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2. d(M ) ¤ δ(M ).
If δ(M ) = ’1, then M = 0 and d(M ) = ’1. Assume M = 0 and r = δ(M ) ≥ 0, and let
a1 , . . . , ar be elements of M such that M/(a1 , . . . , ar )M has ¬nite length. Let I be the
ideal (a1 , . . . , ar ) and let P be the annihilator of M ; set Q = I + P .
We claim that the support of R/Q is {M}. To prove this, ¬rst note that M/IM ∼ =
M —R R/I. (TBGY, subsection S7.1 of the supplement.) By Problem 9 of Chapter 1,
Supp M/IM = Supp M ©Supp R/I, which by (1.5.5) is V (P )©V (I) = V (Q) = Supp R/Q.
(Note that the annihilator of R/I is I and the annihilator of R/Q is Q.) But the support
5.4. CONSEQUENCES OF THE DIMENSION THEOREM 9

of M/IM is {M} by (1.6.9), proving the claim. (If M/IM = 0, then M = 0 by NAK,
contradicting our assumption.)
Again by (1.6.9), AP(R/Q) = {M}, so by (1.3.11), Q is M-primary. By (5.2.5) and
(3.3.10), Q is an ideal of de¬nition of R.
Let R = R/P , Q = Q/P , and consider M as an R-module. Then R is a Noetherian
local ring and Q is an ideal of de¬nition of R generated by a1 , . . . , ar , where ai = ai + P .
By (5.2.6), the degree of the Hilbert-Samuel polynomial sQ (M, n) is at most r. But by
n
the correspondence theorem, lR (M/Q M ) = lR (M/Qn M ), hence sQ (M, n) = sQ (M, n).
It follows that d(M ) ¤ r.
3. δ(M ) ¤ dim M .
If dim M = ’1 then M = 0 and δ(M ) = ’1, so assume M = 0. If dim M = 0, then by
(5.3.2), M has ¬nite length, so δ(M ) = 0.
Now assume dim M > 0. (We have already noted in part 1 that dim M is ¬nite.) Let
P1 , . . . , Pt be the associated primes of M whose coheight is as large as it can be, that is,
coht Pi = dim M for all i = 1, . . . , t. Since the dimension of M is greater than 0, Pi ‚ M
for every i, so by the prime avoidance lemma (0.1.1),
M ⊆ ∪1¤i¤t Pi .
Choose an element a in M such a belongs to none of the Pi , and let N = M/aM . Then
Supp N ⊆ Supp M \ {P1 , . . . , Pt }.
To see this, note that if NP = 0, then MP = 0; also, NPi = 0 for all i because a ∈ Pi ,
/
hence division by a is allowed. Thus dim N < dim M , because dim M = coht Pi and we
are removing all the Pi . Let r = δ(N ), and let a1 , . . . , ar be elements of M such that
N/(a1 , . . . , ar )N has ¬nite length. But
M/(a, a1 , . . . , ar )M ∼ N/(a1 , . . . , ar )N
=
(apply the ¬rst isomorphism theorem) , so M/(a, a1 , . . . , ar )M also has ¬nite length.
Thus δ(M ) ¤ r + 1. By the induction hypothesis, δ(N ) ¤ dim N . In summary,
δ(M ) ¤ r + 1 = δ(N ) + 1 ¤ dim N + 1 ¤ dim M. ™


5.4 Consequences of the Dimension Theorem
In this section we will see many applications of the dimension theorem (5.3.3).

5.4.1 Proposition
Let R be a Noetherian local ring with maximal ideal M. If M is a ¬nitely generated
R-module, then dim M < ∞; in particular, dim R < ∞. Moreover, the dimension of R is
the minimum, over all ideals I of de¬nition of R, of the number of generators of I.
Proof. Finiteness of dimension follows from (5.3.3). The last assertion follows from the
de¬nition of Chevalley dimension in (5.3.2). In more detail, R/I has ¬nite length i¬ (by
the Noetherian hypothesis) R/I is Artinian i¬ [by (5.2.5)] I is an ideal of de¬nition. ™
10 CHAPTER 5. DIMENSION THEORY

5.4.2 Proposition
Let R be a Noetherian local ring with maximal ideal M and residue ¬eld k = R/M.
Then dim R ¤ dimk (M/M2 ).
Proof. Let a1 , . . . , ar be elements of M such that a1 , . . . , ar form a k-basis of M/M2 ,
where ai = ai + M. Then by (0.3.4), a1 , . . . , ar generate M. Since M itself is an ideal
of de¬nition (see (5.2.5)), we have dim R ¤ r by (5.4.1). (Alternatively, by (5.4.5), the
height of M is at most r. Since ht M = dim R, the result follows.) ™


