Notes on Commutative Algebra 1

Notes on Commutative Algebra 1 Dario Portelli December 12, 2008 1 available at http://www.dmi.units.it/ portelli/

Contents 1 RINGS 3 1.1 Basic definitions............................... 3 1.2 Rings of polynomials............................ 5 1.3 Ideals and quotient rings.......................... 11 1.4 Unique factorization domains....................... 19 1.5 Prime ideals.................................. 28 2 MODULES 43 2.1 Generalities on modules.......................... 43 2.2 Integral elements and integral extensions............... 46 2.3 Direct product and direct sum of modules............... 55 2.4 Free modules................................. 60 2.5 Tensor product of modules......................... 66 2.6 Extending the ring of scalars....................... 74 2.7 Categories and functors........................... 78 3 LOCALIZATION 81 3.1 Rings and modules of fractions...................... 82 3.2 Ideals in a ring of fractions........................ 87 3.3 Local rings................................... 91 3.4 Local to global principle........................ 94 3.5 The prime spectrum of a ring....................... 98 4 FINITENESS CONDITIONS 109 4.1 Noetherian and artinian rings and modules.............. 109 4.2 A characterization of artinian rings................... 115 4.3 Examples of non noetherian rings.................... 119 5 PRIMARY DECOMPOSITION 122 5.1 Primary decomposition in noetherian rings.............. 123 5.2 Primary ideals................................ 124 1

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 2 5.3 Primary decomposition in general.................... 126 6 DIMENSION THEORY 134 6.1 Height of ideals and Krull dimension of rings............. 135 6.2 Krull s Principal Ideal Theorem..................... 136 6.3 The Krull dimension of rings of polynomials............. 143 6.4 Dimension theory and integral extensions............... 146 6.5 Dimension of finitely generated algebras over a field........ 149 6.6 The trascendence degree of a field extension............. 151 6.7 Regular local rings.............................. 157 7 A PRIMER IN HOMOLOGICAL ALGEBRA 169 7.1 A first problem to start........................... 169 7.1.1 Complexes............................... 172 7.1.2 Getting new complexes from old ones by applying suitable functors................................ 175 7.1.3 Free resolutions of modules.................... 178 7.1.4 Finitely presented modules.................... 180 7.2 Another problem............................... 180 7.3 Exploration.................................. 184 7.4 Basic homological algebra......................... 186 7.5 Projective and flat modules........................ 199 7.6 The Koszul complex............................. 208 7.7 Fine properties of regular local rings.................. 218 BIBLIOGRAPHY 202

Chapter 1 RINGS The reader is adviced to just quickly browse the first two sections of this chapter. 1.1 Basic definitions A ring is a set R with two binary operations, addition and multipli cation such that R is an abelian group with respect to addition; multiplication is associative; multiplication is distributive over addition, i.e. for every x, y, z R the following equalities hold x(y + z) = xy + xz (y + z)x = yx + zx We shall consider only rings which are commutative, i.e. xy = yx, for every x, y R, and have an identity element (also called unit element), namely there is an element 1 R such that x1 = 1x = x for every x R. Neither of these last properties is a consequence of the other. In fact, given an integer n 2, let A denote the set of the square n n matrices with entries in R. Addition in A is the usual addition of matrices, and multiplication is to be intended the row by column procedure. It is well known that with these operations A is a ring, isomorphic to 3

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 4 the ring of linear endomorphisms of the real vector space R n. It has an identity element, the unit matrix, but it is not commutative ( ) ( ) ( ) ( ) 1 2 3 0 3 0 1 2 (here n = 2) 0 0 4 0 4 0 0 0 On the other hand, let R denote the set of even integers with the usual addition and multiplication. Then R is a commutative ring, but it has no identity element. In these notes the word ring shall always mean a commutative ring with an identity element. A ring homomorphism between two rings R and S is a map f : R S such that f(x + y) = f(x) + f(y) for every x, y R ; f(xy) = f(x)f(y) for every x, y R ; f(1 R ) = 1 S. If f : R S and g : S T are ring homomorphisms, then so is their composition g f : R T. A subring of a ring R is a subset S R such that for every x, y S we have x y, xy S and 1 S. In other words, the inclusion S R is a homomorphism of rings. Given a ring R, we will call an element r R invertible if there exists a s R such that rs = 1. We set R := { r R r is invertible in R } Check that R is a group with respect to the multiplication in R. If f : R S is a ring homomorphism, check that f(r ) S, and f : R S is a group map. Example 1. Product of rings Starting from two rings A and B we can construct a new ring as follows. The underlying set is is the cartesian product A B, and the operations on A B are defined by (a, b) + (a, b ) := (a + a, b + b ) (a, b) (a, b ) := (a a, b b ) It is easy to show that A B with these operations is a ring, where the zero and identity elements are respectively 0 A B = (0 A, 0 B ) 1 A B = (1 A, 1 B )

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 5 It is also immediate to check that the maps p : A B A p : (a, b) a q : A B B q : (a, b) b are surjective homomorphisms of rings. The ring A B is called the product of the rings A and B. It is clear that the above construction can be applied to any family of rings, finite or not. Exercise 2. As usual, we will denote by Z the ring of integers. Compute the group (Z Z) of invertible elements of the ring Z Z. Then compute the group (A B), where A, B are any rings. The ring A B is the simplest ring I know where the following happens. Let a A and b B be non zero elements. Then (a, 0), (0, b) are non zero elements of A B, but (a, 0) (0, b) = (0, 0) = 0 A B! Definition 3. A non zero element r of a ring R is called a zero divisor if there is a s R, with s 0, such that rs = 0. If r 0 is not a zero divisor of R, then r is called a regular element of R. A ring without elements that are zero divisors is called an integral domain. We will denote by Z (R) the set of zero divisors of the ring R, with 0 adjointed. It is easily checked that for x R to be regular is equivalent that the map α : R R defined by setting α(r) := xr is injective. It is also easily checked that every invertible element of a ring R is a regular element, i.e. it is not a zero divisor for R. Hence Corollary 4. Every field is an integral domain. For instance, the ring of integers Z is an integral domain. Every subring of an integral domain (in particular: of a field) is an integral domain. 1.2 Rings of polynomials We assume that the reader has already meet polynomials, and has a working knowledge of the operation with them. However, due to the importance of these rings for our purposes, we will recall in this section the basic facts. Let R be a fixed ring. A polynomial in the indeterminate X with coefficients in R is any formal expressions like (1.1) f = a n X n + a n 1 X n 1 +... + a 1 X + a 0 ( n 0, an integer )

