Math 260Q: Noetherian Rings

Transcription

1 Math 260Q: Noetherian Rings Simon Rubinstein-Salzedo Fall 2005

2 0.1 Introduction These notes are based on a graduate course on noetherian rings I took from Professor Ken Goodearl in the Fall of The textbook was An Introduction to Noncommutative Noetherian Rings by Ken Goodearl and Robert Warfield. The theorem numbers in the notes correspond to those in the book. 0.2 Description Although the rings that one typically first meets in an algebra course (rings of integers, polynomial rings, rings of functions) are commutative, the universe holds just as many noncommutative rings. These often arise as rings of operators of various kinds (think of linear transformations on a vector space). Important examples include rings of differential operators, under which heading one can place the so-called enveloping algebras of Lie algebras; group rings (amounting to rings of operators on vector spaces built from groups); and twisted versions of these, such as the strangely named quantum groups which have been intensely studied in the past two decades. Many of these examples are noetherian rings; that is, ring in which the right and left ideals are finitely generated, and there is a rich general theory of noetherian rings that can be applied to them. This quarter will be mainly devoted to the basic theory of noetherian ring, although a few quantum groups will appear as examples. One key goal will be to prove Goldie s Theorem, which gives necessary and sufficient conditions for a particularly useful kind of ring of fractions to exist. Major concepts such as prime ideals will also be studied, and in order to work effectively with concrete example of noetherian rings, skew polynomial rings will be introduced. 1

3 Chapter 1 A Few Noetherian Rings Let k be a field and R = k[x 1,..., x n ]. An algebraic variety V k n is defined by equations p i (x 1,..., x n ) = 0 for p i R. If we let I = {p R p(x 1,..., x n ) = 0 for all (x 1,..., x n ) V } p i s, then I is an ideal of R generated by finitely many polynomials q 1,..., q t. Then the variety corresponding to I is V = {(x 1,..., x n ) k n q j (x 1,..., x n ) = 0 for all j = 1,..., t}. Analogously, consider linear partial differential equations with polynomial coefficients. There exists a ring R consisting of linear combinations of p(x 1,..., x n ) m 1 x m 1 1 mn x mn n. Corresponding to R is a ring of C functions of x 1,..., x n on R[x 1,..., x n ]. Let R be a ring (which in these notes will always be assumed to have a 1). We write R R for R viewed as a right R-module and R R for R viewed as a left R-module. The ascending chain condition (ACC) for submodules is the following statement: For all chains A 1 A 2 for submodules of A, there exists an N so that A n = A N for all n N. Let {A j j J} be a collection of submodules of A. A maximal element in the collection is some A j0 such that there is no j J with A j A j0. 2

4 Proposition 1.1 For a module A, the following are equivalent: (a) A satisfies the ACC on submodules. (b) Every nonempty collection of submodules of A has a maximal element. (c) Every submodule of A is finitely generated. Proof Sketch. We first show that (a) implies (b). Let A be a nonempty collection of submodules. Suppose there is no maximal element in A. Then there is some A 1 A and some A 2 A so that A 2 A 1, and so forth. We end up with a sequence A 1 A 2, which is a contradiction. We now show that (b) implies (c). Suppose B is a submodule which is not finitely generated. Pick b 1 B, and let B 1 be the submodule generated by b 1. Then there is some b 2 B \ B 1. Let B 2 be the submodule generated by b 1 and b 2. In this manner, find submodules B 1 B 2. The collection {B n n N} has no maximal element. We now show that (c) implies (a). Let A 1 A 2 be an ascending chain. Then i=1 A i is a submodule generated by some x 1,..., x k. Then there is some i so that all the x j s are in A i. Proposition 1.2 Let B A be modules. Then A is noetherian iff B and A/B are both noetherian. Corollary 1.3 If A 1,..., A n are noetherian modules, so is A 1 A n. Corollary 1.4 If R is a right noetherian ring, then all finitely generated right R- modules are noetherian. 3

5 Corollary 1.5 Let S be a subring of a ring R. If S is right noetherian and R S is finitely generated, then R is noetherian. Proposition 1.6 If R is a module-finite algebra over a commutative noetherian ring S, then R is noetherian. (If there is a map S Z(R), then R is an algebra over S; R is module-finite over S means that R is a finitely generated S-module.) In general, suppose A is a finitely generated right module over a ring R, say generated by x 1,..., x n. Then A = x 1 R + + x n R. Set F = RR n. There exists an epimorphism φ : F A given by φ(r 1,..., r n ) = x 1 r x n r n. Thus A = F/K for some submodule K. If R is right noetherian, then R R is noetherian, so RR n is noetherian, so RR n /K is noetherian. If R is a ring, then M n (R) is a ring. Proposition 1.7 If R is a right noetherian ring and n NN and S is a subring of r 0 M n (R) and S... r R, then S is right noetherian. 0 r Consider the ring ( ) aa ab + bc 0 cc. ( ) R R 0 R = {( ) a b a, b, c R}. We have 0 c ( ) ( a b a b 0 c 0 c ) = Example. ( ) Q Q + Qπ is a subring of 0 Z ( ) R R. 0 R Suppose A and C are rings and B is both a left A-module and a right C-module. 4

6 ( ) A B has the standard operations of addition and multiplication. We have 0 C (( ) ( )) ( ) ( ) a 0 0 b (ab)c =, c 0 0 ( ) (( ) ( )) ( ) a 0 0 b a(bc) = c 0 0 Definition. Let A and C be rings. An (A, C)-bimodule is an abelian group B with operations A B B and B C B such that (a) B is a left A-module. (b) B is a right C-module. (c) (ab)c = a(bc) for all a A, b B, and c C. Exercise 1B. If A and C are rings and A B C is a bimodule, then the set with formal matrix operations is a ring. ( ) A B 0 C Proposition 1.8 Let R = ( ) A B be a formal triangular matrix ring. 0 C (a) R is right noetherian iff A and C are right noetherian and B C is finitely generated. (b) R is left noetherian iff A and C are left noetherian and A B is finitely generated. Theorem 1.9 (Hilbert Basis Theorem.) Let S = R[x] be a polynomial ring. If R is right (left) noetherian, then so is S. Idea. Assume R is right noetherian. Let I be a right ideal with S. Without loss of generality, S 0. 5

