Math 260Q: Noetherian Rings

Simon Rubinstein-Salzedo

Fall 2005

0.1 IntroductionThese notes are based on a graduate course on noetherian rings I took from ProfessorKen Goodearl in the Fall of 2005. The textbook was An Introduction to Noncommuta-tive Noetherian Rings by Ken Goodearl and Robert Warfield. The theorem numbersin the notes correspond to those in the book.

0.2 DescriptionAlthough the rings that one typically first meets in an algebra course (rings of inte-gers, polynomial rings, rings of functions) are commutative, the universe holds justas many noncommutative rings. These often arise as rings of operators of variouskinds (think of linear transformations on a vector space). Important examples in-clude rings of differential operators, under which heading one can place the so-calledenveloping algebras of Lie algebras; group rings (amounting to rings of operators onvector spaces built from groups); and “twisted” versions of these, such as the strangelynamed “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 leftideals are finitely generated, and there is a rich general theory of noetherian ringsthat can be applied to them.

This quarter will be mainly devoted to the basic theory of noetherian ring, although afew quantum groups will appear as examples. One key goal will be to prove Goldie’sTheorem, which gives necessary and sufficient conditions for a particularly usefulkind of ring of fractions to exist. Major concepts such as prime ideals will also bestudied, and in order to work effectively with concrete example of noetherian rings,skew polynomial rings will be introduced.

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Chapter 1

A Few Noetherian Rings

Let k be a field and R = k[x1, . . . , xn]. An algebraic variety V ⊆ kn is defined byequations pi(x1, . . . , xn) = 0 for pi ∈ R. If we let I = {p ∈ R | p(x1, . . . , xn) = 0for all (x1, . . . , xn) ∈ V } ⊇ 〈pi’s〉, then I is an ideal of R generated by finitely manypolynomials q1, . . . , qt. Then the variety corresponding to I is V = {(x1, . . . , xn) ∈kn | qj(x1, . . . , xn) = 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(x1, . . . , xn)∂m1

∂xm11

· · · ∂mn

∂xmnn

.

Corresponding to R is a ring of C∞ functions of x1, . . . , xn on R[x1, . . . , xn].

Let R be a ring (which in these notes will always be assumed to have a 1). We writeRR for R viewed as a right R-module and RR for R viewed as a left R-module.

The ascending chain condition (ACC) for submodules is the following statement: Forall chains A1 ⊆ A2 ⊆ · · · for submodules of A, there exists an N so that An = AN

for all n ≥ N .

Let {Aj | j ∈ J} be a collection of submodules of A. A maximal element in thecollection is some Aj0 such that there is no j ∈ J with Aj ) Aj0 .

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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 collec-tion of submodules. Suppose there is no maximal element in A. Then there is someA1 ∈ A and some A2 ∈ A so that A2 ) A1, and so forth. We end up with a sequenceA1 ( A2 ( · · · , which is a contradiction.

We now show that (b) implies (c). Suppose B is a submodule which is not finitelygenerated. Pick b1 ∈ B, and let B1 be the submodule generated by b1. Then there issome b2 ∈ B \B1. Let B2 be the submodule generated by b1 and b2. In this manner,find submodules B1 ( B2 ( · · · . The collection {Bn | n ∈ N} has no maximalelement.

We now show that (c) implies (a). Let A1 ⊆ A2 ⊆ · · · be an ascending chain. Then⋃∞i=1Ai is a submodule generated by some x1, . . . , xk. Then there is some i so that

all the xj’s are in Ai.

Proposition 1.2 Let B ≤ A be modules. Then A is noetherian iff B and A/B areboth noetherian.

Corollary 1.3 If A1, . . . , An are noetherian modules, so is A1 ⊕ · · · ⊕ An.

Corollary 1.4 If R is a right noetherian ring, then all finitely generated right R-modules are noetherian.

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Corollary 1.5 Let S be a subring of a ring R. If S is right noetherian and RS isfinitely generated, then R is noetherian.

Proposition 1.6 If R is a module-finite algebra over a commutative noetherian ringS, 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 generatedby x1, . . . , xn. Then A = x1R+ · · ·+xnR. Set F = Rn

R. There exists an epimorphismφ : F → A given by φ(r1, . . . , rn) = x1r1 + · · · + xnrn. Thus A ∼= F/K for somesubmodule K. If R is right noetherian, then RR is noetherian, so Rn

R is noetherian,so Rn

R/K is noetherian.

If R is a ring, then Mn(R) is a ring.

Proposition 1.7 If R is a right noetherian ring and n ∈ NN and S is a subring of

Mn(R) and S ⊇

r 0

. . .0 r

∣∣∣∣∣∣∣r ∈ R

, then S is right noetherian.

Consider the ring(

R R0 R

)=

{(a b0 c

)∣∣∣∣a, b, c ∈ R}

. We have(a b0 c

)(a′ b′

0 c′

)=(

aa′ ab′ + bc′

0 cc′

).

Example.(

Q Q + Qπ0 Z

)is a subring of

(R R0 R

).

Suppose A and C are rings and B is both a left A-module and a right C-module.

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(A B0 C

)has the standard operations of addition and multiplication. We have

((a 00 0

)(0 b0 0

))(0 00 c

)=

(0 (ab)c0 0

),(

a 00 0

)((0 b0 0

)(0 00 c

))=

(0 a(bc)0 0

).

Definition. Let A and C be rings. An (A,C)-bimodule is an abelian group B withoperations 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 ABC is a bimodule, then the set(A B0 C

)with formal matrix operations is a ring.

Proposition 1.8 Let R =

(A B0 C

)be a formal triangular matrix ring.

(a) R is right noetherian iff A and C are right noetherian and BC is finitely gener-ated.

(b) R is left noetherian iff A and C are left noetherian and AB is finitely generated.

Theorem 1.9 (Hilbert Basis Theorem.) Let S = R[x] be a polynomial ring. If R isright (left) noetherian, then so is S.

Idea. Assume R is right noetherian. Let I be a right ideal with S. Without loss ofgenerality, S 6= 0.

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(1) Set J = {leading coefficients of polynomials in I}∪{0}. Check that J is a rightideal of R.

(2) J is generated by some r1, . . . , rk. Without loss of generality, all ri 6= 0. Supposeri is the leading coefficient of pi ∈ I. Without loss of generality, all pi havedegree n. We’re halfway there: If p ∈ I and deg(p) = d ≥ n, then there area1, . . . , ak ∈ R such that

p1(a1xd−n) + · · ·+ pk(akx

d−n) (∗)

has degree d and the same leading coefficient as p, so p− (∗) ∈ I and has degreeless than n.

(3) Set N = {q ∈ S | q = 0 or deg(q) < n}, a right R-submodule of S generated by1, x, . . . , xn−1. Thus NR is a noetherian module. Thus N∩I is finitely generatedas a right R-module. Say N ∩ I = q1R + · · ·+ qtR.

(4) p1, . . . , pk, q1, . . . , qt generate I.

Corollary 1.10 Let R be a commutative right which is an algebra over a field k. IfR is finitely generated as a k-algebra, then R is noetherian.

Idea. Say R is generated by r1, . . . , rm. Let S = k[x1, . . . , xm] be a polynomial ringover k in m indeterminates. Then there exists a k-algebra homomorphism φ : S → Rsuch that φ(xi) = ri. Then R ∼= S/ kerφ. The Hilbert Basis Theorem implies that Ris noetherian.

Example. Differential operators on R[x]. Let L be the set of linear differentialoperators on R[x], an R-subring of EndR(R[x]) generated by D = d

dxand {µf =left

multiplication by f | f ∈ R[x]}. In fact, L = {µf0 + µf1D + · · ·+ µfnDn | fi ∈ R[x]}.

We have D(µf )(g) = (fg)′ = f ′g + fg′ = (µf ′ + µfD)(f). Thus Dµf= µfD + µf ′ .

Identify each f with µf . Then Df = fD + f ′. In particular, Dx = xD + 1. ThenL = {f0 + f1D + · · ·+ fnD

n | fi ∈ 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 groupalgebra {fmx

m + fm+1xm+1 + · · · + fnx

n | fi ∈ k[y±1],m ≤ n ∈ Z}. Then kG is al-most the Laurent polynomial ring k[y±1][x±1], Since xy = y−1x and xf(y) = f(y−1)x,

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f(y) 7→ f(y−1) is an automorphism α of k[y±1]; xf = α(f)x.

1.1 Automorphism-Twisted Skew Polynomial RingsLet 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 usualpolynomial expressions r0 + r1x + · · · + rnx

n, where each ri ∈ R. Addition is asusual. Multiplication will be derived from the relation xr = α(r)x, so in particularxir = αi(r)xi for all i ≥ 0 and r ∈ R. Hence (rxm)(sxn) = rαm(s)xm+n.

Exercise. This builds a ring.

As an R-module, this new ring is R +Rx+Rx2 + · · · , a free R-module.

Definition. Let R be a ring and α ∈ Aut(R). The statement “S = R[x;α]” meansthat

(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 {xi | 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 thatφ |R= idR and φ(x) = y.

Idea. The only possibility is φ(r0 + r!x+ · · ·+ rnxn) = r0 + r1y + · · ·+ rny

n. Checkthat this works.

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Lemma 1.11 Let S = R[x;α], φ : R→ T a ring homomorphism, and y ∈ T . Assumeyφ(r) = φα(r)y for all r ∈ R. Then there is a unique ring homomorphism ψ : S → Tsuch that ψ |R= φ and ψ(x) = y.

Example. Let k be a field, R = k[y] a polynomial ring, and q ∈ k× (the multiplica-tive group of k). Then there exists a unique k-algebra automorphism α of R such thatα(y) = qy. Look at S = R[x;α]. The key property is that xy = qyx. The elementsare

∑ni=0 fix

i, fi ∈ R, or∑n

i=0

∑mj=0 λijy

jxi, λij ∈ k.

Context. If k is a field, then k2 is the “affine plane over k.” O(k2) is the coefficientring of k2, which is the polynomial ring k[x, y]. This is a k-algebra with generators xand y and relation xy = yx.

Definition. Let k be a field and q ∈ k×. Then corresponding quantized coordinatering of k2 is

Oq(k2) = k〈x, y | xy = qyx〉.

Claim. Oq(k2) is a skew polynomial ring as in the last example.

(a) Keep Oq(k2) = k〈x, y | xy = qyx〉.

(b) R = k[y] (polynomial ring), α ∈ AutR is given by p(y) 7→ p(qy), S = R[x;α].

(c) xy = qyx in S, so there is a unique k-algebra homomorphism φ : Oq(k2) → S

such that φ(x) = x and φ(y) = y.

(d) There is a unique k-algebra homomorphism θ : R→ Oq(k2) such that θ(y) = y.

