2.2.7 Proposition

The following conditions are equivalent, for an arbitrary R-module M .

(1) M = 0;

(2) MP = 0 for all prime ideals P of R;

(3) MP = 0 for all maximal ideals P of R.

Proof. It is immediate that (1) ’ (2) ’ (3). To prove that (3) ’ (1), let m ∈ M . If P is

a maximal ideal of R, then m/1 is 0 in MP , so there exists rP ∈ R \ P such that rP m = 0

in M . Let I(m) be the ideal generated by the rP . Then I(m) cannot be contained in

any maximal ideal M, because rM ∈ M by construction. Thus I(m) must be R, and

/

in particular, 1 ∈ I(m). Thus 1 can be written as a ¬nite sum P aP rP where P is a

maximal ideal of R and aP ∈ R. Consequently,

aP rP m = 0. ™

m = 1m =

P

2.3 Going Down

We will prove a companion result to the going up theorem (2.2.3), but additional hy-

potheses will be needed and the analysis is more complicated.

8 CHAPTER 2. INTEGRAL EXTENSIONS

2.3.1 Lemma

√

Let S be integral over the subring R, with I an ideal of R. Then IS is the set of all

s ∈ S satisfying an equation of integral dependence sm + rm’1 sm’1 + · · · + r1 s + r0 = 0

with the ri ∈ I.

√

Proof. If s satis¬es such an equation, then sm ∈ IS, so s ∈ IS. Conversely, let sn ∈

k

IS, n ≥ 1, so that sn = i=1 ri si for some ri ∈ I and si ∈ S. Then S1 = R[s1 , . . . , sk ]

is a subring of S, and is also a ¬nitely generated R-module by (2.1.5). Now

k k

ri si S1 ⊆ ri S1 ⊆ IS1 .

n

s S1 =

i=1 i=1

Moreover, S1 is a faithful R[sn ]-module, because an element that annihilates S1 annihilates

1 and is therefore 0. By (2.1.2), sn , hence s, satis¬es an equation of integral dependence

with coe¬cients in I. ™

2.3.2 Lemma

Let R be an integral domain with fraction ¬eld K, and assume that R is integrally closed.

Let f and g be monic polynomials in K[x]. If f g ∈ R[x], then both f and g are in R[x].

Proof. In a splitting ¬eld containing K, we have f (x) = i (x’ai ) and g(x) = j (x’bj ).

Since the ai and bj are roots of the monic polynomial f g ∈ R[x], they are integral over R.

The coe¬cients of f and g are in K and are symmetric polynomials in the roots, hence

are integral over R as well. But R is integrally closed, and the result follows. ™

2.3.3 Proposition

Let S be integral over the subring R, where R is an integrally closed domain. Assume

that no nonzero element of R is a zero-divisor of S. (This is automatic if S itself is an

integral domain.) If s ∈ S, de¬ne a homomorphism hs : R[x] ’ S by hs (f ) = f (s); thus

hs is just evaluation at s. Then the kernel I of hs is a principal ideal generated by a

monic polynomial.

Proof. If K is the fraction ¬eld of R, then IK[x] is an ideal of the PID K[x], and IK[x] = 0

because s is integral over R. (If this is unclear, see the argument in Step 1 below.) Thus

IK[x] is generated by a monic polynomial f .

Step 1 : f ∈ R[x].

By hypothesis, s is integral over R, so there is a monic polynomial h ∈ R[x] such that

h(s) = 0. Then h ∈ I ⊆ IK[x], hence h is a multiple of f , say h = f g, with g monic in

K[x]. Since R is integrally closed, we may invoke (2.3.2) to conclude that f and g belong

to R[x].

Step 2 : f ∈ I.

Since f ∈ IK[x], we may clear denominators to produce a nonzero element r ∈ R such

that rf ∈ IR[x] = I. By de¬nition of I we have rf (s) = 0, and by hypothesis, r is not a

zero-divisor of S. Therefore f (s) = 0, so f ∈ I.

Step 3 : f generates I.

Let q ∈ I ⊆ IK[x]. Since f generates IK[x], we can take a common denominator and

2.3. GOING DOWN 9

write q = q1 f /r1 with 0 = r1 ∈ R and q1 ∈ R[x]. Thus r1 q = q1 f , and if we pass to

residue classes in the polynomial ring (R/Rr1 )[x], we have q1 f = 0. Since f is monic, the

leading coe¬cient of q1 must be 0, which means that q1 itself must be 0. Consequently,

r1 divides every coe¬cient of q1 , so q1 /r1 ∈ R[x]. Thus f divides q in R[x]. ™

2.3.4 Going Down Theorem

Let the integral domain S be integral over the integrally closed domain R. Suppose we

have a chain of prime ideals P1 ⊆ · · · ⊆ Pn of R and a chain of prime ideals Qm ⊆ · · · ⊆ Qn

of S, with 1 < m ¤ n. If Qi lies over Pi for i = m, . . . , n, then there are prime ideals

Q1 , . . . , Qm’1 such that Q1 ⊆ · · · ⊆ Qm and Qi lies over Pi for every i = 1, . . . , n.

Proof. By induction, it su¬ces to consider n = m = 2. Let T be the subset of S consisting

of all products rt, r ∈ R \ P1 , t ∈ S \ Q2 . In checking that T is a multiplicative set,

we must make sure that it does not contain 0. If rt = 0 for some r ∈ P1 (hence r = 0)

/

and t ∈ Q2 , then the hypothesis that r is not a zero-divisor of S gives t = 0, which is a

/

contradiction (because 0 ∈ Q2 ). Note that R \ P1 ⊆ T (take t = 1), and S \ Q2 ⊆ T (take

r = 1).

First we prove the theorem under the assumption that T © P1 S = …. Now P1 ST is

a proper ideal of ST , else 1 would belong to T © P1 S. Therefore P1 ST is contained in a

maximal ideal M. By basic localization theory, M corresponds to a prime ideal Q1 of S

that is disjoint from T . Explicitly, s ∈ Q1 i¬ s/1 ∈ M. We refer to Q1 as the contraction

of M to S; it is the preimage of M under the canonical map s ’ s/1. With the aid of

the note at the end of the last paragraph, we have (R \ P1 ) © Q1 = (S \ Q2 ) © Q1 = ….

Thus Q1 © R ⊆ P1 and Q1 = Q1 © S ⊆ Q2 . We must show that P1 ⊆ Q1 © R. We do this

by taking the contraction of both sides of the inclusion P1 ST ⊆ M. Since the contraction

of P1 ST to S is P1 S, we have P1 S ⊆ Q1 , so P1 ⊆ (P1 S) © R ⊆ Q1 © R, as desired.

Finally, we show that T © P1 S is empty. If not, then by de¬nition of T , T © P1 S

contains an element rt with r ∈ R \ P1 and t ∈ S \ Q2 . We apply (2.3.1), with I = P1 and

s replaced by rt, to produce a monic polynomial f (x) = xm + rm’1 xm’1 + · · · + r1 x + r0

with coe¬cients in P1 such that f (rt) = 0. De¬ne

v(x) = rm xm + rm’1 rm’1 xm’1 + · · · + r1 rx + r0 .

Then v(x) ∈ R[x] and v(t) = 0. By (2.3.3), there is a monic polynomial g ∈ R[x] that

generates the kernel of the evaluation map ht : R[x] ’ S. Therefore v = ug for some

u ∈ R[x]. Passing to residue classes in the polynomial ring (R/P1 )[x], we have v = u g.

Since ri ∈ P1 for all i = 0, . . . , m ’ 1, we have v = rm xm . Since R/P1 is an integral

domain and g, hence g, is monic, we must have g = xj for some j with 0 ¤ j ¤ m. (Note

that r ∈ P1 , so v is not the zero polynomial.) Consequently,

/

g(x) = xj + aj’1 xj’1 + · · · + a1 x + a0

with ai ∈ P1 , i = 0, . . . , j ’ 1. But g ∈ ker ht , so g(t) = 0. By (2.3.1), t belongs to the

radical of P1 S, so for some positive integer l, we have tl ∈ P1 S ⊆ P2 S ⊆ Q2 S = Q2 , so

t ∈ Q2 . This contradicts our choice of t (recall that t ∈ S \ Q2 ). ™

Chapter 3

Valuation Rings

The results of this chapter come into play when analyzing the behavior of a rational

function de¬ned in the neighborhood of a point on an algebraic curve.

3.1 Extension Theorems

In Theorem 2.2.4, we generalized a result about ¬eld extensions to rings. Here is another

variation.

3.1.1 Theorem

Let R be a subring of the ¬eld K, and h : R ’ C a ring homomorphism from R into an

algebraically closed ¬eld C. If ± is a nonzero element of K, then either h can be extended

to a ring homomorphism h : R[±] ’ C, or h can be extended to a ring homomorphism

h : R[±’1 ] ’ C.

Proof. Without loss of generality, we may assume that R is a local ring and F = h(R) is

a sub¬eld of C. To see this, let P be the kernel of h. Then P is a prime ideal, and we can

extend h to g : RP ’ C via g(a/b) = h(a)/h(b), h(b) = 0. The kernel of g is P RP , so

by the ¬rst isomorphism theorem, g(RP ) ∼ RP /P RP , a ¬eld (because P RP is a maximal

=

ideal). Thus we may replace (R, h) by (RP , g).

Our ¬rst step is to extend h to a homomorphism of polynomial rings. If f ∈ R[x] with

h(ai )xi ∈ F [x]. Let I = {f ∈ R[x] : f (±) = 0}. Then

ai xi , we take h(f ) =

f (x) =

J = h(I) is an ideal of F [x], necessarily principal. Say J = (j(x)). If j is nonconstant,

it must have a root β in the algebraically closed ¬eld C. We can then extend h to

h : R[±] ’ C via h(±) = β, as desired. To verify that h is well-de¬ned, suppose f ∈ I, so

that f (±) = 0. Then h(f ) ∈ J, hence h(f ) is a multiple of j, and therefore h(f )(β) = 0.

