty set S of ideals in A has a maximal element a, i.e., there is an ideal a in S that is not contained in any other ideal in S.
Proof. (a)⇒(b): Let a= ∪ai; it is an ideal, and hence is finitely generated, say a= (a1, . . . , ar). For some n,an will contain all the ai, and soan =an+1 =· · ·=a. (b)⇒(a): Consider an ideala. If a= (0), thenais generated by the empty set, which is finite. Otherwise there is an elementa1 ∈a, a1 = 0. Ifa= (a1), thenais certainly finitely generated. If not, there is an element a2 ∈ a such that (a1) (a1, a2).
Continuing in this way, we obtain a chain of ideals (a1)(a1, a2)· · · . This process must eventually stop with (a1, . . . , an) =a.
(b)⇒(c): Let a1 ∈ S. If a1 is not a maximal element of S, then there is an a2 ∈ S such that a1 a2. Ifa2 is not maximal, then there is an a3 etc.. From (b) we know that this process will lead to a maximal element after only finitely many steps.
(c)⇒(b): Apply (c) to the set S={a1,a2, . . .}.
A ringAsatisfying the equivalent conditions of the lemma is said to beNoetherian4 A famous theorem of Hilbert states that k[X1, ..., Xn] is Noetherian. In practice, almost all the rings that arise naturally in algebraic number theory or algebraic geom- etry are Noetherian, but not all rings are Noetherian. For example,k[X1, . . . , Xn, . . .] is not Noetherian: X1, . . . , Xnis a minimal set of generators for the ideal (X1, . . . , Xn) in k[X1, . . . , Xn], and X1, . . . , Xn, . . . is a minimal set of generators for the ideal (X1, . . . , Xn, . . .) in k[X1, . . . , Xn, . . .]
Proposition 1.2. Every nonzero nonunit element of a Noetherian integral do- main can be written as a product of irreducible elements.
Proof. We shall need to use that
(a)⊂(b) ⇐⇒ b|a, with equality ⇐⇒ b =a×unit.
The first assertion is obvious. For the second, note that if a = bc and b = ad then a=bc=adc, and so dc= 1. Hence both c and d are units.
Suppose the statement is false, and choose an element a ∈ A which contradicts the statement and is such that (a) is maximal among the ideals generated by such elements (here we use that Ais Noetherian). Sincea can not be written as a product of irreducible elements, it is not itself irreducible, and soa =bcwithbandcnonunits.
Clearly (b)⊃(a), and the ideals can’t be equal for otherwisecwould be a unit. From the maximality of (a), we deduce that b can be written as a product of irreducible elements, and similarly for c. Thus a is a product of irreducible elements, and we have a contradiction.
4After Emmy Noether (1882–1935).
12 1. Preliminaries from Commutative Algebra
Local rings. A ring A is said to local if it has exactly one maximal idealm. In this case, A× =Am(complement of m inA).
Lemma 1.3 (Nakayama’s lemma). Let A be a local Noetherian ring, and let a be a proper ideal in A. Let M be a finitely generated A-module, and define
aM ={
aimi |ai∈a, mi∈M}. (a) If aM =M, then M = 0.
(b) If N is a submodule of M such that N+aM =M, then N =M.
Proof. (a) Suppose M = 0. Among the finite sets of generators for M, choose one {m1, ..., mk} having the fewest elements. From the hypothesis, we know that we can write
mk=a1m1+a2m2+...akmk some ai ∈a. Then
(1−ak)mk=a1m1+a2m2+... +ak−1mk−1.
As 1−akis not inm, it is a unit, and so{m1, ..., mk−1}generates M. This contradicts our choice of{m1, ..., mk}, and so M = 0.
(b) We shall show thata(M/N) = M/N, and then apply the first part of the lemma to deduce that M/N = 0. Consider m+N, m ∈ M. From the assumption, we can write
m =n+
aimi, with ai ∈a, mi ∈M.
Whence
m+N =
aimi+N =
ai(mi+N) (definition of the action of A on M/N), and so m+N ∈a(M/N).
The hypothesis thatM be finitely generated in the lemma is crucial. For example, ifAis a local integral domain with maximal ideal m= 0, thenmM =M for any field M containingA but M = 0.