Then xy = qyx, so xyi = qiyix for all i ≥ 0, so xθ(yi) = θα(yi)x, so xθ(r) =θα(r)x for all r ∈ R. Lemma 1.11 implies that there exists a unique k-algebrahomomorphism ψ : S → Oq(k

2) such that ψ |R= θ and ψ(x) = x. Notice thatψ(y) = y.

(e) φψ(x) = x and φψ(y) = y, so φψ = idS. Similarly ψφ(x) = x and ψφ(y) = y,so ψφ = idOq(k2).

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Proposition 1.13 If k is a field and q ∈ k×, then Oq(k2) is a skew polynomial ring

k[y][x;α], where k[x] is a polynomial ring and α is the k-algebra automorphism ofk[y] such that α(y) = qy.

Definition. Let R be a ring and α ∈ AutR. “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 {xi | 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 Oq((k

×)2) = k〈x, x′, y, y′ | xx′ = x′x = y′y =yy′ = 1, xy = qyx〉.

Exercise 1O. Oq((k×)2) = k[y±1][x±1;α], where α(y) = qy.

Theorem 1.14 Let S = R[x;α] and α ∈ AutR. If R is right (left) noetherian, so isS.

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. Jis closed under addition and subtraction. Let α ∈ J and r ∈ R. There existsp = axn+(lower terms)∈ I. pr = aαn(r)xn+(lower terms)∈ I, so aαn(r) ∈ J .Instead, pα−n(r) = arxn+(lower terms)∈ I, so ar ∈ J . Thus J is a right idealof R.

(2) J is generated by a1, . . . , ak. There exists a pi = aixni+(lower terms)∈ I. Thus

without loss of generality all pi have degree n.

(2’) We’re halfway. Suppose p ∈ I with deg(p) = d ≥ n. Let p = axd+(lowerterms), with a ∈ J . Then a = a1r1 + · · · + akrk for some ri ∈ R. Then

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p1α−n(r1) + · · · + pkα

−n(rk) = (a1r1 + · · · + akrk)xn+(lower terms). Then

p1α−n(r1)x

d−n + · · · + pkα−n(rk)x

d−n = axd+(lower terms), so there is a poly-nomial 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. LetJ denote the set of leading coefficients of elements of I together with 0. Supposea, b ∈ J . There exists a p = axm+(lower terms) and q = bxn+(lower terms)∈ I.What if m < n? Then xn−mp = αn−m(a)xn+(lower terms). Thus we only getαn−m(a) + b ∈ J .

Definition. Let T be a ring. The opposite ring T op has the same set as T and thesame addition as T , but a new multiplication ∗ defined by a ∗ b = ba.

Look at S = R[x;α]. Rop is a subring of Sop, and x ∈ Sop. Also x ∗ r = rx =xα−1(r) = α−1(r) ∗ x. Check that S is a free right R-module with basis {xn | n ≥ 0}.Then Sop is a free left Rop-module with basis {xn | n ≥ 0}. Thus Sop = Rop[x;α−1].

If R is left noetherian, then Rop is right noetherian, so Rop[x;α−1] = Sop is rightnoetherian, and so S is left noetherian.

Corollary 1.15 Let T = R[x±1;α] and α ∈ AutR. If R is right (left) noetherian, sois T .

Definition. A ring R is simple iff

(a) R 6= 0.

(b) The only ideals of R are 0 and R.

Let α ∈ AutR. An α-ideal of R is an ideal I CR such that α(I) = I. R is α-simpleiff

(a) R 6= 0.

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(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 thek-algebra automorphism such that α(y) = y + 1. Then R is α-simple.

Theorem 1.17 Let T = R[x±1;α], α ∈ AutR. Then T is simple iff

(a) R is α-simple.

(b) No positive power of α is inner.

Idea for ⇒(a). If I CR is an α-ideal, then IT C T .

Idea for ⇐. Let 0 6= I C T . Let n be the minimum degree for nonzero elementsof I ∩ S (where S = R[x;α]). Let J be the set of leading coefficients of elementsof I ∩ S of degree n together with 0. Note that 0 6= I ∩ S C S, J C R, andJ 6= 0. J is an α-ideal: if 0 6= a ∈ J , then there exists some p = axn+(lowerterms)∈ I ∩ S, so xpx−1 ∈ I ∩ S and xpx−1 = α(a)xn+(lower terms), so α(a) ∈ J .Thus α(J) ⊆ J . Use xpx−1 to see that α−1(J) ⊆ J . Thus (a) implies thatJ = R, so there is some p = xn + an−1x

n−1 + · · · + a0 ∈ I ∩ S. If a0 = 0, thenp = (xn−1 + an−1x

n−2 + · · · + a1)x, so 0 6= px−1 ∈ I ∩ S, which is a contradic-tion. Thus a0 6= 0. We have xpx−1 = xn + α(an−1)x

n−1 + · · · + α(a0) ∈ I ∩ S.Then p − xpx−1 ∈ I ∩ S and has degree less than n, so p − xpx−1 = 0. Thusα(ai) = ai for all i. For all r ∈ R, pr = αn(r)xn + an−1α

n−1(r)xn−1 + · · · + a0r.αn(r)p = αn(r)xn + αn(r)an−1x

n−1 + · · ·+ αn(r)a0. Then pr− αn(r)p = 0 because ithas degree less than n, so aiα

i(r) = αn(r)ai for all i, and a0r = αn(r)a0 for all r ∈ R.Continuing this way, we see that a0 is invertible in R. Thus αn(r) = a0ra

−10 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 = Oq((k×)2). Then T is simple iff q

is not a root of unity.

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Chapter 2

Skew Polynomial Rings

Example. Consider R[x] with multiplication operators and ddx

= D. Then Df =fD + f ′ for all f ∈ R[x].

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 {xj | j ≥ 0}.

We have x2r = x(rx + δ(r)) = (rx + δ(r))x + δ(r)x + δ2(r) = rx2 + 2δ(r)x + δ2(r).More generally, we have

xnr =n∑

i=0

(n

i

)δn−i(r)xi.

Exercise. R[x; δ] exists.

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We now look at some key examples. Let K be any ring, R = K[y], and δ = ddy

.

Weyl Algebra. A1(K) = K[y][x; d

dy

], so xy = yx+ 1. In general,

An(K) = K[y1]

[x1;

d

dy1

][y2]

[x2;

d

dy2

]· · · [yn]

[xn;

d

dyn

].

Definition. Let R be a ring and δ a derivation on R. A δ-ideal of R is any I C Rsuch that δ(I) ⊆ I. R is δ-simple iff R 6= 0 and the only δ-ideals are 0 and R. δ isinner iff there exists an a ∈ R such that δ(r) = ar − ra for all r ∈ R; otherwise δ isouter.

Proposition 2.1 Let R be a Q-algebra and δ a derivation on R. Then R[x; δ] issimple iff

(a) R is δ-simple.

(b) δ is outer.

Corollary 2.2 If K is a field of characteristic 0, then An(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 {xi | 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 weshould have maps α, δ : R → R such that xr = α(r)x + δ(r) for all r ∈ R. We have

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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 endo-morphism 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 ∈ AutR.

2.1 Building Rings from “Operators”Take an abelian group A (or a vector space over a field k). Label the ring E =EndZ(A) (or algebra E = Endk(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-derivationδ 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 setE = EndZ(R[z]).

(2) Consider the map R → E given by r 7→(lert multiplication by r). Identify Rwith its image, a subring of E.

(3) Define x ∈ E by

x

(∑i

rizi

)=∑

i

(α(ri)z + δ(ri))zi.

Check that xr = α(r)x+ δ(r) for all r ∈ R.

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(4) Set S to be the subring of E generated by R∪{x}. We get that S is a ring andR is a subring of S and x ∈ S for free.

(5) Check that S is a free left R-module with basis {xi | i ≥ 0}.

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Chapter 3

Prime Ideals

Definition. A prime ideal in a ring R is an ideal P C R such that P 6= R and forall ideals I, J CR, if IJ ⊆ P , then I ⊆ P or J ⊆ P .

Proposition 3.1 Let R be a ring, P CR, and P 6= R. The following are equivalent:

(a) P is a prime ideal.

(b) If I, J CR and I, J ) P , then IJ 6⊆ P .

(c) R/P is a prime ring (i.e. if I , J CR/P and I J = 0, then I = 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 CR.

• If x, y ∈ R, then RxR,RyRCR.

Proposition 3.2 All maximal ideals are prime.

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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]/(xy2), the minimalprimes are (x)

(xy2)and (y)

(xy2).

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 workwith (X,≤), where “≤” is “⊇.” Let Y be a nonempty chain in X; we need to showthat it has a lower bound in X. Set Q =

⋂Y C R. We claim that Q is prime. Let

x, y ∈ R such that xRy ⊆ Q and x 6∈ Q. There is some Q′ ∈ Y such that x 6∈ Q′.Thus y ∈ Q′. Look at some arbitrary Q′′ ∈ Y . If Q′ ⊆ Q′′, then y ∈ Q′′. If Q′ ⊇ Q′′,then x 6∈ Q′′, so y ∈ Q′′. Thus y ∈ Q and Q is prime.

Definition. If I C R, a prime P ⊆ I is minimal over I iff P is minimal amongprimes 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 ICR. Then there are onlyfinitely many primes minimal over I, and some product of them (with repetitions) iscontained in I.

Example. R = R[x, y], I = (xy2). Then P1 = (x) and P2 = (y) are the only primesminimal over I, and P1P

22 = I.

Proof. We claim that it suffices to prove

there exist primes P1, . . . , Pn ⊇ I such that P1P2 · · ·Pn ⊆ I. (∗)

To see this, note that each Pi contains a prime Qi minimal over I, and

Q1Q2 · · ·Qn ⊆ P1P2 · · ·Pn ⊆ I.

If Q is a prime minimal over I, then P1 · · ·Pn ⊆ I ⊆ Q, so some Pj ⊆ Q. ThenI ⊆ Pj ⊆ Q, so Q = Pj. Thus {primes minimal over I} ⊆ {P1, . . . , Pn}.

17

Now suppose (∗) fails. The noetherian condition implies that there is an ideal I ⊆ Imaximal such that (∗) fails. If I is prime, take P1 = I, which is a contradiction. ThusI is not prime, so there exist J,K ) I such that JK ⊆ I. Thus (∗) holds for J andK, so (∗) holds for I, which is a contradiction. Hence (∗) always holds.

Definition. An ideal I C R is semiprime iff I is the intersection of prime ideals.(This includes R, which is the empty intersection.)

Theorem 3.7 I C R is semiprime iff for all x ∈ R, xRx ⊆ I implies x ∈ I. (If R iscommutative, x2 ∈ I implies x ∈ I, so I is a radical ideal.)

Proof. For the forward direction, we have I =⋂

α∈A Pα, where Pα are prime ideals.But xRx ⊆ Pα implies x ∈ Pα.