Thus we may assume that j is constant. If the constant is zero, then we may extend h

exactly as above, with β arbitrary. So we can assume that j = 0, and it follows that

1 ∈ J. Consequently, there exists f ∈ I such that h(f ) = 1.

1

2 CHAPTER 3. VALUATION RINGS

This gives a relation of the form

r

1, i = 0

ai ±i = 0 with ai ∈ R and ai = h(ai ) = (1)

0, i > 0

i=0

Choose r as small as possible. We then carry out the same analysis with ± replaced by

±’1 . Assuming that h has no extension to R[±’1 ], we have

s

1, i = 0

bi ±’i = 0 with bi ∈ R and bi = h(bi ) = (2)

0, i > 0

i=0

Take s minimal, and assume (without loss of generality) that r ≥ s. Since h(b0 ) = 1 =

h(1), it follows that b0 ’ 1 ∈ ker h ⊆ M, the unique maximal ideal of the local ring R.

Thus b0 ∈ M (else 1 ∈ M), so b0 is a unit. It is therefore legal to multiply (2) by b’1 ±s

/ 0

to get

±s + b’1 b1 ±s’1 + · · · + b’1 bs = 0 (3)

0 0

Finally, we multiply (3) by ar ±r’s and subtract the result from (1) to contradict the

minimality of r. (The result of multiplying (3) by ar ±r’s cannot be a copy of (1). If so,

r = s (hence ±r’s = 1)and a0 = ar b’1 bs . But h(a0 ) = 1 and h(ar b’1 bs ) = 0.) ™

0 0

It is natural to try to extend h to a larger domain, and this is where valuation rings

enter the picture.

3.1.2 De¬nition

A subring R of a ¬eld K is a valuation ring of K if for every nonzero ± ∈ K, either ± or

±’1 belongs to R.

3.1.3 Examples

The ¬eld K is a valuation ring of K, but there are more interesting examples.

1. Let K = Q, with p a ¬xed prime. Take R to be the set of all rationals of the form

pr m/n, where r ≥ 0 and p divides neither m nor n.

2. Let K = k(x), where k is any ¬eld. Take R to be the set of all rational functions

pr m/n, where r ≥ 0, p is a ¬xed polynomial that is irreducible over k and m and n

are arbitrary polynomials in k[x] not divisible by p. This is essentially the same as the

previous example.

3. Let K = k(x), and let R be the set of all rational functions f /g ∈ k(x) such that

deg f ¤ deg g.

4. Let K be the ¬eld of formal Laurent series over k. Thus a nonzero element of K looks

∞

like f = i=r ai xi with ai ∈ k, r ∈ Z, and ar = 0. We may write f = ar xr g, where

g belongs to the ring R = k[[x]] of formal power series over k. Moreover, the constant

term of g is 1, and therefore g, hence f , can be inverted (by long division). Thus R is a

valuation ring of K.

We now return to the extension problem.

3.2. PROPERTIES OF VALUATION RINGS 3

3.1.4 Theorem

Let R be a subring of the ¬eld K, and h : R ’ C a ring homomorphism from R into an

algebraically closed ¬eld C. Then h has maximal extension (V, h). In other words, V is a

subring of K containing R, h is an extension of h, and there is no extension to a strictly

larger subring. In addition, for any maximal extension, V is a valuation ring of K.

Proof. Let S be the set of all (Ri , hi ), where Ri is a subring of K containing R and hi

is an extension of h to Ri . Partially order S by (Ri , hi ) ¤ (Rj , hj ) if and only if Ri is a

subring of Rj and hj restricted to Ri coincides with hi . A standard application of Zorn™s

lemma produces a maximal extension (V, h). If ± is a nonzero element of K, then by

(3.1.1), h has an extension to either V [±] or V [±’1 ]. By maximality, either V [±] = V or

V [±’1 ] = V . Therefore ± ∈ V or ±’1 ∈ V . ™

3.2 Properties of Valuation Rings

We have a long list of properties to verify, and the statement of each property will be

followed immediately by its proof. The end of proof symbol will only appear at the very

end. Throughout, V is a valuation ring of the ¬eld K.

1. The fraction ¬eld of V is K.

This follows because a nonzero element ± of K can be written as ±/1 or as 1/±’1 .

2. Any subring of K containing V is a valuation ring of K.

This follows from the de¬nition of a valuation ring.

3. V is a local ring.

We will show that the set M of nonunits of V is an ideal. If a and b are nonzero nonunits,

then either a/b or b/a belongs to V . If a/b ∈ V , then a + b = b(1 + a/b) ∈ M (because

if b(1 + a/b) were a unit, then b would be a unit as well). Similarly, if b/a ∈ V , then

a + b ∈ M. If r ∈ V and a ∈ M, then ra ∈ M, else a would be a unit. Thus M is an

ideal.

4. V is integrally closed.

Let ± be a nonzero element of K, with ± integral over V . Then there is an equation of

the form

±n + cn’1 ±n’1 + · · · + c1 ± + c0 = 0

with the ci in V . We must show that ± ∈ V . If not, then ±’1 ∈ V , and if we multiply

the above equation of integral dependence by ±’(n’1) , we get

± = ’cn’1 ’ cn’2 ±’1 ’ · · · ’ c1 ±n’2 ’ c0 ±n’1 ∈ V.

5. If I and J are ideals of V , then either I ⊆ J or J ⊆ I. Thus the ideals of V are totally

ordered by inclusion.

Suppose that I is not contained in J, and pick a ∈ I \ J (hence a = 0). If b ∈ J, we

must show that b ∈ I. If b = 0 we are ¬nished, so assume b = 0. We have b/a ∈ V (else

a/b ∈ V , so a = (a/b)b ∈ J, a contradiction). Therefore b = (b/a)a ∈ I.

4 CHAPTER 3. VALUATION RINGS

6. Conversely, let V be an integral domain with fraction ¬eld K. If the ideals of V are

partially ordered by inclusion, then V is a valuation ring of K.

If ± is a nonzero element of K, then ± = a/b with a and b nonzero elements of V . By

hypothesis, either (a) ⊆ (b), in which case a/b ∈ V , or (b) ⊆ (a), in which case b/a ∈ V .

7. If P is a prime ideal of the valuation ring V , then VP and V /P are valuation rings.

First note that if K is the fraction ¬eld of V , it is also the fraction ¬eld of VP . Also, V /P

is an integral domain, hence has a fraction ¬eld. Now by Property 5, the ideals of V are

totally ordered by inclusion, so the same is true of VP and V /P . The result follows from

Property 6.

8. If V is a Noetherian valuation ring, then V is a PID. Moreover, for some prime p ∈ V ,

every ideal is of the form (pm ), m ≥ 0. For any such p, ©∞ (pm ) = 0.

m=1

Since V is Noetherian, an ideal I of V is ¬nitely generated, say by a1 , . . . , an . By Property

5, we may renumber the ai so that (a1 ) ⊆ (a2 ) · · · ⊆ (an ). But then I ⊆ (an ) ⊆ I, so

I = (an ). In particular, the maximal ideal M of V is (p) for some p, and p is prime

because M is a prime ideal. If (a) is an arbitrary ideal, then (a) = V if a is a unit, so

assume a is a nonunit, that is, a ∈ M. But then p divides a, so a = pb. If b is a nonunit,

then p divides b, and we get a = p2 c. Continuing inductively and using the fact that V

is a PID, hence a UFD, we have a = pm u for some positive integer m and unit u. Thus

(a) = (pm ). Finally, if a belongs to (pm ) for every m ≥ 1, then pm divides a for all m ≥ 1.

Again using unique factorization, we must have a = 0. (Note that if a is a unit, so is p, a

contradiction.)

9. Let R be a subring of the ¬eld K. The integral closure R of R in K is the intersection

of all valuation rings V of K such that V ⊇ R.

If a ∈ R, then a is integral over R, hence over any valuation ring V ⊇ R. But V is

integrally closed by Property 4, so a ∈ V . Conversely, assume a ∈ R. Then a fails to

/

’1 ’1

belong to the ring R = R[a ]. (If a is a polynomial in a , multiply by a su¬ciently

high power of a to get a monic equation satis¬ed by a.) Thus a’1 cannot be a unit in

R . (If ba’1 = 1 with b ∈ R , then a = a1 = aa’1 b = b ∈ R , a contradiction.) It follows

that a’1 belongs to a maximal ideal M of R . Let C be an algebraic closure of the ¬eld

k = R /M , and let h be the composition of the canonical map R ’ R /M = k and

the inclusion k ’ C. By (3.1.4), h has a maximal extension to h : V ’ C for some

valuation ring V of K containing R ⊇ R. Now h(a’1 ) = h(a’1 ) since a’1 ∈ M ⊆ R ,

and h(a’1 ) = 0 by de¬nition of h. Consequently a ∈ V , for if a ∈ V , then

/

1 = h(1) = h(aa’1 ) = h(a)h(a’1 ) = 0,

a contradiction. The result follows.

10. Let R be an integral domain with fraction ¬eld K. Then R is integrally closed if and

only if R = ©± V± , the intersection of some (not necessarily all) valuation rings of K.

The “only if” part follows from Property 9. For the “if” part, note that each V± is

integrally closed by Property 4, hence so is R. (If a is integral over R, then a is integral

over each V± , hence a belongs to each V± , so a ∈ R.) ™

3.3. DISCRETE VALUATION RINGS 5

3.3 Discrete Valuation Rings

3.3.1 De¬nitions and Comments

An absolute value on a ¬eld K is a mapping x ’ |x| from K to the real numbers, such

that for every x, y ∈ K,

1. |x| ≥ 0, with equality if and only if x = 0;

2. |xy| = |x| |y|;

3. |x + y| ¤ |x| + |y|.