Rings of fractions. LetA be an integral domain; there is a fieldK ⊃A, called the field of fractions ofA, with the property that everyc∈K can be written in the form c= ab−1, a, b∈ A, b = 0. For example, Q is the field of fractions of Z, and k(X) is the field of fractions of k[X].
Let A be an integral domain with field of fractions K. A subset S of A is said to be multiplicative if 0 ∈/ S, 1 ∈ S, and S is closed under multiplication. If S is a multiplicative subset, then we define
S−1A={a/b ∈K |b ∈S}. It is obviously a subring ofK.
Example 1.4. (a) Lett be a nonzero element ofA; then St=df {1,t,t2,...}
1. Preliminaries from Commutative Algebra 13
is a multiplicative subset ofA, and we (sometimes) writeAt forSt−1A. For example, if d is a nonzero integer,
Zd ={a/dn ∈Q|a∈Z, n≥0}.
It consists of those elements of Qwhose denominator divides some power of d.
(b) If p is a prime ideal, then Sp =Ap is a multiplicative set (if neithera nor b belongs to p, thenab does not belong top). We write Ap for Sp−1A. For example,
Z(p) ={m/n∈Q|n is not divisible by p}.
Proposition 1.5. LetAbe an integral domain, and letS be a multiplicative subset of A. The map
p→S−1p=df {a/s|a∈p, s∈S}
is a bijection from the set of prime ideals in Asuch thatp∩S =∅ to the set of prime ideals in S−1A; the inverse map is q→q∩A.
Proof. It is easy to see that
p a prime ideal disjoint from S ⇒S−1p is a prime ideal,
q a prime ideal in S−1A⇒q∩A is a prime ideal disjoint from S, and so we only have to show that the two maps are inverse, i.e.,
(S−1p)∩A=pand S−1(q∩A) = q.
(S−1p)∩A =p: Clearly (S−1p)∩A⊃p. For the reverse inclusion, leta/s∈(S−1p)∩A, a ∈p, s∈ S. Consider the equation as ·s=a ∈ p. Both a/s and s are in A, and so at least one of a/s or s is in p (because it is prime); but s /∈ p(by assumption), and so a/s∈p.
S−1(q∩A) = q : Clearly S−1(q∩A) ⊂ q because q∩A ⊂ q and q is an ideal in S−1A. For the reverse inclusion, letb∈q. We can write itb =a/switha∈A,s ∈S.
Then a=s·(a/s)∈q∩A, and so a/s= (s·(a/s))/s∈S−1(q∩A).
Example 1.6. (a) If p is a prime ideal in A, then Ap is a local ring (because p contains every prime ideal disjoint from Sp).
(b) We list the prime ideals in some rings:
Z: (2),(3),(5),(7),(11), . . . ,(0);
Z2: (3),(5),(7),(11), . . . ,(0);
Z(2): (2),(0);
Z42: (5),(11),(13), . . . ,(0);
Z/(42): (2),(3),(7).
Note that in general, for t a nonzero element of an integral domain,
{prime ideals of At} ↔ {prime ideals ofA not containing t} {prime ideals of A/(t)} ↔ {prime ideals of A containing t}.
14 1. Preliminaries from Commutative Algebra
The Chinese remainder theorem. Recall the classical form of the theorem: let d1, ..., dn be integers, relatively prime in pairs; then for any integers x1, ..., xn, the equations
x≡xi (mod di)
have a simultaneous solution x∈Z; if x is one solution, then the other solutions are the integers of the form x+md, m∈Z, whered=
di.
We want to translate this in terms of ideals. Integersm and n are relatively prime if and only if (m, n) = Z, i.e., if and only if (m) + (n) = Z. This suggests defining ideals a and b in a ring A to be relatively prime if a+b=A.
Ifm1, ..., mkare integers, then∩(mi) = (m) wheremis the least common multiple of the mi. Thus ∩(mi) ⊃ (
mi) =
(mi). If the mi are relatively prime in pairs, then m=
mi, and so we have ∩(mi) =
(mi). Note that in general, a1·a2· · ·an⊂a1∩a2∩...∩an.
These remarks suggest the following statement.