5.4.3 Proposition
ˆ
Let R be Noetherian local ring with maximal ideal M, and R its M-adic completion.
ˆ
Then dim R = dim R.
Proof. By (4.2.9), R/Mn ∼ R/Mn . By (5.2.6), sM (R, n) = sM (R, n). In particular, the
=ˆ ˆ ˆ
ˆ
ˆ
two Hilbert-Samuel polynomials must have the same degree. Therefore d(R) = d(R), and
the result follows from (5.3.3). ™


5.4.4 Theorem
If R is a Noetherian ring, then the prime ideals of R satisfy the descending chain condition.
Proof. We may assume without loss of generality that R is a local ring. Explicitly, if P0 is
a prime ideal of R, let A be the localized ring RP0 . Then the chain P0 ⊃ P1 ⊃ P2 ⊃ . . . of
prime ideals of R will stabilize if and only if the chain AP0 ⊃ AP1 ⊃ AP2 ⊃ · · · of prime
ideals of A stabilizes. But if R is local as well as Noetherian, the result is immediate
because dim R < ∞. ™


5.4.5 Generalization of Krull™s Principal Ideal Theorem
Let P be a prime ideal of the Noetherian ring R. The following conditions are equivalent:
(a) ht P ¤ n;
(b) There is an ideal I of R that is generated by n elements, such that P is a minimal
prime ideal over I. (In other words, P is minimal subject to P ⊇ I.)
Proof. If (b) holds, then IRP is an ideal of de¬nition of RP√that is generated by n
elements. (See (3.3.10), and note that if P is minimal over I i¬ I = P .) By (5.3.1) and
(5.4.1), ht P = dim RP ¤ n. Conversely, if (a) holds then dim RP ¤ n, so by (5.4.1) there
is an ideal of de¬nition J of RP generated by n elements a1 /s, . . . , an /s with s ∈ R \ P .
The elements ai must belong to P , else the ai /s will generate RP , which is a contradiction
because J must be proper; see (5.2.5). Take I to be the ideal of R generated by a1 , . . . , an .
Invoking (3.3.10) as in the ¬rst part of the proof, we conclude that I satis¬es (b). ™


5.4.6 Krull™s Principal Ideal Theorem
Let a be a nonzero element of the Noetherian ring R. If a is neither a unit nor a zero-
divisor, then every minimal prime ideal P over (a) has height 1.
5.4. CONSEQUENCES OF THE DIMENSION THEOREM 11

Proof. It follows from (5.4.5) that ht P ¤ 1. Thus assume ht P = dim RP = 0. We claim
that RP = 0, hence P ∈ Supp R. For if a/1 = 0, then for some s ∈ R \ P we have sa = 0,
which contradicts the hypothesis that a is not a zero-divisor. We may assume that P is
minimal in the support of R, because otherwise P has height 1 and we are ¬nished. By
(1.5.9), P is an associated prime of R, so by (1.3.6), P consists entirely of zero-divisors,
a contradiction. ™
The hypothesis that a is not a unit is never used, but nothing is gained by dropping
it. If a is a unit, then a cannot belong to the prime ideal P and the theorem is vacuously
true.


5.4.7 Theorem
Let R be a Noetherian local ring with maximal ideal M, and let a ∈ M be a non zero-
divisor. Then dim R/(a) = dim R ’ 1.
Proof. We have dim R > 0, for if dim R = 0, then M is the only prime ideal, and as
in the proof of (5.4.6), M consists entirely of zero-divisors, a contradiction. In the proof
of part 3 of the dimension theorem (5.3.3), take M = R and N = R/(a) to conclude
that dim R/(a) < dim R, hence dim R/(a) ¤ dim R ’ 1. To prove equality, we appeal
to part 2 of the proof of (5.3.3). This allows us to ¬nd elements a1 , . . . , as ∈ M, with
s = dim R/(a), such that the images ai in R/(a) generate an M/(a)-primary ideal of
R/(a). Then a, a1 , . . . , as generate an M-primary ideal of R, so by (5.4.1) and (3.3.10),
dim R ¤ 1 + s = 1 + dim R/(a). ™


5.4.8 Corollary
Let a be a non zero-divisor belonging to the prime ideal P of the Noetherian ring R. Then
ht P/(a) = ht P ’ 1.
Proof. In (5.4.7), replace R by RP and R/(a) by (RP )Q , where Q is a minimal prime
ideal over (a). ™


5.4.9 Theorem
Let R = k[[X1 , . . . , Xn ]] be a formal power series ring in n variables over the ¬eld k.
Then dim R = n.
Proof. The unique maximal ideal is (X1 , . . . , Xn ), so the dimension of R is at most n.
On the other hand, the dimension is at least n because of the chain

(0) ‚ (X1 ) ‚ (X1 , X2 ) ‚ · · · ‚ (X1 , . . . , Xn )

of prime ideals. ™
12 CHAPTER 5. DIMENSION THEORY

5.5 Strengthening of Noether™s Normalization Lemma
5.5.1 De¬nition
An a¬ne k-algebra is an integral domain that is also a ¬nite-dimensional algebra over a
¬eld k.
A¬ne algebras are of great interest in algebraic geometry because they are the coor-
dinate rings of a¬ne algebraic varieties. To study them we will need a stronger version
of Noether™s normalization lemma. In this section we will give the statement and proof,
following Serre™s Local Algebra, page 42.