The absolute value is nonarchimedean if the third condition is replaced by a stronger

version:

3 . |x + y| ¤ max(|x|, |y|).

As expected, archimedean means not nonarchimedean.

The familiar absolute values on the reals and the complex numbers are archimedean.

However, our interest will be in nonarchimedean absolute values. Here is where most of

them come from.

A discrete valuation on K is a surjective map v : K ’ Z ∪ {∞}, such that for every

x, y ∈ K,

(a) v(x) = ∞ if and only if x = 0;

(b) v(xy) = v(x) + v(y);

(c) v(x + y) ≥ min(v(x), v(y)).

A discrete valuation induces a nonarchimedean absolute value via |x| = cv(x) , where c

is a constant with 0 < c < 1.

3.3.2 Examples

We can place a discrete valuation on all of the ¬elds of Subsection 3.1.3. In Examples

1 and 2, we take v(pr m/n) = r. In Example 3, v(f /g) = deg g ’ deg f . In Example 4,

∞

v( i=r ai xi ) = r (if ar = 0).

3.3.3 Proposition

If v is a discrete valuation on the ¬eld K, then V = {a ∈ K : v(a) ≥ 0} is a valuation

ring with maximal ideal M = {a ∈ K : v(a) ≥ 1}.

Proof. The de¬ning properties (a), (b) and (c) of 3.3.1 show that V is a ring. If a ∈ V ,

/

’1 ’1

then v(a) < 0, so v(a ) = v(1) ’ v(a) = 0 ’ v(a) > 0, so a ∈ V , proving that V is a

valuation ring. Since a is a unit of V i¬ both a and a’1 belong to V i¬ v(a) = 0, M is

the ideal of nonunits and is therefore the maximal ideal of the valuation ring V . ™

3.3.4 De¬nitions and Comments

Discrete valuations do not determine all valuation rings. An arbitrary valuation ring

corresponds to a generalized absolute value mapping into an ordered group rather than

the real numbers. We will not consider the general situation, as discrete valuations will

be entirely adequate for us. A valuation ring V arising from a discrete valuation v as in

6 CHAPTER 3. VALUATION RINGS

(3.3.3) is said to be a discrete valuation ring, abbreviated DVR. An element t ∈ V with

v(t) = 1 is called a uniformizer or prime element. A uniformizer tells us a lot about the

DVR V and the ¬eld K.

3.3.5 Proposition

Let t be a uniformizer in the discrete valuation ring V . Then t generates the maximal

ideal M of V , in particular, M is principal. Conversely, if t is any generator of M, then

t is a uniformizer.

Proof. Since M is the unique maximal ideal, (t) ⊆ M. If a ∈ M, then v(a) ≥ 1, so

v(at’1 ) = v(a) ’ v(t) ≥ 1 ’ 1 = 0, so at’1 ∈ V , and consequently a ∈ (t). Now suppose

M = (t ). Since t ∈ M, we have t = ct for some c ∈ V . Thus

1 = v(t) = v(c) + v(t ) ≥ 0 + 1 = 1,

which forces v(t ) = 1. ™

3.3.6 Proposition

If t is a uniformizer, then every nonzero element a ∈ K can be expressed uniquely as

a = utn where u is a unit of V and n ∈ Z. Also, K = Vt , that is, K = S ’1 V where

S = {1, t, t2 , . . . }.

Proof. Let n = v(a), so that v(at’n ) = 0 and therefore at’n is a unit u. To prove

uniqueness, note that if a = utn , then v(a) = v(u) + nv(t) = 0 + n = n, so that n, and

hence u, is determined by a. The last statement follows by Property 1 of Section 3.2 and

the observation that the elements of V are those with valuation n ≥ 0. ™

A similar result holds for ideals.

3.3.7 Proposition

Every nonzero ideal I of the DVR V is of the form Mn , where M is the maximal ideal

of V and n is a unique nonnegative integer. We write v(I) = n; by convention, M0 = V .

Proof. Choose a ∈ I such that n = v(a) is as small as possible. By (3.3.6), a = utn , so

tn = u’1 a ∈ I. By (3.3.5), M = (t), and therefore Mn ⊆ I. Conversely, let b ∈ I, with

v(b) = k ≥ n by minimality of n. As in the proof of (3.3.6), bt’k is a unit u , so b = u tk .

Since k ≥ n we have b ∈ (tn ) = Mn , proving that I ⊆ Mn . The uniqueness of n is a

consequence of Nakayama™s lemma. If Mr = Ms with r < s, then Mr = Mr+1 = MMr .

Thus Mr , hence M, is 0, contradicting the hypothesis that I is nonzero. ™

We may interpret v(I) as the length of a composition series.

3.3.8 Proposition

Let I be a nonzero ideal of the discrete valuation ring R. Then v(I) = lR (R/I), the

composition length of the R-module R/I.

3.3. DISCRETE VALUATION RINGS 7

Proof. By (3.3.7), we have R ⊃ M ⊃ M2 ⊃ · · · ⊃ Mn = I, hence

R/I ⊃ M/I ⊃ M2 /I ⊃ · · · ⊃ Mn /I = 0.

By basic properties of composition length, we have, with l = lR ,

M/I

R/I

) + l(M/I) = l(R/M) + l( 2 ) + l(M2 /I).

l(R/I) = l(

M/I M /I

Continuing in this fashion, we get

n’1

l(Mi /Mi+1 ).

l(R/I) =

i=0

Since M is generated by a uniformizer t, it follows that ti + Mi+1 generates Mi /Mi+1 .

Since Mi /Mi+1 is annihilated by M, it is an R/M-module, that is, a vector space, over

the ¬eld R/M. The vector space is one-dimensional because the Mi , i = 0, 1, . . . , n, are

distinct [see the proof of (3.3.7)]. Consequently, l(R/I) = n. ™

We are going to prove a characterization theorem for DVR™s, and some preliminary

results will be needed.

3.3.9 Proposition

Let I be an ideal of the Noetherian ring R. Then for some positive integer m, we have

√

( I)m ⊆ I. In particular (take I = 0), the nilradical of R is nilpotent.

√

Since R is Noetherian, I is ¬nitely generated, say by a1 , . . . , at , with ani ∈ I.

Proof. √ i

t

Then ( I)m is generated by all products ar1 · · · art with i=1 ri = m. Our choice of m

t

1

is

t

(ni ’ 1).

m=1+

i=1

We claim that ri ≥ ni for some i. If not, then ri ¤ ni ’ 1 for all i, and

t t

(ni ’ 1) = m,

m= ri < 1 +

i=1 i=1

√

a contradiction. But then each product ar1 · · · art is in I, hence ( I)m ⊆ I. ™

t

1

3.3.10 Proposition

Let M be a maximal ideal of the Noetherian ring R, and let Q be any ideal of R. The

following conditions are equivalent:

1. Q is M-primary.

√

2. Q = M.

3. For some positive integer n, we have Mn ⊆ Q ⊆ M.

8 CHAPTER 3. VALUATION RINGS

Proof. We have (1) implies (2) by de¬nition of M-primary; see (1.1.1). The implication

(2) ’ (1) follows from (1.1.2). To prove that (2) implies (3), apply (3.3.9) with I = Q to

get, for some positive integer n,

Mn ⊆ Q ⊆ Q = M.

To prove that (3) implies (2), observe that by (1.1.1),

√ √

M = Mn ⊆ Q ⊆ M = M. ™

Now we can characterize discrete valuation rings.

3.3.11 Theorem

Let R be a Noetherian local domain with fraction ¬eld K and unique maximal ideal

M = 0. (Thus R is not a ¬eld.) The following conditions are equivalent:

1. R is a discrete valuation ring.

2. R is a principal ideal domain.

3. M is principal.

4. R is integrally closed and every nonzero prime ideal is maximal.

5. Every nonzero ideal is a power of M.

6. The dimension of M/M2 as a vector space over R/M is 1.

Proof.

(1) ’ (2): This follows from (3.3.7) and (3.3.5).

(2) ’ (4): This holds because a PID is integrally closed, and a PID is a UFD in which

every nonzero prime ideal is maximal.

(4) ’ (3): Let t be a nonzero element of M. By hypothesis, M is the only nonzero

prime ideal, so the radical of (t), which is the intersection of all prime ideals containing

t, coincides with M. By (3.3.10), for some n ≥ 1 we have Mn ⊆ (t) ⊆ M, and we

may assume that (t) ‚ M, for otherwise we are ¬nished. Thus for some n ≥ 2 we have

Mn ⊆ (t) but Mn’1 ⊆ (t). Choose a ∈ Mn’1 with a ∈ (t), and let β = t/a ∈ K. If

/

’1

= a/t ∈ R, then a ∈ Rt = (t), contradicting the choice of a. Therefore β ’1 ∈ R.

β /

’1 ’1

is not integral over R. But then β M ⊆ M, for if

Since R is integrally closed, β

’1 ’1

β M ⊆ M, then β stabilizes a ¬nitely generated R-module, and we conclude from

the implication (4) ’ (1) in (2.1.4) that β ’1 is integral over R, a contradiction.

Now β ’1 M ⊆ R, because β ’1 M = (a/t)M ⊆ (1/t)Mn ⊆ R. (Note that a ∈ Mn’1

and Mn ⊆ (t).) Thus β ’1 M is an ideal of R, and if it were proper, it would be contained

in M, contradicting β ’1 M ⊆ M. Consequently, β ’1 M = R, hence M is the principal

ideal (β).