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 6 where the a i s are in R and are called the coefficients of f. If a n 0, then a n and n are called respectively the leading coefficient and the degree of f ; usually we will denote the degree of f by f. In particular, if a n = 1 then f is called a monic polynomial. But this is just terminology, so let us consider a more substantial point. We said above that a polynomial is a formal expressions. Formal because of the presence of X and its powers, whereas it is perfectly clear what the coefficients are, they are elements of the ring R. The basic idea to make X something concrete is that all the information carried by a polynomial is encoded in the sequence of its coefficients. This is usually expressed by saying that two polynomials are equal if they have equal coefficients. To exploit this idea we have simply to identify the polynomial f as written in (1.1), with the following sequence of elements of R, indexed by the natural numbers (1.2) (... 0, 0, 0, a n, a n 1,...,a 1, a 0 ) It is then convenient to define a polynomial as being a sequence of elements of R as in (1.2), i.e. for which there is an integer n N such that a m = 0 for every m > n. The set of all polynomials will be denoted by R [ X ]. Let us consider now the algebraic operations in R [ X ]. The sum of two polynomials is defined componentwise (..., a n, a n 1,..., a 2, a 1, a 0 ) + (..., b m, b m 1,..., b 2, b 1, b 0 ) := (......, a 2 + b 2, a 1 + b 1, a 0 + b 0 ) In a similar way we can define the multiplication of a polynomial by an element r of R r (..., a n, a n 1,..., a 1, a 0 ) := (..., ra n, ra n 1,..., ra 1, ra 0 ) It is easily checked that the formal properties of R [ X ] with respect to these two operations are exactly those of a vector space over a field; the only difference is that here in general we have just a ring, instead of a field. This new perspective suggests that we can write any polynomial also in the following way (..., a n, a n 1,..., a 1, a 0 ) = = (..., 0, a n, 0, 0,..., 0, 0 ) +... + (..., 0, 0, 0, a 1, 0 ) + (..., 0, 0, 0, a 0 ) = = a n (..., 0, 1, 0, 0,..., 0, 0 )+ +a n 1 (..., 0, 0, 1, 0,..., 0, 0 ) +...... + a 1 (..., 0, 0,..., 0, 1, 0 )+ +a 0 (..., 0, 0,..., 0, 0, 1 )

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 7 ( we will formalize all this in the next chapter by saying that R [X] is a free R-module). The comparison of (1.1) with the above computation shows that the powers of X play the same role as the sequences with only one entry different from zero, and equal to 1. This suggest to define X := (..., 0, 0,..., 0, 1, 0) By the same reasons, we can set X 2 := (..., 0, 0,..., 0, 1, 0, 0), and so on. Of course, this will be satisfactory only if the polynomial X 2 written above is the square of the polynomial X previously defined for a multiplication on R [ X ] that still remains to be defined. To do this we borrow from our highschool experience with polynomials the idea that a multiplying a polynomial by X simply means to shift the sequence of coefficients one place to the left, and define (1.3) X(... 0, 0, 0, a n, a n 1,..., a 1, a 0 ) = (... 0, 0, a n, a n 1,...,a 1, a 0, 0) It is easily seen that (1.3) is all we need to define a distributive product on the set R [X]. It turns out that this product is also commutative and associative, then R [X] becomes a ring. So nothing formal is left, everything is quite concrete. The identity element of R [X] is the polynomial 1 R [X] = (..., 0, 0,..., 0, 0, 1) of degree zero. It is also easily seen that R is a subring of R [X]. The first important property of the ring of polynomials we will need is the Euclidean division algorithm. Let f, g R [X] and assume that the leading coefficient of g is an invertible element of R (note that this implies g 0). Then there exist two polynomials q, r R [X] such that f = q g + r where r = 0 or 0 deg(r) < deg(g) Moreover, the polynomials q, r are unique. I hope that the reader has been sufficiently trained at high school in the task to divide one polynomial by another to readily convince him/herself that the well known algorithm works over an arbitrary ring of coefficients because of the assumption that the leading coefficient of g is an invertible element of R. This for the existence part of the statement. The uniqueness is left as an Exercise, as well as the construction of an example which shows that the assumption that the leading coefficient of g is in R is essential for the Euclidean division algorithm.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 8 Up to now we have only considered polynomials in one indeterminate. But already at highschool we were confronted with things like X 2 4Y 6 and so on. How can we formalize polynomials in two or more indeterminates? Let us consider first the case of two indeterminates X and Y. It is rather clear that we have to replace the usual sequences of elements of the ring of coefficients R, with double sequences, i.e. sequences indexed over N 2, like.............................. (1.4) F(X, Y ) =...... a 22 a 21 a 20...... a 12 a 11 a 10... a 03 a 02 a 01 a 00 where e.g. a 21 is meant to be the coefficient of XY 2. It is undestood that for every i, j sufficiently large we have a ij = 0. Starting from here we can construct the ring of polynomials in two variables R [X, Y ] by following the same path used above to construct R [X]. For instance, we will have................................. 0 0 0 0... 0 0 0 0 X =... 0 0 0 0 Y =... 0 0 0 0... 0 0 0 1... 0 0 0 0... 0 0 0 0... 0 0 1 0 On the other hand, this way is impracticable to construct rings of polynomials in three or more indeterminates. An elegant solution of this problem is as follows. Let us consider again the matrix (1.4). The elements in the third column (from the right) of this matrix are the coefficients of the various monomials X n Y 2. The factor Y 2 appears in all these terms, so that the column can be identified with Y 2 times a polynomial A 2 in the only indeterminate X. All the information on the polynomial F(X, Y ) given by the third column of (1.4) is encoded into A 2. Every column of (1.4) can be treated similarly. Therefore, a convenient way to rewrite (1.4) is (1.5) (... 0, 0, 0, A m, A m 1,...,A 1, A 0 ), where the A h are elements of R [X]. Note that the convention we made above about (1.4), namely that for every i, j sufficiently large a ij = 0, translates nicely into A h = 0 for every h >> 0. Hence (1.5) represents a polynomial in the indeterminate Y, with coefficients in the ring R [X]. Then R [X, Y ] = (R [X]) [Y ] This idea works for any number of indeterminates R [X, Y, Z] = (R [X, Y ]) [Z],...