For the reverse direction, it suffices to prove that if x ∈ R \ I, there exists a primeP ⊇ I such that x 6∈ P . Set x0 = x. Since x0 6∈ I, x0Rx0 6⊆ I, so there exists anx1 = x0r0x0 6∈ I. Then x1 6∈ I, so there is an x2 = x1r1x1 6∈ I, and so forth. Zorn’sLemma implies that there is an ideal P ⊇ I which is maximal with respect to thecondition P ∩ {x0, x1, x2, . . .} = ∅. We now claim that P is prime. Suppose P is notprime. Then there exist J,K ) P such that JK ⊆ P . The maximality of P impliesthat there exist xj ∈ J and xk ∈ K. Let m = max{j, k}; note that xm ∈ J ∩ K.Then xm+1 ∈ xmRxm ⊆ JK ⊆ P , which is a contradiction. Thus P is prime.

Corollary 3.8 The following are equivalent for I CR:

(a) I is semiprime.

(b) If J CR and J2 ⊆ I, then J ⊆ I.

(c) If J is a right ideal of R and J2 ⊆ I, then J ⊆ I.

(d) If J is a left ideal of R and J2 ⊆ I, then J ⊆ I.

(e) If J CR and J ) I, then J2 6⊆ I.

18

Definition. A right or left ideal J in a ring is nilpotent iff there is some n ∈ N suchthat Jn = 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, thesmallest semiprime ideal.

Example. Let R = Z/nZ, where n = pm11 · · · pmk

k . The prime ideals in R are piZ/nZ.The prime radical is (p1 · · · pk)Z/nZ.

Theorem 3.11 Let N be the prime radical of a right or left noetherian ring R. ThenN is nilpotent and N contains all nilpotent right or left ideals.

Definition. Let A = AR. For X ⊆ A, the annihilator of X in R is ann(X) =annR(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 annA(Y ) = {a ∈ A | ay = 0 for all y ∈ Y }. For X ⊆ R,`. ann(X) = {r ∈ R | rx = 0 for all x ∈ X} and r. ann(X) = {r ∈ R | xr = 0 for allx ∈ X}.

Example. Let R = Z and A = Z/6Z. Then annR(2 + 6Z) = 3Z C Z andannA(2) = {3 + 6Z, 6Z}C A.

If B is a submodule of A, then annR(B) is an ideal of R. If I CR, then annA(I) is asubmodule of A.

Definition. A module AR is faithful iff annR(A) = 0. A is fully faithful iff A isfaithful and all its nonzero submodules are faithful.

Example. Let R = Z. Then nZ (n 6= 0), Z, Q, and R are fully faithful.⊕∞

i=1 Z/nZis faithful but not fully faithful.

19

If AR 6= 0 and I = annR(A) C R, then A is also an R/I-module and is faithful.Possibly there is some 0 6= B ≤ A with annR(B) > I.

Proposition 3.12 Let AR 6= 0. Suppose there is a P C R maximal among annihi-lators of nonzero submodules of A. Then P is a prime ideal and annA(P ) is a fullyfaithful (R/P )-module.

Idea. Suppose I, J C R with I, J ) R. Then annR(A) cannot properly containP , so annR(A) ) I. Thus AI 6= 0. Now annR(AI) ) P , so it does not containJ , and AIJ 6= 0. Thus IJ 6⊆ annR(A). There exists a nonzero B ≤ A such thatP = annR(B). Since I ) annR(B), BI 6= 0. Now annR(BI) 6⊇ J , so BIJ 6= 0. ThusIJ 6⊆ P . Thus P is prime.

Suppose R is right noetherian and AR is finitely generated, and A 6= 0.

(1) Since R is right noetherian, there exists a P1 CR maximal among annihilatorsof nonzero submodules of A. Proposition 3.12 implies that P1 is prime, andA1 = annA(P1) is a fully faithful (R/P1)-module.

(2) If A1 6= A, repeat for A/A1: there exists a P2 C R and a submodule A2 ≤ Awith A2 > A1 such that A2/A1 is a fully faithful (R/P2)-module.

(3) Continue. Since AR is noetherian, there is some n so that An = A.

Proposition 3.13 Suppose R is right noetherian and AR is finitely generated. Thenthere exist submodules

A0 = 0 < a! < · · · < An = A

and prime ideals P1, . . . , Pn CR such that for all i, Ai/Ai−1 is a fully faithful (R/Pi)-module.

Proposition 3.14 Suppose R,A,Ai, Pi are as in Proposition 3.13. Set I = annR(A).If P is any prime minimal over I, then P is some Pj.

20

Proof. (Ai/Ai−1)Pi = 0 implies that AiPi ⊆ Ai−1. So APnPn−1 · · ·P1 = 0. ThusPnPn−1 · · ·P1 ⊆ I ⊆ P . Thus there exists a j such that Pj ⊆ P . But I ⊆ Pj ⊆ P . Sothe minimality of P implies that P = Pj.

Definition. A module A is simple (or irreducible) iff A 6= 0 and the only submod-ules of A are 0 and A.

If A is simple, then ann(A) is the only ideal which is an annihilator of 0 6= a ∈ A. It ismaximal and hence prime. If A = AR is simple, then annR(A) is a right primitiveideal of R.

Proposition 3.16 In any ring R, the intersection of left primitive ideals, the in-tersection of right primitive ideals, the intersection of maximal left ideals, and theintersection of maximal right ideals all coincide.

Definition. This intersection is the Jacobson radical of R, denoted J(R). R isa semiprimitive ring iff J(R) = 0. I C R is a semiprimitive ideal iff R/I is asemiprimitive ring.

Note. Primitive ideals are prime. Thus J(R) is semiprime and contains the primeradical.

Theorem 3.22 Let S = R[x; δ], where R is a commutative noetherian Q-algebra andδ is a derivation.

(a) If P C 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 C S, QS is prime, and QS ∩R = Q.

(ii) If δ(R) 6⊆ Q, then QS is the unique prime of S whose intersection with Ris equal to Q.

21

(iii) If δ(R) ⊆ Q, then there exist infinitely many primes P C S such thatP ∩ R = Q. They are inverse images of primes in K[x], where K is thequotient field of R/Q under the map

Squot7−→ S

QS= (R/Q)[x] ⊆ K[x].

Example. Let k be an algebraically closed field of characteristic 0. Let R = k[y],δ = (y2 − 1) d

dy, and S = R[x; δ]. The prime ideals in R are 0 and 〈y − α〉 (α ∈ k).

The prime δ-ideals are 0, 〈y − 1〉, 〈y + 1〉. If Q = 0, then δ(R) 6⊆ Q, so only P = 0 isin 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:

NNNNNNNNNNNNNN · · · 〈y − 1, x− λ〉 · · ·

pppppppppppppp

NNNNNNNNNNNNNN · · · 〈y + 1, x+ λ〉 · · ·

pppppppppppppp

〈y − 1〉

TTTTTTTTTTTTTTTTTTT 〈y + 1〉

jjjjjjjjjjjjjjjjjjj

0

22

Chapter 4

Semisimple Modules, ArtinianModules, and Torsionfree Modules

Commutative domains are contained in quotient fields. The modules of these quo-tient fields are vector spaces, which are direct sums of 1-dimensional submodules.Noncommutative domains are contained in division rings of fractions. The modulesof these rings of fractions are direct sums of 1-dimensional submodules. We wouldlike to contain noetherian prime rings in some sort of quotient object whose modulesare direct sums of copies of some particular simple module.

Definition. For a module A, the socle of A is the sum of all simple submodulesof A. It is denoted by soc(A). A is semisimple (or completely reducible) iffsoc(A) = A.

Examples.

(1) Vector spaces.

(2) A ∈ Z-mod. soc(A) = {a ∈ A | the order of a is prime}. So if n =pm1

1 pm22 · · · pmk

k , where the pi are distinct primes and mi > 0, then soc(Z/nZ) =n

p1p2···pkZ/nZ.

Say {Ai | i ∈ I} is a collection of R-modules. The direct product has the Cartesianproduct set

∏i∈I Ai with natural operations. The (external) direct sum

⊕i∈I Ai is

the set of a ∈∏

i∈I Ai with ai = 0 for all but finitely many i. Now suppose all Ai are

23

submodules of some A. Then∑

i∈I Ai is the submodule generated by⋃

i∈I Ai. Thereexists a homomorphism (“sum map”) S :

⊕i∈I Ai →

∑i∈I Ai given by a 7→

∑i∈I ai.

Check that s is injective iff for all j ∈ I, Aj ∩(∑

i∈Ii6=j

Ai

)= 0. In this case, we say

{Ai | i ∈ I} is independent. If this happens, we identify∑

i∈I Ai (the internal directsum) with

⊕i∈I Ai.

Proposition 4.1 If A is a module, soc(A) is the direct sum of some family of simplesubmodules 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. SupposeY = {Bi | i ∈ I}, where the Bi are simple and B =

∑i∈I Bi =

⊕i∈I Bi ≤ soc(A).

Suppose B 6= soc(A). Then there exists a simple submodule S ≤ A such that S 6⊆ 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 directsummand.

Proof. We first do the forward direction. Let B ≤ A. Zorn’s Lemma implies thatthere 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 6⊆ B ⊕C,so {B,C, S} is independent, so B ∩ (C ⊕ S) = 0, which is a contradiction. ThusB ⊕ C = A.

For the reverse direction, we have A = soc(A) ⊕ B for some B ≤ A. SupposeB 6= 0. Choose 0 6= b ∈ B and set C = bR or Rb. Zorn’s Lemma implies thatthere is some submodule M ≤ C maximal with respect to b 6∈ M . Then C/Mis a simple module. The hypothesis implies that there is some N ≤ A such thatA = M ⊕ N ,so C = M ⊕ (N ∩ C). (This follows from the Modular Law, whichstates that if M ≤ C ≤ A and N ≤ A, then C ∩ (N + M) = (C ∩ N) + M .) NowN ∩C ∼= C/M is simple. But N ∩C ≤ soc(A), which is a contradiction. Thus B = 0and A = soc(A).

24

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) RR is semisimple.

(d) RR is semisimple.

(e) R = 0 or R ∼= Mn1(D1)⊕Mn2(D2)⊕· · ·⊕Mnk(Dk) for some ni ∈ N and division

rings Di.

Proof. We first show that (c) implies (a). Let A = AR. There exists a free moduleFR with a basis {xa | a ∈ A}. There is a unique homomorphism f : F → A such thatf(xa) = a for all a ∈ A. So if K = ker(f), A ∼= F/K. Since RR is semisimple andF =

⊕a∈A πaR is isomorphic to a direct sum of copies of RR, 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 6= 0.

(1) Since RR is semisimple, RR =⊕

i∈I Si, where Si are simple right ideals. RR iscyclic, so I is finite.

(2) Group the Si by isomorphism class and relabel. Then

RR =k⊕

i=1

ni⊕j=1

Sij

such that Sij are simple right ideals, k, ni ∈ N. Furthermore, Sij∼= Stn iff i = t.

(3) HomR(Sij, Stn) = 0 for i 6= t.