(3) ’ (2): By hypothesis, M is a principal ideal (t), and we claim that ©∞ Mn = 0.

n=0

Suppose that a belongs to Mn for all n, with a = bn tn for some bn ∈ R. Then bn tn =

bn+1 tn+1 , hence bn = bn+1 t. Thus (bn ) ⊆ (bn+1 ) for all n, and in fact (bn ) = (bn+1 ) for

su¬ciently large n because R is Noetherian. Therefore bn = bn+1 t = ctbn for some c ∈ R,

so (1 ’ ct)bn = 0. But t ∈ M, so t is not a unit, and consequently ct = 1. Thus bn must

be 0, and we have a = bn tn = 0, proving the claim.

Now let I be any nonzero ideal of R. Then I ⊆ M, but by the above claim we

have I ⊆ ©n=0 Mn . Thus there exists n ≥ 0 such that I ⊆ Mn and I ⊆ Mn+1 . Choose

3.3. DISCRETE VALUATION RINGS 9

a ∈ I \Mn+1 ; since Mn = (t)n = (tn ), we have a = utn with u ∈ M (because a ∈ Mn+1 ).

/ /

’1

But then u is a unit, so t = u a ∈ I. To summarize, I ⊆ M = (t ) ⊆ I, proving that

n n n

I is principal.

(2) ’ (1): By hypothesis, M is a principal ideal (t), and by the proof of (3) ’ (2),

©∞ Mn = 0. Let a be any nonzero element of R. Then (a) ⊆ M, and since ©∞ Mn = 0,

n=0 n=0

we will have a ∈ (t ) but a ∈ (t ) for some n. Thus a = ut with u ∈ M, in other

n n+1 n

/ /

words, u is a unit. For ¬xed a, both u and n are unique (because t, a member of M, is a

nonunit). It follows that if β is a nonzero element of the fraction ¬eld K, then β = utm

uniquely, where u is a unit of R and m is an integer, possibly negative. If we de¬ne

v(β) = m, then v is a discrete valuation on K with valuation ring R.

(1) ’ (5): This follows from (3.3.7).

(5) ’ (3): As in the proof of (3.3.7), M = M2 . Choose t ∈ M \ M2 . By hypothesis,

(t) = Mn for some n ≥ 0. We cannot have n = 0 because (t) ⊆ M ‚ R, and we cannot

have n ≥ 2 by choice of t. The only possibility is n = 1, hence M = (t).

(1) ’ (6): This follows from the proof of (3.3.8).

(6) ’ (3): By hypothesis, M = M2 , so we may choose t ∈ M \ M2 . But then t + M2 is a

generator of the vector space M/M2 over the ¬eld R/M. Thus R(t+M2 )/M2 = M/M2 .

By the correspondence theorem, t + M2 = M. Now M(M/(t)) = (M2 + (t))/(t) =

M/(t), so by NAK, M/(t) = 0, that is, M = (t). ™.

Let us agree to exclude the trivial valuation v(a) = 0 for every a = 0.

3.3.12 Corollary

The ring R is a discrete valuation ring if and only if R is a local PID that is not a ¬eld.

In particular, since R is a PID, it is Noetherian.

Proof. The “if” part follows from (2) implies (1) in (3.3.11). For the “only if” part, note

that a discrete valuation ring R is a PID by (1) implies (2) of (3.3.11); the Noetherian

hypothesis is not used here. Moreover, R is a local ring by Property 3 of Section 3.2. If R

is a ¬eld, then every nonzero element a ∈ R is a unit, hence v(a) = 0. Thus the valuation

v is trivial, contradicting our convention. ™

3.3.13 Corollary

Let R be a DVR with maximal ideal M. If t ∈ M \ M2 , then t is a uniformizer.

Proof. This follows from the proof of (5) implies (3) in (3.3.11). ™

Chapter 4

Completion

The set R of real numbers is a complete metric space in which the set Q of rationals

is dense. In fact any metric space can be embedded as a dense subset of a complete

metric space. The construction is a familiar one involving equivalence classes of Cauchy

sequences. We will see that under appropriate conditions, this procedure can be general-

ized to modules.

4.1 Graded Rings and Modules

4.1.1 De¬nitions and Comments

A graded ring is a ring R that is expressible as •n≥0 Rn where the Rn are additive

subgroups such that Rm Rn ⊆ Rm+n . Sometimes, Rn is referred to as the nth graded

piece and elements of Rn are said to be homogeneous of degree n. The prototype is a

polynomial ring in several variables, with Rd consisting of all homogeneous polynomials

of degree d (along with the zero polynomial). A graded module over a graded ring R is a

module M expressible as •n≥0 Mn , where Rm Mn ⊆ Mm+n .

Note that the identity element of a graded ring R must belong to R0 . For if 1 has a

component a of maximum degree n > 0, then 1a = a forces the degree of a to exceed n,

a contradiction.

Now suppose that {Rn } is a ¬ltration of the ring R, in other words, the Rn are additive

subgroups such that

R = R0 ⊇ R 1 ⊇ · · · ⊇ R n ⊇ · · ·

with Rm Rn ⊆ Rm+n . We call R a ¬ltered ring. A ¬ltered module

M = M0 ⊇ M 1 ⊇ · · · ⊇ · · ·

over the ¬ltered ring R may be de¬ned similarly. In this case, each Mn is a submodule

and we require that Rm Mn ⊆ Mm+n .

If I is an ideal of the ring R and M is an R-module, we will be interested in the I-adic

¬ltrations of R and of M , given respectively by Rn = I n and Mn = I n M . (Take I 0 = R,

so that M0 = M .)

1

2 CHAPTER 4. COMPLETION

4.1.2 Associated Graded Rings and Modules

If {Rn } is a ¬ltration of R, the associated graded ring of R is de¬ned as

gr(R) = grn (R)

n≥0

where grn (R) = Rn /Rn+1 . We must be careful in de¬ning multiplication in gr(R). If

a ∈ Rm and b ∈ Rn , then a + Rm+1 ∈ Rm /Rm+1 and b + Rn+1 ∈ Rn /Rn+1 . We take

(a + Rm+1 )(b + Rn+1 ) = ab + Rm+n+1

so that the product of an element of grm (R) and an element of grn (R) will belong to

grm+n (R). If a ∈ Rm+1 and b ∈ Rn , then ab ∈ Rm+n+1 , so multiplication is well-de¬ned.

If M is a ¬ltered module over a ¬ltered ring R, we de¬ne the associated graded module

of M as

gr(M ) = grn (M )

n≥0

where grn (M ) = Mn /Mn+1 . If a ∈ Rm and x ∈ Mn , we de¬ne scalar multiplication by

(a + Rm+1 )(x + Mn+1 ) = ax + Mm+n+1

and it follows that

(Rm /Rm+1 )(Mn /Mn+1 ) ⊆ Mm+n /Mm+n+1 .

Thus gr(M ) is a graded module over the graded ring gr(R).

It is natural to ask for conditions under which a graded ring will be Noetherian, and

the behavior of the subring R0 is critical.

4.1.3 Proposition

Let R = •d≥0 Rd be a graded ring. Then R is Noetherian if and only if R0 is Noetherian

and R is a ¬nitely generated R0 -algebra.

Proof. If the condition on R0 holds, then R is a quotient of a polynomial ring R0 [X1 , . . . , Xn ],

hence R is Noetherian by the Hilbert Basis Theorem. Conversely, if R is Noetherian, then

so is R0 , because R0 ∼ R/I where I is the ideal •d≥1 Rd . By hypothesis, I is ¬nitely

=

generated, say by homogeneous elements a1 , . . . , ar of degree n1 , . . . , nr respectively. Let

R = R0 [a1 , . . . , ar ] be the R0 -subalgebra of R generated by the ai . It su¬ces to show

that Rn ⊆ R for all n ≥ 0 (and therefore R = R ). We have R0 ⊆ R by de¬nition of

R , so assume as an induction hypothesis that Rd ⊆ R for d ¤ n ’ 1, where n > 0. If

a ∈ Rn , then a can be expressed as c1 a1 + · · · + cr ar , where ci (i = 1, . . . , r) must be a

homogeneous element of degree n ’ ni < n = deg a. By induction hypothesis, ci ∈ R ,

and since ai ∈ R we have a ∈ R . ™

We now prepare for the basic Artin-Rees lemma.

4.1. GRADED RINGS AND MODULES 3

4.1.4 De¬nitions and Comments

Let M be a ¬ltered R-module with ¬ltration {Mn }, I an ideal of R. We say that {Mn }

is an I-¬ltration if IMn ⊆ Mn+1 for all n. An I-¬ltration with IMn = Mn+1 for all

su¬ciently large n is said to be I-stable. Note that the I-adic ¬ltration is I-stable.

4.1.5 Proposition

Let M be a ¬nitely generated module over a Noetherian ring R, and suppose that {Mn }

is an I-¬ltration of M . The following conditions are equivalent.

1. {Mn } is I-stable.

2. De¬ne a graded ring R— and a graded R— -module M — by

R— = M— =

I n, Mn .

n≥0 n≥0

Then M — is ¬nitely generated.

Proof. Let Nn = •n Mi , and de¬ne

i=0

—

Mn = M0 • · · · • Mn • IMn • I 2 Mn • · · ·

Since Nn is ¬nitely generated over R, it follows that Mn is a ¬nitely generated R— -module.

—

By de¬nition, M — is the union of the Mn over all n ≥ 0. Therefore M — is ¬nitely generated

—

over R— if and only if M — = Mm for some m, in other words, Mm+k = I k Mm for all k ≥ 1.