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 9 hence it allows us to define inductively the ring of polynomials in n indeterminates by the formula (1.6) R [X 1,...,X n ] := ( R [X 1,...,X n 1 ] ) [X n ] The other relevant property of the ring of polynomials we will need is Theorem 5. Universal property of the ring of polynomials. Let R be a ring, ϕ : R S a homomorphism of rings, and ξ 1,...,ξ n arbitrary elements of S. Then there exists one and only one homomorphism of rings Φ : R[X 1,...,X n ] S such that (1.7) Φ R = ϕ and Φ(X i ) = ξ i for every i = 1,...,n. Proof. Because of (1.6), it is sufficient to prove the case n = 1. Assume that there is a homomorphism Φ : R [X] S satisfiyng (1.7) (with n = 1). Then, for every polynomial f R [X] we have Φ(f) = Φ(a n X n + a n 1 X n 1 +... + a 1 X + a 0 ) = = Φ(a n )(Φ(X)) n + Φ(a n 1 )(Φ(X)) n 1 +... + Φ(a 1 )Φ(X) + Φ(a 0 ) = = ϕ(a n )ξ n + ϕ(a n 1 )ξ n 1 +... + ϕ(a 1 )ξ + ϕ(a 0 ) where the last equality is a consequence of (1.7). This shows that, if a homomorphism Φ : R [ X ] S satisfiyng (1.7) exists, then it is unique. On the other hand, if we set for any f = a n X n + a n 1 X n 1 +... + a 1 X + a 0 Φ(f) := ϕ(a n )ξ n + ϕ(a n 1 )ξ n 1 +... + ϕ(a 1 )ξ + ϕ(a 0 ) then it is easily seen that such a map Φ : R [X] S satisfies (1.7) and preserves the addition of polynomials. To prove that Φ preserves also the multiplication, let f = i a ix i and g = j b jx j be two polynomials. Then ( ) Φ(fg) = Φ a i b j X i+j = ϕ(a i )ϕ(b j ) ξ i+j = i,j i,j ( )( ) ϕ(a i ) ξ i ϕ(b j ) ξ j = Φ(f) Φ(g) i Hence, also the existence part of our statement is proved. j The name Universal property of the ring of polynomials of this theorem deserves some explanation. First of all we prove the

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 10 Proposition 6. Assume that R is a subring of a ring P and that there exist elements X 1,..., X n of P such that the following property holds true. UP Let ϕ : R S be a ring map, and let ξ 1,...,ξ n be arbitrary elements of S. Then there exists one and only one ring map Ψ : P S such that (1.8) Ψ R = ϕ and Ψ(X i ) = ξ i for every i = 1,...,n From these assumptions it follows the existence of a canonical isomorphism of rings ι : R[X 1,..., X n ] P such that (1.9) ι R = id R and ι(x i ) = X i for every i = 1,..., n Proof. By the Universal property of the ring of polynomials the inclusion R P extends to a ring map Φ : R[X 1,...,X n ] P such that Φ R is the inclusion R P and Φ(X i ) = X i for every i = 1,...,n Similarly, if we apply UP with the inclusion R R[X 1,...,X n ] playing the role of ϕ, we will get a ring map Ψ : P R[X 1,..., X n ] such that Ψ R is R R[X 1,...,X n ] and Ψ(X i ) = X i for any i = 1,..., n Then, from the uniqueness parts of the universal property of R[X 1,...,X n ] and of UP, it follows at once that the compositions Ψ Φ and Φ Ψ are necessarily the identity maps of R[X 1,...,X n ] and P respectively. Then Φ has the properties required for ι. The meaning of the proposition is that the universal property UP determines uniquely the ring R[X 1,...,X n ], up to canonical isomorphisms. Therefore, we can take UP as the definition of the ring of polynomials with coefficients in R, in n indeterminates (up to canonical isomorphisms). There are several problems that can be formalized by a suitable universal property, as we will see. Usually, the proof of the existence of a solution for them requires an adhoc construction in each case. But the proof of the uniqueness (again, up to canonical isomorphisms) for such a solution is completely standard. As an Exercise the student will try to prove the uniqueness for any universal property we will state in the sequel. Proposition 7. If R is an integral domain, then R[X] is also an integral domain, and conversely. Proof. Consider two non zero polynomials f = a n X n + a n 1 X n 1 +... + a 1 X + a 0 g = b m X m + b m 1 X m 1 +... + b 1 X + b 0 with a n 0 and b m 0. Then f g = a n b m X n+m + terms of lower degree where a n b m 0 because R is an integral domain. Hence fg 0. The converse is trivial because R is a subring of R [X]. and