25

(4) RR = R1⊕· · ·⊕Rk, whereRi =⊕ni

j=1 Sij is a right ideal. By (3), HomR(Ri, Rt) =0 for i 6= t, so Ri CR for all i.

(5) Now each Ri is a ring, and R ∼= R1 × · · · ×Rk. Thus without loss of generality,R ∼= R1 =

⊕n1

j=1 Tj, all Tj∼= T1 (as right ideals). Thus RR

∼= T n1 . Write

T n1 =

x

1

...xn

∣∣∣∣∣∣∣xi ∈ T1

. Then D1 = EndR(T1), a division ring (by Schur’s

Lemma). There is a map θ : Mn1(D1) → EndR(T n1 ). Then

[fij] 7→

x1

...xn1

7→ [fij] ·

x1...xn1

.

Exercise: θ is a ring isomorphism. Thus Mn1(D1) ∼= EndR(T n1 ) ∼= EndR(RR) ∼=

R.

We now outline the proof that (e) implies (c).

(1) R = 0 is easy, so without loss of generality assume R 6= 0. Without loss ofgenerality, R = R1×· · ·×Rk, where Ri = Mni

(Di), ni ∈ N, and Di is a divisionring.

(2) soc(RR) = soc(R1)× · · · × soc(Rk). Thus without loss of generality, R = R1 =Mn1(D1).

(3) For i = 1, . . . , ni, set Si = {x ∈ R | all nonzero entries in x are in row i}. Exer-cise: Si is a simple right ideal of R. Since RR = S1⊕· · ·⊕Sn, RR 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 A0 > A1 >A2 > · · · . A ring R is right (left) artinian if RR (RR) is artinian.

26

Proposition. Every semisimple ring R is artinian and noetherian.

Proof. RR = S1 ⊕ · · · ⊕ Sn, where the Si are simple right ideals. But Si is artinianand noetherian for all i. Thus RR is artinian and noetherian. So is RR.

Theorem. (Noether.) Let R be a ring. The following are equivalent.

(a) R ∼= Mn(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 6= 0. Then R ∼= Mn1(D1) × · · · ×Mnk

(Dk). 0×Mn2(D2)× · · · ×Mnk(Dk) 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 toshow that if D is a simple ring, then Mn(D) is also simple.

(b) implies (c) is clear. To see that (c) implies (b), let R 6= 0. Since R is rightartinian, there is a right ideal S minimal among nonzero right ideals, so S is simple.Thus soc(RR) 6= 0. We leave it as an exercise to show that soc(RR) C R. Thussoc(RR) = 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

Then A/T (A) is torsionfree.

Let R = Z and A be a torsion Z-module. Then A =⊕

p prime Tp(A), where Tp(A) ={a ∈ A | pna = 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 adomain, rs 6= 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 andX is closed under multiplication. An R-module A is X-torsion iff every element ofA is annihilated by an element of X. A is X-torsionfree iff no nonzero element ofA is annihilated by a nonzero element of X.

Question. Let A = AR, X ⊆ R a multiplicative set, and tX(A) = {a ∈ A | ax = 0for some x ∈ X}. Is (when is) this a submodule of A?

Example. Let A = R/xR, where x ∈ X. Then 1x = 0, so 1 ∈ tX(A). So if tX(A)is a submodule, it equals A. So for all r ∈ R, we need ry = 0 for some y ∈ X, sory ∈ xR, so ry = xs for some s ∈ R.

Definition. Let R be a ring and X ⊆ R a multiplicative set. X satisfies the rightOre condition iff for all r ∈ R and x ∈ X, there exists s ∈ R and y ∈ X such thatry = xs. Then X is a right Ore set.

Lemma 4.21 Let X ⊆ R be right Ore.

(a) For all x1, . . . , xn ∈ X, there exists s1, . . . , sn ∈ R such that x1s1 = · · · = xnsn ∈X.

(b) For all AR, tX(A) is a submodule of A.

Proof.

28

(a) It is enough to do n = 2. Given x1, x2 ∈ X, the right Ore condition impliesthat there exist s ∈ R and y ∈ X such that x1y = x2s.

(b) Let a1, a2 ∈ tX(A). There exist x1, x2 ∈ X such that aixi = 0. There ex-ist s1, s2 ∈ R such that x1s1 = x2s2 = y ∈ X. Thus aiy = aixisi = 0, so(a1±a2)y = 0. If r ∈ R, there exist s ∈ R and z ∈ X such that rz = x1s. Thus(a1r)z = a1x1s = 0.

Lemma 4.22 Let R be a ring, X ⊆ R a right Ore set, and A = AR.

(a) tX(A) is an X-torsion module, and A/tX(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 thesame 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 = AR is a noetherianX-torsionfree module, and f ∈ EndR(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 thatker(fn) = ker(fn+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/fn(A) is X-torsion. Look at a ∈ ker(f). Thereis some x ∈ X such that ax = 0 in A/fn(A), so ax = fn(b) for some b ∈ A.fn+1(b) = f(a)x = 0, so fn(b) = 0. Thus ax = 0, and so a = 0.

Corollary. If A = AR is noetherian and f ∈ EndR(A) is epic, then f ∈ AutR(A).

Proof. X = {1}.

Corollary. If R is right noetherian and x, y ∈ R with xy = 1, then yx = 1.

29

Proof. Let f ∈ EndR(RR) be left multiplication by x. If xy = 1, then f is epic, sof is an automorphism. Thus r. annR(x) = ker(f) = 0. x(yx− 1) = 0, so yx− 1 = 0.

30

Chapter 5

Injective Hulls

A commutative domain R is contained in its quotient field F . Torsionfree R-modulesbecome vector spaces over F .

For every vector v and 0 6= r ∈ R, there is a vector 1rv so that r

(1rv)

= v.

Definition. Let R be a commutative domain. A = RA is divisible iff for all a ∈ Aand 0 6= r ∈ R, there is some b ∈ A such that rb = a.

Exercise. Let R be a commutative domain with quotient field F . A = RA is theR-module of a vector space over F iff A is torsionfree and divisible.

Say A = RA, a ∈ A, and 0 6= 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 suchthat f(sr) = sa for all s ∈ R. This is defined irrespective of whether there is such ana′. If there is such such an a′, then there is a homomorphism f : R → A such thatg(s) = sa′ for all s ∈ R. g(sr) = sra′ = sa = f(sr), so g |(r)= f .

31

Proposition. Let R be a commutative domain and A = RA. Then A is divisibleiff for every principal ideal I C R and every homomorphism f : I → A, there is anextension g : R→ A.

Proof. We show only the forward direction; the reverse direction is similar. If ICR isprincipal, say I = (r), then r = 0 implies that f is the zero map R→ A. Thus with-out loss of generality, r 6= 0. Let f : I → A be a homomorphism, f(r) = a ∈ A. Bydivisibility, there is an a′ ∈ A such that ra′ = a. Define a homomorphism g : R→ Asuch 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 = AR is injective iff for every right R-modules B ≤ C, all homomor-phisms 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. ZZ is not injective.

2Z⊆ //

2n7→n��

Z

6∃~~}}

}}

Z

Z(Z/nZ) is not injective.

Proposition 5.1 (Baer’s Criterion.) A = AR is injective iff for any right ideal I ofR, all homomorphism I → A extend to homomorphisms R→ A.

32

Proof. The forward direction is obvious. For the reverse direction, let B ≤ C be rightR-modules and f : B → A a homomorphism. Let X = {(D, g) | B ≤ D ≤ C, g ∈HomR(D,A), g extends f}. Then (B, f) ∈ X. Define ≤ on X: (D1, g1) ≤ (D2, g2) iffD1 ≤ D2 and g@ extends g1. Check that ≤ is a partial order, and nonempty chainshave upper bounds: A chain {(Di, gi) | i ∈ I} has an upper bound

(⋃i∈I Di,

⋃i∈I Bi

).

Zorn’s Lemma implies that there is a maximal element (D∗, g∗) ∈ X. If D∗ = C,we’re done. Suppose D∗ 6= C, and choose c ∈ C \ D∗. Set D′ = D∗ + cR. LetI = {r ∈ R | cr ∈ D∗}, a right ideal of R. Define h : I → A by h(r) = g∗(cr). Checkthat 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 awell-defined homomorphism g′ : D∗ + cR→ A given by g′(d+ cr) = g∗(d) +xr for alld ∈ D∗ and r ∈ R.

Corollary. Let R be a commutative PID. Then an R-module A is injective iff it isdivisible.

Example. The Z-modules Q and Q/Z are injective. So is

Z(p∞) =

[∞⋃

n=0

1

pnZ

]/Z,

the p-primary part of Q/Z.

Proposition. If R is a commutative PID, every R-module A is a submodule of aninjective R-module.

Proof. Write A ∼= F/K for some free R-module F and some K ≤ F . F is a directsum of copies of RR. Set D to be the direct sum of the quotient field (over someindex set). Then D is a divisible R-module. F ⊆ D. Now A ∼= F/K ≤ D/K, andD/K is divisible. Identify A with its image in D/K.

General Problem. Let P be some property of modules. Suppose A = AR and weknow that A is isomorphic to a submodule of an R-module B satisfying P . Why isA equal to a submodule of such a module?

(1) We are given a monomorphism f : A→ B.

33

(2) Choose a set B1 such that B1 ∩ A = ∅ and there is a bijection β : B → B1.Use β to transfer the R-module operations from B to B1: for all x, y ∈ B1,define x + y = β(β−1(x) + β−1(y)), and for all r ∈ R, define xr = β(β−1(x)r).Now B1 is an R-module, and β : B → B1 is a module isomorphism. Alsog = βf : A→ B1 is monic.

(3) Define B2 = A t (B1 \ g(A)). There is a bijection γ : B1 → B2 such that{γ(g(a)) = a for all a ∈ A,γ(b) = b for all b ∈ B1 \ g(A).

Use γ to make B2 into an R-module; now γ is a module isomorphism. γg : A→B2 is a monomorphism, and γg(a) = a for all a ∈ A. Thus A is a submodule ofB2. B ∼= B1

∼= B2, so B2 has P .

Analogue. Constructing an algebraic closure of a field k.

(1) Given p ∈ k[x], p 6∈ 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 k1 ⊇ k such that all nonconstant polynomials of k[x] haveroots in k1.

(4) Repeat: There exist fields k ⊆ k1 ⊆ k2 ⊆ · · · such that all nonconstant polyno-mials in ki[x] have roots in ki+1.

(5) Thus k =⋃∞

n=1 kn 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 ∈ HomR(B,A). Then there is amodule F ≥ A such that f extends to a homomorphism C → F .

Proof. The aim is to fill in the diagram

B⊆ //

f��

C

∃�����

A⊆ //___ F

34

as follows:B

⊆ //

f

��

C

ir

��

πir��

Aπi` //

i` ++

A⊕CD

A⊕ C

πccHHHHHHHHH

Define D = {(f(b),−b) | b ∈ B}, a submodule of A ⊕ C. If a ∈ ker(πi`), then(a, 0) ∈ kerπ, so (a, 0) = (f(b),−b) for some b ∈ B, so a = 0. Thus πi` is monic.Identify A with the image of πi`.