7 (1) Set J = {leading coefficients of polynomials in I} {0}. Check that J is a right ideal of R. (2) J is generated by some r 1,..., r k. Without loss of generality, all r i 0. Suppose r i is the leading coefficient of p i I. Without loss of generality, all p i have degree n. We re halfway there: If p I and deg(p) = d n, then there are a 1,..., a k R such that p 1 (a 1 x d n ) + + p k (a k x d n ) ( ) has degree d and the same leading coefficient as p, so p ( ) I and has degree less than n. (3) Set N = {q S q = 0 or deg(q) < n}, a right R-submodule of S generated by 1, x,..., x n 1. Thus N R is a noetherian module. Thus N I is finitely generated as a right R-module. Say N I = q 1 R + + q t R. (4) p 1,..., p k, q 1,..., q t generate I. Corollary 1.10 Let R be a commutative right which is an algebra over a field k. If R is finitely generated as a k-algebra, then R is noetherian. Idea. Say R is generated by r 1,..., r m. Let S = k[x 1,..., x m ] be a polynomial ring over k in m indeterminates. Then there exists a k-algebra homomorphism φ : S R such that φ(x i ) = r i. Then R = S/ ker φ. The Hilbert Basis Theorem implies that R is noetherian. Example. Differential operators on R[x]. Let L be the set of linear differential operators on R[x], an R-subring of End R (R[x]) generated by D = d and {µ dx f =left multiplication by f f R[x]}. In fact, L = {µ f0 + µ f1 D + + µ fn D n f i R[x]}. We have D(µ f )(g) = (fg) = f g + fg = (µ f + µ f D)(f). Thus D µf = µ f D + µ f. Identify each f with µ f. Then Df = fd + f. In particular, Dx = xd + 1. Then L = {f 0 + f 1 D + + f n D n f i R[x]} is almost the polynomial ring R[x][D]. Example. Let k be a field, G the group x, y xyx 1 = y 1, and kg the group algebra {f m x m + f m+1 x m f n x n f i k[y ±1 ], m n Z}. Then kg is almost the Laurent polynomial ring k[y ±1 ][x ±1 ], Since xy = y 1 x and xf(y) = f(y 1 )x, 6

8 f(y) f(y 1 ) is an automorphism α of k[y ±1 ]; xf = α(f)x. 1.1 Automorphism-Twisted Skew Polynomial Rings Let R be a ring and α Aut(R). We wish to build a skew polynomial ring with indeterminate x such that xr = α(r)x for all r R. The elements are the usual polynomial expressions r 0 + r 1 x + + r n x n, where each r i R. Addition is as usual. Multiplication will be derived from the relation xr = α(r)x, so in particular x i r = α i (r)x i for all i 0 and r R. Hence (rx m )(sx n ) = rα m (s)x m+n. Exercise. This builds a ring. As an R-module, this new ring is R + Rx + Rx 2 +, a free R-module. Definition. Let R be a ring and α Aut(R). The statement S = R[x; α] means that (a) S is a ring containing R as a subring. (b) x is an element of S, and xr = α(r)x for all r R. (c) S is a free left R-module with basis {x i i 0}. Such a ring exists and is unique. Lemma. Let R be a ring and α Aut(R). Suppose S = R[x; α] and T = R[y; α]. Then S = T ; more precisely, there exists an isomorphism φ : S T such that φ R = id R and φ(x) = y. Idea. The only possibility is φ(r 0 + r! x + + r n x n ) = r 0 + r 1 y + + r n y n. Check that this works. 7

9 Lemma 1.11 Let S = R[x; α], φ : R T a ring homomorphism, and y T. Assume yφ(r) = φα(r)y for all r R. Then there is a unique ring homomorphism ψ : S T such that ψ R = φ and ψ(x) = y. Example. Let k be a field, R = k[y] a polynomial ring, and q k (the multiplicative group of k). Then there exists a unique k-algebra automorphism α of R such that α(y) = qy. Look at S = R[x; α]. The key property is that xy = qyx. The elements are n i=0 f ix i, f i R, or n m i=0 j=0 λ ijy j x i, λ ij k. Context. If k is a field, then k 2 is the affine plane over k. O(k 2 ) is the coefficient ring of k 2, which is the polynomial ring k[x, y]. This is a k-algebra with generators x and y and relation xy = yx. Definition. Let k be a field and q k. Then corresponding quantized coordinate ring of k 2 is O q (k 2 ) = k x, y xy = qyx. Claim. O q (k 2 ) is a skew polynomial ring as in the last example. (a) Keep O q (k 2 ) = k x, y xy = qyx. (b) R = k[ŷ] (polynomial ring), α Aut R is given by p(ŷ) p(qŷ), S = R[ˆx; α]. (c) ˆxŷ = qŷˆx in S, so there is a unique k-algebra homomorphism φ : O q (k 2 ) S such that φ(x) = ˆx and φ(y) = ŷ. (d) There is a unique k-algebra homomorphism θ : R O q (k 2 ) such that θ(ŷ) = y. Then xy = qyx, so xy i = q i y i x for all i 0, so xθ(ŷ i ) = θα(ŷ i )x, so xθ(r) = θα(r)x for all r R. Lemma 1.11 implies that there exists a unique k-algebra homomorphism ψ : S O q (k 2 ) such that ψ R = θ and ψ(ˆx) = x. Notice that ψ(ŷ) = y. (e) φψ(ˆx) = ˆx and φψ(ŷ) = ŷ, so φψ = id S. Similarly ψφ(x) = x and ψφ(y) = y, so ψφ = id Oq(k 2 ). 8

10 Proposition 1.13 If k is a field and q k, then O q (k 2 ) is a skew polynomial ring k[y][x; α], where k[x] is a polynomial ring and α is the k-algebra automorphism of k[y] such that α(y) = qy. Definition. Let R be a ring and α Aut R. T = R[x ±1 ; α] means (a) T is a ring with R as a subring. (b) x is an invertible element of T, and xr = α(r)x for all r R. (c) T is a free left R-module with basis {x i i Z}. Example. Let k be a field and q k. The algebraic torus is (k ) 2. O((k ) 2 ) = k[x ±1, y ±1 ]. The quantized version is O q ((k ) 2 ) = k x, x, y, y xx = x x = y y = yy = 1, xy = qyx. Exercise 1O. O q ((k ) 2 ) = k[y ±1 ][x ±1 ; α], where α(y) = qy. Theorem 1.14 Let S = R[x; α] and α Aut R. If R is right (left) noetherian, so is S. Idea for the right noetherian case. Let I be a nonzero right ideal of S. (1) Let J denote the set of leading coefficients of elements of I together with 0. J is closed under addition and subtraction. Let α J and r R. There exists p = ax n +(lower terms) I. pr = aα n (r)x n +(lower terms) I, so aα n (r) J. Instead, pα n (r) = arx n +(lower terms) I, so ar J. Thus J is a right ideal of R. (2) J is generated by a 1,..., a k. There exists a p i = a i x n i +(lower terms) I. Thus without loss of generality all p i have degree n. (2 ) We re halfway. Suppose p I with deg(p) = d n. Let p = ax d +(lower terms), with a J. Then a = a 1 r a k r k for some r i R. Then 9

