<< стр. 2(всего 4)СОДЕРЖАНИЕ >>
diп¬Ђerent proof is given here.)

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) в‡’ (2) в‡’ (3). To prove that (3) в‡’ (1), let m в€€ M . If P is
a maximal ideal of R, then m/1 is 0 in MP , so there exists rP в€€ 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 в€€ M by construction. Thus I(m) must be R, and
/
in particular, 1 в€€ I(m). Thus 1 can be written as a п¬Ѓnite sum P aP rP where P is a
maximal ideal of R and aP в€€ R. Consequently,

aP rP m = 0. в™Ј
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
в€љ
Let S be integral over the subring R, with I an ideal of R. Then IS is the set of all
s в€€ S satisfying an equation of integral dependence sm + rmв€’1 smв€’1 + В· В· В· + r1 s + r0 = 0
with the ri в€€ I.
в€љ
Proof. If s satisп¬Ѓes such an equation, then sm в€€ IS, so s в€€ IS. Conversely, let sn в€€
k
IS, n в‰Ґ 1, so that sn = i=1 ri si for some ri в€€ I and si в€€ S. Then S1 = R[s1 , . . . , sk ]
is a subring of S, and is also a п¬Ѓnitely generated R-module by (2.1.5). Now
k k
ri si S1 вЉ† ri S1 вЉ† 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, satisп¬Ѓes an equation of integral dependence
with coeп¬ѓcients in I. в™Ј

2.3.2 Lemma
Let R be an integral domain with fraction п¬Ѓeld K, and assume that R is integrally closed.
Let f and g be monic polynomials in K[x]. If f g в€€ R[x], then both f and g are in R[x].
Proof. In a splitting п¬Ѓeld containing K, we have f (x) = i (xв€’ai ) and g(x) = j (xв€’bj ).
Since the ai and bj are roots of the monic polynomial f g в€€ R[x], they are integral over R.
The coeп¬ѓcients 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. в™Ј

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 в€€ S, deп¬Ѓne a homomorphism hs : R[x] в†’ 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 п¬Ѓeld 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 в€€ R[x].
By hypothesis, s is integral over R, so there is a monic polynomial h в€€ R[x] such that
h(s) = 0. Then h в€€ I вЉ† 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 в€€ I.
Since f в€€ IK[x], we may clear denominators to produce a nonzero element r в€€ R such
that rf в€€ IR[x] = I. By deп¬Ѓnition of I we have rf (s) = 0, and by hypothesis, r is not a
zero-divisor of S. Therefore f (s) = 0, so f в€€ I.
Step 3 : f generates I.
Let q в€€ I вЉ† 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 в€€ R and q1 в€€ 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 coeп¬ѓcient of q1 must be 0, which means that q1 itself must be 0. Consequently,
r1 divides every coeп¬ѓcient of q1 , so q1 /r1 в€€ R[x]. Thus f divides q in R[x]. в™Ј

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 вЉ† В· В· В· вЉ† Pn of R and a chain of prime ideals Qm вЉ† В· В· В· вЉ† Qn
of S, with 1 < m в‰¤ n. If Qi lies over Pi for i = m, . . . , n, then there are prime ideals
Q1 , . . . , Qmв€’1 such that Q1 вЉ† В· В· В· вЉ† Qm and Qi lies over Pi for every i = 1, . . . , n.
Proof. By induction, it suп¬ѓces to consider n = m = 2. Let T be the subset of S consisting
of all products rt, r в€€ R \ P1 , t в€€ 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 в€€ P1 (hence r = 0)
/
and t в€€ Q2 , then the hypothesis that r is not a zero-divisor of S gives t = 0, which is a
/
contradiction (because 0 в€€ Q2 ). Note that R \ P1 вЉ† T (take t = 1), and S \ Q2 вЉ† T (take
r = 1).
First we prove the theorem under the assumption that T в€© P1 S = в€…. Now P1 ST is
a proper ideal of ST , else 1 would belong to T в€© 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 в€€ Q1 iп¬Ђ s/1 в€€ M. We refer to Q1 as the contraction
of M to S; it is the preimage of M under the canonical map s в†’ s/1. With the aid of
the note at the end of the last paragraph, we have (R \ P1 ) в€© Q1 = (S \ Q2 ) в€© Q1 = в€….
Thus Q1 в€© R вЉ† P1 and Q1 = Q1 в€© S вЉ† Q2 . We must show that P1 вЉ† Q1 в€© R. We do this
by taking the contraction of both sides of the inclusion P1 ST вЉ† M. Since the contraction
of P1 ST to S is P1 S, we have P1 S вЉ† Q1 , so P1 вЉ† (P1 S) в€© R вЉ† Q1 в€© R, as desired.
Finally, we show that T в€© P1 S is empty. If not, then by deп¬Ѓnition of T , T в€© P1 S
contains an element rt with r в€€ R \ P1 and t в€€ S \ Q2 . We apply (2.3.1), with I = P1 and
s replaced by rt, to produce a monic polynomial f (x) = xm + rmв€’1 xmв€’1 + В· В· В· + r1 x + r0
with coeп¬ѓcients in P1 such that f (rt) = 0. Deп¬Ѓne

v(x) = rm xm + rmв€’1 rmв€’1 xmв€’1 + В· В· В· + r1 rx + r0 .

Then v(x) в€€ R[x] and v(t) = 0. By (2.3.3), there is a monic polynomial g в€€ R[x] that
generates the kernel of the evaluation map ht : R[x] в†’ S. Therefore v = ug for some
u в€€ R[x]. Passing to residue classes in the polynomial ring (R/P1 )[x], we have v = u g.
Since ri в€€ P1 for all i = 0, . . . , m в€’ 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 в‰¤ j в‰¤ m. (Note
that r в€€ P1 , so v is not the zero polynomial.) Consequently,
/

g(x) = xj + ajв€’1 xjв€’1 + В· В· В· + a1 x + a0

with ai в€€ P1 , i = 0, . . . , j в€’ 1. But g в€€ 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 в€€ P1 S вЉ† P2 S вЉ† Q2 S = Q2 , so
t в€€ Q2 . This contradicts our choice of t (recall that t в€€ S \ Q2 ). в™Ј
Chapter 3

Valuation Rings

The results of this chapter come into play when analyzing the behavior of a rational
function deп¬Ѓned in the neighborhood of a point on an algebraic curve.

3.1 Extension Theorems
In Theorem 2.2.4, we generalized a result about п¬Ѓeld extensions to rings. Here is another
variation.

3.1.1 Theorem
Let R be a subring of the п¬Ѓeld K, and h : R в†’ C a ring homomorphism from R into an
algebraically closed п¬Ѓeld C. If О± is a nonzero element of K, then either h can be extended
to a ring homomorphism h : R[О±] в†’ C, or h can be extended to a ring homomorphism
h : R[О±в€’1 ] в†’ C.
Proof. Without loss of generality, we may assume that R is a local ring and F = h(R) is
a subп¬Ѓeld 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 в†’ C via g(a/b) = h(a)/h(b), h(b) = 0. The kernel of g is P RP , so
by the п¬Ѓrst isomorphism theorem, g(RP ) в€ј RP /P RP , a п¬Ѓeld (because P RP is a maximal
=
ideal). Thus we may replace (R, h) by (RP , g).
Our п¬Ѓrst step is to extend h to a homomorphism of polynomial rings. If f в€€ R[x] with
h(ai )xi в€€ F [x]. Let I = {f в€€ R[x] : f (О±) = 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 ОІ in the algebraically closed п¬Ѓeld C. We can then extend h to
h : R[О±] в†’ C via h(О±) = ОІ, as desired. To verify that h is well-deп¬Ѓned, suppose f в€€ I, so
that f (О±) = 0. Then h(f ) в€€ J, hence h(f ) is a multiple of j, and therefore h(f )(ОІ) = 0.
Thus we may assume that j is constant. If the constant is zero, then we may extend h
exactly as above, with ОІ arbitrary. So we can assume that j = 0, and it follows that
1 в€€ J. Consequently, there exists f в€€ I such that h(f ) = 1.