Lemma. Let A and Ci (i ∈ I) be R-modules, Bi ≤ Ci, and fi ∈ HomR(Bi, A) for alli. Then there is an R-module F ≥ A such that for all i, fi extends to a homomor-phism Ci → F .

Proof. Set B =⊕

i∈I Bi ≤ C =⊕

i∈I Ci. Define f =∑

i∈I fi : B → A. Labelnatural injections ji : Bi → B and qi : Ci → C. The previous Lemma implies thatthere is an R-module F ≥ A and a homomorphism g : C → F which extends f .

Bi⊆ //

ji

@@@

@@@@

fi

��

Ciqi

~~~~~~

~~~

gqi

��

B⊆ //

f

��

C

g

��A

⊆ // F

Corollary. For any A = AR, there is a module F1(A) ≥ A such that for any rightideal I of R, all homomorphisms I → A extend to homomorphisms R→ F1(A).

So we make A ≤ F1(A) ≤ F2(A) such that I → Fn(A) extends to R→ Fn+1(A).

Suppose I is a right ideal of R and f a homomorphism I →⋃∞

n=1 Fn(A). If I isfinitely generated, say I =

∑tj=1 xjR, then there is some n such that f(xj) ∈ Fn(A)

35

for all j, so f(I) ≤ Fn(A), so f extends to R → Fn+1(A) ≤ U . Thus if R is rightnoetherian,

⋃∞n=1 Fn(A) is injective.

Theorem 5.4 Any A = AR 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 Aα for all α ≤ γ such that A ≤ A1 = F1(A).When Aα is defined, set Aα+1 = F1(Aα). When β is a limit ordinal less than or equalto γ, set Aβ =

⋃α<β Aα.

We claim that Aγ is injective. By Baer’s Criterion, we only need to show that forany right ideal I of R, any f ∈ HomR(I, Aγ) extends to a homomorphism R → Aγ.f(I) ≤ Aγ, and card(f(I)) ≤ ℵ. γ is a limit ordinal, so Aγ =

⋃α<γ Aα. Thus for

all x ∈ I, there is an αx < γ such that f(x) ∈ Aαx and card({αx | x ∈ I}) ≤ ℵ, sosup{αx | x ∈ I} = σ has cardinality at most γ, so σ < γ. Thus f(I) ≤ Aσ, so fextends to a homomorphism R→ Aσ+1 ≤ Aγ.

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 ahomomorphism f : B → A.

A⊆ //

id��

B

∃f��~~

~~

A

Thus B = A⊕ ker(f).

For the reverse direction, by Theorem 5.4, there is an injective module B ≥ A. ThenB = A⊕A′ for some A′ ≤ B. We observe that A1 ⊕A2 is injective iff A1 and A2 areboth injective.

Definition. A submodule A ≤ B is essential iff for every nonzero submodule C of

36

B, A ∩ C 6= 0. We write A ≤e B. A homomorphism f : A → B is an essentialmonomorphism iff f is monic and f(A) ≤e B.

Given A ≤ B, A ≤e B iff A ∩ C 6= 0 for all 0 6= C ≤ B iff A ∩ C 6= 0 for all nonzerocyclic submodules C of B iff A ∩ bR 6= 0 for all 0 6= b ∈ B iff for all 0 6= b ∈ B, thereis some r ∈ R such that 0 6= br ∈ A.

Proposition 5.6

(a) Let A ≤ B ≤ C. Then A ≤e C iff A ≤e B ≤e C.

(b) If Ai ≤e Bi ≤e C (i = 1, 2), then A1 ∩ A2 ≤e B1 ∩B2.

If A ≤ C is not essential, there is some 0 6= B ≤ C such that A ∩B = 0. Thus

A⊆ //

monic22C

quot// C/B.

Proposition 5.7 Let A,B ≤ C. Assume B is maximal with respect to A ∩ B = 0.Then

A+B

B≤e

C

B,

and A⊕B ≤e C.

Proof. Suppose D/B ≤ C/B and A+BB∩ D

B= 0. Then D ≥ B and (A+B)∩D = B,

so A ∩ D ≤ A ∩ B = 0. Maximality of B implies that D = B, so D/B = 0. ThusA+B

B≤e

CB

. The other part is similar.

Corollary 5.8 If A ≤ C, then A is a direct summand of some essential submoduleof C.

Proof. Zorn’s Lemma implies that there is a submodule B ≤ C maximal with re-spect to A ∩B = 0. Thus A⊕B ≤e C by Proposition 5.7.

37

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 someB. Now A ∩ B = 0, so B = 0, so A = C. For the reverse direction, if A ≤ C, thenCorollary 5.8 implies that there is some B ≤ C so that A⊕B ≤e C. The hypothesisimplies that A⊕B = C.

Definition. An essential extension of a module A is any module B such thatA ≤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 isa direct summand of any C ≥ A. Zorn’s Lemma implies that there is some B ≤ Cmaximal with respect to A ∩ B = 0. Proposition 5.7 implies that A⊕B

B≤e

CB

. Thusf : A

⊆→ Cquot→ C/B is an essential monomorphism. There is an injective module

E ≥ A.

A⊆ //

� _

f��

E

C/B∃g

==zz

zz

Then ker(g) ∩ f(A) = 0, so ker g = 0. Now g maps C/B isomorphically into E, soA = gf(A) ≤e g(C/B). The hypothesis implies that A = g(C/B), so f(A) = C/B,so A⊕B

B= C

B, so A⊕B = C.

Example. Let 0 6= n ∈ Z. Then Z/nZ ∼= nZ/n2Z ≤e Z/n2Z, so Z/nZ is not injec-tive as a Z-module.

38

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 iffA is injective.

Proof. We leave the reverse direction as an exercise. For the forward direction, it isenough 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 isessentially 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 Eis injective and A ≤e E.

Proof. Zorn’s Lemma implies that there is a submodule E ≤ F maximal with re-spect 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 in-jective module E such that A ≤e E.

Injective hulls exist b Theorem 5.4 and 5.12.

Proposition 5.13 Suppose E and E ′ are injective hulls for A and A′. If there is an

39

isomorphism f : A→ A′, then f extends to an isomorphism E → E ′.

E∃∼= //___ C ′

A

⊆

OO

f

∼=// A′

⊆

OO

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 isdivisible as a Z-module and hence injective. Look at E, the subspace of V spannedby A. The elements of E are s1

t1a1 + · · ·+ sk

tkak = 1

ta, where si, ti ∈ Z, ti 6= 0, ai ∈ A,

t = t1 · · · tk 6= 0 ∈ Z, and a ∈ A. Thus E = {1/t a | 0 6= t ∈ Z, a ∈ A}, which isdivisible and hence injective. If 0 6= x ∈ E, then x = 1

ta, a 6= 0, so tx = a ∈ A. Thus

A ≤e E, the injective hull of A. dimQ(E) is a measure of size. Thus the measure ofthe size of A is the rank of A.

Suppose V is a vector space over a field k. Then V =⊕

i∈I Vi, dimk(Vi) = 1.dimk(V ) = card(I). The Vi are simple k-modules; they are also not the direct sum oftwo nonzero subspaces.

Definition. A module A is indecomposable iff A is not the direct sum of twononzero submodules.

Definition. A module A is uniform iff A 6= 0 and B1 ∩ B2 6= 0 for all nonzerosubmodule B1, B2 ≤ A (i.e. all nonzero submodules of A are essential).

Examples. Q, Z, Z/pnZ, and Z(p∞) (p prime) are uniform Z-modules.

Lemma 5.14 A nonzero module A is uniform iff E(A) is indecomposable.

40

Proof. For the forward direction, suppose E(A) = E1⊕E2. Then (E1∩A)∩(E2∩A) =0, implies that some Ei ∩ A = 0, so Ei = 0.

For the reverse direction, let 0 6= B ≤ A. Then B ≤ A ≤ E(A), so there is a sub-module 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), soB ≤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 uniformsubmodules.

Proposition 5.15 A module A has finite rank iff there is a finite direct sum of uni-form submodules which is an essential submodule of A.

Lemma 5.16 Suppose E = E1 ⊕ · · · ⊕ En with all Ei uniform. Then E does notcontain 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 = 1follows because then E is uniform. Suppose n > 1 and A1 ⊕ An+1 ≤ E, where eachAj 6= 0. Set A = A1 ⊕ · · · ⊕ An. By induction, A does not embed in a direct sum ofn− 1 uniform modules. Thus A does not embed in

n⊕`=1` 6=i

E`

for all i = 1, . . . , n, so A ∩ Ei 6= 0 for all i, so A ∩ Ei ≤ Ei for all i.

Definition. The rank of a module A is a nonnegative integer n if there is a directsum of n nonzero submodules in A but not a direct sum of n+1 nonzero submodules,and it is infinite otherwise.

41

Proposition 5.20 tells us that the rank of A is n <∞ iff E(A) is the direct sum of nuniforms.

Example. If A is a torsionfree Z-module, then E(A) is a vector space over Q, andthe rank of A is equal to dimQE(A).

Theorem 5.17 (Goldie.) For a module A, the rank of A is finite iff there is no infinitedirect sum of nonzero submodules in A.

Proof. The reverse direction is clear. For the forward direction, suppose A 6= 0, andE(A) is not a finite direct sum of indecomposable submodules.

(1) E(A) = C0 is not indecomposable. Thus E(A) = M ⊕ N with M,N 6= 0. Atleast one of M and N is not a finite direct sum of indecomposables. Now C0 =C1⊕B1 such that B1 6= 0 and C1 is not a finite direct sum of indecomposables.

(2) Repeat: C1 = C2 ⊕ B2 such that B2 6= 0 and C2 is not a finite direct sum ofindecomposables. Continue this process.

Check that B1, B2, . . . are independent. Thus we have⊕∞

n=1Bn ≤ E(A). Thus⊕∞n=1(Bn ∩ A) ≤ A, and all Bn ∩ A 6= 0.

Corollary 5.18 Every noetherian module has finite rank.

Proof. Suppose that the rank of A is infinite. Then Theorem 5.17 implies thatthere is an infinite direct sum

⊕∞n=1Bn ≤ A with all Bn 6= 0. So B1 < B1 ⊕ B2 <

B1 ⊕B2 ⊕B3 < · · · .

Corollary 5.19 Every nonzero noetherian module contains a uniform submodule.

Proof. If A 6= 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 A1 ⊕ · · · ⊕An ≤e A, and

42

all Ai are uniform.

Corollary 5.21 Let B ≤ A.

(a) rank(B) ≤ rank(A).

(b) Suppose rank(A) <∞. Then rank(B) = rank(A) iff B ≤e A.