5.5.2 Theorem
Let A be a ¬nitely generated k-algebra, and I1 ‚ · · · ‚ Ir a chain of nonzero proper ideals
of A. There exists a nonnegative integer n and elements x1 , . . . , xn ∈ A algebraically
independent over k such that the following conditions are satis¬ed.
1. A is integral over B = k[x1 , . . . , xn ]. (This is the standard normalization lemma.)
2. For each i = 1, . . . , r, there is a positive integer h(i) such that Ii © B is generated (as
an ideal of B) by x1 , . . . , xh(i) .
Proof. It su¬ces to let A be a polynomial ring k[Y1 , . . . , Ym ]. For we may write A = A /I0
where A = k[Y1 , . . . , Ym ]. If Ii is the preimage of Ii under the canonical map A ’ A /I0 ,
and we ¬nd elements x1 , . . . , xn ∈ A , relative to the ideals I0 ‚ I1 ‚ · · · ‚ Ir , then the
images of xi’h(0) in A, i > h(0), satisfy the required conditions. The proof is by induction
on r.
Assume r = 1. We ¬rst consider the case in which I1 is a principal ideal (x1 ) = x1 A
with x1 ∈ k. By our assumption that A is a polynomial ring, we have x1 = g(Y1 , . . . , Ym )
/
for some nonconstant polynomial g with coe¬cients in k. We claim that there are positive
integers ri (i = 2, . . . , m) such that A is integral over B = k[x1 , . . . , xm ], where
xi = Yi ’ Y1ri , i = 2, . . . , m.
If we can show that Y1 is integral over B, then (since the xi belong to B, hence are integral
over B) all Yi are integral over B, and therefore A is integral over B. Now Y1 satis¬es
the equation x1 = g(Y1 , . . . , Ym ), so
g(Y1 , x2 + Y1r2 , . . . , xm + Y1rm ) ’ x1 = 0.
c± Y ± , ± = (a1 , . . . , am ), c± = 0,
If we write the polynomial g as a sum of monomials
the above equation becomes

c± Y1a1 (x2 + Y1r2 )a2 · · · (xm + Y1rm )am ’ x1 = 0.

To produce the desired ri , let f (±) = a1 + a2 r2 + · · · + am rm , and pick the ri so that all
the f (±) are distinct. For example, take ri = si , where s is greater than the maximum of
the aj . Then there will be a unique ± that maximizes f , say ± = β, and we have
f (β)
pj (x1 , . . . , xm )Y1j = 0
cβ Y1 +
j<f (β)
5.6. PROPERTIES OF AFFINE K-ALGEBRAS 13

so Y1 is integral over B, and as we noted above, A = k[Y1 , . . . , Ym ] is integral over
B = k[x1 , . . . , xm ]. Since A has transcendence degree m over k and an integral extension
must be algebraic, it follows that x1 , . . . , xm are algebraically independent over k. Thus
the ¬rst assertion of the theorem holds (in this ¬rst case, where I1 is principal). If we
can show that I1 © B = (x1 ) = x1 B, the second assertion will also hold. The right-to-left
inclusion follows from our assumptions about x1 , so let t belong to I1 © B. Then t = x1 u
with u ∈ A, hence, dividing by x1 , u ∈ A © k(x1 , . . . , xm ). Since B is isomorphic to a
polynomial ring, it is a unique factorization domain and therefore integrally closed. Since
A is integral over B, we have u ∈ B. Thus x1 A © B = x1 B, and the proof of the ¬rst case
is complete. Note that we have also shown that A © k(x1 , . . . , xm ) = B = k[x1 , . . . , xm ].
Still assuming r = 1, we now consider the general case by induction on m. If m =
0 there is nothing to prove, and we have already taken care of m = 1 (because A is
then a PID). Let x1 be a nonzero element of I1 , and note that x1 ∈ k because I1 is/
proper. By what we have just proved, there are elements t2 , . . . , tm ∈ A such that
x1 , t2 , . . . , tm are algebraically independent over k, A is integral over the polynomial ring
C = k[x1 , t2 , . . . , tm ], and x1 A © C = x1 C. By the induction hypothesis, there are
elements x2 , . . . , xm satisfying the conditions of the theorem for k[t2 , . . . , tm ] and the
ideal I1 © k[t2 , . . . , tm ]. Then x1 , . . . , xm satisfy the desired conditions.
Finally, we take the inductive step from r ’ 1 to r. let t1 , . . . , tm satisfy the conditions
of the theorem for the chain of ideals I1 ‚ · · · ‚ Ir’1 , and let s = h(r ’ 1). By the
argument of the previous paragraph, there are elements xs+1 , . . . , xm ∈ k[ts+1 , . . . , tm ]
satisfying the conditions for the ideal Ir © k[ts+1 , . . . , tm ]. Take xi = ti for 1 ¤ i ¤ s. ™