11 p 1 α n (r 1 ) + + p k α n (r k ) = (a 1 r a k r k )x n +(lower terms). Then p 1 α n (r 1 )x d n + + p k α n (r k )x d n = ax d +(lower terms), so there is a polynomial of degree less than d in I. Finish as before. There are problems with the left noetherian case. Let I be a left ideal of S. Let J denote the set of leading coefficients of elements of I together with 0. Suppose a, b J. There exists a p = ax m +(lower terms) and q = bx n +(lower terms) I. What if m < n? Then x n m p = α n m (a)x n +(lower terms). Thus we only get α n m (a) + b J. Definition. Let T be a ring. The opposite ring T op has the same set as T and the same addition as T, but a new multiplication defined by a b = ba. Look at S = R[x; α]. R op is a subring of S op, and x S op. Also x r = rx = xα 1 (r) = α 1 (r) x. Check that S is a free right R-module with basis {x n n 0}. Then S op is a free left R op -module with basis {x n n 0}. Thus S op = R op [x; α 1 ]. If R is left noetherian, then R op is right noetherian, so R op [x; α 1 ] = S op is right noetherian, and so S is left noetherian. Corollary 1.15 Let T = R[x ±1 ; α] and α Aut R. If R is right (left) noetherian, so is T. Definition. A ring R is simple iff (a) R 0. (b) The only ideals of R are 0 and R. Let α Aut R. An α-ideal of R is an ideal I R such that α(i) = I. R is α-simple iff (a) R 0. 10

12 (b) The only α-ideals are 0 and R. Exercise 1V. Let R = k[x], where K is a field of characteristic 0, and let α be the k-algebra automorphism such that α(y) = y + 1. Then R is α-simple. Theorem 1.17 Let T = R[x ±1 ; α], α Aut R. Then T is simple iff (a) R is α-simple. (b) No positive power of α is inner. Idea for (a). If I R is an α-ideal, then IT T. Idea for. Let 0 I T. Let n be the minimum degree for nonzero elements of I S (where S = R[x; α]). Let J be the set of leading coefficients of elements of I S of degree n together with 0. Note that 0 I S S, J R, and J 0. J is an α-ideal: if 0 a J, then there exists some p = ax n +(lower terms) I S, so xpx 1 I S and xpx 1 = α(a)x n +(lower terms), so α(a) J. Thus α(j) J. Use xpx 1 to see that α 1 (J) J. Thus (a) implies that J = R, so there is some p = x n + a n 1 x n a 0 I S. If a 0 = 0, then p = (x n 1 + a n 1 x n a 1 )x, so 0 px 1 I S, which is a contradiction. Thus a 0 0. We have xpx 1 = x n + α(a n 1 )x n α(a 0 ) I S. Then p xpx 1 I S and has degree less than n, so p xpx 1 = 0. Thus α(a i ) = a i for all i. For all r R, pr = α n (r)x n + a n 1 α n 1 (r)x n a 0 r. α n (r)p = α n (r)x n + α n (r)a n 1 x n α n (r)a 0. Then pr α n (r)p = 0 because it has degree less than n, so a i α i (r) = α n (r)a i for all i, and a 0 r = α n (r)a 0 for all r R. Continuing this way, we see that a 0 is invertible in R. Thus α n (r) = a 0 ra 1 0 for all r R. (b) implies that n = 0. Thus p = 1. Finally, 1 I S I, so I = T. Corollary 1.18 Let k be a field, q k, and T = O q ((k ) 2 ). Then T is simple iff q is not a root of unity. 11

13 Chapter 2 Skew Polynomial Rings Example. Consider R[x] with multiplication operators and d fd + f for all f R[x]. dx = D. Then Df = Definition. Let R be a ring. A map δ : R R is a derivation iff δ(r + s) = δ(r) + δ(s) and δ(rs) = δ(r)s + rδ(s) for all r, s R. Definition. S = R[x; δ] means (a) S is a ring containing R as a subring. (b) x S and xr = rx + δ(r) for all r R. (c) S is a free left R-module with basis {x j j 0}. We have x 2 r = x(rx + δ(r)) = (rx + δ(r))x + δ(r)x + δ 2 (r) = rx 2 + 2δ(r)x + δ 2 (r). More generally, we have n ( ) n x n r = δ n i (r)x i. i i=0 Exercise. R[x; δ] exists. 12

14 We now look at some key examples. Let K be any ring, R = K[y], and δ = d dy. [ ] Weyl Algebra. A 1 (K) = K[y] x; d, so xy = yx + 1. In general, dy [ A n (K) = K[y 1 ] x 1 ; d dy 1 ] [ [y 2 ] x 2 ; d dy 2 ] [ [y n ] x n ; d dy n ]. Definition. Let R be a ring and δ a derivation on R. A δ-ideal of R is any I R such that δ(i) I. R is δ-simple iff R 0 and the only δ-ideals are 0 and R. δ is inner iff there exists an a R such that δ(r) = ar ra for all r R; otherwise δ is outer. Proposition 2.1 Let R be a Q-algebra and δ a derivation on R. Then R[x; δ] is simple iff (a) R is δ-simple. (b) δ is outer. Corollary 2.2 If K is a field of characteristic 0, then A n (K) is simple for all n. We wish to form a skew polynomial ring with coefficient ring R and indeterminate x. We require the following: It is a ring S with R as a subring. x S and S is a free left R-module with basis {x i i 0}. deg(st) deg s + deg t for all s, t S. Consequences of the above wish list are that deg(x) = 1 and deg(r) = 0 for all r R. Then xr should have degree at most 1, so xr = r x + r for some r, r R. Hence we should have maps α, δ : R R such that xr = α(r)x + δ(r) for all r R. We have 13

15 x(r + s) = xr + xs for all r, s R, so α(r + s) = α(r) + α(s), δ(r + s) = δ(r) + δ(s). Also x1 = x so α(1) = 1 and δ(1) = 0. Furthermore, (xr)s = x(rs) = α(rs)x + δ(rs). Thus α(rs) = α(r)α(s) and δ(rs) = α(r)δ(s) + δ(r)s. Hence α must be a ring endomorphism of R and δ a left α-derivation of R. The next job is to show that there exists a ring R[x; α, δ] with some properties: Universal property. Uniqueness up to isomorphism. Show the Hilbert Basis Theorem if α Aut R. 2.1 Building Rings from Operators Take an abelian group A (or a vector space over a field k). Label the ring E = End Z (A) (or algebra E = End k (A)). Take some elements of E and form the ring (algebra) they generate. Proposition 2.3 Given a ring R, a ring endomorphism α or R, and an α-derivation δ on R, there exists a ring R[x; α, δ]. Sketch. (1) Form the ordinary polynomial ring R[z]. View it as an abelian group, and set E = End Z (R[z]). (2) Consider the map R E given by r (lert multiplication by r). Identify R with its image, a subring of E. (3) Define x E by ( ) x r i z i = i Check that xr = α(r)x + δ(r) for all r R. i (α(r i )z + δ(r i ))z i. 14

16 (4) Set S to be the subring of E generated by R {x}. We get that S is a ring and R is a subring of S and x S for free. (5) Check that S is a free left R-module with basis {x i i 0}. 15

17 Chapter 3 Prime Ideals Definition. A prime ideal in a ring R is an ideal P R such that P R and for all ideals I, J R, if IJ P, then I P or J P. Proposition 3.1 Let R be a ring, P R, and P R. The following are equivalent: (a) P is a prime ideal. (b) If I, J R and I, J P, then IJ P. (c) R/P is a prime ring (i.e. if Ī, J R/P and Ī J = 0, then Ī = 0 or J = 0). (d) If I and J are right ideals of R and IJ P, then I P or J P. (e) If I and J are left ideals of R and IJ P, then I P or J P. (f) If x, y R and xry P, then either x P or y P. If I and J are right ideals, then RI, RJ R. If x, y R, then RxR, RyR R. Proposition 3.2 All maximal ideals are prime. 16