(c) rank(A) ≤ rank(B) + rank(A/B).

Corollary 5.22 Suppose rank(A) < ∞, f ∈ EndR(A), and f is monic. Thenf(A) ≤e A.

Proof. f(A) ∼= A, so they have the same rank.

Observation. If (Ei)i∈I is an infinite collection of injective modules, then∏

i∈I Ei isinjective.

B⊆ //

��

A

{{vv

vv

v

∏i∈I Ei.

Theorem 5.23 (Papp, Bass.) A ring R is right noetherian iff any direct sum ofinjective right modules is injective.

Proof. For the forward direction, let (Ei)i∈I be a family of injective right R-modulesand E =

⊕i∈I Ei. Let J ≤ RR and f ∈ HomR(J,E). J is finitely generated, so

f(J) is finitely generated, so there is a finite K ⊆ J such that f(J) ⊆⊕

k∈K Ek. But⊕k∈K Ek =

∏k∈K|Ek is injective, so f extends to a homomorphism

R→⊕k∈K

Ek ≤⊕i∈I

Ei.

43

Thus by Baer’s Criterion,⊕

i∈I Ei is injective.

For the reverse direction, let I1 ≤ I2 ≤ · · · be a chain of right ideals of R, soI =

⋃∞n=1 In is a right ideal. By hypothesis, E =

⊕∞n=1E(R/In) is injective. For any

n, there is a homomorphism

I⊆→ T

quot→ R/In⊆→ E(R/In).

These induce a homomorphism f : I →∏∞

n=1E(R/In). Observe that f(I) ≤ E.Thus f extends to a homomorphism g : R→ E. Let z = g(1) ∈

⊕tn=1E(R/In). For

all x ∈ I, f(x) = g(x) = zx, so f(x)n = znx = 0 for all n > t, so x ∈ In for all n > t.Thus I = It+1, so our chain stops.

Corollary 5.24 If R is right noetherian, then every injective right R-module is adirect sum of uniform submodules.

Proof. Let E = ER be injective. Zorn’s Lemma implies that there is an independentfamily (Ai)i∈i of uniform submodules, maximal among such families. Thus

⊕i∈I Ai ≤

E. For all i ∈ I, E contains an injective hull Ei for Ai. Since {Ai} is independent,{Ei} is independent, so ⊕

i∈I

Ai ≤e

⊕i∈I

Ei ≤ E.

Since R is right noetherian,⊕

i∈I Ei is injective, so

E =

(⊕i∈I

Ei

)⊕ E ′

for some E ′. If E ′ 6= 0, there is a uniform B ≤ E ′. But then {Ai | i ∈ I} ∪ {B} isindependent, which is a contradiction. Thus E ′ = 0 and E =

⊕i∈I Ei.

Example. Let R = Z. Let E be a uniform injective Z-module. Pick 0 6= x ∈ E.

• If x has infinite order, Zx ∼= Z. But Zx ≤e E, so E = E(Zx) ∼= E(Z) ∼= Q.

• If x has order m < ∞, m = m′p with p prime and y = m′x has order p.Zy ∼= Z/pZ. Thus E ∼= E(Z/pZ) ∼= Z(p∞). Thus all injective Z-modules aredirect sums of copies of Q, Z(2∞), Z(3∞), Z(5∞), and so forth.

44

Chapter 6

Semisimple Rings of Fractions

We know that commutative domains embed in quotient fields. We want noncommu-tative domains to embed in quotient division rings, prime rings to embed in quotientsimple artinian rings, and semiprime rings to embed in quotient semisimple rings.

Suppose R ⊆ S are rings, and X ⊆ R is such that elements of S are fractions withnumerators from R and denominators from X.

(1) rx

is the product of r and 1x. We have two choices: r · 1

xor 1

x· r.

(a) rx

is ambiguous when S is noncommutative.(b) Therefore we instead use rx−1 or x−1r.(c) We make a choice — for example, assume all elements of S are fractions

rx−1 for r ∈ R and x ∈ X.

(2) For x ∈ X, x−1 = 1x−1 ∈ S, so if ax = 0 in R, then a = axx−1 = 0 in S. Thusa = 0 in R. Thus we need all elements of X to be non-zero-divisors.

(3) For x, y ∈ X, we have x−1y−1 ∈ S. So it is helpful to have X closed undermultiplication and 1 ∈ X.

(4) Suppose a, b ∈ R and x, y ∈ X. Look at (ax−1)(by−1). We need x−1b to be somefraction with a right denominator: x−1b = cz−1 for some c ∈ R and z ∈ X.Thus bz = xc. Thus we need to have X be a right Ore set.

(5) To add ax−1 + by−1, we need common denominators. This is okay if the Orecondition is satisfied.

45

Definition. A regular element in a ring R is any non-zero-divisor: r. annR(x) =`. annR(x) = 0.

Let R be a ring and X ⊆ R a multiplicative set of regular elements. A right ring offractions with respect to X is a ring S ⊇ R such that all elements of X are invertiblein S and every element of S can be written as rx−1 for some r ∈ R and x ∈ X.

Lemma 6.1 Suppose there is a right ring of fractions S for R with respect to X.

(a) X is a right Ore set.

(b) Given s1, . . . , sn ∈ S, there exist ai ∈ R and x ∈ X such that si = aix−1 for all

i.

(c) Let a, b ∈ R and x, y ∈ X. Then ax−1 = by−1 iff there exist c, d ∈ R such thatac = bd and xc = yd ∈ X.

Proof.

(a) Done.

(b) Suppose si = bix−1i for some bi ∈ R and xi ∈ X. By Lemma 4.21, there exist

c1, . . . , cn ∈ R such that x1c1 = · · · = xncn = x ∈ X. xi and x are invertible inS, so ci is invertible, and si = bix

−1i = bicic

−1i x−1

i = (bici)x−1.

(c) First assume that ax−1 = by−1. By Lemma 4.21, there exist c, d ∈ R such thatxc = yd ∈ X. Then

bd(yd)−1 = by−1 = ax−1 = ac(xc)−1 = ac(yd)−1,

so ac = bd. For the converse, suppose ac = bc. Then xc = yd ∈ X for somec, d ∈ R. Then

ax−1 = (ac)(xc)−1 = (bd)(yd)−1 = by−1.

Grunge Method. Assume R is a ring and X ⊆ R a right Ore set of regular elements.

46

(1) Define ∼ on R × X by (c) of Lemma 6.1. Check that ∼ is an equivalencerelation.

(2) Set [a, x] to be the equivalence class of (a, x). Let S = {[a, x] | (a, x) ∈ R×X}.

(3) Define an addition on S with common denominators:

[a, x] + [b, y] = [a′, z] + [b′, z] = [a′ + b′, z].

Check well-definedness, associativity, and so forth. Then R ∼= {[a, 1] | a ∈ R},and x = [x, 1] with inverse [1, x].

Suppose R is a domain, A = AR, a ∈ A, and 0 6= x ∈ R. We wish to “divide a byx.” If A is contained in a module with an element b such that bx = a, then there isa homomorphism R → L given by r 7→ br, x 7→ a, and xr 7→ ar ∈ A. Start witha homomorphism f : xR → A, f(xr) = ar. This is well-defined: if xr = xr′, thenr = r′, so ar = ar′. f extends to g : R → E(A), g(1)x = g(x) = f(x) = a. We justneed x to be not a left zero divisor.

Summary. Suppose x ∈ R is regular, A = AR, and a ∈ A. Then there is a b ∈ E(A)such that bx = a. If A = RR, we get b ∈ E(RR) such that bx = 1. Furthermore,suppose X ⊆ R is a multiplicative set of regular elements and B = {b ∈ E(RR) |bx ∈ R for some x ∈ X}.

(1) For all a ∈ R, there exist b ∈ B and x ∈ X such that bx = a.

(2) If X is right Ore, then B/R = tX(E(RR)/R), a submodule. Thus B ≤ E(RR).

We have R ∼= EndR(RR). We would like B ∼= EndB(BB) ⊆ EndR(BR).

Theorem 6.2 (Ore, Asano.) Let R be a ring and X ⊆ R a multiplicative set ofregular elements. Then there is a right ring of fractions for R with respect to X iffX is right Ore.

Proof. The forward direction is done. For the reverse direction, pick E = E(RR).Then tX(E/R) is a submodule of E/R, or A/R for some A ≤ E. Then A = {a ∈ E |ax ∈ R for some x ∈ X}. Observe that tX(E) ∩R = tX(R) = 0, so tX(E) = 0. Thus

47

E and A are X-torsionfree. Set S = EndR(A).

We claim that for all a ∈ A and x ∈ X, there is a b ∈ A such that bx = a. Thereis a homomorphism f : xR → A such that f(xr) = ar for all r ∈ R. f extends tog : R → E, so b = g(1) satisfies bx = a. Since bx ∈ A, there is a y ∈ X such thatbxy ∈ R, so b ∈ A.

We now claim that for all r ∈ R, there is a unique φr ∈ S such that φr(1) = r. (Soφr(r

′) = rr′ for all r′ ∈ R.) There is a homomorphism f : RR → RR given by leftmultiplication by r. This extends to a homomorphism g : E → E. For all a ∈ A,there is an x ∈ X such that ax ∈ R, so

g(a)x = g(ax) = f(ax) = r(ax) ∈ R,

so g(a) ∈ A. Thus g(A) ⊆ A, so define φr = g |A∈ S. Suppose also that φ′ ∈ S suchthat φ′(1) = r. So h = φ′ − φr ∈ S has h(1) = 0. For a ∈ A, there is an x ∈ X suchthat ax ∈ R and h(a)x = h(ax) = h(1)ax = 0, so h(a) = 0. Thus h = 0, so φ′ = φr.

Proposition 6.5 Let X ⊆ R be a right and left Ore set of regular elements. ThenRX−1 = X−1R.

Proof. Look at S = RX−1. Then R is a subring of S. Elements of X are invertiblein S. Let s ∈ S. Then s = ax−1 for some a ∈ R and x ∈ X. Since X is left Ore, thereis a b ∈ R and a y ∈ X such that ya = bx. Then s = ax−1 = yb−1. Thus S = X−1R.The other way is symmetric.

Definition. A classical right (left) quotient ring for a ring R is a right (left)ring of fractions with respect to the set of regular elements of R.

Definition. A right (left) Ore domain is a domain R such that R \ {0} is a right(left) Ore set.

For all a, x ∈ R, x 6= 0, there exist b, y ∈ R, y 6= 0, such that ay = xb iff for allnonzero u, v ⊆ R, there are nonzero elements r, s ∈ R such that ur = vx 6= 0.

48

Lemma 6.6 Let R be a nonzero domain. The following are equivalent.

(a) R is a right Ore domain.

(b) RR is a uniform module.

(c) RR has finite rank.