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 11 1.3 Ideals and quotient rings Definition 8. A (non empty) subset I of a ring R is called an ideal if the following properties hold for every x, y I, we have x y I ; for every x I and for every y R, we have also xy I. For every ring R, the subsets {0} and R itself are ideals of R, the so called trivial ideals. Exercise 9. Let I be an ideal of the ring R. Check that, if I contains an invertible element of R, then I = R. Then the only ideals of a field are {0} and the field itself. Prove that also the converse is true, namely if the only ideals of a ring R are the trivial ones, then R is a field. Example 10. Let R be a ring and let r R. Then ( r ) := { sr s R } is an ideal of R ; ideals of this kind are called principal. It shoud be well known to the reader that every ideal in Z or in K[X], where K is a field, is principal. We will see that this is no more true for general rings. Exercise 11. Let r be a fixed element in a ring R. Is it possible that the principal ideal ( rx 1 ) of R [X] equals R [X]? Let f : R S be a homomorphism of rings. We define the kernel of f as Ker(f) := { r R f(r) = 0 } = f 1 (0) Then Ker(f) is an ideal of R. Check that, more generally, for every ideal J of S we have that f 1 (J) is an ideal of R, containing Ker(f). When the homomorphism is clear we will use J c (to be read the contraction of J or J contracted ) instead of f 1 (J). We will review now some standard way of constructing new ideals. The starting point is the straightforward Exercise 12. Let { I α } α A be an arbitrary family of ideals in a given ring R. Show that α A I α is still an ideal in R.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 12 Example 13. Let U R be any subset, and consider the family F of all the ideals I R containing U. Note that at least the trivial ideal R belongs to this family. By the previous exercise, the intersection of all the ideals of F is a new ideal J of R. It is called the ideal generated by U, and it is the smallest ideal of R containing U. We will also say that the elements of U are a system of generators for J. If U is finite set, say U = { x 1, x 2,..., x n }, then the notation commonly used for the ideal generate by the x u s is ( x 1, x 2,..., x n ). Such an ideal will be called finitely generated. We will learn their importance in the next chapters. In fact, we will work almost exclusively with noetherian rings, i.e. rings such that every ideal is finitely generated. Exercise 14. Let U be as above. Prove that the ideal generated by U is { a1 x 1 +... + a m x m m N, xh U, a h R } i.e. the set of all finite linear combinations of elements of U, with coefficients in R. Let us consider now some important particular cases of the above construction. For given ideals I, J in the ring R, let U denote the set of all the elements of R of the kind i + j, where i I, and j J. The ideal generated by U will be denoted by I + J, and called the sum of I and J (check that, actually, U is already an ideal). Similarly, the product of the ideals I and J will be the ideal IJ generated by the set V of all the elements of the kind ij, where i I, and j J. Note that IJ I J, but in general the two ideals are different. Both the sum and the product of ideals can be performed on any finite set of ideals; formal definitions for them can be given either by induction on the number of ideals involved, or by suitably redefine the sets U and V. A particular case of product ideal is when the ideals we multiply are all equal to the same ideal I. If we consider r > 0 copies of I, their product ideal will be denoted by I r. We will set I 0 := R. Exercise 15. For R = Z, compute (6) + (16), (6) (16) and (4) 3. Find out an example of two ideals (m) and (n) of Z such that (m)(n) (m) (n). Exercise 16. Let r s be two positive integers. Check that I r I s, for every ideal I of R. If I = (x) is a principal ideal, check that I r = (x r ). Compute r 0 (x)r, where R = Z, and x Z is > 1 (what happens if x = 1?). An important instance where IJ = I J is when the ideals I, J are coprime, i.e. when I + J = R. In fact, in this case there are elements i I and j J such that i + j = 1 Multiplying the above relation by an arbitrary r I J we get r = r i + r j

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 13 and both terms in the R.H.S. belong to IJ because of r I J. For the easy of reference we summarize Proposition 17. If I, J are coprime ideals in a ring R, then IJ = I J. Let f : R S be a homomorphism of rings, and let I R be an ideal. We will denote by I e the ideal of S generated by f(i) (to be read the extension of I or I extended ; this notation will be used only when the homomorphism f is clear from the context.). Exercise 18. If I, J are ideals in R, prove that I e J e = ( I J ) e. Here is a very concrete example of an ideal generated by some element of R. Example 19. Let K be a fixed ring. Consider the rings of polynomials K[X, Y, Z] and S := K[T]. By Theorem 5 we get a homomorphism of rings Φ : K[X, Y, Z] K[T] such that the restriction of Φ to K is the inclusion K K[T], and Φ(X) = T, Φ(Y ) = T 2 and Φ(Z) = T 3. Then the map Φ is Φ : F(X, Y, Z) F( T, T 2, T 3 ) for every F K[X, Y, Z] We want to compute Ker(Φ). Since Φ(X 2 ) = T 2 = Φ(Y ), we have Φ(X 2 Y ) = 0, hence X 2 Y Ker(Φ). In the same way we get also X 3 Z Ker(Φ), Y 3 Z 2 Ker(Φ). But note that and also Y 2 (X 2 Y ) + Z(X 3 Z) = Y 3 Z 2 + X 2 (XZ Y 2 ) XZ Y 2 = X(X 3 Z) + (X 2 + Y )(X 2 Y ) Then Y 3 Z 2 is a linear combinations of X 2 Y and X 3 Z. More trials does not produce new interesting elements in Ker(Φ), i.e. elements which are not linear combinations of X 2 Y and X 3 Z. Therefore, a natural guess is that the inclusion (1.10) ( X 2 Y, X 3 Z ) Ker(Φ) is actually an equality. In fact, take an arbitrary f Ker(Φ). If we consider X 3 Z as a polynomial in the indeterminate Z, with coefficients in K[X, Y ] by (1.6), then by Euclidean division algorithm we have f = q(x 3 Z) + r where r = 0, or Z (r) < 1 Here Z denotes the degree w.r.t. the indeterminate Z. Note that the conditions on r can be summarized as r K[X, Y ]. Moreover, from the first relation above we get r Ker(Φ). Now we divide r by X 2 Y in the ring K[X, Y ] r = h(x 2 Y ) + s where s = 0, or Y (s) < 1.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 14 Again, the first of the above relations implies that s Ker(Φ), whereas the second means s K[X]. But it is clear from the definition of Φ that the restriction of this map to K[X] is an isomorphism K[X] K[T], hence s = 0 and we conclude f = q(x 3 Z) + h(x 2 Y ) So, also the inclusion Ker(Φ) ( X 2 Y, X 3 Z ) holds, and in (1.10) we have equality. Finally, we have computed Ker(Φ) in the sense that we have find out an explicit system of generators for such ideal. Exercise 20. Compute Ker(Φ), where Φ : K[X, Y, Z, W] K[s, t] is the homomorphism between rings of polynomials such that Φ restricted to K is the inclusion K K[s, t], and Φ : X s 3 Φ : Y s 2 t Φ : Z st 2 Φ : W t 3 Let us briefly recall the definition of quotient ring and its main properties. Let R be a ring and let I R be an ideal. To grasp the idea of quotient ring R/I (or quotient group, vector space, module,...) you have just to put = 0 everything is in I. We can transform this slogan into something concrete by remarking that, if every element of I must be considered as zero, then two elements x, y of R whose difference x y is in I must be considered as equal. This will produce a new ring, namely the quotient ring. Let us formalize all this. We define the following equivalence relation in R x y if and only if x y I x, y R Usual notations for the equivalence class of r R are r + I or [r] I or r The important feature of this equivalence relation is that it is compatible with the operations on R, namely if x x and y y, then x + y x + y and xy x y Therefore we can define canonically on the quotient set R/ =: R/I two operations, and R/I becomes a ring w.r.t. these operations, the quotient ring. Note that R/I consists of the only element 0 if and only if I = R. To avoid trivialities, we will always assume that I R, when the quotient ring has to be considered. Now we have a new ring, R/I, which hopefully is better suited to study the situation we are interested to. We have to translate the informations encoded into the old ring