1
2 CHAPTER 3. VALUATION RINGS

This gives a relation of the form
r
1, i = 0
ai О±i = 0 with ai в€€ 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 О± replaced by
О±в€’1 . Assuming that h has no extension to R[О±в€’1 ], we have
s
1, i = 0
bi О±в€’i = 0 with bi в€€ R and bi = h(bi ) = (2)
0, i > 0
i=0

Take s minimal, and assume (without loss of generality) that r в‰Ґ s. Since h(b0 ) = 1 =
h(1), it follows that b0 в€’ 1 в€€ ker h вЉ† M, the unique maximal ideal of the local ring R.
Thus b0 в€€ M (else 1 в€€ M), so b0 is a unit. It is therefore legal to multiply (2) by bв€’1 О±s
/ 0
to get

О±s + bв€’1 b1 О±sв€’1 + В· В· В· + bв€’1 bs = 0 (3)
0 0

Finally, we multiply (3) by ar О±rв€’s and subtract the result from (1) to contradict the
minimality of r. (The result of multiplying (3) by ar О±rв€’s cannot be a copy of (1). If so,
r = s (hence О±rв€’s = 1)and a0 = ar bв€’1 bs . But h(a0 ) = 1 and h(ar bв€’1 bs ) = 0.) в™Ј
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 Deп¬Ѓnition
A subring R of a п¬Ѓeld K is a valuation ring of K if for every nonzero О± в€€ K, either О± or
О±в€’1 belongs to R.

3.1.3 Examples
The п¬Ѓeld K is a valuation ring of K, but there are more interesting examples.
1. Let K = Q, with p a п¬Ѓxed prime. Take R to be the set of all rationals of the form
pr m/n, where r в‰Ґ 0 and p divides neither m nor n.
2. Let K = k(x), where k is any п¬Ѓeld. Take R to be the set of all rational functions
pr m/n, where r в‰Ґ 0, p is a п¬Ѓxed 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 в€€ k(x) such that
deg f в‰¤ deg g.
4. Let K be the п¬Ѓeld of formal Laurent series over k. Thus a nonzero element of K looks
в€ћ
like f = i=r ai xi with ai в€€ k, r в€€ 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.
3.2. PROPERTIES OF VALUATION RINGS 3

3.1.4 Theorem
Let R be a subring of the п¬Ѓeld K, and h : R в†’ C a ring homomorphism from R into an
algebraically closed п¬Ѓeld 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 ) в‰¤ (Rj , hj ) if and only if Ri is a
subring of Rj and hj restricted to Ri coincides with hi . A standard application of ZornвЂ™s
lemma produces a maximal extension (V, h). If О± is a nonzero element of K, then by
(3.1.1), h has an extension to either V [О±] or V [О±в€’1 ]. By maximality, either V [О±] = V or
V [О±в€’1 ] = V . Therefore О± в€€ V or О±в€’1 в€€ V . в™Ј

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 п¬Ѓeld K.
1. The fraction п¬Ѓeld of V is K.
This follows because a nonzero element О± of K can be written as О±/1 or as 1/О±в€’1 .
2. Any subring of K containing V is a valuation ring of K.
This follows from the deп¬Ѓnition 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 в€€ V , then a + b = b(1 + a/b) в€€ M (because
if b(1 + a/b) were a unit, then b would be a unit as well). Similarly, if b/a в€€ V , then
a + b в€€ M. If r в€€ V and a в€€ M, then ra в€€ M, else a would be a unit. Thus M is an
ideal.
4. V is integrally closed.
Let О± be a nonzero element of K, with О± integral over V . Then there is an equation of
the form

О±n + cnв€’1 О±nв€’1 + В· В· В· + c1 О± + c0 = 0

with the ci in V . We must show that О± в€€ V . If not, then О±в€’1 в€€ V , and if we multiply
the above equation of integral dependence by О±в€’(nв€’1) , we get

О± = в€’cnв€’1 в€’ cnв€’2 О±в€’1 в€’ В· В· В· в€’ c1 О±nв€’2 в€’ c0 О±nв€’1 в€€ V.

5. If I and J are ideals of V , then either I вЉ† J or J вЉ† I. Thus the ideals of V are totally
ordered by inclusion.
Suppose that I is not contained in J, and pick a в€€ I \ J (hence a = 0). If b в€€ J, we
must show that b в€€ I. If b = 0 we are п¬Ѓnished, so assume b = 0. We have b/a в€€ V (else
a/b в€€ V , so a = (a/b)b в€€ J, a contradiction). Therefore b = (b/a)a в€€ I.
4 CHAPTER 3. VALUATION RINGS

6. Conversely, let V be an integral domain with fraction п¬Ѓeld K. If the ideals of V are
partially ordered by inclusion, then V is a valuation ring of K.
If О± is a nonzero element of K, then О± = a/b with a and b nonzero elements of V . By
hypothesis, either (a) вЉ† (b), in which case a/b в€€ V , or (b) вЉ† (a), in which case b/a в€€ 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 п¬Ѓeld of V , it is also the fraction п¬Ѓeld of VP . Also, V /P
is an integral domain, hence has a fraction п¬Ѓeld. 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 в€€ V ,
every ideal is of the form (pm ), m в‰Ґ 0. For any such p, в€©в€ћ (pm ) = 0.
m=1

Since V is Noetherian, an ideal I of V is п¬Ѓnitely generated, say by a1 , . . . , an . By Property
5, we may renumber the ai so that (a1 ) вЉ† (a2 ) В· В· В· вЉ† (an ). But then I вЉ† (an ) вЉ† 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 в€€ 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 в‰Ґ 1, then pm divides a for all m в‰Ґ 1.
Again using unique factorization, we must have a = 0. (Note that if a is a unit, so is p, a
9. Let R be a subring of the п¬Ѓeld K. The integral closure R of R in K is the intersection
of all valuation rings V of K such that V вЉ‡ R.
If a в€€ R, then a is integral over R, hence over any valuation ring V вЉ‡ R. But V is
integrally closed by Property 4, so a в€€ V . Conversely, assume a в€€ R. Then a fails to
/
в€’1 в€’1
belong to the ring R = R[a ]. (If a is a polynomial in a , multiply by a suп¬ѓciently
high power of a to get a monic equation satisп¬Ѓed by a.) Thus aв€’1 cannot be a unit in
R . (If baв€’1 = 1 with b в€€ R , then a = a1 = aaв€’1 b = b в€€ R , a contradiction.) It follows
that aв€’1 belongs to a maximal ideal M of R . Let C be an algebraic closure of the п¬Ѓeld
k = R /M , and let h be the composition of the canonical map R в†’ R /M = k and
the inclusion k в†’ C. By (3.1.4), h has a maximal extension to h : V в†’ C for some
valuation ring V of K containing R вЉ‡ R. Now h(aв€’1 ) = h(aв€’1 ) since aв€’1 в€€ M вЉ† R ,
and h(aв€’1 ) = 0 by deп¬Ѓnition of h. Consequently a в€€ V , for if a в€€ V , then
/