—

Equivalently, the ¬ltration {Mn } is I-stable. ™

4.1.6 Induced Filtrations

If {Mn } is a ¬ltration of the R-module M , and N is a submodule of M , then we have

¬ltrations induced on N and M/N , given by Nn = N © Mn and (M/N )n = (Mn + N )/N

respectively.

4.1.7 Artin-Rees Lemma

Let M be a ¬nitely generated module over the Noetherian ring R, and assume that M

has an I-stable ¬ltration {Mn }, where I is an ideal of R. Let N be a submodule of M .

Then the ¬ltration {Nn = N © Mn } induced by M on N is also I-stable.

Proof. As in (4.1.5), let R— = •n≥0 I n , M — = •n≥0 Mn , and N — = •n≥0 Nn . Since R

is Noetherian, I is ¬nitely generated, so R— is a ¬nitely generated R-algebra. (Elements

of R— can be expressed as polynomials in a ¬nite set of generators of I.) By (4.1.3), R— is

a Noetherian ring. Now by hypothesis, M is ¬nitely generated over the Noetherian ring

R and {Mn } is I-stable, so by (4.1.5), M — is ¬nitely generated over R— . Therefore the

submodule N — is also ¬nitely generated over R— . Again using (4.1.5), we conclude that

{Nn } is I-stable. ™

4 CHAPTER 4. COMPLETION

4.1.8 Applications

Let M be a ¬nitely generated module over the Noetherian ring R, with N a submodule of

M . The ¬ltration on N induced by the I-adic ¬ltration on M is given by Nm = (I m M )©N .

By Artin-Rees, for large enough m we have

I k ((I m M ) © N ) = (I m+k M ) © N

for all k ≥ 0.

There is a basic topological interpretation of this result. We can make M into a

topological abelian group in which the module operations are continuous. The sets I m M

are a base for the neighborhoods of 0, and the translations x + I m M form a basis for the

neighborhoods of an arbitrary point x ∈ M . The resulting topology is called the I-adic

topology on M . The above equation says that the I-adic topology on N coincides with

the topology induced on N by the I-adic topology on M .

4.2 Completion of a Module

4.2.1 Inverse Limits

Suppose we have countably many R-modules M0 , M1 , . . . , with R-module homomor-

phisms θn : Mn ’ Mn’1 , n ≥ 1. (We are restricting to the countable case to simplify

the notation, but the ideas carry over to an arbitrary family of modules, indexed by a

directed set. If i ¤ j, we have a homomorphism fij from Mj to Mi . We assume that the

maps can be composed consistently, in other words, if i ¤ j ¤ k, then fij —¦ fjk = fik .)The

collection of modules and maps is called an inverse system.

A sequence (xi ) in the direct product Mi is said to be coherent if it respects the

maps θn in the sense that for every i we have θi+1 (xi+1 ) = xi . The collection M of all

coherent sequences is called the inverse limit of the inverse system. The inverse limit is

denoted by

lim Mn .

←’

Note that M becomes an R-module with componentwise addition and scalar multiplica-

tion of coherent sequences, in other words, (xi ) + (yi ) = (xi + yi ) and r(xi ) = (rxi ).

Now suppose that we have homomorphisms gi from an R-module M to Mi , i =

0, 1, . . . . Call the gi coherent if θi+1 —¦ gi+1 = gi for all i. Then the gi can be lifted to a

homomorphism g from M to M . Explicitly, g(x) = (gi (x)), and the coherence of the gi

forces the sequence (gi (x)) to be coherent.

An inverse limit of an inverse system of rings can be constructed in a similar fashion,

as coherent sequences can be multiplied componentwise, that is, (xi )(yi ) = (xi yi ).

4.2.2 Examples

1. Take R = Z, and let I be the ideal (p) where p is a ¬xed prime. Take Mn = Z/I n and

θn+1 (a + I n+1 ) = a + I n . The inverse limit of the Mn is the ring Zp of p-adic integers.

4.2. COMPLETION OF A MODULE 5

2. Let R = A[x1 , . . . , xn ] be a polynomial ring in n variables, and I the maximal ideal

(x1 , . . . , xn ). Let Mn = R/I n and θn (f + I n ) = f + I n’1 , n = 1, 2, . . . . An element of

Mn is represented by a polynomial f of degree at most n ’ 1. (We take the degree of f

to be the maximum degree of a monomial in f .) The image of f in I n’1 is represented

by the same polynomial with the terms of degree n ’ 1 deleted. Thus the inverse limit

can be identi¬ed with the ring A[[x1 , . . . , xn ]] of formal power series.

Now let M be a ¬ltered R-module with ¬ltration {Mn }. The ¬ltration determines a

topology on M as in (4.1.8), with the Mn forming a base for the neighborhoods of 0. We

have the following result.

4.2.3 Proposition

If N is a submodule of M , then the closure of N is given by N = ©∞ (N + Mn ).

n=0

Proof. Let x be an element of M . Then x fails to belong to N i¬ some neighborhood of x

is disjoint from N , in other words, (x+Mn )©N = … for some n. Equivalently, x ∈ N +Mn

/

for some n, and the result follows. To justify the last step, note that if x ∈ N + Mn ,

then x = y + z, y ∈ N, z ∈ Mn . Thus y = x ’ z ∈ (x + Mn ) © N . Conversely, if

y ∈ (x + Mn ) © N , then for some z ∈ Mn we have y = x ’ z, so x = y + z ∈ N + Mn . ™

4.2.4 Corollary

The topology is Hausdor¬ if and only if ©∞ Mn = {0}.

n=0

Proof. By (4.2.3), ©∞ Mn = {0}, so we are asserting that the Hausdor¬ property is

n=0

equivalent to points being closed, that is, the T1 condition. This holds because separating

distinct points x and y by disjoint open sets is equivalent to separating x ’ y from 0. ™

4.2.5 De¬nition of the Completion

Let {Mn } be a ¬ltration of the R-module M . Recalling the construction of the reals from

the rationals, or the process of completing an arbitrary metric space, let us try to come up

with something similar in this case. If we go far out in a Cauchy sequence, the di¬erence

between terms becomes small. Thus we can de¬ne a Cauchy sequence {xn } in M by

the requirement that for every positive integer r there is a positive integer N such that

xn ’ xm ∈ Mr for n, m ≥ N . We identify the Cauchy sequences {xn } and {yn } if they

get close to each other for large n. More precisely, given a positive integer r there exists

a positive integer N such that xn ’ yn ∈ Mr for all n ≥ N . Notice that the condition

xn ’ xm ∈ Mr is equivalent to xn + Mr = xm + Mr . This suggests that the essential

feature of the Cauchy condition is that the sequence is coherent with respect to the maps

θn : M/Mn ’ M/Mn’1 . Motivated by this observation, we de¬ne the completion of M

as

ˆ

M = lim(M/Mn ).

←’

The functor that assigns the inverse limit to an inverse system of modules is left exact,

and becomes exact under certain conditions.

6 CHAPTER 4. COMPLETION

4.2.6 Theorem

Let {Mn , θn }, {Mn , θn }, and {Mn , θn } be inverse systems of modules, and assume that

the diagram below is commutative with exact rows.

fn+1 gn+1

G Mn+1 G Mn+1 G Mn+1 G0

0

θn+1 θn+1

θn+1

G Mn G Mn G Mn G0

fn gn

0

Then the sequence

0 ’ lim Mn ’ lim Mn ’ lim Mn

←’ ←’ ←’

is exact. If θn is surjective for all n, then

0 ’ lim Mn ’ lim Mn ’ lim Mn ’ 0

←’ ←’ ←’

is exact.

Mn and de¬ne an R- homomorphism dM : M ’ M by dM (xn ) =

Proof. Let M =

(xn ’ θn+1 (xn+1 )). The kernel of dM is the inverse limit of the Mn . Now the maps (fn )

and (gn ) induce f = fn : M = Mn ’ M and g = gn : M ’ M = Mn . We

have the following commutative diagram with exact rows.

GM GM GM G0

f g

0

dM dM

dM

GM GM GM G0

f g

0

We now apply the snake lemma, which is discussed in detail in TBGY (Section S2 of the

supplement). The result is an exact sequence

0 ’ ker dM ’ ker dM ’ ker dM ’ coker dM ,

proving the ¬rst assertion. If θn is surjective for all n, then dM is surjective, and conse-

quently the cokernel of dM is 0. The second assertion follows. ™

4.2.7 Corollary

Suppose that the sequence

GM GM GM G0

f g

0

is exact. Let {Mn } be a ¬ltration of M , so that {Mn } induces ¬ltrations {M © f ’1 (Mn )}

and {g(Mn )} on M and M respectively. Then the sequence

ˆ

0 ’ (M )

ˆ’ M ’ (M )

ˆ’ 0

4.2. COMPLETION OF A MODULE 7

is exact.

Proof. Exactness of the given sequence implies that the diagram below is commutative

with exact rows.

G M /(M © f ’1 (Mn+1 )) G M/Mn+1 G M /g(Mn+1 ) G0

0

θn+1 θn+1

θn+1

G M /(M © f ’1 (Mn )) G M/Mn G M /g(Mn ) G0

0

Since θn is surjective for all n, (4.2.6) allows us to pass to the inverse limit. ™

4.2.8 Remark

A ¬ltration {Mn } of an R-module M induces in a natural way a ¬ltration {N © Mn }

on a given submodule N , and a ¬ltration {(N + Mn )/N } on the quotient module M/N .

We have already noted this in (4.2.7) (with f the inclusion map and g the canonical

epimorphism), but the point is worth emphasizing.

4.2.9 Corollary

ˆ

Let {Mn } be a ¬ltration of the R-module M . Let Mn be the completion of Mn with

ˆ ˆ

respect to the induced ¬ltration on Mn [see (4.2.8)]. Then Mn is a submodule of M and

M /Mn ∼ M/Mn for all n.