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 15 R into the corresponding ones in the ring R/I. To do this, we need a homomorphism of rings R R/I. We are lucky! In fact, the canonical map π : R R/I given by r [r] = r is a surjective homomorphism of rings, the canonical epimorphism. Finally, we have (1.11) Ker(π) = I. This is the exact meaning of put = 0 everything is in I! Namely, when you change the ring from R to R/I by using the map π, then everything in I goes to zero. Example 21. We want to understand the quotient ring R[X]/(X). Let f = a n X n + a n 1 X n 1 +...+a 1 X +a 0 be a polynomial. Since the kernel of the canonical epimorphism π : R[X] R[X]/(X) is the principal ideal (X), we have simply π(f) = a 0! This amounts to putting = 0 every term of f which is in (X). Hence R[X]/(X) R. Similarly, we have R[X, Y ]/(X) R[Y ], and so on. Needless to say, things are almost never so easy. So to deal with quotient rings we need some theorem. Theorem 22. Let f : R S be a homomorphism of rings. Then the map f : R/Ker(f) S given by f([ r ]) = f(r) for every [ r ] R/Ker(f), is well defined, and it is an injective homomorphism. Moreover, we have f = f π where π : R R/Ker(f) is the canonical epimorphism. Example 23. Let I 1,..., I n be ideals in a ring R. The map ϕ : R R/I 1... R/I n given by ϕ : r ( r + I 1,..., r + I n ) is easily seen to be a ring homomorphism. We have Ker(ϕ) = I 1... I n Finally, you can prove as an exercise that ϕ is surjective if and only if I i + I j = R for every i j, namely if the ideals I i are pairwise coprime. This result is often called the Chinese Remainder Theorem. Example 24. Let I be an ideal of the ring R. Check that IR[X] := { a n X n + a n 1 X n 1 +... + a 1 X + a 0 R[X] a h I, for every h = 0,...,n} is an ideal of R[X].

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 16 We want to compute the quotient ring R[X]/IR[X]. First of all, let us try to guess the correct answer. Two polynomials a m X m + a m 1 X m 1 +... + a 1 X + a 0 and b n X n + b n 1 X n 1 +... + b 1 X + b 0 will have the same image into R[X]/IR[X] if and only if their difference is in IR[X], hence if and only if a h b h I for every h 0. This suggests that possibly the ring (R/I) [X] will be isomorphic to R[X]/IR[X]. By the universal property of the ring of polynomials (Theorem 5), we can extend the canonical epimorphism π : R R/I to a homomorphism Π : R[X] (R/I) [X] such that Π(X) = X. Now, it is clear that Π is surjective, and it is also clear that Ker(Π) = IR[X]. We conclude by Theorem 22 that (1.12) R[X] / I R[X] ( R / I ) [X] Example 25. Let us consider again the homomorphism Φ : K[X, Y, Z] K[T] of Example 19. It is clear that it is surjective (already its restriction to the subring K [X] is surjective). Then by Theorem 22 we have the isomorphism of rings Φ : K[X, Y, Z]/ ( X 2 Y, X 3 Z ) K [T] Maybe it is useful to get an intuitive idea of the quotient ring in the L.H.S. above; let us denote it by A. The residue class of the polynomial X 2 Y in A is zero, hence Ȳ = X 2 holds true in A. Therefore, when we compute the residue class of any polynomial F K [X, Y, Z] we have to replace everywhere Y by X 2 (and, of course, X by X ). Similarly, we have to replace everywhere Z by X 3. Then the residue class of F in A is F = F( X, X2, X3 ) a polynomial expression with coefficients in K, in the indeterminate X. If we identify X with T we get the isomorphism Φ showed above. Remark 26. From Theorem 22 it follows that, for every homomorphism f : R S, we have the isomorphism f : R/Ker(f) f(r) So in this case we have the concrete ring f(r) which can be used instead of the abstract ring R/Ker(f). The following theorem is the universal property of the quotient ring. Theorem 27. Let f : R S be a homomorphism of rings, and let I R be an ideal. If I Ker(f), then there is one and only one homomorphism g : R/I S such that (as usual, π denotes the canonical map R R/I ) We will say that f factorizes through π. f = g π