18 Exercise. In a ring R, every proper ideal is contained in a maximal ideal. Example. In Z, 0 is the unique minimal prime. In R[x, y]/(xy 2 ), the minimal primes are (x) and (y). (xy 2 ) (xy 2 ) Proposition 3.3 Every prime ideal P in a ring R contains a minimal prime. Proof. The idea is as follows: Let X = {prime ideals contained in P }. We work with (X, ), where is. Let Y be a nonempty chain in X; we need to show that it has a lower bound in X. Set Q = Y R. We claim that Q is prime. Let x, y R such that xry Q and x Q. There is some Q Y such that x Q. Thus y Q. Look at some arbitrary Q Y. If Q Q, then y Q. If Q Q, then x Q, so y Q. Thus y Q and Q is prime. Definition. If I R, a prime P I is minimal over I iff P is minimal among primes containing I. Equivalently, P/I is a minimal prime in R/I. Theorem 3.4 Let R be a right or left noetherian ring and I R. Then there are only finitely many primes minimal over I, and some product of them (with repetitions) is contained in I. Example. R = R[x, y], I = (xy 2 ). Then P 1 = (x) and P 2 = (y) are the only primes minimal over I, and P 1 P 2 2 = I. Proof. We claim that it suffices to prove there exist primes P 1,..., P n I such that P 1 P 2 P n I. ( ) To see this, note that each P i contains a prime Q i minimal over I, and Q 1 Q 2 Q n P 1 P 2 P n I. If Q is a prime minimal over I, then P 1 P n I Q, so some P j I P j Q, so Q = P j. Thus {primes minimal over I} {P 1,..., P n }. Q. Then 17

19 Now suppose ( ) fails. The noetherian condition implies that there is an ideal Ī I maximal such that ( ) fails. If Ī is prime, take P 1 = Ī, which is a contradiction. Thus Ī is not prime, so there exist J, K Ī such that JK Ī. Thus ( ) holds for J and K, so ( ) holds for Ī, which is a contradiction. Hence ( ) always holds. Definition. An ideal I R is semiprime iff I is the intersection of prime ideals. (This includes R, which is the empty intersection.) Theorem 3.7 I R is semiprime iff for all x R, xrx I implies x I. (If R is commutative, x 2 I implies x I, so I is a radical ideal.) Proof. For the forward direction, we have I = α A P α, where P α are prime ideals. But xrx P α implies x P α. For the reverse direction, it suffices to prove that if x R \ I, there exists a prime P I such that x P. Set x 0 = x. Since x 0 I, x 0 Rx 0 I, so there exists an x 1 = x 0 r 0 x 0 I. Then x 1 I, so there is an x 2 = x 1 r 1 x 1 I, and so forth. Zorn s Lemma implies that there is an ideal P I which is maximal with respect to the condition P {x 0, x 1, x 2,...} =. We now claim that P is prime. Suppose P is not prime. Then there exist J, K P such that JK P. The maximality of P implies that there exist x j J and x k K. Let m = max{j, k}; note that x m J K. Then x m+1 x m Rx m JK P, which is a contradiction. Thus P is prime. Corollary 3.8 The following are equivalent for I R: (a) I is semiprime. (b) If J R and J 2 I, then J I. (c) If J is a right ideal of R and J 2 I, then J I. (d) If J is a left ideal of R and J 2 I, then J I. (e) If J R and J I, then J 2 I. 18

20 Definition. A right or left ideal J in a ring is nilpotent iff there is some n N such that J n = 0. J is a nil ideal iff every element of J is nilpotent. Definition. The prime radical of R is the intersection of all prime ideals of R, the smallest semiprime ideal. Example. Let R = Z/nZ, where n = p m 1 1 p m k k. The prime ideals in R are p iz/nz. The prime radical is (p 1 p k )Z/nZ. Theorem 3.11 Let N be the prime radical of a right or left noetherian ring R. Then N is nilpotent and N contains all nilpotent right or left ideals. Definition. Let A = A R. For X A, the annihilator of X in R is ann(x) = ann R (X) = {r R xr = 0 for all x X}, which is a right ideal of R. For Y R, the annihilator of Y in A is ann A (Y ) = {a A ay = 0 for all y Y }. For X R, l. ann(x) = {r R rx = 0 for all x X} and r. ann(x) = {r R xr = 0 for all x X}. Example. Let R = Z and A = Z/6Z. Then ann R (2 + 6Z) = 3Z Z and ann A (2) = {3 + 6Z, 6Z} A. If B is a submodule of A, then ann R (B) is an ideal of R. If I R, then ann A (I) is a submodule of A. Definition. A module A R is faithful iff ann R (A) = 0. A is fully faithful iff A is faithful and all its nonzero submodules are faithful. Example. Let R = Z. Then nz (n 0), Z, Q, and R are fully faithful. i=1 Z/nZ is faithful but not fully faithful. 19

21 If A R 0 and I = ann R (A) R, then A is also an R/I-module and is faithful. Possibly there is some 0 B A with ann R (B) > I. Proposition 3.12 Let A R 0. Suppose there is a P R maximal among annihilators of nonzero submodules of A. Then P is a prime ideal and ann A (P ) is a fully faithful (R/P )-module. Idea. Suppose I, J R with I, J R. Then ann R (A) cannot properly contain P, so ann R (A) I. Thus AI 0. Now ann R (AI) P, so it does not contain J, and AIJ 0. Thus IJ ann R (A). There exists a nonzero B A such that P = ann R (B). Since I ann R (B), BI 0. Now ann R (BI) J, so BIJ 0. Thus IJ P. Thus P is prime. Suppose R is right noetherian and A R is finitely generated, and A 0. (1) Since R is right noetherian, there exists a P 1 R maximal among annihilators of nonzero submodules of A. Proposition 3.12 implies that P 1 is prime, and A 1 = ann A (P 1 ) is a fully faithful (R/P 1 )-module. (2) If A 1 A, repeat for A/A 1 : there exists a P 2 R and a submodule A 2 A with A 2 > A 1 such that A 2 /A 1 is a fully faithful (R/P 2 )-module. (3) Continue. Since A R is noetherian, there is some n so that A n = A. Proposition 3.13 Suppose R is right noetherian and A R is finitely generated. Then there exist submodules A 0 = 0 < a! < < A n = A and prime ideals P 1,..., P n R such that for all i, A i /A i 1 is a fully faithful (R/P i )- module. Proposition 3.14 Suppose R, A, A i, P i are as in Proposition Set I = ann R (A). If P is any prime minimal over I, then P is some P j. 20