1 = h(1) = h(aaв€’1 ) = h(a)h(aв€’1 ) = 0,

10. Let R be an integral domain with fraction п¬Ѓeld K. Then R is integrally closed if and
only if R = в€©О± VО± , the intersection of some (not necessarily all) valuation rings of K.
The вЂњonly ifвЂќ part follows from Property 9. For the вЂњifвЂќ part, note that each VО± is
integrally closed by Property 4, hence so is R. (If a is integral over R, then a is integral
over each VО± , hence a belongs to each VО± , so a в€€ R.) в™Ј
3.3. DISCRETE VALUATION RINGS 5

3.3 Discrete Valuation Rings
An absolute value on a п¬Ѓeld K is a mapping x в†’ |x| from K to the real numbers, such
that for every x, y в€€ K,
1. |x| в‰Ґ 0, with equality if and only if x = 0;
2. |xy| = |x| |y|;
3. |x + y| в‰¤ |x| + |y|.
The absolute value is nonarchimedean if the third condition is replaced by a stronger
version:
3 . |x + y| в‰¤ 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 в†’ Z в€Є {в€ћ}, such that for every
x, y в€€ K,
(a) v(x) = в€ћ if and only if x = 0;
(b) v(xy) = v(x) + v(y);
(c) v(x + y) в‰Ґ 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 п¬Ѓelds 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 в€’ deg f . In Example 4,
в€ћ
v( i=r ai xi ) = r (if ar = 0).

3.3.3 Proposition
If v is a discrete valuation on the п¬Ѓeld K, then V = {a в€€ K : v(a) в‰Ґ 0} is a valuation
ring with maximal ideal M = {a в€€ K : v(a) в‰Ґ 1}.
Proof. The deп¬Ѓning properties (a), (b) and (c) of 3.3.1 show that V is a ring. If a в€€ V ,
/
в€’1 в€’1
then v(a) < 0, so v(a ) = v(1) в€’ v(a) = 0 в€’ v(a) > 0, so a в€€ V , proving that V is a
valuation ring. Since a is a unit of V iп¬Ђ both a and aв€’1 belong to V iп¬Ђ v(a) = 0, M is
the ideal of nonunits and is therefore the maximal ideal of the valuation ring V . в™Ј

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 в€€ V with
v(t) = 1 is called a uniformizer or prime element. A uniformizer tells us a lot about the
DVR V and the п¬Ѓeld 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) вЉ† M. If a в€€ M, then v(a) в‰Ґ 1, so
v(atв€’1 ) = v(a) в€’ v(t) в‰Ґ 1 в€’ 1 = 0, so atв€’1 в€€ V , and consequently a в€€ (t). Now suppose
M = (t ). Since t в€€ M, we have t = ct for some c в€€ V . Thus

1 = v(t) = v(c) + v(t ) в‰Ґ 0 + 1 = 1,

which forces v(t ) = 1. в™Ј

3.3.6 Proposition
If t is a uniformizer, then every nonzero element a в€€ K can be expressed uniquely as
a = utn where u is a unit of V and n в€€ Z. Also, K = Vt , that is, K = S в€’1 V where
S = {1, t, t2 , . . . }.
Proof. Let n = v(a), so that v(atв€’n ) = 0 and therefore atв€’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 в‰Ґ 0. в™Ј
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 в€€ I such that n = v(a) is as small as possible. By (3.3.6), a = utn , so
tn = uв€’1 a в€€ I. By (3.3.5), M = (t), and therefore Mn вЉ† I. Conversely, let b в€€ I, with
v(b) = k в‰Ґ n by minimality of n. As in the proof of (3.3.6), btв€’k is a unit u , so b = u tk .
Since k в‰Ґ n we have b в€€ (tn ) = Mn , proving that I вЉ† Mn . The uniqueness of n is a
consequence of NakayamaвЂ™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. в™Ј
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 вЉѓ M вЉѓ M2 вЉѓ В· В· В· вЉѓ Mn = I, hence

R/I вЉѓ M/I вЉѓ M2 /I вЉѓ В· В· В· вЉѓ 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
nв€’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 п¬Ѓeld 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. в™Ј
We are going to prove a characterization theorem for DVRвЂ™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
в€љ
( I)m вЉ† I. In particular (take I = 0), the nilradical of R is nilpotent.
в€љ
Since R is Noetherian, I is п¬Ѓnitely generated, say by a1 , . . . , at , with ani в€€ I.
Proof. в€љ i
t
Then ( I)m is generated by all products ar1 В· В· В· art with i=1 ri = m. Our choice of m
t
1
is
t
(ni в€’ 1).
m=1+
i=1

We claim that ri в‰Ґ ni for some i. If not, then ri в‰¤ ni в€’ 1 for all i, and
t t
(ni в€’ 1) = m,
m= ri < 1 +
i=1 i=1
в€љ
a contradiction. But then each product ar1 В· В· В· art is in I, hence ( I)m вЉ† I. в™Ј
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.
в€љ
2. Q = M.
3. For some positive integer n, we have Mn вЉ† Q вЉ† M.
8 CHAPTER 3. VALUATION RINGS

Proof. We have (1) implies (2) by deп¬Ѓnition of M-primary; see (1.1.1). The implication
(2) в‡’ (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 вЉ† Q вЉ† Q = M.
To prove that (3) implies (2), observe that by (1.1.1),
в€љ в€љ
M = Mn вЉ† Q вЉ† M = M. в™Ј
Now we can characterize discrete valuation rings.

3.3.11 Theorem
Let R be a Noetherian local domain with fraction п¬Ѓeld K and unique maximal ideal
M = 0. (Thus R is not a п¬Ѓeld.) 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) в‡’ (2): This follows from (3.3.7) and (3.3.5).
(2) в‡’ (4): This holds because a PID is integrally closed, and a PID is a UFD in which
every nonzero prime ideal is maximal.
(4) в‡’ (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 в‰Ґ 1 we have Mn вЉ† (t) вЉ† M, and we
may assume that (t) вЉ‚ M, for otherwise we are п¬Ѓnished. Thus for some n в‰Ґ 2 we have
Mn вЉ† (t) but Mnв€’1 вЉ† (t). Choose a в€€ Mnв€’1 with a в€€ (t), and let ОІ = t/a в€€ K. If
/
в€’1
= a/t в€€ R, then a в€€ Rt = (t), contradicting the choice of a. Therefore ОІ в€’1 в€€ R.
ОІ /
в€’1 в€’1
is not integral over R. But then ОІ M вЉ† M, for if
Since R is integrally closed, ОІ
в€’1 в€’1
ОІ M вЉ† M, then ОІ stabilizes a п¬Ѓnitely generated R-module, and we conclude from
the implication (4) в‡’ (1) in (2.1.4) that ОІ в€’1 is integral over R, a contradiction.
Now ОІ в€’1 M вЉ† R, because ОІ в€’1 M = (a/t)M вЉ† (1/t)Mn вЉ† R. (Note that a в€€ Mnв€’1
and Mn вЉ† (t).) Thus ОІ в€’1 M is an ideal of R, and if it were proper, it would be contained
in M, contradicting ОІ в€’1 M вЉ† M. Consequently, ОІ в€’1 M = R, hence M is the principal
ideal (ОІ).
(3) в‡’ (2): By hypothesis, M is a principal ideal (t), and we claim that в€©в€ћ Mn = 0.
n=0
Suppose that a belongs to Mn for all n, with a = bn tn for some bn в€€ R. Then bn tn =
bn+1 tn+1 , hence bn = bn+1 t. Thus (bn ) вЉ† (bn+1 ) for all n, and in fact (bn ) = (bn+1 ) for
suп¬ѓciently large n because R is Noetherian. Therefore bn = bn+1 t = ctbn for some c в€€ R,
so (1 в€’ ct)bn = 0. But t в€€ 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 вЉ† M, but by the above claim we
have I вЉ† в€©n=0 Mn . Thus there exists n в‰Ґ 0 such that I вЉ† Mn and I вЉ† Mn+1 . Choose
3.3. DISCRETE VALUATION RINGS 9