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 17 The last theorem on quotient rings is an essential part of the dictionary which allows the translation of properties of R to properties of R/I, and conversely. Here we are concerned with the ideals of the two rings. Theorem 28. Let R be a ring, and let I R be a fixed ideal. If π : R R/I denotes the canonical epimophism, then there is a ono-to-one onto, order preserving correspondence between the set of ideals J of R containing I, and the set of ideals J of R/I given by J J = π 1 ( J). The inverse correspondence is given by J π(j) =: J/I. A natural question arises from this last statement. Let R, I, J be as above. Then J/I is an ideal of R/I. Is there a reasonable way to understand the (horrible) quotient ring R/I? J/I Yes! Just compose the canonical epimorphisms R R/I R/I to get the epimorphism f : R (R/I)/(J/I). It is easily seen that Ker(f) = J (do it as an Exercise!). J/I By Theorem 22 we conclude that f is a (canonical) isomorphism (1.13) R J R/I J/I Exercise 29. With the same notations of Theorem 28, check that if H R is any ideal, then π(h) is still an ideal R/I. Moreover, check that π 1 ( π(h) ) = H + I. Example 30. Assume R = Z. We want to study the ideals of the quotient ring Z/(6) =: Z 6. By Theorem 28 they are of the form (m)/(6), where (6) (m). This last condition is equivalent to require that m divides 6 ; we will use for this the notation m 6. So there are four possibilities for m, namely m = 1, 2, 3, 6. The first and the last cases lead respectively to the ideals Z 6 and 0, the trivial ideals. The other ideals are ( 2) = { 2, 4, 0 } and ( 3) = { 3, 0 } What about 5? We have 5 2 = 25 = 4 6 + 1, hence 5 2 = 1, because 4 6 (6). Then we have computed (Z 6 ) = { 1, 5 }, the cyclic group with two elements. Note that 2 and 3 are zero divisors of Z 6 because 2 3 = 6 = 0, but 2 + 3 = 5 is not a zero divisor. Hence the sum of two zero divisors of a ring is not a zero divisor, in general. In particular, the set of zero divisors is not an ideal. We compute now the quotient ring Z 6 /( 3) = Z/(6) (3)/(6) Z (3) = Z 3

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 18 a field. The isomorphism here is that of (1.13). The ideals in the rings Z m can be studied in a similar way. Before ending this example, we want to consider briefly Z 8. You can compute the group (Z 8 ) as an exercise. Moreover, the element 2 is a zero divisor in Z 8 of a peculiar type, because 2 3 = 0. Definition 31. An element r of a ring R is called nilpotent if there is an integer m > 0 such that r m = 0. We remarked in the last example that the zero divisors in a ring do not form an ideal, in general. On the other hand, nilpotents are peculiar zero divisors, so maybe it is not so silly to ask if the set of nilpotent elements of a given ring is an ideal. Proposition 32. Let R be a ring. Then the set N (R) := { r R r m = 0 for some integer m > 0 } is an ideal of R, called the nilradical of R. Proof. It is clear that for r N (R) and for any a R, then a r N (R). The crucial point is to show that for r, s N (R) we have r s N (R). Let m, n be two positive integers such that r m = 0 and s n = 0. We are looking for a positive integer N such that (r s) N = 0. By the binomial coefficient theorem (which holds true in any ring), (r s) N will be the sum of terms of the form ( ) N ± r u s v with u + v = N u So we have just to choose N large enough so that either r u = 0 or s v = 0. The smallest integer that works in general is N = m + n 1. Definition 33. Given an ideal I R, set I := π 1 ( N ( R/I ) ) By Proposition 32 it is clear that I is an ideal of R, called the radical of I. It is also clear that I = { r R r m I for some integer m > 0 } Definition 34. A ring R is called reduced if its nilradical is zero N (R) = 0. Exercise 35. Check that Z 6 is reduced. For which integers m the ring Z m is reduced? Exercise 36. Consider the ideal I = ( 243 ) in the ring Z. Compute I. The same for I = ( 1944 ).