Theorem 1.7. Let a1, ...,an be ideals in a ring A, relatively prime in pairs. Then for any elements x1, ..., xn of A, the equations
x≡xi (modai)
have a simultaneous solution x∈A; if x is one solution, then the other solutions are the elements of the form x+a with a ∈ ∩ai; moreover, ∩ai =
ai. In other words, the natural maps give an exact sequence
0→a→A→ n
i=1
A/ai →0 with a=∩ai =
ai.
Proof. Suppose first that n = 2. Asa1 +a2 =A, there are elementsai∈ai such that a1+a2 = 1. The elementx=df a1x2+a2x1 has the required property.
For each i we can find elementsai ∈a1 and bi ∈ai such that ai+bi = 1, alli≥2.
The product
i≥2(ai+bi) = 1, and lies in a1+
i≥2ai, and so a1+
i≥2
ai =A.
We can now apply the theorem in the case n = 2 to obtain an element y1 of A such that
y1 ≡1 mod a1, y1 ≡0 mod
i≥2
ai.
These conditions imply
y1 ≡1 mod a1, y1 ≡0 mod aj, all j >1.
Similarly, there exist elementsy2, ..., ynsuch that
yi ≡1 mod ai, yi ≡0 mod aj for j =i.
The elementx=
xiyi now satisfies the requirements.
1. Preliminaries from Commutative Algebra 15
It remains to prove that∩ai =
ai. We have already noted that∩ai ⊃
ai. First suppose that n = 2, and leta1 +a2= 1, as before. For c∈a1 ∩a2, we have
c=a1c+a2c∈a1·a2
which proves thata1∩a2 =a1a2. We complete the proof by induction. This allows us to assume that
i≥2ai =∩i≥2ai. We showed above that a1 and
i≥2ai are relatively prime, and so
a1·(
i≥2
ai) =a1∩(
i≥2
ai) =∩ai.
The theorem extends to A-modules.
Theorem 1.8. Let a1, ...,anbe ideals in A, relatively prime in pairs, and let M be an A-module. There is an exact sequence:
0→aM →M →
i
M/aiM →0 with a=
ai =∩ai.
This has an elementary proof (see Janusz 1996, p. 9), but I prefer to use tensor products, which I now review.
Review of tensor products. LetM,N, andP beA-modules. A mappingf: M× N →P is said to be A-bilinear if
f(m+m, n) = f(m, n) +f(m, n); f(m, n+n) =f(m, n) +f(m, n) f(am, n) = af(m, n) =f(m, an), a∈A, m, m ∈M, n, n ∈N, i.e., if it is linear in each variable. A pair (Q, f) consisting of an A-module Q and an A-bilinear map f : M ×N → Q is called the tensor product of M and N if any other A-bilinear map f : M ×N → P factors uniquely into f = α◦f with α : Q → P A-linear. The tensor product exists, and is unique (up to a unique isomorphism). We denote it by M ⊗A N, and we write (m, n) → m⊗ n for f.
The pair (M ⊗AN,(m, n) → m⊗n) is characterized by each of the following two conditions:
(a) The map M × N → M ⊗A N is A-bilinear, and any other A-bilinear map M×N →P is of the form (m, n)→α(m⊗n) for a uniqueA-linear mapα: M⊗AN → P; thus
BilinA(M×N, P) = HomA(M ⊗AN, P).
(b) As an A-module, M ⊗AN generated by the symbols m⊗n, m ∈ M, n ∈ N, which satisfy the relations
(m+m)⊗n = m⊗n+m⊗n; m⊗(n+n) =m⊗n+m⊗n am⊗n = a(m⊗n) =m⊗an.
Tensor products commute with direct sums: there is a canonical isomorphism (⊕iMi)⊗A(⊕jNj)→ ⊕≈ i,jMi⊗ANj, (
mi)⊗(
nj)→
mi⊗nj.
16 1. Preliminaries from Commutative Algebra
It follows that ifM and N are free A-modules5 with bases (ei) and (fj) respectively, then M ⊗AN is a freeA-module with basis (ei⊗fj). In particular, ifV and W are vector spaces over a field k of dimensions m and n respectively, then V ⊗kW is a vector space overk of dimensionmn.
Letα:M →N and β: M →N be A-linear maps. Then (m, n)→α(m)⊗β(n) : M ×N →M⊗AN
is A-bilinear, and therefore factors through M ×N → M ⊗AN. Thus there is an A-linear map α⊗β :M ⊗AN →M ⊗AN such that
(α⊗β)(m⊗n) =α(m)⊗β(n).