Proof. The equivalence of (a) and (b) is done, as is (b) implies (c). We need only show(c) implies (b). Suppose 0 < rank(RR) <∞. Then there is a uniform U ⊆ RR. Pick0 6= u ∈ U . The map RR → uR given by r 7→ ur is an isomorphism. 0 6= uR ≤ U , souR is uniform, so RR is uniform.

Corollary 6.7 Any right (left) noetherian domain is a right (left) Ore domain.

Corollary 5.18 tells us that RR has finite rank.

Theorem 6.8 (Ore’s Theorem.) Let R be a ring. The following are equivalent.

(a) There is a right Ore set X ⊆ R of regular elements such that RX−1 is a divisionring.

(b) R has a classical right quotient ring D, and D is a division ring.

(c) R is a right Ore domain.

Proof. We first show that (a) implies (b). R is a subring of RX−1. Now RX−1 is adivision ring, so R is a domain. Look at 0 6= u ∈ R. There is a u−1 ∈ RX−1 becauseRX−1 is a division ring. All elements of RX−1 have the form ax−1 for some a, x ∈ R,x 6= 0. Thus RX−1 is a classical right quotient ring of R. We have already shown (b)implies (a).

49

For (b) implies (c), note that R is a domain, so the set of regular elements is R \ {0}.For (c) implies (b), there is a ring D = RX−1, X = R \ {0}, so D is a classical rightquotient ring of R. If 0 6= d ∈ D, d = ax−1 for some a, x ∈ R, a, x 6= 0. So there issome e = xa−1 ∈ D and de = ed = 1. Thus D is a division ring.

The Weyl algebra A1(C) is contained in its quotient division ring D1(C). The centerof D1(C) is C, and D1(C) is transcendental over C.

Definition. A right annihilator in a ring R is any right ideal I = r. annR(X) forsome X ⊆ R.

Observe that if I = r. ann(X), then XI = 0, so X ⊆ `. ann(I), so r. ann(`. ann(I)) ⊆r. ann(X) = I, so I = r. ann(`. ann(I)).

Proposition 6.9 Suppose R has a classical right quotient ring Q.

(a) If Q is right noetherian, then RR is finite rank.

(b) If Q is right noetherian, then R satisfies the ACC on right annihilators.

(c) If Q is semisimple, then R is semiprime.

Proof.

(a) If rank(RR) = ∞, then there is some⊕∞

n=1An ≤ RR for some nonzero rightideals An. Choose 0 6= an ∈ An for all n. Look at the right ideals anQ in Q.∑∞

n=1 anQ is not direct because Q is right noetherian, so there are q1, . . . , qk ∈ Qsuch that a1q1 + · · · + akqk = 0 but akqk 6= 0. There exist b1, . . . , bk, x ∈R, x regular, such that qi = bix

−1. Then a1b1x−1 + · · · + akbkx

−1 = 0, soa1b1 + · · · + akbk = 0 (in R). aibi ∈ A and the Ai are independent, so aibi = 0for i = 1, . . . , k. Thus akqk = akbkx

−1 = 0, which is a contradiction. Thusrank(RR) <∞.

(b) Let I1 ≤ I2 ≤ · · · be a chain of right annihilators in R. Set Jn = `. ann(In) sothat In = r. annR(Jn). Then J1 ≥ J2 ≥ · · · . Now r. annQ(J1) ≤ r. annQ(J2) ≤· · · . These are right ideals of Q, and Q is right noetherian. Thus there is somem

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such that r. annQ(Jn) = r. annQ(Jm) for all n ≥ m. Note that r. annQ(Jn)∩R =r. annR(Jn) = In. Thus In = Im for all n ≥ m.

(c) Assume Q is semisimple. Suppose N C R and N2 = 0. Let L = `. annR(N).Exercise 5ZA implies that LR ≤e RR. We claim that LQ ≤e QQ. Let 0 6= q ∈ Q.Write q = ax−1, where a, x ∈ R and x is regular. Then qx = a 6= 0. There issome r ∈ R such that 0 6= ar ∈ L. Thus 0 6= q(xr) ∈ LQ. Thus LQ ≤e QQ.Now QQ is semisimple, so LQ = Q. Thus 1 = `1q1 + · · ·+ `nqn for some `i ∈ Land qi ∈ Q. There exist ai, x ∈ R with x regular such that qi = aix

−1, sox = `1a1 + · · ·+ `nan ∈ L, so xN = 0, so N = 0. Thus R is semiprime.

Definition. A ring R is a right Goldie ring iff rank(RR) <∞ and R satisfies theACC on right annihilators.

A ring which has a semisimple classical right quotient ring is a semiprime right Goldiering.

Suppose R is a semiprime right Goldie ring and X is the set of regular elementsof R. We want the right Ore condition: given a ∈ R and x ∈ X, there shouldbe some b ∈ R and y ∈ X such that ay = xb, or equivalently, ay ∈ xR, ory ∈ J := {r ∈ R | ar ∈ xR}, a right ideal of R. Since x is regular, xR ∼= RR,so rank(xR) = rank(RR) <∞, so xR ≤e RR, so J ≤e RR.

Our new goal is to show that every essential right ideal of R contains a regular ele-ment.

Ideas.

• Show r. annR(x) = 0 so that x is regular.

• We get this from xR ≤e RR by proving that `. annR(I) = 0 for all I ≤e RR.

• If J ≤e RR, we want x ∈ J such that r. annR(x) = 0, so we try for r. annR(x)as small as possible.

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• We need the DCC on right annihilators.

Proposition 6.10 If R is semiprime right Goldie and K ≤e RR, then `. annR(K) = 0.

Proof. SetJ =

⋃I≤eRR

`. annR(I).

We leave it as an exercise to show that J C R. Then we are done if J is nilpotent.We have J ≥ J2 ≥ J3 ≥ · · · , so r. annR(J) ≤ r. annR(J2) ≤ · · · , so there is some k sothat r. annR(Jk) = r. annR(Jk+1).

We claim that Jk = 0. Suppose not; then Jk 6= 0. Then r. annR(Jk) 6= R. Amongx ∈ R \ r. annR(Jk), pick x with r. annR(x) maximal. If a ∈ J , then a ∈ `. annR(I),and I ≤e RR. There is some r ∈ R such that 0 6= xr ∈ I, so axr = 0, sor. annR(x) < r. annR(ax). Maximality of r. annR(x) implies that ax ∈ r. annR(Jk),so Jkax = 0. Thus Jk+1x = 0, so x ∈ r. annR(Jk+1) = r. annR(Jk), which is a contra-diction. Thus Jk = 0. Since R is semiprime, J = 0.

Lemma 6.11 Let R be semiprime right Goldie. Then the following are equivalent.

(a) x is regular.

(b) r. annR(x) = 0.

(c) xR ≤e RR.

Proof. (a) implies (b) is clear, and (b) implies (c) follows from Corollary 5.22 (RR

has finite rank). We therefore need only show that (c) implies (a). By Proposition6.10, `. annR(xR) = 0, so `. annR(x) = 0. Look at I = r. annR(x). Proposition 5.7tells us that there is some J ≤ RR such that I ∩ J = 0 and I ⊕ J ≤e RR.

We claim that xJ ≤e RR. If 0 6= a ∈ xR, then a = xr for some r ∈ R. SetK = {k ∈ R | rk ∈ I ⊕ J}. Then K ≤e RR. By Proposition 6.10, `. annR(K) = 0,

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so aK 6= 0. Then aK = xrK ≤ x(I ⊕ J) = xJ , so 0 6= aK ≤ aR ∩ xJ . NowxJ ≤e RR Because of the finite rank condition, rank(xJ) = rank(xR) = rank(RR).Then J ∩ r. annR(x) = J ∩ I = 0, so J ∼= xJ , so rank(J) = rank(RR), so J ≤e RR, soI = 0.

Lemma 6.12 If R is semiprime right Goldie, then R satisfies the DCC on right an-nihilators.

Proof. Let I1 ≥ I2 ≥ · · · be a descending chain of right annihilators. SupposeIn = r. annR(Xn) for some Xn ⊆ R. Then rank(I1) ≥ rank(I2) ≥ · · · , so there issome m so that rank(In) = rank(Im) for all n ≥ m. Then In ≤e Im for all n ≥ m.

We claim that In = Im for all n ≥ m. Let n ≥ m and x ∈ Im. Then J = {r ∈ R |xr ∈ In} ≤e RR. Then XnxJ ⊆ XnIn = 0. Proposition 6.10 tells us that Xnx = 0,so x ∈ In. Thus In = Im.

Proposition 6.13 (Goldie’s Regular Element Lemma.) Let R be semiprime rightGoldie. Then I ≤e RR iff there is a regular element in I.

Proof. The reverse direction is Lemma 6.11. For the forward direction, we use theDCC on right annihilators. There is some x ∈ I so that A = r. annR(x) is minimalamong right annihilators of single elements of I.

We claim that xR ≤e I. Then we’re done, because this implies that xR ≤e RR; useLemma 6.11. Let B ≤ I be such that B ∩ xR = 0. We must show that B = 0.

Our first goal is to show that BA = 0. If b ∈ B, then bR ∩ xR = 0, so r. annR(b +x) = r. annR(b) ∩ r. annR(x). [If (b + x)r = 0, then br = −xr, so br = xr = 0.]b+x ∈ B+xR ≤ I, so minimality implies that r. annR(b+x) = A, so A ≤ r. annR(b),so bA = 0.

Our next goal is to show that RB ∩ A = 0. We have (RB ∩ A)2 ≤ (RB)A = 0. By

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semiprimeness, RB ∩ A = 0.

The next goal is to show that xRB ≤e (RB)R. RB C R, so xRB ⊆ RB. Leftmultiplication by x defines a homomorphism f : (RB)R → (RB)R. ker(f) =RB ∩ r. annR(x) = RB ∩ A = 0, so xRB ∼= (RB)R, so they have the same finiterank, so xRB ≤e (RB)R.

Finally we are ready to show that B = 0. B ∩ xRB ≤ B ∩ xR = 0 and B ≤ RB.Thus xRB ≤e (RB)R, so B = 0. Thus xR ≤e I ≤e RR, so Lemma 6.11 implies thatx is regular.

Corollary 6.14 If R is prime right Goldie and 0 6= I C R, then there is a regularelement in I.

Proof. By Exercise 5A, IR ≤e RR.

Theorem 6.15 (Goldie’s Theorem.) R has a semisimple classical right quotient ringiff R is semiprime right Goldie.

Proof. The forward direction is Proposition 6.9. For the reverse direction, let Xdenote the set of regular elements of R. Let a ∈ R and x ∈ X. Since x is regular,xR ≤e RR (by Lemma 6.11), so

I = {r ∈ R | ar ∈ xR} ≤e RR.

Proposition 5.6 implies that there is some y ∈ X ∩ I, so ay = xb for some b ∈ R.Thus X is right Ore. Thus there is a classical right quotient ring Q = RX−1.