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 19 Exercise 37. Prove that for any ring R, the quotient ring R/N (R) is reduced. Exercise 38. Prove that if x N (R), then 1 + x R. More generally, prove that for every e R and x N (R), we have e + x R. Exercise 39. Let R be a ring, and let f = a n X n + a n 1 X n 1 +... + a 1 X + a 0 be a polynomial with coefficients in R. Prove that f (R [X]) if and only if a 0 is invertible in R, and a 1,...,a n are nilpotent (hint: if g = b m X m + b m 1 X m 1 +... + b 0 is the inverse of f, prove by induction on r that a r+1 n b m r = 0. Hence show that a n is nilpotent, and then use Exercise 38). In particular, if N (R) = 0 (e.g. if R is an integral domain) then (1.14) ( R [X] ) = R f is nilpotent if and only if a 0, a 1...,a n are nilpotent. f is a zero divisor if and only if ther is a r R, with r 0, such that rf = 0 (hint: let g = b m X m + b m 1 X m 1 +... + b 0 of least degree m such that fg = 0. Then a n b m = 0, hence a n g = 0 because a n g annihilates f and has degree < m. Now show by induction that a n r g = 0, for r = 0,..., n). Check that the indeterminate X is a regular element of R [X]. 1.4 Unique factorization domains Our insight on algebraic properties is based essentially on all the computations we performed at hight school with integers and polynomials. These rings are of a very special kind, they are unique factorization domains. In these rings a very precise control is possible on every property concerning multiplication of elements. Needless to say, most of these good properties are lost if we work in more general rings. But we will see that some essential features survive in rather general rings if we replace elements by ideals. This will be the content of Chapter 5, devoted to the primary decomposition of an ideal. In this section we review the basic definitions about divisibility, and show how the concept of irreducible element arise naturally when we perform iterated factorizations in Z, or in the polynomials with coefficients in a given field. Moreover, we explore thoroughly the fact that suitable uniqueness properties of such factorizations can be obtained only when all the irreducible elements of the ring under consideration enjoy a stronger property, that to be prime elements. This leads naturally to the notion of prime ideal, which will be formally introduced and studied in the next section.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 20 To start, let us work in Z. The basic idea is quite simple: given any integer m, we can check whether it is possible to get it as a product of two other integers, both having modulus < m. For instance, if m = 180, then 180 = 9 20 or 180 = 3 ( 60 ) or 180 = 10 18 or... Then, if we succeded in factorize m, we can apply the same procedure to the single factors. And so on. In our example 180 = 9 20 = 3 3 2 10 = 3 3 2 2 5 We can iterate until each factor is divisible only by itself and by ±1, namely it is a prime number up to the sign. This to the sign is related to the invertible elements of Z, because Z = { 1, 1 }. The above procedure certainly stops after finitely many steps because at each step the modulus of some factor is replaced by two smaller ones. In the ring K [X], where K is a field, we can do the same. First of all, recall that by (1.14) we have K[X] = K. Now, let f 0 be a polynomial of positive degree. By the above remark, this is equivalent to require that f is not a unit in K [X]. Then a non trivial factorization of f will be f = g h, where g < f and h < f. This rules out the possibility that g = 0, namely that g K[X], and similarly for h. If f does not admit a non trivial factorization it is called irreducible; this means explicitly that f is divisible only by any a K, and by its associate elements a f, where again a K. On the other hand, if f = g h is a non trivial factorization, we can check whether g and h admit non trivial factorizations or not. Then iterate until a product of irreducible polynomials is obtained. At each step the degree of some factor decrease, hence the whole procedure stops. Note that the degree of a polynomial plays here the role of the modulus for the integers ( you are led to the same remark by comparing the euclidean division algorithm of these two rings). Let us formalize things by giving the basic definitions. From now on in this section R will denote an integral domain, and we will deal only with non zero elements of R. Definition 40. For a, b R we will say that a divides b (or that a is a divisor of b) if b = ac, for some c R. We will denote this by a b. If a b and b a, then a = ub with u R. In this case a and b are called associates. To be associates is an equivalence relation. If a b and a and b are not associates, we will say that a divides b properly, or that it is a proper divisor of b. Finally, a R is called irreducible if it has no proper divisors, i.e. if from b a it follows either that b R, or that a and b are associates.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 21 These properties translate nicely in terms of ideals. In fact a b is equivalent to b (a), which in turn is equivalent to (b) (a). a and b are associates if and only if (a) = (b). (1.15) If a is a proper divisor of b, then (b) (a), and conversely. a is irreducible if and only if (a) is maximal (w.r.t. ) among the set of proper, principal ideals of R. After all these preparations, let us introduce the core of the subject by an example. Example 41. Consider the subring of C Z[ 3 ] := { a + b 3 C a, b Z } C The standard tool for studing this ring is the modulus of a complex number. Recall that if z = x + iy (x, y R) is any complex number, its modulus is z := x 2 + y 2 = z z This second way to compute z shows immediately that, for any z, w C we have zw = z w. It will be clear in a moment that it is better for us to work with the square of the modulus; hence we set N(x) := x x = a 2 + 3 b 2 for every x = a + b 3 Z[ 3 ] The main properties of the norm N(x) of x are 1. N(x) N for every x Z[ 3 ] ; (1.16) 2. N(x) = 0 if and only if x = 0 ; 3. N(xy) = N(x) N(y) for every x, y Z[ 3 ] ; 4. N(x) = 1 if and only if x ( Z[ 3 ] ). Only this last item is not completely trivial. If xy = 1, then N(x) N(y) = N(xy) = N(1) = 1. But N(x), N(y) are natural numbers, hence they are both 1 (note that it is impossible to conclude in this way if we use the modulus instead of the norm).

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 22 On the other hand, if N(x) = 1, then x x = 1. Since x = a b 3 Z[ 3 ], we conclude that x is invertible in Z[ 3 ]. By the way, we have proved that the complex conjugation induces an automorphism of Z[ 3 ]. Finally, from 4. in (1.16) it follows at once that ( Z[ 3 ] ) = { 1, 1 }. Our interest for this example is due to the fact that in Z[ 3 ] the integer 4 can be factorized in two ways (1.17) ( 1 + 3 ) ( 1 3) = 4 = 2 2 It is useful for what follows to use the Argand-Gauss plane to get an intuitive understanding of what is going on. The elements of Z[ 3 ] are the points of a lattice inside this plane (some of them are represented in the figure below) Im(z) 1+ 3 1 2 4 Re(z) 1 3 Also the circle of all complex numbers of unitary modulus is represented in the figure, and we see that it intersects the lattice Z[ 3 ] only in the points 1 and 1, as computed above. Finally, it is clear how the factorizations (1.17) were conceived. The element 1 + 3 is irreducible in Z[ 3 ]. In fact, assume that in this ring there is a factorization (1.18) 1 + 3 = xy By taking norms we get 4 = N(1 + 3) = N(x) N(y) and the factorization (1.18) is non trivial only if N(x) = 2 and N(y) = 2. But, if say x = a + b 3, this would mean that the equation a 2 + 3 b 2 = 2