22 Proof. (A i /A i 1 )P i = 0 implies that A i P i A i 1. So AP n P n 1 P 1 = 0. Thus P n P n 1 P 1 I P. Thus there exists a j such that P j P. But I P j P. So the minimality of P implies that P = P j. Definition. A module A is simple (or irreducible) iff A 0 and the only submodules of A are 0 and A. If A is simple, then ann(a) is the only ideal which is an annihilator of 0 a A. It is maximal and hence prime. If A = A R is simple, then ann R (A) is a right primitive ideal of R. Proposition 3.16 In any ring R, the intersection of left primitive ideals, the intersection of right primitive ideals, the intersection of maximal left ideals, and the intersection of maximal right ideals all coincide. Definition. This intersection is the Jacobson radical of R, denoted J(R). R is a semiprimitive ring iff J(R) = 0. I R is a semiprimitive ideal iff R/I is a semiprimitive ring. Note. Primitive ideals are prime. Thus J(R) is semiprime and contains the prime radical. Theorem 3.22 Let S = R[x; δ], where R is a commutative noetherian Q-algebra and δ is a derivation. (a) If P S is prime, then P R is a prime δ-ideal of R. (b) Suppose Q is a prime δ-ideal of R. (i) QS S, QS is prime, and QS R = Q. (ii) If δ(r) Q, then QS is the unique prime of S whose intersection with R is equal to Q. 21

23 (iii) If δ(r) Q, then there exist infinitely many primes P S such that P R = Q. They are inverse images of primes in K[x], where K is the quotient field of R/Q under the map S quot S QS = (R/Q)[x] K[x]. Example. Let k be an algebraically closed field of characteristic 0. Let R = k[y], δ = (y 2 1) d, and S = R[x; δ]. The prime ideals in R are 0 and y α (α k). dy The prime δ-ideals are 0, y 1, y + 1. If Q = 0, then δ(r) Q, so only P = 0 is in S. If Q = y 1, then δ(r) Q, so S/QS = k[x]. The primes are 0 and x λ (λ k), so P = y 1, x λ (λ k) or y 1. The case Q = y + 1 is similar. Hence the prime ideals in S have the following structure: y 1, x λ y 1 y + 1, x + λ y

24 Chapter 4 Semisimple Modules, Artinian Modules, and Torsionfree Modules Commutative domains are contained in quotient fields. The modules of these quotient fields are vector spaces, which are direct sums of 1-dimensional submodules. Noncommutative domains are contained in division rings of fractions. The modules of these rings of fractions are direct sums of 1-dimensional submodules. We would like to contain noetherian prime rings in some sort of quotient object whose modules are direct sums of copies of some particular simple module. Definition. For a module A, the socle of A is the sum of all simple submodules of A. It is denoted by soc(a). A is semisimple (or completely reducible) iff soc(a) = A. Examples. (1) Vector spaces. (2) A Z-mod. soc(a) = {a A the order of a is prime}. So if n = p m 1 1 p m 2 2 p m k k, where the p i are distinct primes and m i > 0, then soc(z/nz) = n p 1 p 2 p k Z/nZ. Say {A i i I} is a collection of R-modules. The direct product has the Cartesian product set i I A i with natural operations. The (external) direct sum i I A i is the set of a i I A i with a i = 0 for all but finitely many i. Now suppose all A i are 23

25 submodules of some A. Then i I A i is the submodule generated by i I A i. There exists a homomorphism ( sum map ) S : i I A i i I A i given by a i I a i. ( ) Check that s is injective iff for all j I, A j i I A i = 0. In this case, we say i j {A i i I} is independent. If this happens, we identify i I A i (the internal direct sum) with i I A i. Proposition 4.1 If A is a module, soc(a) is the direct sum of some family of simple submodules of A. Proof. Let X = {all independent collections of simple submodules of A}. Then X. Zorn s Lemma implies that there is a maximal element Y X. Suppose Y = {B i i I}, where the B i are simple and B = i I B i = i I B i soc(a). Suppose B soc(a). Then there exists a simple submodule S A such that S B, so S B = 0 and Y {S} is independent, which is a contradiction. Thus B = soc(a). Proposition 4.2 A module A is semisimple iff every submodule of A is a direct summand. Proof. We first do the forward direction. Let B A. Zorn s Lemma implies that there is a submodule C A maximal with respect to B C = 0. Thus B+C = B C. Suppose B C < a. Then there is a simple submodule S A such that S B C, so {B, C, S} is independent, so B (C S) = 0, which is a contradiction. Thus B C = A. For the reverse direction, we have A = soc(a) B for some B A. Suppose B 0. Choose 0 b B and set C = br or Rb. Zorn s Lemma implies that there is some submodule M C maximal with respect to b M. Then C/M is a simple module. The hypothesis implies that there is some N A such that A = M N,so C = M (N C). (This follows from the Modular Law, which states that if M C A and N A, then C (N + M) = (C N) + M.) Now N C = C/M is simple. But N C soc(a), which is a contradiction. Thus B = 0 and A = soc(a). 24

26 Corollary 4.3 Any submodule of a semisimple module is semisimple. For which R are all modules semisimple? Theorem 4.4 For a ring R, the following are equivalent. (a) All right R-modules are semisimple. (b) All left R-modules are semisimple. (c) R R is semisimple. (d) R R is semisimple. (e) R = 0 or R = M n1 (D 1 ) M n2 (D 2 ) M nk (D k ) for some n i N and division rings D i. Proof. We first show that (c) implies (a). Let A = A R. There exists a free module F R with a basis {x a a A}. There is a unique homomorphism f : F A such that f(x a ) = a for all a A. So if K = ker(f), A = F/K. Since R R is semisimple and F = a A π ar is isomorphic to a direct sum of copies of R R, F is semisimple, and so A is semisimple. It now suffices to show that (c) and (e) are equivalent. We outline the proof that (c) implies (e) first. Without loss of generality, R 0. (1) Since R R is semisimple, R R = i I S i, where S i are simple right ideals. R R is cyclic, so I is finite. (2) Group the S i by isomorphism class and relabel. Then R R = k n i i=1 j=1 S ij such that S ij are simple right ideals, k, n i N. Furthermore, S ij = Stn iff i = t. (3) Hom R (S ij, S tn ) = 0 for i t. 25

27 (4) R R = R 1 R k, where R i = n i j=1 S ij is a right ideal. By (3), Hom R (R i, R t ) = 0 for i t, so R i R for all i. (5) Now each R i is a ring, and R = R 1 R k. Thus without loss of generality, R = R 1 = n 1 j=1 T j, all T j = T1 (as right ideals). Thus R R = T n 1. Write x 1 T1 n =. x i T 1 x n. Then D 1 = End R (T 1 ), a division ring (by Schur s Lemma). There is a map θ : M n1 (D 1 ) End R (T1 n ). Then [f ij ] x 1. [f ij ] x 1.. x n1 x n1 Exercise: θ is a ring isomorphism. Thus M n1 (D 1 ) = End R (T n 1 ) = End R (R R ) = R. We now outline the proof that (e) implies (c). (1) R = 0 is easy, so without loss of generality assume R 0. Without loss of generality, R = R 1 R k, where R i = M ni (D i ), n i N, and D i is a division ring. (2) soc(r R ) = soc(r 1 ) soc(r k ). Thus without loss of generality, R = R 1 = M n1 (D 1 ). (3) For i = 1,..., n i, set S i = {x R all nonzero entries in x are in row i}. Exercise: S i is a simple right ideal of R. Since R R = S 1 S n, R R is semisimple. Such rings are called semisimple rings (or semisimple artinian rings). Definition. A module A is artinian if it satisfies the descending chain condition (DCC), i.e. there is no infinite descending chain of proper submodules A 0 > A 1 > A 2 >. A ring R is right (left) artinian if R R ( R R) is artinian. 26