ˆ ˆ=

Proof. We apply (4.2.7) with M = Mn and M = M/Mn , to obtain the exact sequence

ˆ ˆ

0 ’ Mn ’ M ’ (M/Mn )

ˆ’ 0.

ˆ ˆ

Thus we may identify Mn with a submodule of M , and

M /Mn ∼ (M/Mn ) (M ).

ˆ ˆ= ˆ= ˆ

Now the mth term of the induced ¬ltration on M is

Mm = (Mn + Mm )/Mn = Mn /Mn = 0

for m ≥ n. Thus M has the discrete topology, so Cauchy sequences (and coherent

sequences) can be identi¬ed with single points. Therefore M is isomorphic to its com-

pletion, and we have M /Mn ∼ M/Mn for every n. ™

ˆ ˆ=

4.2.10 Remarks

Two ¬ltrations {Mn } and {Mn } of a given R-module are said to be equivalent if they

induce the same topology. For example, under the hypothesis of (4.1.8), the ¬ltrations

{I n N } and {N © I n M } of the submodule N are equivalent (Problem 5). Since equivalent

¬ltrations give rise to the same set of Cauchy sequences, it follows that completions of a

given module with respect to equivalent ¬ltrations are isomorphic.

8 CHAPTER 4. COMPLETION

4.3 The Krull Intersection Theorem

4.3.1 De¬nitions and Comments

Recall from (4.1.1) and (4.1.8) that the I-adic topology on the R-module M is the topology

induced on M by the I-adic ¬ltration Mn = I n M . The completion of M with respect to

the I-adic ¬ltration is called the I-adic completion.

ˆ

There is a natural map from a ¬ltered module M to its completion M given by

x ’ {x + Mn }. The kernel of this map is ©∞ Mn , which is ©∞ I n M if the ¬ltration is

n=0 n=0

I-adic. The Krull intersection theorem (4.3.2) gives precise information about this kernel.

4.3.2 Theorem

Let M be a ¬nitely generated module over the Noetherian ring R, I an ideal of R, and

ˆ ˆ

M the I-adic completion of M . Let N be the kernel of the natural map M ’ M . Then

N is the set of elements x ∈ M such that x is annihilated by some element of 1 + I. In

fact, we can ¬nd a single element of 1 + I that works for the entire kernel.

Proof. Suppose that a ∈ I, x ∈ M , and (1 + a)x = 0. Then

x = ’ax = ’a(’ax) = a2 x = a2 (’ax) = ’a3 x = a4 x = · · · ,

hence x ∈ I n M for all n ≥ 0. Conversely, we must show that for some a ∈ I, 1 + a

annihilates everything in the kernel N . By (4.1.8), for some n we have, for all k ≥ 0,

I k ((I n M ) © N ) = (I n+k M ) © N.

Set k = 1 to get

I((I n M ) © N ) = (I n+1 M ) © N.

But N ⊆ I n+1 M ⊆ I n M , so the above equation says that IN = N . By (0.3.1), there

exists a ∈ I such that (1 + a)N = 0. ™

4.3.3 Corollary

If I is a proper ideal of the Noetherian integral domain R, then ©∞ I n = 0.

n=0

ˆ

n

Proof. The intersection of the I is the kernel N of the natural map from R to R. By

(4.3.2), 1 + a annihilates N for some a ∈ I. If 0 = x ∈ N then (1 + a)x = 0, and since R

is a domain, 1 + a = 0. But then ’1, hence 1, belongs to I, contradicting the hypothesis

that I is proper. ™

4.3.4 Corollary

Let M be a ¬nitely generated module over the Noetherian ring R. If the ideal I of R is

contained in the Jacobson radical J(R), then ©∞ I n M = 0. Thus by (4.2.4), the I-adic

n=0

topology on M is Hausdor¬.

Proof. Let a ∈ I ⊆ J(R) be such that (1 + a) annihilates the kernel N = ©∞ I n M n=0

ˆ . By (0.2.1), 1 + a is a unit of R, so if x ∈ N (hence

of the natural map from M to M

(1 + a)x = 0), we have x = 0. ™

4.4. HENSEL™S LEMMA 9

4.3.5 Corollary

Let R be a Noetherian local ring with maximal ideal M. If M is a ¬nitely generated

R-module, then ©∞ Mn M = 0. Thus the M-adic topology on M , in particular the

n=0

M-adic topology on R, is Hausdor¬.

Proof. Since M = J(R), this follows from (4.3.4). ™

4.4 Hensel™s Lemma

Let A be a local ring with maximal ideal P , and let k = A/P be the residue ¬eld.

Assume that A is complete with respect to the P -adic topology, in other words, every

Cauchy sequence converges. In algebraic number theory, where this result is most often

applied, A is a discrete valuation ring of a local ¬eld. But the statement and proof of the

algebraic number theory result can be copied, as follows.

If a ∈ A, then the coset a + P ∈ k will be denoted by a. If f is a polynomial in A[X],

then reduction of the coe¬cients of f mod P yields a polynomial f in k[X]. Thus

d d

ai X ∈ A[X], f (X) = ai X i ∈ k[X].

i

f (X) =

i=0 i=0

Hensel™s lemma is about lifting a factorization of f from k[X] to A[X]. Here is the precise

statement.

4.4.1 Hensel™s Lemma

Assume that f is a monic polynomial of degree d in A[X], and that the corresponding

polynomial F = f factors as the product of relatively prime monic polynomials G and H

in k[X]. Then there are monic polynomials g and h in A[X] such that g = G, h = H and

f = gh.

Proof. Let r be the degree of G, so that deg H = d ’ r. We will inductively construct

gn , hn ∈ A[X], n = 1, 2, . . . , such that deg gn = r, deg hn = d ’ r, g n = G, hn = H, and

f (X) ’ gn (X)hn (X) ∈ P n [X].

Thus the coe¬cients of f ’ gn hn belong to P n .

The basis step: Let n = 1. Choose monic g1 , h1 ∈ A[X] such that g 1 = G and h1 = H.

Then deg g1 = r and deg h1 = d ’ r. Since f = g 1 h1 , we have f ’ g1 h1 ∈ P [X].

The inductive step: Assume that gn and hn have been constructed. Let f (X)’gn (X)hn (X) =

d

i=0 ci X with the ci ∈ P . Since G = g n and H = hn are relatively prime, for each

i n

i = 0, . . . , d there are polynomials v i and wi in k[X] such that

X i = v i (X)g n (X) + wi (X)hn (X).

Since g n has degree r, the degree of v i is at most d ’ r, and similarly the degree of wi is

at most r. Moreover,

X i ’ vi (X)gn (X) ’ wi (X)hn (X) ∈ P [X]. (1)

10 CHAPTER 4. COMPLETION

We de¬ne

d d

gn+1 (X) = gn (X) + ci wi (X), hn+1 (X) = hn (X) + ci vi (X).

i=0 i=0

Since the ci belong to P n ⊆ P , it follows that g n+1 = g n = G and hn+1 = hn = H. Since

the degree of gn+1 is at most r, it must be exactly r, and similarly the degree of hn+1 is

d ’ r. To check the remaining condition,

f ’ gn+1 hn+1 = f ’ (gn + ci wi )(hn + ci vi )

i i

= (f ’ gn hn ’ ci (X i ’ gn vi ’ hn wi ) ’

ci X i ) + ci cj wi vj .

i i i,j

By the induction hypothesis, the ¬rst grouped term on the right is zero, and, with the

aid of Equation (1) above, the second grouped term belongs to P n P [X] = P n+1 [X]. The

¬nal term belongs to P 2n [X] ⊆ P n+1 [X], completing the induction.

Finishing the proof. By de¬nition of gn+1 , we have gn+1 ’ gn ∈ P n [X], so for any

¬xed i, the sequence of coe¬cients of X i in gn (X) is Cauchy and therefore converges.

To simplify the notation we write gn (X) ’ g(X), and similarly hn (X) ’ h(X), with

g(X), h(X) ∈ A[X]. By construction, f ’ gn hn ∈ P n [X], and we may let n ’ ∞ to get

f = gh. Since g n = G and hn = H for all n, we must have g = G and h = H. Since

f, G and H are monic, the highest degree terms of g and h are of the form (1 + a)X r and

(1 + a)’1 X d’r respectively, with a ∈ P . (Note that 1 + a must reduce to 1 mod P .) By

replacing g and h by (1 + a)’1 g and (1 + a)h, respectively, we can make g and h monic

without disturbing the other conditions. The proof is complete. ™

Chapter 5

Dimension Theory

The geometric notion of the dimension of an a¬ne algebraic variety V is closely related

to algebraic properties of the coordinate ring of the variety, that is, the ring of polynomial

functions on V . This relationship suggests that we look for various ways of de¬ning

the dimension of an arbitrary commutative ring. We will see that under appropriate

hypotheses, several concepts of dimension are equivalent. Later, we will connect the

algebraic and geometric ideas.

5.1 The Calculus of Finite Di¬erences

Regrettably, this charming subject is rarely taught these days, except in actuarial pro-

grams. It turns out to be needed in studying Hilbert and Hilbert-Samuel polynomials in

the next section.

5.1.1 Lemma

Let g and G be real-valued functions on the nonnegative integers, and assume that ∆G =

g, that is, G(k + 1) ’ G(k) = g(k) for all k ≥ 0. (We call ∆G the di¬erence of G.) Then

b

g(k) = G(k)|b+1 = G(b + 1) ’ G(a).

a

k=a

Proof. Add the equations G(a + 1) ’ G(a) = g(a), G(a + 2) ’ G(a + 1) = g(a + 1), . . . ,

G(b + 1) ’ G(b) = g(b). ™

5.1.2 Lemma

If r is a positive integer, de¬ne k (r) = k(k ’ 1)(k ’ 2) · · · (k ’ r + 1). Then ∆k (r) = rk (r’1) .