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 23 has a solution in integers, which is clearly impossible. The same argument shows that also 1 3 and 2 are irreducible in Z[ 3 ]. Finally, since the units in Z[ 3 ] are ±1, no two of 2, 1 + 3 and 1 3 are associates. We will express this by saying that the two factorizations of 4 in irreducible factors given in (1.17) are essentially distinct. ( For the reader interested in these rings of arithmetic flavour, the books [13] and [16] are recommended to start.) On the other hand, it is common knowledge that in Z and in K [X] the final factorizations we obtain are essentially independent of the path we followed to get them, namely two such factorizations can differ only by the order of the factors and by invertible elements of the ring. For example 180 = 3 3 2 2 5 = 2 ( 2 ) 3 2 5 = 3 ( 3 ) 2 2 ( 5 ) =... Hence both Z and K [X] enjoy a particular property, which is worth of a formal definition. Definition 42. An integral domain R is called a unique factorization domain (UFD for short) if the following conditions hold true (UFD 1) every non zero, non invertible element of R is a product of irreducible elements; (UFD 2) this factorization is essentially unique, where by this we will mean that, if p 1 p 2... p m = q 1 q 2... q n and all the p i s and the q j s are irreducible, then m = n and, after eventually rearranging the indices, p i and q i are associates for any i = 1,..., m. Therefore the ring Z[ 3 ] is not a UFD. The real difference between this ring and Z or K [X] is the content of the next theorem. To state it we have to recall that in both these two last rings we have the euclidean division algorithm available, and from this it follows easily that every ideal is principal. Definition 43. An integral domain such that every ideal is principal is called a PID ( Principal Ideal Domain ). Euclid s Theorem 1 Let R be a PID, and let x R be an irreducible element. If x divides the product yz, then x divides y, or x divides z. Proof. Since x is irreducible, by (1.15) the ideal (x) is maximal among the set of all proper, principal ideals of R. But R is a PID, hence (x) is maximal for the set of all proper ideals. 1 See Prop. 30 in Elements, Book VII, for the case of integers. Actually, I have not checked this reference. For a more recent proof for R = Z, which demands almost nothing to ring theory, cfr. Gauss Disquisitiones Arithmeticæ, Sectio Secunda, Theorema 13.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 24 Assume that x does not divide neither y, nor z. Then by (1.15) we have y / (x), and by the maximality of (x) we conclude (y, x) = R. Hence 1 = a x + by for suitable a, b R ; similarly we get 1 = c x + d z. Therefore 1 = ( a x + by ) ( c x + d z ) = acx 2 + adzx + bcyx + bdyz But yz (x) by assumption, hence 1 (x). Contradiction. Euclid s Theorem is concerned with a property of peculiar elements of a Principal Ideal Domain R. Thanks to our dictionary (1.15), we can translate it in terms of ideals of R as follows Assume that R is a PID, and that x R is irreducible. If yz (x), then y (x), or z (x). This suggests to define in general Definition 44. An ideal P of a ring R is called a prime ideal if P R and from xy P it follows that x P or y P. The next section is devoted to the study of these ideals. The following definition is also convenient Definition 45. An element x of an integral domain R, different from 0, which generates a prime ideal is called a prime element of R. The relations between irreducible and prime elements in a general integral domain is Proposition 46. Every prime element is irreducible. Proof. In fact, from x = y z it follows y z (x), hence y (x), say. Now, from y = r x we get x = r xz, then 1 = r z because R is an integral domain and x 0. Therefore z R. Example 47. The converse of Proposition 46 is false in general. In fact, the irreducible element 1+ 3 of Z[ 3 ] is not a prime element because 2 / ( 1+ 3 ), but (1.17) shows that 2 2 ( 1 + 3 ) On the other hand, Euclid s Theorem holds true in any UFD because of Proposition 48. Every irreducible element of a UFD is prime.

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 25 Proof. Assume that R is a UFD and x R is irreducible. Then x / R, and therefore (x) R. If y z (x), decompose both sides of the resulting relation y z = r x into irreducibles p 1 p 2... p }{{ m q } 1 q 2... q }{{ n } = y = z = u 1 u 2... u }{{} t x = r This decomposition is essentially unique because R is a UFD. Then either x is associate to some of the p i s, hence x divides y, or x is associate to some of the q j s and divides z. The proof is complete. The results from Euclid s Theorem up to the last proposition indicate a possible relation between prime elements and uniqueness of factorizations. This is made explicit in the next lemma, whose proof is a recommended Exercise, based on the definitions. Lemma 49. Let R be any domain, and let a R a non zero, non invertible element. Assume that a = p 1 p 2... p m, where all the p i s are prime elements of R. Then such a decomposition is essentially unique. From Proposition 48 and Lemma 49 it follows at once a characterization of the UFD s, namely Proposition 50. Let R be an integral domain such that every non zero, non invertible element of R is product of irreducible elements ( i.e. R satisfies condition (UFD 1) of Definition 42). Then R is a UFD if and only if every irreducible element of R is prime. We will use it now to prove that Proposition 51. Every PID is a UFD. Proof. Let R be a PID. By Euclid s Theorem every irreducible element of R is a prime element. Hence, it remains to prove that any non zero, non invertible element x R factorizes in a product of irreducible elements. If x is irreducible there is nothing to prove. Then assume that x is not irreducible, and consider a non trivial factorization x = y 1 y 2. If y 1 is irreducible, consider y 2. Otherwise, there is a non trivial factorization y 1 = z 1 z 2. If z 1 is irreducible, consider z 2. Otherwise, there is a non trivial factorization... In this way we construct a tree. Since the above factorizations are non trivial we get strictly ascending chains of principal ideals of R (x) (y 1 ) (z 1 )... (x) (y 1 ) (z 2 )...

NOTES ON COMMUTATIVE ALGEBRA by DARIO PORTELLI 26 and so on. Then it is sufficient to prove that such a chain in a PID cannot be infinite. Let us simplify a little bit the notations by considering (x 1 ) (x 2 ) (x 3 )... Set I := i 1 (x i ) I is an ideal of R, as it is easily checked. Hence I = (y). Therefore, by definition of I there is a suitable index i N such that y (x i ). It follows that (y) (x i ) (y), hence I = (y) = (x i ). Remark 52. We will introduce in one of the next chapters an important, wide class of rings, the noetherian rings. They are defined as those rings for which every ideal is finitely generated. In particular, any PID is a noetherian ring. It is interesting to note that the final argument in the proof of Proposition 51 still holds true for a noetherian integral domain. In fact, if I = (y 1, y 2...y r ), then by definition of I there is a suitable index i N such that y h (x i ) for every h = 1,...r. It follows that I (x i ) I, hence I = (x i ). Therefore, in a noetherian integral domain R any non zero, non invertible element factorizes into product of irreducible elements, namely R satisfies condition (UFD 1) of Definition 42. Remark 53. There are UFD which are not PID. In fact, we will prove that if R is a UFD, then R[X] is also a UFD (see Theorem 170). In particular, if K is a field, the integral domain K[X, Y ] is a UFD. Assume that it is a PID. The polinomial X is irreducible by a degree argument. Hence the ideal (X) is maximal in K[X, Y ], as we have seen. But this is equivalent K[X, Y ]/(X) being a field, whereas which is not a field. Contradiction. K[X, Y ]/(X) K[Y ] Let us conclude this section by reconsidering the Example 41. Example 54. The two factorizations of 4 in (1.17) are essentially different because the irreducible elements involved are not associates. A possible reason for this is that the ring Z [ 3 ] has very few invertible elements. We will try now to construct a new subring of C, containing Z [ 3 ], with some more invertible elements. In the following figure the circle of complex numbers with unitary modulus intersects the vector 1 + 3 in a point that we will denote by η. It is clear that η is a primitive 6 th root of the unity. Consider the set A := { a + bη a, b Z } C