a в€€ I \Mn+1 ; since Mn = (t)n = (tn ), we have a = utn with u в€€ M (because a в€€ Mn+1 ).
/ /
в€’1
But then u is a unit, so t = u a в€€ I. To summarize, I вЉ† M = (t ) вЉ† I, proving that
n n n

I is principal.
(2) в‡’ (1): By hypothesis, M is a principal ideal (t), and by the proof of (3) в‡’ (2),
в€©в€ћ Mn = 0. Let a be any nonzero element of R. Then (a) вЉ† M, and since в€©в€ћ Mn = 0,
n=0 n=0
we will have a в€€ (t ) but a в€€ (t ) for some n. Thus a = ut with u в€€ M, in other
n n+1 n
/ /
words, u is a unit. For п¬Ѓxed a, both u and n are unique (because t, a member of M, is a
nonunit). It follows that if ОІ is a nonzero element of the fraction п¬Ѓeld K, then ОІ = utm
uniquely, where u is a unit of R and m is an integer, possibly negative. If we deп¬Ѓne
v(ОІ) = m, then v is a discrete valuation on K with valuation ring R.
(1) в‡’ (5): This follows from (3.3.7).
(5) в‡’ (3): As in the proof of (3.3.7), M = M2 . Choose t в€€ M \ M2 . By hypothesis,
(t) = Mn for some n в‰Ґ 0. We cannot have n = 0 because (t) вЉ† M вЉ‚ R, and we cannot
have n в‰Ґ 2 by choice of t. The only possibility is n = 1, hence M = (t).
(1) в‡’ (6): This follows from the proof of (3.3.8).
(6) в‡’ (3): By hypothesis, M = M2 , so we may choose t в€€ M \ M2 . But then t + M2 is a
generator of the vector space M/M2 over the п¬Ѓeld 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). в™Ј.
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 п¬Ѓeld.
In particular, since R is a PID, it is Noetherian.
Proof. The вЂњifвЂќ part follows from (2) implies (1) in (3.3.11). For the вЂњonly ifвЂќ 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 п¬Ѓeld, then every nonzero element a в€€ R is a unit, hence v(a) = 0. Thus the valuation
v is trivial, contradicting our convention. в™Ј

3.3.13 Corollary
Let R be a DVR with maximal ideal M. If t в€€ M \ M2 , then t is a uniformizer.
Proof. This follows from the proof of (5) implies (3) in (3.3.11). в™Ј
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.

A graded ring is a ring R that is expressible as вЉ•nв‰Ґ0 Rn where the Rn are additive
subgroups such that Rm Rn вЉ† 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 вЉ•nв‰Ґ0 Mn , where Rm Mn вЉ† 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,
Now suppose that {Rn } is a п¬Ѓltration of the ring R, in other words, the Rn are additive
subgroups such that
R = R0 вЉ‡ R 1 вЉ‡ В· В· В· вЉ‡ R n вЉ‡ В· В· В·
with Rm Rn вЉ† Rm+n . We call R a п¬Ѓltered ring. A п¬Ѓltered module
M = M0 вЉ‡ M 1 вЉ‡ В· В· В· вЉ‡ В· В· В·
over the п¬Ѓltered ring R may be deп¬Ѓned similarly. In this case, each Mn is a submodule
and we require that Rm Mn вЉ† 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
п¬Ѓltrations 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 п¬Ѓltration of R, the associated graded ring of R is deп¬Ѓned as

gr(R) = grn (R)
nв‰Ґ0

where grn (R) = Rn /Rn+1 . We must be careful in deп¬Ѓning multiplication in gr(R). If
a в€€ Rm and b в€€ Rn , then a + Rm+1 в€€ Rm /Rm+1 and b + Rn+1 в€€ 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 в€€ Rm+1 and b в€€ Rn , then ab в€€ Rm+n+1 , so multiplication is well-deп¬Ѓned.
If M is a п¬Ѓltered module over a п¬Ѓltered ring R, we deп¬Ѓne the associated graded module
of M as

gr(M ) = grn (M )
nв‰Ґ0

where grn (M ) = Mn /Mn+1 . If a в€€ Rm and x в€€ Mn , we deп¬Ѓne scalar multiplication by

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

and it follows that

(Rm /Rm+1 )(Mn /Mn+1 ) вЉ† 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 = вЉ•dв‰Ґ0 Rd be a graded ring. Then R is Noetherian if and only if R0 is Noetherian
and R is a п¬Ѓnitely 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 в€ј R/I where I is the ideal вЉ•dв‰Ґ1 Rd . By hypothesis, I is п¬Ѓnitely
=
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 suп¬ѓces to show
that Rn вЉ† R for all n в‰Ґ 0 (and therefore R = R ). We have R0 вЉ† R by deп¬Ѓnition of
R , so assume as an induction hypothesis that Rd вЉ† R for d в‰¤ n в€’ 1, where n > 0. If
a в€€ Rn , then a can be expressed as c1 a1 + В· В· В· + cr ar , where ci (i = 1, . . . , r) must be a
homogeneous element of degree n в€’ ni < n = deg a. By induction hypothesis, ci в€€ R ,
and since ai в€€ R we have a в€€ R . в™Ј
We now prepare for the basic Artin-Rees lemma.
4.1. GRADED RINGS AND MODULES 3

Let M be a п¬Ѓltered R-module with п¬Ѓltration {Mn }, I an ideal of R. We say that {Mn }
is an I-п¬Ѓltration if IMn вЉ† Mn+1 for all n. An I-п¬Ѓltration with IMn = Mn+1 for all
suп¬ѓciently large n is said to be I-stable. Note that the I-adic п¬Ѓltration is I-stable.

4.1.5 Proposition
Let M be a п¬Ѓnitely generated module over a Noetherian ring R, and suppose that {Mn }
is an I-п¬Ѓltration of M . The following conditions are equivalent.

1. {Mn } is I-stable.
2. Deп¬Ѓne a graded ring Rв€— and a graded Rв€— -module M в€— by

Rв€— = Mв€— =
I n, Mn .
nв‰Ґ0 nв‰Ґ0

Then M в€— is п¬Ѓnitely generated.