28 Proposition. Every semisimple ring R is artinian and noetherian. Proof. R R = S 1 S n, where the S i are simple right ideals. But S i is artinian and noetherian for all i. Thus R R is artinian and noetherian. So is R R. Theorem. (Noether.) Let R be a ring. The following are equivalent. (a) R = M n (D) for some n N and division ring D. (b) R is semisimple and simple. (c) R is simple and right artinian. (d) R is simple and left artinian. We call these rings simple artinian rings. Proof. We first show that (b) implies (a). Let R 0. Then R = M n1 (D 1 ) M nk (D k ). 0 M n2 (D 2 ) M nk (D k ) is a proper ideal and hence 0. Thus k = 1. Now we show that (a) implies (b). R is semisimple. We leave it as an exercise to show that if D is a simple ring, then M n (D) is also simple. (b) implies (c) is clear. To see that (c) implies (b), let R 0. Since R is right artinian, there is a right ideal S minimal among nonzero right ideals, so S is simple. Thus soc(r R ) 0. We leave it as an exercise to show that soc(r R ) R. Thus soc(r R ) = R. A commutative domain R is embedded in its quotient field F. Then torsionfree R- modules extend to vector spaces over F. Definition. If A is an R-module, there is a torsion submodule T (A) = {a A ra = 0 for some nonzero r R}. 27

29 Then A/T (A) is torsionfree. Let R = Z and A be a torsion Z-module. Then A = p prime T p(a), where T p (A) = {a A p n a = 0 for some n N}. If a, b T (A), there are nonzero r, s R so that ra = sb = 0. Then since R is a domain, rs 0, so (rs)(a + b) = 0, so a + b T (A). Definition. A multiplicative set in a ring R is any X R such that 1 X and X is closed under multiplication. An R-module A is X-torsion iff every element of A is annihilated by an element of X. A is X-torsionfree iff no nonzero element of A is annihilated by a nonzero element of X. Question. Let A = A R, X R a multiplicative set, and t X (A) = {a A ax = 0 for some x X}. Is (when is) this a submodule of A? Example. Let A = R/xR, where x X. Then 1x = 0, so 1 t X (A). So if t X (A) is a submodule, it equals A. So for all r R, we need ry = 0 for some y X, so ry xr, so ry = xs for some s R. Definition. Let R be a ring and X R a multiplicative set. X satisfies the right Ore condition iff for all r R and x X, there exists s R and y X such that ry = xs. Then X is a right Ore set. Lemma 4.21 Let X R be right Ore. (a) For all x 1,..., x n X, there exists s 1,..., s n R such that x 1 s 1 = = x n s n X. (b) For all A R, t X (A) is a submodule of A. Proof. 28

30 (a) It is enough to do n = 2. Given x 1, x 2 X, the right Ore condition implies that there exist s R and y X such that x 1 y = x 2 s. (b) Let a 1, a 2 t X (A). There exist x 1, x 2 X such that a i x i = 0. There exist s 1, s 2 R such that x 1 s 1 = x 2 s 2 = y X. Thus a i y = a i x i s i = 0, so (a 1 ± a 2 )y = 0. If r R, there exist s R and z X such that rz = x 1 s. Thus (a 1 r)z = a 1 x 1 s = 0. Lemma 4.22 Let R be a ring, X R a right Ore set, and A = A R. (a) t X (A) is an X-torsion module, and A/t X (A) is X-torsionfree. (b) Submodules, factor modules, and sums of X-torsion modules are X-torsion. (c) If B A, and B and B/A are both X-torsion or both X-torsionfree, then the same is true for A. (d) Submodules and direct products of X-torsionfree modules are X-torsionfree. Proposition 4.23 Suppose X R is a multiplicative set, A = A R is a noetherian X-torsionfree module, and f End R (A). If A/f(A) is X-torsion, then f is injective. Proof. ker(f) ker(f 2 ). Since A is noetherian, there is some n such that ker(f n ) = ker(f n+1 ). For all i N, A/f(A) f i (A)/f i+1 (A). Thus f i (A)/f i+1 (A) is X-torsion. This implies that A/f n (A) is X-torsion. Look at a ker(f). There is some x X such that āx = 0 in A/f n (A), so ax = f n (b) for some b A. f n+1 (b) = f(a)x = 0, so f n (b) = 0. Thus ax = 0, and so a = 0. Corollary. If A = A R is noetherian and f End R (A) is epic, then f Aut R (A). Proof. X = {1}. Corollary. If R is right noetherian and x, y R with xy = 1, then yx = 1. 29

31 Proof. Let f End R (R R ) be left multiplication by x. If xy = 1, then f is epic, so f is an automorphism. Thus r. ann R (x) = ker(f) = 0. x(yx 1) = 0, so yx 1 = 0. 30

32 Chapter 5 Injective Hulls A commutative domain R is contained in its quotient field F. Torsionfree R-modules become vector spaces over F. For every vector v and 0 r R, there is a vector 1 r v so that r ( 1 r v) = v. Definition. Let R be a commutative domain. A = R A is divisible iff for all a A and 0 r R, there is some b A such that rb = a. Exercise. Let R be a commutative domain with quotient field F. A = R A is the R-module of a vector space over F iff A is torsionfree and divisible. Say A = R A, a A, and 0 r R. Question. When can we solve r? = a inside A? If ra = a, then sra = sa for all s R. We get a homomorphism f : (r) A such that f(sr) = sa for all s R. This is defined irrespective of whether there is such an a. If there is such such an a, then there is a homomorphism f : R A such that g(s) = sa for all s R. g(sr) = sra = sa = f(sr), so g (r) = f. 31

33 Proposition. Let R be a commutative domain and A = R A. Then A is divisible iff for every principal ideal I R and every homomorphism f : I A, there is an extension g : R A. Proof. We show only the forward direction; the reverse direction is similar. If I R is principal, say I = (r), then r = 0 implies that f is the zero map R A. Thus without loss of generality, r 0. Let f : I A be a homomorphism, f(r) = a A. By divisibility, there is an a A such that ra = a. Define a homomorphism g : R A such that g(s) = sa for all s R. g(sr) = sra = sa = sf(r) = f(sr) for all s R, so g extends f. Definition. A = A R is injective iff for every right R-modules B C, all homomorphisms B A extend to homomorphisms C A. B C A (Note that C A is not necessarily unique.) Corollary. Over a commutative domain, all injective modules are divisible. Example. Z Z is not injective. Z(Z/nZ) is not injective. 2n n 2Z Z Z Proposition 5.1 (Baer s Criterion.) A = A R is injective iff for any right ideal I of R, all homomorphism I A extend to homomorphisms R A. 32