Remark 1.9. Let α: km → km and β: kn → kn be two matrices, regarded as a linear maps. Then α⊗β is a linear map kmn → kmn. Its matrix with respect to the canonical basis is called the Kronecker product of the two matrices. (Kronecker products of matrices pre-date tensor products by about 70 years.)
Lemma 1.10. If α: M →N and β: M →N are surjective, then so also is α⊗β: M ⊗AN →M ⊗AN.
Proof. Recall that M⊗N is generated as an A-module by the elementsm⊗n, m ∈ M, n ∈ N. By assumption m = α(m) for some m ∈ M and n = β(n) for some n∈N, and som⊗n =α(m)⊗β(n) = (α⊗β)(m⊗n). Therefore Im(α⊗β) contains a set of generators for M ⊗AN and so it is equal to it.
One can also show that if
M →M →M →0 is exact, then so also is
M⊗AP →M ⊗AP →M⊗AP →0.
For example, if we tensor the exact sequence
0→a→A →A/a→0 with M, we obtain an exact sequence
a⊗AM →M →(A/a)⊗AM →0 The image of a⊗M in M is
aM =df {
aimi |ai ∈a, mi∈M}, and so we obtain from the exact sequence that
M/aM ∼= (A/aA)⊗AM (1.11).
By way of contrast, if M → N is injective, then M ⊗A P → N ⊗AP need not be injective. For example, take A = Z, and note that (Z →m Z)⊗Z(Z/mZ) equals Z/mZ→m Z/mZ, which is the zero map.
5Let M b e an A-module. Elements e1, . . . , em form a basis for M if every element of M can be expressed uniquely as a linear combination of the ei’s with coefficients in A. Then Am → M, (a1, . . . , am)→
aiei, is an isomorphism ofA-modules, andM is said to be a free A-module of rank m.
1. Preliminaries from Commutative Algebra 17
Proof of Theorem 1.8. Return to the situation of the theorem. When we ten- sor the isomorphism
A/a→∼= A/ai
with M, we get an isomorphism
M/aM ∼= (A/a)⊗AM →∼=
(A/ai)⊗AM ∼=
M/aiM, as required.
Extension of scalars. IfA→B is an A-algebra andM is an A-module, thenB⊗AM has a natural structure of a B-module for which
b(b⊗m) = bb⊗m, b, b ∈B, m∈M.
We say thatB⊗AM is theB-module obtained from M byextension of scalars. The map m → 1⊗m: M → B⊗AM is uniquely determined by the following universal property: it isA-linear, and for anyA-linear mapα: M →N fromMinto aB-module N, there is a unique B-linear map α: B ⊗AM → N such that α(1⊗m) = α(m).
Thus α→α defines an isomorphism
HomA(M, N)→HomB(B⊗AM, N), N a B-module).
For example, A ⊗AM = M. If M is a free A-module with basis e1, . . . , em, then B⊗AM is a free B-module with basis 1⊗e1, . . . ,1⊗em.
Tensor products of algebras. Iff: A→B andg: A→CareA-algebras, thenB⊗AC has a natural structure of an A-algebra: the product structure is determined by the rule
(b⊗c)(b⊗c) =bb⊗cc and the mapA →B ⊗AC isa→f(a)⊗1 = 1⊗g(a).
For example, there is a canonical isomorphism
a⊗f →af :K ⊗kk[X1, . . . , Xm]→K[X1, . . . , Xm] (1.12).
Tensor products of fields. We are now able to computeK⊗kΩ ifKis a finite separable field extension of k and Ω is an arbitrary field extension of k. According to the primitive element theorem (FT, 5.1), K = k[α] for some α ∈ K. Let f(X) be the minimum polynomial of α. By definition this means that the map g(X) → g(α) determines an isomorphism
k[X]/(f(X))→K.
Hence
K⊗kΩ∼= (k[X]/(f(X)))⊗Ω∼= Ω[X]/(f(X))
by (1.11) and (1.12). Because K is separable overk,f(X) has distinct roots. There- fore f(X) factors in Ω[X] into monic irreducible polynomials
f(X) =f1(X)· · ·fr(X)