Proof. Let Nn = вЉ•n Mi , and deп¬Ѓne
i=0

в€—
Mn = M0 вЉ• В· В· В· вЉ• Mn вЉ• IMn вЉ• I 2 Mn вЉ• В· В· В·

Since Nn is п¬Ѓnitely generated over R, it follows that Mn is a п¬Ѓnitely generated Rв€— -module.
в€—

By deп¬Ѓnition, M в€— is the union of the Mn over all n в‰Ґ 0. Therefore M в€— is п¬Ѓnitely generated
в€—

over Rв€— if and only if M в€— = Mm for some m, in other words, Mm+k = I k Mm for all k в‰Ґ 1.
в€—

Equivalently, the п¬Ѓltration {Mn } is I-stable. в™Ј

4.1.6 Induced Filtrations
If {Mn } is a п¬Ѓltration of the R-module M , and N is a submodule of M , then we have
п¬Ѓltrations induced on N and M/N , given by Nn = N в€© Mn and (M/N )n = (Mn + N )/N
respectively.

4.1.7 Artin-Rees Lemma
Let M be a п¬Ѓnitely generated module over the Noetherian ring R, and assume that M
has an I-stable п¬Ѓltration {Mn }, where I is an ideal of R. Let N be a submodule of M .
Then the п¬Ѓltration {Nn = N в€© Mn } induced by M on N is also I-stable.
Proof. As in (4.1.5), let Rв€— = вЉ•nв‰Ґ0 I n , M в€— = вЉ•nв‰Ґ0 Mn , and N в€— = вЉ•nв‰Ґ0 Nn . Since R
is Noetherian, I is п¬Ѓnitely generated, so Rв€— is a п¬Ѓnitely generated R-algebra. (Elements
of Rв€— can be expressed as polynomials in a п¬Ѓnite set of generators of I.) By (4.1.3), Rв€— is
a Noetherian ring. Now by hypothesis, M is п¬Ѓnitely generated over the Noetherian ring
R and {Mn } is I-stable, so by (4.1.5), M в€— is п¬Ѓnitely generated over Rв€— . Therefore the
submodule N в€— is also п¬Ѓnitely generated over Rв€— . Again using (4.1.5), we conclude that
{Nn } is I-stable. в™Ј
4 CHAPTER 4. COMPLETION

4.1.8 Applications
Let M be a п¬Ѓnitely generated module over the Noetherian ring R, with N a submodule of
M . The п¬Ѓltration on N induced by the I-adic п¬Ѓltration on M is given by Nm = (I m M )в€©N .
By Artin-Rees, for large enough m we have

I k ((I m M ) в€© N ) = (I m+k M ) в€© N

for all k в‰Ґ 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 в€€ 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 Оёn : Mn в†’ Mnв€’1 , n в‰Ґ 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 в‰¤ j, we have a homomorphism fij from Mj to Mi . We assume that the
maps can be composed consistently, in other words, if i в‰¤ j в‰¤ k, then fij в—¦ 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 Оёn in the sense that for every i we have Оёi+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 .
в†ђв€’

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 Оёi+1 в—¦ 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 п¬Ѓxed prime. Take Mn = Z/I n and
Оёn+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 Оёn (f + I n ) = f + I nв€’1 , n = 1, 2, . . . . An element of
Mn is represented by a polynomial f of degree at most n в€’ 1. (We take the degree of f
to be the maximum degree of a monomial in f .) The image of f in I nв€’1 is represented
by the same polynomial with the terms of degree n в€’ 1 deleted. Thus the inverse limit
can be identiп¬Ѓed with the ring A[[x1 , . . . , xn ]] of formal power series.

Now let M be a п¬Ѓltered R-module with п¬Ѓltration {Mn }. The п¬Ѓltration 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 = в€©в€ћ (N + Mn ).
n=0
Proof. Let x be an element of M . Then x fails to belong to N iп¬Ђ some neighborhood of x
is disjoint from N , in other words, (x+Mn )в€©N = в€… for some n. Equivalently, x в€€ N +Mn
/
for some n, and the result follows. To justify the last step, note that if x в€€ N + Mn ,
then x = y + z, y в€€ N, z в€€ Mn . Thus y = x в€’ z в€€ (x + Mn ) в€© N . Conversely, if
y в€€ (x + Mn ) в€© N , then for some z в€€ Mn we have y = x в€’ z, so x = y + z в€€ N + Mn . в™Ј

4.2.4 Corollary
The topology is Hausdorп¬Ђ if and only if в€©в€ћ Mn = {0}.
n=0
Proof. By (4.2.3), в€©в€ћ Mn = {0}, so we are asserting that the Hausdorп¬Ђ 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 в€’ y from 0. в™Ј

4.2.5 Deп¬Ѓnition of the Completion
Let {Mn } be a п¬Ѓltration 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 diп¬Ђerence
between terms becomes small. Thus we can deп¬Ѓne a Cauchy sequence {xn } in M by
the requirement that for every positive integer r there is a positive integer N such that
xn в€’ xm в€€ Mr for n, m в‰Ґ 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 в€’ yn в€€ Mr for all n в‰Ґ N . Notice that the condition
xn в€’ xm в€€ 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
Оёn : M/Mn в†’ M/Mnв€’1 . Motivated by this observation, we deп¬Ѓne the completion of M
as
Л†
M = lim(M/Mn ).
в†ђв€’

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 , Оёn }, {Mn , Оёn }, and {Mn , Оёn } 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
Оёn+1 Оёn+1
Оёn+1
 

G Mn G Mn G Mn G0
fn gn
0

Then the sequence
0 в†’ lim Mn в†’ lim Mn в†’ lim Mn
в†ђв€’ в†ђв€’ в†ђв€’

is exact. If Оёn is surjective for all n, then
0 в†’ lim Mn в†’ lim Mn в†’ lim Mn в†’ 0
в†ђв€’ в†ђв€’ в†ђв€’

is exact.
Mn and deп¬Ѓne an R- homomorphism dM : M в†’ M by dM (xn ) =
Proof. Let M =
(xn в€’ Оёn+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 в†’ M and g = gn : M в†’ 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 в†’ ker dM в†’ ker dM в†’ ker dM в†’ coker dM ,
proving the п¬Ѓrst assertion. If Оёn is surjective for all n, then dM is surjective, and conse-
quently the cokernel of dM is 0. The second assertion follows. в™Ј

4.2.7 Corollary
Suppose that the sequence

GM GM GM G0
f g
0

is exact. Let {Mn } be a п¬Ѓltration of M , so that {Mn } induces п¬Ѓltrations {M в€© f в€’1 (Mn )}
and {g(Mn )} on M and M respectively. Then the sequence
Л†
0 в†’ (M )
Л†в†’ M в†’ (M )
Л†в†’ 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 в€© f в€’1 (Mn+1 )) G M/Mn+1 G M /g(Mn+1 ) G0
0
Оёn+1 Оёn+1
Оёn+1
 

G M /(M в€© f в€’1 (Mn )) G M/Mn G M /g(Mn ) G0
0
Since Оёn is surjective for all n, (4.2.6) allows us to pass to the inverse limit. в™Ј

4.2.8 Remark
A п¬Ѓltration {Mn } of an R-module M induces in a natural way a п¬Ѓltration {N в€© Mn }
on a given submodule N , and a п¬Ѓltration {(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
Л†
Let {Mn } be a п¬Ѓltration of the R-module M . Let Mn be the completion of Mn with
Л† Л†
respect to the induced п¬Ѓltration on Mn [see (4.2.8)]. Then Mn is a submodule of M and
M /Mn в€ј M/Mn for all n.
Л† Л†=
Proof. We apply (4.2.7) with M = Mn and M = M/Mn , to obtain the exact sequence

Л† Л†
0 в†’ Mn в†’ M в†’ (M/Mn )
Л†в†’ 0.