Proof. Just compute:

∆k (r) = (k + 1)(r) ’ k (r) = (k + 1)k(k ’ 1) · · · (k ’ r + 2) ’ k(k ’ 1) · · · (k ’ r + 1)

= k(k ’ 1) · · · (k ’ r + 2)[k + 1 ’ (k ’ r + 1)] = rk (r’1) . ™

1

2 CHAPTER 5. DIMENSION THEORY

5.1.3 Examples

n

= [k (2) /2]|n+1 = (n + 1)n/2.

∆k (2) = 2k (1) , so k=1 k 1

k 2 = k(k ’ 1) + k = k (2)

+ k (1) , so

n

k 2 = [k (3) /3]|n+1 + (n + 1)n/2 = (n + 1)n(n ’ 1)/3 + (n + 1)n/2

1

k=1

= n(n + 1)(2n + 1)/6.

k (3) = k(k ’ 1)(k ’ 2) = k 3 ’ 3k 2 + 2k, so k 3 = k (3) + 3k 2 ’ 2k. Therefore

n

k 3 = [k (4) /4]|n+1 + 3n(n + 1)(2n + 1)/6 ’ 2n(n + 1)/2.

1

k=1

The ¬rst term on the right is (n + 1)n(n ’ 1)(n ’ 2)/4, so the sum of the ¬rst n cubes is

[n(n + 1)/4][n2 ’ 3n + 2 + 2(2n + 1) ’ 4]

which simpli¬es to [n(n + 1)/2]2 .

n

k s for any positive integer s.

In a similar fashion we can ¬nd k=1

5.1.4 De¬nitions and Comments

A polynomial-like function is a function f from the natural numbers (nonnegative integers)

N to the rational numbers Q, such that f eventually agrees with a polynomial g ∈ Q[X].

In other words, f (n) = g(n) for all su¬ciently large n (abbreviated n >> 0). The degree

of f is taken to be the degree of g.

5.1.5 Lemma

Let f : N ’ Q. Then f is a polynomial-like function of degree r if and only if ∆f is a

polynomial-like function of degree r ’ 1. (We can allow r = 0 if we take the degree of the

zero polynomial to be -1.)

Proof. This follows from (5.1.1) and (5.1.2), along with the observation that a function

whose di¬erence is zero is constant. (The analogous result from di¬erential calculus that

a function with zero derivative is constant is harder to prove, and is usually accomplished

via the mean value theorem.) ™

5.2 Hilbert and Hilbert-Samuel Polynomials

There will be two polynomial-like functions of interest, and we begin preparing for their

arrival.

5.2. HILBERT AND HILBERT-SAMUEL POLYNOMIALS 3

5.2.1 Proposition

Let R = •n≥0 Rn be a graded ring. Assume that R0 is Artinian and R is ¬nitely generated

as an algebra over R0 . If M = •n≥0 Mn is a ¬nitely generated graded R-module, then

each Mn is a ¬nitely generated R0 -module.

Proof. By (4.1.3) and (1.6.13), R is a Noetherian ring, hence M is a Noetherian R-

module. Let Nn be the direct sum of the Mm , m ≥ n. Since M is Noetherian, Nn is

¬nitely generated over R, say by x1 , . . . , xt . Since Nn = Mn • •m>n Mm , we can write

xi = yi + zi with yi ∈ Mn and zi ∈ •m>n Mm . It su¬ces to prove that y1 , . . . , yt generate

t

Mn over R0 . If y ∈ Mn , then y is of the form i=1 ai xi with ai ∈ R. But just as we

decomposed xi above, we can write ai = bi + ci where bi ∈ R0 and ci ∈ •j>0 Rj . Thus

t t

y= (bi + ci )(yi + zi ) = bi y i

i=1 i=1

because the elements bi zi , ci yi and ci zi must belong to •m>n Mm . ™

5.2.2 Corollary

In (5.2.1), the length lRo (Mn ) of the R0 -module Mn is ¬nite for all n ≥ 0.

Proof. Apply (5.2.1) and (1.6.14). ™

We will need the following basic property of composition length.

5.2.3 Additivity of Length

Suppose we have an exact sequence of R-modules 0 ’ A1 ’ A2 ’ · · · ’ An ’ 0, all

with ¬nite length. Then we have additivity of length, that is,

l(A1 ) ’ l(A2 ) + · · · + (’1)n’1 l(An ) = 0.

This is probably familiar for a short exact sequence 0 ’ N ’ M ’ M/N ’ 0, where the

additivity property can be expressed as l(M ) = l(N ) + l(M/N ). (See TBGY, Section 7.5,

Problem 5.) The general result is accomplished by decomposing a long exact sequence

into short exact sequences. (“Long” means longer than short.) To see how the process

works, consider an exact sequence

GA f GB GC GD GE G 0.

g h i

0

Our ¬rst short exact sequence is

0 ’ A ’ B ’ coker f ’ 0.

Now coker f = B/ im f = B/ ker g ∼ im g (= ker h), so our second short exact sequence is

=

0 ’ im g ’ C ’ coker g ’ 0.

As above, coker g ∼ im h (= ker i), so the third short exact sequence is

=

0 ’ im h ’ D ’ coker h ’ 0.

4 CHAPTER 5. DIMENSION THEORY

But coker h ∼ im i = E, so we may replace the third short exact sequence by

=

0 ’ im h ’ D ’ E ’ 0.

Applying additivity for short exact sequences, we have

l(A) ’ l(B) + l(coker f ) ’ l(im g) + l(C) ’ l(coker g) + l(im h) ’ l(D) + l(E) = 0.

After cancellation, this becomes

l(A) ’ l(B) + l(C) ’ l(D) + l(E) = 0

as desired.

5.2.4 Theorem

Let R = •n≥0 Rn be a graded ring. Assume that R0 is Artinian and R is ¬nitely generated

as an algebra over R0 , with all generators a1 , . . . , ar belonging to R1 . If M is a ¬nitely

generated graded R-module, de¬ne h(M, n) = lR0 (Mn ), n ∈ N. Then h, as a function of

n with M ¬xed, is polynomial-like of degree at most r ’ 1. Using slightly loose language,

we call h the Hilbert polynomial of M .

Proof. We argue by induction on r. If r = 0, then R = R0 . Choose a ¬nite set

of homogeneous generators for M over R. If d is the maximum of the degrees of the

generators, then Mn = 0 for n > d, and therefore h(M, n) = 0 for n >> 0. Now

assume r > 0, and let »r be the endomorphism of M given by multiplication by ar . By

hypothesis, ar ∈ R1 , so »r (Mn ) ⊆ Mn+1 . if Kn is the kernel, and Cn the cokernel, of

»r : Mn ’ Mn+1 , we have the exact sequence

G Kn G Mn »r G Mn+1 G Cn G 0.

0

Let K be the direct sum of the Kn and C the direct sum of the Cn , n ≥ 0. Then

K is a submodule of M and C a quotient of M . Thus K and C are ¬nitely generated

Noetherian graded R-modules, so by (5.2.1) and (5.2.2), h(K, n) and h(C, n) are de¬ned

and ¬nite. By (5.2.3),

h(K, n) ’ h(M, n) + h(M, n + 1) ’ h(C, n) = 0

hence ∆h(M, n) = h(C, n) ’ h(K, n). Now ar annihilates K and C, so K and C are

¬nitely generated T -modules, where T is the graded subring of R generated over R0 by

a1 , . . . , ar’1 . (If an ideal I annihilates an R-module M , then M is an R/I-module;

see TBGY, Section 4.2, Problem 6.) By induction hypothesis, h(K, n) and h(C, n) are

polynomial-like of degree at most r ’ 2, hence so is ∆h(M, n). By (5.1.5), h(M, n) is

polynomial-like of degree at most r ’ 1. ™

5.2.5 De¬nitions and Comments

Let R be any Noetherian local ring with maximal ideal M. An ideal I of R is said to be

an ideal of de¬nition if Mn ⊆ I ⊆ M for some n ≥ 1. Equivalently, R/I is an Artinian

√

ring. [See (3.3.10), and note that I = M if and only if every prime ideal containing I

is maximal, so (1.6.11) applies.]

5.2. HILBERT AND HILBERT-SAMUEL POLYNOMIALS 5

5.2.6 The Hilbert-Samuel Polynomial

Let I be an ideal of de¬nition of the Noetherian local ring R. If M is a ¬nitely generated

R-module, then M/IM is a ¬nitely generated module over the Artinian ring R/I. Thus

M/IM is Artinian (as well as Noetherian), hence has ¬nite length over R/I. With the I-

adic ¬ltrations on R and M , the associated graded ring and the associated graded module

[see (4.1.2)] are given by

grI (R) = •n≥0 (I n /I n+1 ), grI (M ) = •n≥0 (I n M/I n+1 M ).

If I is generated over R by a1 , . . . , ar , then the images a1 , . . . , ar in I/I 2 generate grI (R)

over R/I. (Note that by de¬nition of a graded ring, Ri Rj ⊆ Ri+j , which allows us to

produce elements in Rt for arbitrarily large t.) By (5.2.4),

h(grI (M ), n) = lR/I (I n M/I n+1 M ) < ∞.

Again recall that if N is an R-module and the ideal I annihilates N , then N becomes an

R/I-module via (a + I)x = ax. It follows that we may replace lR/I by lR in the above

formula. We de¬ne the Hilbert-Samuel polynomial by

sI (M, n) = lR (M/I n M ).