We claim that the only I ≤e QQ is I = Q. The goal is to show that I ∩ R ≤e RR.Let 0 6= b ∈ R. IF I ≤e QQ, then there is some q ∈ Q such that 0 6= bq ∈ I. Supposeq = ax−1, a ∈ R and x ∈ X. Then 0 6= bax−1 ∈ I, so 0 6= ba ∈ I ∩ R. Proposition6.13 implies that there is some y ∈ (I ∩ R) ∩X. Then y ∈ I and y−1 ∈ Q, so 1 ∈ I,so I = Q. Corollary 5.9 then tells us that QQ is semisimple.

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Corollary 6.16 If R is semiprime right noetherian, then R has a semisimple classicalright quotient ring.

Proposition 6.20 If R is a semiprime right and left Goldie ring, then every right(left) Goldie quotient ring of R is also a left (right) Goldie quotient ring.

Lemma 6.17 Let R be semiprime right Goldie, and let Q be the right Goldie quotientring of R. Then Q is simple iff R is prime.

Proof. Observe that RR ≤e QR. If 0 6= q ∈ Q, then q = ax−1 for 0 6= a ∈ R andx ∈ R regular, so 0 6= qx = a ∈ R. For the reverse direction, let 0 6= I C Q. ThenRR ≤e QR, so I ∩ R 6= 0; also I ∩ R C R. Exercise 5A tells us that (I ∩Q)Q ≤e RR.By Proposition 6.13, there is a regular element x ∈ I ∩R, so 1 ∈ I, so I = Q.

For the forward direction, suppose A,B C R such that AB = 0 and Q 6= 0. Then0 6= QAQ C Q, so QAQ = Q, so 1 = p1a1q1 + · · · + pnanqn for some pi, qi ∈ Qand ai ∈ A. There exist bi ∈ R and x ∈ R regular such that all qi = bix

−1. Thusx = p1a1b1 + · · ·+ pnanbn ∈ QA. Then xB ⊆ QAB = 0 in R, so B = 0.

Theorem 6.18 R has a simple artinian classical right quotient ring iff R is primeright Goldie.

55

Chapter 7

Modules over Semiprime GoldieRings

Proposition 7.19 Let R be a semiprime right and left Goldie ring. Then everyfinitely generated torsionfree right R-module A is isomorphic to a submodule of afinitely generated free module.

Proof. We claim that every nonzero submodule B ≤ A contains a uniform submod-ule isomorphic to a right ideal of R. Pick 0 6= b ∈ B, and let I = r. annR(b) ≤ RR.Since A is torsionfree, there is no regular element in I, so I 6≤e RR, so there is some0 6= J ≤ RR such that J ∩ I = 0. Now J ∼= bJ ≤ B. rank(K) < rank(RR) < ∞, sothere is some uniform U ≤ J . Thus bU ∼= U is a uniform submodule of B.

Zorn’s Lemma and the claim imply that there is some⊕

i∈I Ui ≤e A such that Ui isuniform and isomorphic to Vi ≤ RR. Let Q be the right Goldie quotient ring of R.Let Qi be a copy of QR for all i ∈ I. There exists a monomorphism

f :⊕i∈I

Ui

∼=→⊕i∈I

Vi ≤⊕i∈I

Qi,

i.e.⊕

i∈I Ui is isomorphic to a submodule of a free right Q-module. Then⊕

i∈I Qi istorsionfree and divisible as a right R-module and is therefore injective (Proposition7.11). Thus f extends to a homomorphism g : A→

⊕i∈I Qi. Then

(ker g) ∩

(⊕i∈I

Ui

)= ker f = 0,

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so ker g = 0. Since A is finitely generated, g(A) is finitely generated, so g(A) ⊆⊕i∈I0

Qi for some finite I0 ⊆ I. Thus A is isomorphic to a submodule of QnR for some

n ∈ N. Thus without loss of generality, A ≤ QnR. Choose generators a1, . . . , am for A

and write ai = (ai1, . . . , ain) for some aij ∈ Q. There exist bij ∈ R and a regular x ∈ Rsuch that aij = x−1bij for all i, j. Set bi = (bi1, . . . , bin) ∈ Rn for all i. Notice thatxai = bi for all i, so xA ≤ Rn. Thus left multiplication by x defines a homomorphismh : A→ Rn. Since x is regular, h is monic.

57

Chapter 8

Bimodules and Affiliated Prime Ideals

Example. Let

R =

(Z Q0 Q

),

the subring of M2(Q). R is right noetherian but not left noetherian. Let

I =

(0 Q0 0

)CR.

Then IR is simple (artinian and noetherian), but RI is not noetherian.

Lemmas 8.10 and 8.11 Let RAS be a bimodule such that

(a) AS is finitely generated.

(b) RA is artinian and noetherian.

(c) S is semiprime right noetherian.

(d) AS is torsionfree and faithful.

Then AS is semisimple and artinian.

Proof. We first claim that Ac = A for all regular c ∈ S. A ≥ Ac ≥ Ac2 ≥ · · · areleft submodules of A. By (b), there is some m ∈ N such that Acm = Acm+1. If a ∈ A,then acm = bcm+1 for some b ∈ A. (d) implies that a = bc.

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We now claim that the right Goldie quotient ring Q of S is noetherian as a rightS-module. By (b), A = Ra1 + · · · + Ram for some ai ∈ A. (d) implies that0 = r. annS(A) = r. annS({a1, . . . , am}), so the homomorphism f : SS → Am

S given bys 7→ (a1s, . . . , ams) is monic. (d) and the first claim imply that AS is injective, so fextends to a homomorphism g : QS → Am

S . SS ≤e QS, so g is monic. Now (a) and(c) imply that Am

S is noetherian, so QS is noetherian.

Finally, we claim thatQ = S. Let c be a regular element of S. S ≤ c−1S ≤ c−2S ≤ · · ·are right S-submodules of Q. The second claim implies that there is some k so thatc−kS = c−k−1S, so c−1 ∈ S.

Now AS is a finitely generated module over a semisimple ring.

Theorem 8.12 (Lenagan’s Theorem.) Let RAS be a bimodule such that

(a) AS is finitely generated.

(b) RA is artinian and noetherian.

(c) S is right noetherian.

Then AS is artinian.

Corollary. Suppose R is right and left noetherian, I C R, and RI is artinian. ThenIR is artinian.

Proof. Suppose not. (a) and (c) imply that AS is noetherian, so it satisfies the ACCon sub-bimodules. Use noetherian induction: without loss of generality, (A/B)S isartinian for all nonzero sub-bimodules B. Choose P CS maximal among annihilatorsof nonzero submodules of AS. Proposition 3.12 implies that P is a prime ideal andB = `. annA(P ) is a fully faithful (S/P )-module.

Observe that B is a nonzero sub-bimodule of A. Then (A/B)S is artinian. Look atB = RBS/P .

• BS/P is finitely generated and faithful.

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• RB is artinian and noetherian.

• S/P is semiprime right noetherian.

Let T be the torsion submodule of BS/P . Observe that T is a sub-bimodule of B anda left R-module. Thus T = Rt1 + · · ·+Rtn for some ti. There exists a regular x ∈ Ssuch that tix = 0 for all i, so Tx = 0, so TS/P is not faithful, so T = 0. Thus BS/P istorsionfree. The Lemma implies that BS is artinian. Thus AS is artinian.

60

Chapter 9

Problems

(1) Let A be the algebra over a field k presented by two elements x and y andthe relation yx − xy = x. Show that A = k[y][x;α], where α is the k-algebraautomorphism of the polynomial ring k[y] such that α(y) = y − 1.

(2) Here is another way to obtain Corollary 1.15 from Theorem 1.14. Let S =R[x;α] and T = [x±1;α], where α is an automorphism of R. Show that therule β(s) = x−1sx defines an automorphism β of S, and then use the universalmapping property of S[y; β] to show that the inclusion map S → T extends toa ring homomorphism φ : S[y; β] → T such that φ(y) = x−1. Conclude thatT ∼= S[y; β]/ ker(φ). Thus, if R is, say, right noetherian, then two applicationsof Theorem 1.14 show that S[y; β] is right noetherian, and therefore T is rightnoetherian.

(3) Let S be the algebra over a field k presented by two generators x and y and onerelation xy − yx = y. Show that S = k[y][x; δ], where k[y] is a polynomial ringover k and δ = y(d/dy).

(4) (a) Let R = k[t] be a polynomial ring over a field k, and let α be the k-algebraendomorphism of R given by the rule α(f(t)) = f(t2). Show that R[x;α]is neither right nor left noetherian.

(b) Now let R = k(t) be the quotient field of k[t] and extend α to the k-algebra endomorphism of R given by the same rule α(f(t)) = f(t2). Showthat R[x;α] is not right noetherian. In this case, however, R[x;α] is leftnoetherian, as will follow from Theorem 2.8.

(5) Let R[x;α.δ] be a skew polynomial ring, and assume that α is an automorphismof R. Show that α−1 is an automorphism of the opposite ring Rop, that −δα−1

is an α−1-derivation of Rop, and that R[x;α, δ]op = Rop[x;α−1,−δα−1].

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(6) Let S = R[x;α, δ] be a skew polynomial ring and I an ideal of R such thatα(I) ⊆ I and δ(I) ⊆ I. Let α and δ denote the ring endomorphism and skewderivation on R/I induced by α and δ. Show that IS is a two-sided ideal of Ssuch that IS ∩ R = I, and that IS = SI in case α is an automorphism andα(I) = I. Then show that S/IS ∼= (R/I)[x; α, δ].

(7) Let R = Oq(k2), where k is a field and q ∈ k× is not a root of unity. Show that

R/R(xy − 1) is a faithful simple left R-module.

(8) Let R be a semiprime ring.

(a) Show that any simple right or left ideal of R is generated by an idempotent.

(b) Given an idempotent e ∈ R, show that eR is a simple right ideal if andonly if Re is a simple left ideal, if and only if eRe is a division ring.

(c) Show that soc(RR) = soc(RR).

(9) Let R be a right noetherian domain.

(a) Show that the intersection of any two nonzero right ideals of R is nonzero.

(b) Show that R \ {0} is a right Ore set in R.

(10) If I is a nonzero ideal in a prime ring R, show that I is both an essential rightideal and an essential left ideal of R.

(11) If A is a right module over a ring R, show that the set

Z(A) = {x ∈ A | xI = 0 for some I ≤e RR} = {x ∈ A | annR(x) ≤e RR}

is a submodule of A. It is called the singular submodule of A.

(12) Let S = k[x, y] be a polynomial ring over a field k, and let R be the ringS/〈x, y〉2. Show that RR has finite rank but is not a direct sum of uniformsubmodules.

(13) If N is a nilpotent ideal in a ring R, show that `. annR(N) is an essential rightideal of R.

(14) If R is a ring such that RR has finite rank, show that any right or left invertibleelement in R is invertible.

(15) Let R be a ring and X ⊆ R a right Ore set of regular elements. If R is rightnoetherian, show that RX−1 is right noetherian.

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