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

M /Mn в€ј (M/Mn ) (M ).
Л† Л†= Л†= Л†

Now the mth term of the induced п¬Ѓltration on M is

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

for m в‰Ґ n. Thus M has the discrete topology, so Cauchy sequences (and coherent
sequences) can be identiп¬Ѓed with single points. Therefore M is isomorphic to its com-
pletion, and we have M /Mn в€ј M/Mn for every n. в™Ј
Л† Л†=

4.2.10 Remarks
Two п¬Ѓltrations {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 п¬Ѓltrations
{I n N } and {N в€© I n M } of the submodule N are equivalent (Problem 5). Since equivalent
п¬Ѓltrations give rise to the same set of Cauchy sequences, it follows that completions of a
given module with respect to equivalent п¬Ѓltrations are isomorphic.
8 CHAPTER 4. COMPLETION

4.3 The Krull Intersection Theorem
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 п¬Ѓltration Mn = I n M . The completion of M with respect to
Л†
There is a natural map from a п¬Ѓltered module M to its completion M given by
x в†’ {x + Mn }. The kernel of this map is в€©в€ћ Mn , which is в€©в€ћ I n M if the п¬Ѓltration is
n=0 n=0

4.3.2 Theorem
Let M be a п¬Ѓnitely generated module over the Noetherian ring R, I an ideal of R, and
Л† Л†
M the I-adic completion of M . Let N be the kernel of the natural map M в†’ M . Then
N is the set of elements x в€€ M such that x is annihilated by some element of 1 + I. In
fact, we can п¬Ѓnd a single element of 1 + I that works for the entire kernel.
Proof. Suppose that a в€€ I, x в€€ M , and (1 + a)x = 0. Then
x = в€’ax = в€’a(в€’ax) = a2 x = a2 (в€’ax) = в€’a3 x = a4 x = В· В· В· ,
hence x в€€ I n M for all n в‰Ґ 0. Conversely, we must show that for some a в€€ I, 1 + a
annihilates everything in the kernel N . By (4.1.8), for some n we have, for all k в‰Ґ 0,
I k ((I n M ) в€© N ) = (I n+k M ) в€© N.
Set k = 1 to get
I((I n M ) в€© N ) = (I n+1 M ) в€© N.
But N вЉ† I n+1 M вЉ† I n M , so the above equation says that IN = N . By (0.3.1), there
exists a в€€ I such that (1 + a)N = 0. в™Ј

4.3.3 Corollary
If I is a proper ideal of the Noetherian integral domain R, then в€©в€ћ I n = 0.
n=0
Л†
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 в€€ I. If 0 = x в€€ N then (1 + a)x = 0, and since R
is a domain, 1 + a = 0. But then в€’1, hence 1, belongs to I, contradicting the hypothesis
that I is proper. в™Ј

4.3.4 Corollary
Let M be a п¬Ѓnitely generated module over the Noetherian ring R. If the ideal I of R is
contained in the Jacobson radical J(R), then в€©в€ћ I n M = 0. Thus by (4.2.4), the I-adic
n=0
topology on M is Hausdorп¬Ђ.
Proof. Let a в€€ I вЉ† J(R) be such that (1 + a) annihilates the kernel N = в€©в€ћ I n M n=0
Л† . By (0.2.1), 1 + a is a unit of R, so if x в€€ N (hence
of the natural map from M to M
(1 + a)x = 0), we have x = 0. в™Ј
4.4. HENSELвЂ™S LEMMA 9

4.3.5 Corollary
Let R be a Noetherian local ring with maximal ideal M. If M is a п¬Ѓnitely generated
R-module, then в€©в€ћ Mn M = 0. Thus the M-adic topology on M , in particular the
n=0
M-adic topology on R, is Hausdorп¬Ђ.
Proof. Since M = J(R), this follows from (4.3.4). в™Ј

4.4 HenselвЂ™s Lemma
Let A be a local ring with maximal ideal P , and let k = A/P be the residue п¬Ѓeld.
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 п¬Ѓeld. But the statement and proof of the
algebraic number theory result can be copied, as follows.
If a в€€ A, then the coset a + P в€€ k will be denoted by a. If f is a polynomial in A[X],
then reduction of the coeп¬ѓcients of f mod P yields a polynomial f in k[X]. Thus
d d
ai X в€€ A[X], f (X) = ai X i в€€ k[X].
i
f (X) =
i=0 i=0

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

4.4.1 HenselвЂ™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 в€’ r. We will inductively construct
gn , hn в€€ A[X], n = 1, 2, . . . , such that deg gn = r, deg hn = d в€’ r, g n = G, hn = H, and
f (X) в€’ gn (X)hn (X) в€€ P n [X].
Thus the coeп¬ѓcients of f в€’ gn hn belong to P n .
The basis step: Let n = 1. Choose monic g1 , h1 в€€ A[X] such that g 1 = G and h1 = H.
Then deg g1 = r and deg h1 = d в€’ r. Since f = g 1 h1 , we have f в€’ g1 h1 в€€ P [X].
The inductive step: Assume that gn and hn have been constructed. Let f (X)в€’gn (X)hn (X) =
d
i=0 ci X with the ci в€€ 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 в€’ r, and similarly the degree of wi is
at most r. Moreover,
X i в€’ vi (X)gn (X) в€’ wi (X)hn (X) в€€ P [X]. (1)
10 CHAPTER 4. COMPLETION

We deп¬Ѓne
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 вЉ† 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 в€’ r. To check the remaining condition,

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

= (f в€’ gn hn в€’ ci (X i в€’ gn vi в€’ hn wi ) в€’
ci X i ) + ci cj wi vj .
i i i,j

By the induction hypothesis, the п¬Ѓrst 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
п¬Ѓnal term belongs to P 2n [X] вЉ† P n+1 [X], completing the induction.
Finishing the proof. By deп¬Ѓnition of gn+1 , we have gn+1 в€’ gn в€€ P n [X], so for any
п¬Ѓxed i, the sequence of coeп¬ѓcients of X i in gn (X) is Cauchy and therefore converges.
To simplify the notation we write gn (X) в†’ g(X), and similarly hn (X) в†’ h(X), with
g(X), h(X) в€€ A[X]. By construction, f в€’ gn hn в€€ P n [X], and we may let n в†’ в€ћ 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)в€’1 X dв€’r respectively, with a в€€ P . (Note that 1 + a must reduce to 1 mod P .) By
replacing g and h by (1 + a)в€’1 g and (1 + a)h, respectively, we can make g and h monic
without disturbing the other conditions. The proof is complete. в™Ј
Chapter 5