34 Proof. The forward direction is obvious. For the reverse direction, let B C be right R-modules and f : B A a homomorphism. Let X = {(D, g) B D C, g Hom R (D, A), g extends f}. Then (B, f) X. Define on X: (D 1, g 1 ) (D 2, g 2 ) iff D 1 D 2 and extends g 1. Check that is a partial order, and nonempty chains have upper bounds: A chain {(D i, g i ) i I} has an upper bound ( i I D i, i I B i). Zorn s Lemma implies that there is a maximal element (D, g ) X. If D = C, we re done. Suppose D C, and choose c C \ D. Set D = D + cr. Let I = {r R cr D }, a right ideal of R. Define h : I A by h(r) = g (cr). Check that h is a homomorphism. By hypothesis, h extends to a homomorphism h : R A. Set x = h (1) so that xr = h (r) = g (cr) for all r I. We claim that there is a well-defined homomorphism g : D + cr A given by g (d + cr) = g (d) + xr for all d D and r R. Corollary. Let R be a commutative PID. Then an R-module A is injective iff it is divisible. Example. The Z-modules Q and Q/Z are injective. So is [ ] Z(p 1 ) = p Z /Z, n the p-primary part of Q/Z. n=0 Proposition. If R is a commutative PID, every R-module A is a submodule of an injective R-module. Proof. Write A = F/K for some free R-module F and some K F. F is a direct sum of copies of R R. Set D to be the direct sum of the quotient field (over some index set). Then D is a divisible R-module. F D. Now A = F/K D/K, and D/K is divisible. Identify A with its image in D/K. General Problem. Let P be some property of modules. Suppose A = A R and we know that A is isomorphic to a submodule of an R-module B satisfying P. Why is A equal to a submodule of such a module? (1) We are given a monomorphism f : A B. 33

35 (2) Choose a set B 1 such that B 1 A = and there is a bijection β : B B 1. Use β to transfer the R-module operations from B to B 1 : for all x, y B 1, define x + y = β(β 1 (x) + β 1 (y)), and for all r R, define xr = β(β 1 (x)r). Now B 1 is an R-module, and β : B B 1 is a module isomorphism. Also g = βf : A B 1 is monic. (3) Define B 2 = A (B 1 \ g(a)). There is a bijection γ : B 1 B 2 such that { γ(g(a)) = a for all a A, γ(b) = b for all b B 1 \ g(a). Use γ to make B 2 into an R-module; now γ is a module isomorphism. γg : A B 2 is a monomorphism, and γg(a) = a for all a A. Thus A is a submodule of B 2. B = B 1 = B2, so B 2 has P. Analogue. Constructing an algebraic closure of a field k. (1) Given p k[x], p k, there is a field k k in which p has a root. (2) Likewise for any set of polynomials. (3) Thus there is a field k 1 k such that all nonconstant polynomials of k[x] have roots in k 1. (4) Repeat: There exist fields k k 1 k 2 such that all nonconstant polynomials in k i [x] have roots in k i+1. (5) Thus k = n=1 k n is a field containing k and is algebraically closed. (6) Set k = {α k α is algebraic over k}, which is an algebraic closure of k. Lemma. Let A and B C be R-modules and f Hom R (B, A). Then there is a module F A such that f extends to a homomorphism C F. Proof. The aim is to fill in the diagram B C f A 34 F

36 as follows: B C f A πi r πi l A C D i l i r π A C Define D = {(f(b), b) b B}, a submodule of A C. If a ker(πi l ), then (a, 0) ker π, so (a, 0) = (f(b), b) for some b B, so a = 0. Thus πi l is monic. Identify A with the image of πi l. Lemma. Let A and C i (i I) be R-modules, B i C i, and f i Hom R (B i, A) for all i. Then there is an R-module F A such that for all i, f i extends to a homomorphism C i F. Proof. Set B = i I B i C = i I C i. Define f = i I f i : B A. Label natural injections j i : B i B and q i : C i C. The previous Lemma implies that there is an R-module F A and a homomorphism g : C F which extends f. B i j i B f i f q i C i C gq i g A F Corollary. For any A = A R, there is a module F 1 (A) A such that for any right ideal I of R, all homomorphisms I A extend to homomorphisms R F 1 (A). So we make A F 1 (A) F 2 (A) such that I F n (A) extends to R F n+1 (A). Suppose I is a right ideal of R and f a homomorphism I n=1 F n(a). If I is finitely generated, say I = t j=1 x jr, then there is some n such that f(x j ) F n (A) 35

37 for all j, so f(i) F n (A), so f extends to R F n+1 (A) U. Thus if R is right noetherian, n=1 F n(a) is injective. Theorem 5.4 Any A = A R is contained in an injective module. Proof. Let ℵ = card(r), and let γ be the first infinite ordinal such that card(γ) > ℵ. Build a transfinite sequence of modules A α for all α γ such that A A 1 = F 1 (A). When A α is defined, set A α+1 = F 1 (A α ). When β is a limit ordinal less than or equal to γ, set A β = α<β A α. We claim that A γ is injective. By Baer s Criterion, we only need to show that for any right ideal I of R, any f Hom R (I, A γ ) extends to a homomorphism R A γ. f(i) A γ, and card(f(i)) ℵ. γ is a limit ordinal, so A γ = α<γ A α. Thus for all x I, there is an α x < γ such that f(x) A αx and card({α x x I}) ℵ, so sup{α x x I} = σ has cardinality at most γ, so σ < γ. Thus f(i) A σ, so f extends to a homomorphism R A σ+1 A γ. Corollary 5.5 A module A is injective iff A is a direct summand of every overmodule. Proof. We first do the forward direction. Given A B, id : A A extends to a homomorphism f : B A. Thus B = A ker(f). A B id f A For the reverse direction, by Theorem 5.4, there is an injective module B A. Then B = A A for some A B. We observe that A 1 A 2 is injective iff A 1 and A 2 are both injective. Definition. A submodule A B is essential iff for every nonzero submodule C of 36

38 B, A C 0. We write A e B. A homomorphism f : A B is an essential monomorphism iff f is monic and f(a) e B. Given A B, A e B iff A C 0 for all 0 C B iff A C 0 for all nonzero cyclic submodules C of B iff A br 0 for all 0 b B iff for all 0 b B, there is some r R such that 0 br A. Proposition 5.6 (a) Let A B C. Then A e C iff A e B e C. (b) If A i e B i e C (i = 1, 2), then A 1 A 2 e B 1 B 2. If A C is not essential, there is some 0 B C such that A B = 0. Thus A C quot C/B. monic Proposition 5.7 Let A, B C. Assume B is maximal with respect to A B = 0. Then A + B B C e B, and A B e C. Proof. Suppose D/B C/B and A+B D = 0. Then D B and (A + B) D = B, B B so A D A B = 0. Maximality of B implies that D = B, so D/B = 0. Thus. The other part is similar. A+B B e C B Corollary 5.8 If A C, then A is a direct summand of some essential submodule of C. Proof. Zorn s Lemma implies that there is a submodule B C maximal with respect to A B = 0. Thus A B e C by Proposition