Now the sequence

0 ’ I n M/I n+1 M ’ M/I n+1 M ’ M/I n M ’ 0

is exact by the third isomorphism theorem. An induction argument using additivity of

length shows that sI (M, n) is ¬nite. Consequently

∆sI (M, n) = sI (M, n + 1) ’ sI (M, n) = h(grI (M ), n).

By (5.2.4), h(grI (M ), n) is polynomial-like of degree at most r ’ 1, so by (5.1.5), sI (M, n)

is polynomial like of degree at most r.

The Hilbert-Samuel polynomial sI (M, n) depends on the particular ideal of de¬nition

I, but the degree d(M ) of sI (M, n) is the same for all possible choices. To see this, let t be a

positive integer such that Mt ⊆ I ⊆ M. Then for every n ≥ 1 we have Mtn ⊆ I n ⊆ Mn ,

so sM (M, tn) ≥ sI (M, n) ≥ sM (M, n). If the degrees of these polynomial are, from right

to left, d1 , d2 and d3 , we have O(dn ) ¤ O(dn ) ¤ O(dn ), with d3 = d1 . Therefore all three

1 2 3

degrees coincide.

The Hilbert-Samuel polynomial satis¬es a property analogous to (4.1.7), the additivity

of length.

5.2.7 Theorem

Let I be an ideal of de¬nition of the Noetherian local ring R, and suppose we have an

exact sequence 0 ’ M ’ M ’ M ’ 0 of ¬nitely generated R-modules. Then

sI (M , n) + sI (M , n) = sI (M, n) + r(n)

6 CHAPTER 5. DIMENSION THEORY

where r(n) is polynomial-like of degree less than d(M ), and the leading coe¬cient of r(n)

is nonnegative.

Proof. The following sequence is exact:

0 ’ M /(M © I n M ) ’ M/I n M ’ M /I n M ’ 0.

Set Mn = M © I n M . Then by additivity of length,

sI (M, n) ’ sI (M , n) = lR (M /Mn )

hence lR (M /Mn ) is polynomial-like. By the Artin-Rees lemma (4.1.7), the ¬ltration

{Mn } is I-stable, so IMn = Mn+1 for su¬ciently large n, say, n ≥ m. Thus for every

n ≥ 0 we have Mn+m = M © I n+m M ⊇ I n+m M , and consequently

I n+m M ⊆ Mn+m = I n Mm ⊆ I n M ,

which implies that

lR (M /I n+m M ) ≥ lR (M /Mn+m ) ≥ lR (M /I n M ).

The left and right hand terms of this inequality are sI (M , n + m) and sI (M , n) respec-

tively, and it follows that sI (M , n) and lR (M /Mn ) have the same degree and the same

leading coe¬cient. Moreover, sI (M , n) ’ lR (M /Mn ) = r(n) is polynomial-like of degree

less than deg lR (M /Mn ) ¤ deg sI (M, n), as well as nonnegative for n >> 0. The result

now follows upon adding the equations

sI (M, n) ’ sI (M , n) = lR (M /Mn )

and

r(n) = sI (M , n) ’ lR (M /Mn ). ™

5.2.8 Corollary

Let M be a submodule of M , where M is a ¬nitely generated module over the Noetherian

local ring R. Then d(M ) ¤ d(M ).

Proof. Apply (5.2.7), noting that we can ignore r(n) because it is of lower degree than

sI (M, n). ™

5.3 The Dimension Theorem

5.3.1 De¬nitions and Comments

The dimension of a ring R, denoted by dim R, will be taken as its Krull dimension, the

maximum length n of a chain P0 ‚ P1 ‚ · · · ‚ Pn of prime ideals of R. If there is no

upper bound on the length of such a chain, we take n = ∞. An example of an in¬nite-

dimensional ring is the non-Noetherian ring k[X1 , X2 , . . . ], where k is a ¬eld. We have

5.3. THE DIMENSION THEOREM 7

the in¬nite chain of prime ideals (X1 ) ‚ (X1 , X2 ) ‚ (X1 , X2 , X3 ) ‚ · · · . At the other

extreme, a ¬eld, and more generally an Artinian ring, has dimension 0 by (1.6.4).

A Dedekind domain is a Noetherian, integrally closed integral domain in which every

nonzero prime ideal is maximal. A Dedekind domain that is not a ¬eld has dimension 1.

Algebraic number theory provides many examples, because the ring of algebraic integers

of a number ¬eld is a Dedekind domain.

There are several other ideas that arise from the study of chains of prime ideals. We

de¬ne the height of a prime ideal P (notation ht P ) as the maximum length n of a chain

of prime ideals P0 ‚ P1 ‚ · · · ‚ Pn = P . By (0.4.6), the height of P is the dimension of

the localized ring RP .

The coheight of the prime ideal P (notation coht P ) is the maximum length n of a

chain of prime ideals P = P0 ‚ P1 ‚ · · · ‚ Pn . It follows from the correspondence

theorem and the third isomorphism theorem that the coheight of P is the dimension of

the quotient ring R/P . (If I and J are ideals of R with I ⊆ J, and S = (R/I)/(J/I),

then S ∼ R/J, so S is an integral domain i¬ R/J is an integral domain, and J/I is a

=

prime ideal of R/I i¬ J is a prime ideal of R.)

If I is an arbitrary ideal of R, we de¬ne the height of I as the in¬mum of the heights

of prime ideals P ⊇ I, and the coheight of I as the supremum of the coheights of prime

ideals P ⊇ I.

5.3.2 The Dimension of a Module

Intuitively, the dimension of an R-module M , denoted by dim M , will be measured by

length of chains of prime ideals, provided that the prime ideals in the chain contribute

to M in the sense that they belong to the support of M . Formally, we de¬ne dim M =

dim(R/ ann M ) if M = 0, and we take the dimension of the zero module to be ’1.

We now assume that M is nonzero and ¬nitely generated over the Noetherian ring R.

By (1.3.3), M has at least one associated prime. By (1.5.5), P ⊇ ann M i¬ P ∈ Supp M ,

and by (1.5.9), the minimal elements of AP(M ) and Supp M are the same. Thus

dim M = sup{coht P : P ∈ Supp M } = sup{coht P : P ∈ AP(M )}.

By (1.6.9), the following conditions are equivalent.

1. dim M = 0;

2. Every prime ideal in Supp M is maximal;

3. Every associated prime ideal of M is maximal;

4. The length of M as an R-module is ¬nite.

We make the additional assumption that R is a local ring with maximal ideal M.

Then by (1.5.5),

Supp(M/MM ) = V (ann(M/MM ))

which coincides with {M} by Problem 2. By the above equivalent conditions, lR (M/MM )

is ¬nite. Since M is ¬nitely generated, we can assert that there is a smallest positive

integer r, called the Chevalley dimension δ(M ), such that for some elements a1 , . . . , ar

belonging to M we have lR (M/(a1 , . . . , ar )M < ∞. If M = 0 we take δ(M ) = ’1.

8 CHAPTER 5. DIMENSION THEORY

5.3.3 Dimension Theorem

Let M be a ¬nitely generated module over the Noetherian local ring R. The following

quantities are equal:

1. The dimension dim M of the module M ;

2. The degree d(M ) of the Hilbert-Samuel polynomial sI (M, n), where I is any ideal of

de¬nition of R. (For convenience we take I = M, the maximal ideal of R, and we specify

that the degree of the zero polynomial is -1.);

3. The Chevalley dimension δ(M ).

Proof. We divide the proof into three parts.

1. dim M ¤ d(M ), hence dim M is ¬nite.

If d(M ) = ’1, then sM (M, n) = lR (M/Mn M ) = 0 for n >> 0. By NAK, M = 0 so

dim M = ’1. Thus assume d(M ) ≥ 0. By (1.3.9) or (1.5.10), M has only ¬nitely many

associated primes. It follows from (5.3.2) and (5.3.1) that for some associated prime P

we have dim M = coht P = dim R/P . By (1.3.2) there is an injective homomorphism

from R/P to M , so by (5.2.8) we have d(R/P ) ¤ d(M ). If we can show that dim R/P ¤

d(R/P ), it will follow that dim M = dim R/P ¤ d(R/P ) ¤ d(M ).

It su¬ces to show that for any chain of prime ideals P = P0 ‚ · · · ‚ Pt in R, the

length t of the chain is at most d(R/P ). If t = 0, then R/P = 0 (because P is prime),

hence d(R/P ) = ’1 and we are ¬nished. Thus assume t ≥ 1, and assume that the

result holds up to t ’ 1. Choose a ∈ P1 \ P , and consider prime ideals Q such that

Ra + P ⊆ Q ⊆ P1 . We can pick a Q belonging to AP(R/(Ra + P )). [Since Ra + P ⊆ Q,

we have (R/(Ra+P ))Q = 0; see Problem 3. Then choose Q to be a minimal element in the

support of R/(Ra+P ), and apply (1.5.9).] By (1.3.2) there is an injective homomorphism

from R/Q to R/(Ra+P ), so by (5.2.8) we have d(R/Q) ¤ d(R/(Ra+P )). Since the chain

Q ‚ P2 ‚ · · · ‚ Qr is of length t’1, the induction hypothesis implies that t’1 ¤ d(R/Q),

hence t ’ 1 ¤ d(R/(Ra + P )). Now the sequence

0 ’ R/P ’ R/P ’ R/(Ra + P ) ’ 0

is exact, where the map from R/P to itself is multiplication by a. (The image of the map

is Ra + P .) By (5.2.7),

sM (R/P, n) + sM (R/(Ra + P ), n) = sM (R/P, n) + r(n)

where r(n) is polynomial-like of degree less than d(R/P ). Thus d(R/(Ra+P )) < d(R/P ),

and consequently t ’ 1 < d(R/P ). Therefore t ¤ d(R/P ), as desired.