Dimension Theory

The geometric notion of the dimension of an aп¬ѓne 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 deп¬Ѓning
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 Diп¬Ђerences
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 в€†G =
g, that is, G(k + 1) в€’ G(k) = g(k) for all k в‰Ґ 0. (We call в€†G the diп¬Ђerence of G.) Then
b
g(k) = G(k)|b+1 = G(b + 1) в€’ G(a).
a
k=a

Proof. Add the equations G(a + 1) в€’ G(a) = g(a), G(a + 2) в€’ G(a + 1) = g(a + 1), . . . ,
G(b + 1) в€’ G(b) = g(b). в™Ј

5.1.2 Lemma
If r is a positive integer, deп¬Ѓne k (r) = k(k в€’ 1)(k в€’ 2) В· В· В· (k в€’ r + 1). Then в€†k (r) = rk (rв€’1) .
Proof. Just compute:
в€†k (r) = (k + 1)(r) в€’ k (r) = (k + 1)k(k в€’ 1) В· В· В· (k в€’ r + 2) в€’ k(k в€’ 1) В· В· В· (k в€’ r + 1)
= k(k в€’ 1) В· В· В· (k в€’ r + 2)[k + 1 в€’ (k в€’ r + 1)] = rk (rв€’1) . в™Ј

1
2 CHAPTER 5. DIMENSION THEORY

5.1.3 Examples
n
= [k (2) /2]|n+1 = (n + 1)n/2.
в€†k (2) = 2k (1) , so k=1 k 1

k 2 = k(k в€’ 1) + k = k (2)
+ k (1) , so
n
k 2 = [k (3) /3]|n+1 + (n + 1)n/2 = (n + 1)n(n в€’ 1)/3 + (n + 1)n/2
1
k=1
= n(n + 1)(2n + 1)/6.

k (3) = k(k в€’ 1)(k в€’ 2) = k 3 в€’ 3k 2 + 2k, so k 3 = k (3) + 3k 2 в€’ 2k. Therefore
n
k 3 = [k (4) /4]|n+1 + 3n(n + 1)(2n + 1)/6 в€’ 2n(n + 1)/2.
1
k=1

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

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

which simpliп¬Ѓes to [n(n + 1)/2]2 .
n
k s for any positive integer s.
In a similar fashion we can п¬Ѓnd k=1

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 в€€ Q[X].
In other words, f (n) = g(n) for all suп¬ѓciently large n (abbreviated n >> 0). The degree
of f is taken to be the degree of g.

5.1.5 Lemma
Let f : N в†’ Q. Then f is a polynomial-like function of degree r if and only if в€†f is a
polynomial-like function of degree r в€’ 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 diп¬Ђerence is zero is constant. (The analogous result from diп¬Ђerential calculus that
a function with zero derivative is constant is harder to prove, and is usually accomplished
via the mean value theorem.) в™Ј

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 = вЉ•nв‰Ґ0 Rn be a graded ring. Assume that R0 is Artinian and R is п¬Ѓnitely generated
as an algebra over R0 . If M = вЉ•nв‰Ґ0 Mn is a п¬Ѓnitely generated graded R-module, then
each Mn is a п¬Ѓnitely 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 в‰Ґ n. Since M is Noetherian, Nn is
п¬Ѓnitely generated over R, say by x1 , . . . , xt . Since Nn = Mn вЉ• вЉ•m>n Mm , we can write
xi = yi + zi with yi в€€ Mn and zi в€€ вЉ•m>n Mm . It suп¬ѓces to prove that y1 , . . . , yt generate
t
Mn over R0 . If y в€€ Mn , then y is of the form i=1 ai xi with ai в€€ R. But just as we
decomposed xi above, we can write ai = bi + ci where bi в€€ R0 and ci в€€ вЉ•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 вЉ•m>n Mm . в™Ј

5.2.2 Corollary
In (5.2.1), the length lRo (Mn ) of the R0 -module Mn is п¬Ѓnite for all n в‰Ґ 0.
Proof. Apply (5.2.1) and (1.6.14). в™Ј
We will need the following basic property of composition length.

Suppose we have an exact sequence of R-modules 0 в†’ A1 в†’ A2 в†’ В· В· В· в†’ An в†’ 0, all
with п¬Ѓnite length. Then we have additivity of length, that is,

l(A1 ) в€’ l(A2 ) + В· В· В· + (в€’1)nв€’1 l(An ) = 0.

This is probably familiar for a short exact sequence 0 в†’ N в†’ M в†’ M/N в†’ 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. (вЂњLongвЂќ 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 п¬Ѓrst short exact sequence is

0 в†’ A в†’ B в†’ coker f в†’ 0.

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

0 в†’ im g в†’ C в†’ coker g в†’ 0.

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

0 в†’ im h в†’ D в†’ coker h в†’ 0.
4 CHAPTER 5. DIMENSION THEORY

But coker h в€ј im i = E, so we may replace the third short exact sequence by
=
0 в†’ im h в†’ D в†’ E в†’ 0.
Applying additivity for short exact sequences, we have
l(A) в€’ l(B) + l(coker f ) в€’ l(im g) + l(C) в€’ l(coker g) + l(im h) в€’ l(D) + l(E) = 0.
After cancellation, this becomes
l(A) в€’ l(B) + l(C) в€’ l(D) + l(E) = 0
as desired.

5.2.4 Theorem
Let R = вЉ•nв‰Ґ0 Rn be a graded ring. Assume that R0 is Artinian and R is п¬Ѓnitely generated
as an algebra over R0 , with all generators a1 , . . . , ar belonging to R1 . If M is a п¬Ѓnitely
generated graded R-module, deп¬Ѓne h(M, n) = lR0 (Mn ), n в€€ N. Then h, as a function of
n with M п¬Ѓxed, is polynomial-like of degree at most r в€’ 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 п¬Ѓnite 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 О»r be the endomorphism of M given by multiplication by ar . By
hypothesis, ar в€€ R1 , so О»r (Mn ) вЉ† Mn+1 . if Kn is the kernel, and Cn the cokernel, of
О»r : Mn в†’ Mn+1 , we have the exact sequence
G Kn G Mn О»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 в‰Ґ 0. Then
K is a submodule of M and C a quotient of M . Thus K and C are п¬Ѓnitely generated
Noetherian graded R-modules, so by (5.2.1) and (5.2.2), h(K, n) and h(C, n) are deп¬Ѓned
and п¬Ѓnite. By (5.2.3),
h(K, n) в€’ h(M, n) + h(M, n + 1) в€’ h(C, n) = 0
hence в€†h(M, n) = h(C, n) в€’ h(K, n). Now ar annihilates K and C, so K and C are
п¬Ѓnitely generated T -modules, where T is the graded subring of R generated over R0 by
a1 , . . . , arв€’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 в€’ 2, hence so is в€†h(M, n). By (5.1.5), h(M, n) is
polynomial-like of degree at most r в€’ 1. в™Ј

Let R be any Noetherian local ring with maximal ideal M. An ideal I of R is said to be
an ideal of deп¬Ѓnition if Mn вЉ† I вЉ† M for some n в‰Ґ 1. Equivalently, R/I is an Artinian
в€љ
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 deп¬Ѓnition of the Noetherian local ring R. If M is a п¬Ѓnitely generated
R-module, then M/IM is a п¬Ѓnitely generated module over the Artinian ring R/I. Thus
M/IM is Artinian (as well as Noetherian), hence has п¬Ѓnite length over R/I. With the I-
[see (4.1.2)] are given by

grI (R) = вЉ•nв‰Ґ0 (I n /I n+1 ), grI (M ) = вЉ•nв‰Ґ0 (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 deп¬Ѓnition of a graded ring, Ri Rj вЉ† 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 ) < в€ћ.