39 Corollary 5.9 A module C is semisimple iff the only essential submodule of C is C. Proof. For the forward direction, let A e C. By semisimplicity, C = A B for some B. Now A B = 0, so B = 0, so A = C. For the reverse direction, if A C, then Corollary 5.8 implies that there is some B C so that A B e C. The hypothesis implies that A B = C. Definition. An essential extension of a module A is any module B such that A e B. It is a proper essential extension iff B > A. Proposition 5.10 A module A is injective iff A has no proper essential extensions. Proof. For the forward direction, let A e B. Injectivity implies that B = A A for some A. Then A A = 0, so A = 0, so B = A. Now we do the reverse direction. By Corollary 5.5, it is enough to show that A is a direct summand of any C A. Zorn s Lemma implies that there is some B C maximal with respect to A B = 0. Proposition 5.7 implies that A B B e C. Thus B f : A C quot C/B is an essential monomorphism. There is an injective module E A. A E f g C/B Then ker(g) f(a) = 0, so ker g = 0. Now g maps C/B isomorphically into E, so A = gf(a) e g(c/b). The hypothesis implies that A = g(c/b), so f(a) = C/B, so A B = C, so A B = C. B B Example. Let 0 n Z. Then Z/nZ = nz/n 2 Z e Z/n 2 Z, so Z/nZ is not injective as a Z-module. 38

40 Definition. A submodule A C is essentially closed iff A e B C iff A = B. Exercise 5E If C = A A, then A is essentially closed in C. Proposition 5.11 Let A E with E injective. Then A is essentially closed in E iff A is injective. Proof. We leave the reverse direction as an exercise. For the forward direction, it is enough to show that A has no proper essential extensions. So suppose A e B. A B f E As before, A e B implies that ker f = 0, so A = f(a) e f(b) E. Thus A is essentially closed in E, so f(b) = A = f(a), so B = A. Theorem 5.12 Let A F with F injective. Then there is some E F such that E is injective and A e E. Proof. Zorn s Lemma implies that there is a submodule E F maximal with respect to A e E. If E e E F, then A e E, so E = E by maximality of E. Thus E is essentially closed in F, so Proposition 5.11 implies that E is injective. Definition. An injective hull (or injective envelope) for a module A is any injective module E such that A e E. Injective hulls exist b Theorem 5.4 and Proposition 5.13 Suppose E and E are injective hulls for A and A. If there is an 39

41 isomorphism f : A A, then f extends to an isomorphism E E. = E C A f = A Notation. E(A) is an injective hull for a module A. Suppose A is a torsionfree Z-module. A embeds in a vector space V over Q. V is divisible as a Z-module and hence injective. Look at E, the subspace of V spanned by A. The elements of E are s 1 t1 a s k tk a k = 1a, where s t i, t i Z, t i 0, a i A, t = t 1 t k 0 Z, and a A. Thus E = {1/t a 0 t Z, a A}, which is divisible and hence injective. If 0 x E, then x = 1 a, a 0, so tx = a A. Thus t A e E, the injective hull of A. dim Q (E) is a measure of size. Thus the measure of the size of A is the rank of A. Suppose V is a vector space over a field k. Then V = i I V i, dim k (V i ) = 1. dim k (V ) = card(i). The V i are simple k-modules; they are also not the direct sum of two nonzero subspaces. Definition. A module A is indecomposable iff A is not the direct sum of two nonzero submodules. Definition. A module A is uniform iff A 0 and B 1 B 2 0 for all nonzero submodule B 1, B 2 A (i.e. all nonzero submodules of A are essential). Examples. Q, Z, Z/p n Z, and Z(p ) (p prime) are uniform Z-modules. Lemma 5.14 A nonzero module A is uniform iff E(A) is indecomposable. 40

42 Proof. For the forward direction, suppose E(A) = E 1 E 2. Then (E 1 A) (E 2 A) = 0, implies that some E i A = 0, so E i = 0. For the reverse direction, let 0 B A. Then B A E(A), so there is a submodule F E(A) which is an injective hull for B. F is injective, so E(A) = F F for some F. Indecomposability of E(A) implies that F = 0, so F = E(A), so B e E(A), so B e A. Thus A is uniform. Corollary. The nonzero indecomposables are exactly the uniform injectives. Definition. A module A has finite rank iff E(A) is a finite direct sum of uniform submodules. Proposition 5.15 A module A has finite rank iff there is a finite direct sum of uniform submodules which is an essential submodule of A. Lemma 5.16 Suppose E = E 1 E n with all E i uniform. Then E does not contain a direct sum of n + 1 nonzero submodules. Proof. We proceed by induction on n. The case n = 0 is clear, and the case n = 1 follows because then E is uniform. Suppose n > 1 and A 1 A n+1 E, where each A j 0. Set A = A 1 A n. By induction, A does not embed in a direct sum of n 1 uniform modules. Thus A does not embed in for all i = 1,..., n, so A E i 0 for all i, so A E i E i for all i. n l=1 l i E l Definition. The rank of a module A is a nonnegative integer n if there is a direct sum of n nonzero submodules in A but not a direct sum of n+1 nonzero submodules, and it is infinite otherwise. 41

43 Proposition 5.20 tells us that the rank of A is n < iff E(A) is the direct sum of n uniforms. Example. If A is a torsionfree Z-module, then E(A) is a vector space over Q, and the rank of A is equal to dim Q E(A). Theorem 5.17 (Goldie.) For a module A, the rank of A is finite iff there is no infinite direct sum of nonzero submodules in A. Proof. The reverse direction is clear. For the forward direction, suppose A 0, and E(A) is not a finite direct sum of indecomposable submodules. (1) E(A) = C 0 is not indecomposable. Thus E(A) = M N with M, N 0. At least one of M and N is not a finite direct sum of indecomposables. Now C 0 = C 1 B 1 such that B 1 0 and C 1 is not a finite direct sum of indecomposables. (2) Repeat: C 1 = C 2 B 2 such that B 2 0 and C 2 is not a finite direct sum of indecomposables. Continue this process. Check that B 1, B 2,... are independent. n=1 (B n A) A, and all B n A 0. Thus we have n=1 B n E(A). Thus Corollary 5.18 Every noetherian module has finite rank. Proof. Suppose that the rank of A is infinite. Then Theorem 5.17 implies that there is an infinite direct sum n=1 B n A with all B n 0. So B 1 < B 1 B 2 < B 1 B 2 B 3 <. Corollary 5.19 Every nonzero noetherian module contains a uniform submodule. Proof. If A 0 is noetherian, Corollary 5.18 implies that the rank of A is n < and is nonzero. Proposition 5.20 implies that there is some A 1 A n e A, and 42