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 deп¬Ѓne the Hilbert-Samuel polynomial by

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

Now the sequence

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

is exact by the third isomorphism theorem. An induction argument using additivity of
length shows that sI (M, n) is п¬Ѓnite. Consequently

в€†sI (M, n) = sI (M, n + 1) в€’ sI (M, n) = h(grI (M ), n).

By (5.2.4), h(grI (M ), n) is polynomial-like of degree at most r в€’ 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 deп¬Ѓnition
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 вЉ† I вЉ† M. Then for every n в‰Ґ 1 we have Mtn вЉ† I n вЉ† Mn ,
so sM (M, tn) в‰Ґ sI (M, n) в‰Ґ sM (M, n). If the degrees of these polynomial are, from right
to left, d1 , d2 and d3 , we have O(dn ) в‰¤ O(dn ) в‰¤ O(dn ), with d3 = d1 . Therefore all three
1 2 3
degrees coincide.
The Hilbert-Samuel polynomial satisп¬Ѓes a property analogous to (4.1.7), the additivity
of length.

5.2.7 Theorem
Let I be an ideal of deп¬Ѓnition of the Noetherian local ring R, and suppose we have an
exact sequence 0 в†’ M в†’ M в†’ M в†’ 0 of п¬Ѓnitely 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 coeп¬ѓcient of r(n)
is nonnegative.
Proof. The following sequence is exact:

0 в†’ M /(M в€© I n M ) в†’ M/I n M в†’ M /I n M в†’ 0.

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

sI (M, n) в€’ sI (M , n) = lR (M /Mn )

hence lR (M /Mn ) is polynomial-like. By the Artin-Rees lemma (4.1.7), the п¬Ѓltration
{Mn } is I-stable, so IMn = Mn+1 for suп¬ѓciently large n, say, n в‰Ґ m. Thus for every
n в‰Ґ 0 we have Mn+m = M в€© I n+m M вЉ‡ I n+m M , and consequently

I n+m M вЉ† Mn+m = I n Mm вЉ† I n M ,

which implies that

lR (M /I n+m M ) в‰Ґ lR (M /Mn+m ) в‰Ґ 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 coeп¬ѓcient. Moreover, sI (M , n) в€’ lR (M /Mn ) = r(n) is polynomial-like of degree
less than deg lR (M /Mn ) в‰¤ deg sI (M, n), as well as nonnegative for n >> 0. The result
now follows upon adding the equations

sI (M, n) в€’ sI (M , n) = lR (M /Mn )

and

r(n) = sI (M , n) в€’ lR (M /Mn ). в™Ј

5.2.8 Corollary
Let M be a submodule of M , where M is a п¬Ѓnitely generated module over the Noetherian
local ring R. Then d(M ) в‰¤ d(M ).
Proof. Apply (5.2.7), noting that we can ignore r(n) because it is of lower degree than
sI (M, n). в™Ј

5.3 The Dimension Theorem
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 вЉ‚ P1 вЉ‚ В· В· В· вЉ‚ Pn of prime ideals of R. If there is no
upper bound on the length of such a chain, we take n = в€ћ. An example of an inп¬Ѓnite-
dimensional ring is the non-Noetherian ring k[X1 , X2 , . . . ], where k is a п¬Ѓeld. We have
5.3. THE DIMENSION THEOREM 7

the inп¬Ѓnite chain of prime ideals (X1 ) вЉ‚ (X1 , X2 ) вЉ‚ (X1 , X2 , X3 ) вЉ‚ В· В· В· . At the other
extreme, a п¬Ѓeld, 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 п¬Ѓeld has dimension 1.
Algebraic number theory provides many examples, because the ring of algebraic integers
of a number п¬Ѓeld is a Dedekind domain.
There are several other ideas that arise from the study of chains of prime ideals. We
deп¬Ѓne the height of a prime ideal P (notation ht P ) as the maximum length n of a chain
of prime ideals P0 вЉ‚ P1 вЉ‚ В· В· В· вЉ‚ 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 вЉ‚ P1 вЉ‚ В· В· В· вЉ‚ 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 вЉ† J, and S = (R/I)/(J/I),
then S в€ј R/J, so S is an integral domain iп¬Ђ R/J is an integral domain, and J/I is a
=
prime ideal of R/I iп¬Ђ J is a prime ideal of R.)
If I is an arbitrary ideal of R, we deп¬Ѓne the height of I as the inп¬Ѓmum of the heights
of prime ideals P вЉ‡ I, and the coheight of I as the supremum of the coheights of prime
ideals P вЉ‡ 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 deп¬Ѓne dim M =
dim(R/ ann M ) if M = 0, and we take the dimension of the zero module to be в€’1.
We now assume that M is nonzero and п¬Ѓnitely generated over the Noetherian ring R.
By (1.3.3), M has at least one associated prime. By (1.5.5), P вЉ‡ ann M iп¬Ђ P в€€ 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 в€€ Supp M } = sup{coht P : P в€€ 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 п¬Ѓnite.
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 п¬Ѓnite. Since M is п¬Ѓnitely generated, we can assert that there is a smallest positive
integer r, called the Chevalley dimension Оґ(M ), such that for some elements a1 , . . . , ar
belonging to M we have lR (M/(a1 , . . . , ar )M < в€ћ. If M = 0 we take Оґ(M ) = в€’1.
8 CHAPTER 5. DIMENSION THEORY

5.3.3 Dimension Theorem
Let M be a п¬Ѓnitely 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
deп¬Ѓnition 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 Оґ(M ).
Proof. We divide the proof into three parts.
1. dim M в‰¤ d(M ), hence dim M is п¬Ѓnite.
If d(M ) = в€’1, then sM (M, n) = lR (M/Mn M ) = 0 for n >> 0. By NAK, M = 0 so
dim M = в€’1. Thus assume d(M ) в‰Ґ 0. By (1.3.9) or (1.5.10), M has only п¬Ѓnitely 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 ) в‰¤ d(M ). If we can show that dim R/P в‰¤
d(R/P ), it will follow that dim M = dim R/P в‰¤ d(R/P ) в‰¤ d(M ).
It suп¬ѓces to show that for any chain of prime ideals P = P0 вЉ‚ В· В· В· вЉ‚ 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 ) = в€’1 and we are п¬Ѓnished. Thus assume t в‰Ґ 1, and assume that the
result holds up to t в€’ 1. Choose a в€€ P1 \ P , and consider prime ideals Q such that
Ra + P вЉ† Q вЉ† P1 . We can pick a Q belonging to AP(R/(Ra + P )). [Since Ra + P вЉ† 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) в‰¤ d(R/(Ra+P )). Since the chain
Q вЉ‚ P2 вЉ‚ В· В· В· вЉ‚ Qr is of length tв€’1, the induction hypothesis implies that tв€’1 в‰¤ d(R/Q),
hence t в€’ 1 в‰¤ d(R/(Ra + P )). Now the sequence

0 в†’ R/P в†’ R/P в†’ R/(Ra + P ) в†’ 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 в€’ 1 < d(R/P ). Therefore t в‰¤ d(R/P ), as desired.
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