5.6 Properties of A¬ne k-algebras
We will look at height, coheight and dimension of a¬ne algebras.

5.6.1 Proposition
Let S = R[X] where R is an arbitrary ring. If Q ‚ Q , where Q and Q are prime ideals
of S both lying above the same prime ideal P of R, then Q = P S.
Proof. Since R/P can be regarded as a subring of S/Q, we may assume without loss of
generality that P = 0. By localizing with respect to the multiplicative set R \ {0}, we
may assume that R is a ¬eld. But then every nonzero prime ideal of S is maximal, hence
Q = 0. Since P S is also 0, the result follows. ™

5.6.2 Corollary
Let I be an ideal of the Noetherian ring R, and let P be a prime ideal of R with I ⊆ P .
Let S be the polynomial ring R[X], and take J = IS and Q = P S. If P is a minimal
prime ideal over I, then Q is a minimal prime ideal over J.
Proof. To verify that Q is prime, note that R[X]/P R[X] ∼ R[X]/P [X] ∼ (R/P )[X], an
= =
integral domain. By modding out I, we may assume that I = 0. Suppose that the prime
ideal Q1 of S is properly contained in Q. Then Q1 © R ⊆ Q © R = P S © R = P . (A
14 CHAPTER 5. DIMENSION THEORY

polynomial belonging to R coincides with its constant term.) By minimality, Q1 © R = P .
By (5.6.1), Q1 = P S = Q, a contradiction. ™

5.6.3 Proposition
As above, let S = R[X], R Noetherian, P a prime ideal of R, Q = P S. Then ht P = ht Q.
Proof. Let n be the height of P . By the generalized Krull principal ideal theorem
(5.4.5), there is an ideal I of R generated by n elements such that P is a minimal prime
ideal over I. By (5.6.2), Q = P S is a minimal prime ideal over J = IS. But J is
also generated over S by n elements, so again by (5.4.5), ht Q ¤ ht P . Conversely, if
P0 ‚ P1 ‚ · · · ‚ Pn = P ‚ R and Qi = Pi [X], then Q0 ‚ Q1 ‚ · · · ‚ Qn = Q, so
ht Q ≥ ht P . ™
We may now prove a major result on the dimension of a polynomial ring.

5.6.4 Theorem
Let S = R[X], where R is a Noetherian ring. Then dim S = 1 + dim R.
Proof. Let P0 ‚ P1 ‚ · · · ‚ Pn be a chain of prime ideals of R. If Qn = Pn S, then by
(5.6.3), ht Qn = ht Pn . But the Q sequence can be extended via Qn ‚ Qn+1 = Qn + (X).
(Note that X cannot belong to Qn because 1 ∈ Pn .) It follows that dim S ≥ 1 + dim R.
/
Now consider a chain Q0 ‚ Q1 ‚ · · · ‚ Qn of prime ideals of S, and let Pi = Qi © R
for every i = 0, 1, . . . , n. We may assume that the Pi are not all distinct (otherwise
dim R ≥ dim S ≥ dim S ’ 1). Let j be the largest index i such that Pi = Pi+1 . By (5.6.1),
Qj = Pj S, and by (5.6.3), ht Pj = ht Qj ≥ j. But by choice of j,
Pj = Pj+1 ‚ Pj+2 ‚ · · · ‚ Pn
so ht Pj + n ’ j ’ 1 ¤ dim R. Since the height of Pj is at least j, we have n ’ 1 ¤ dim R,
hence dim S ¤ 1 + dim R. ™

5.6.5 Corollary
If R is a Noetherian ring, then dim R[X1 , . . . , Xn ] = n + dim R. In particular, if K is a
¬eld then dim K[X1 , . . . , Xn ] = n.
Proof. This follows from (5.6.4) by induction. ™

5.6.6 Corollary
Let R = K[X1 , . . . , Xn ], where K is a ¬eld. Then ht(X1 , . . . , Xi ) = i, 1 ¤ i ¤ n.
Proof. First consider i = n. The height of (X1 , . . . , Xn ) is at most n, the dimension of R,
and in fact the height is n, in view of the chain (X1 ) ‚ (X1 , X2 ) ‚ · · · ‚ (X1 , . . . , Xn ).
The general result now follows by induction, using (5.4.8). ™
If X is an a¬ne algebraic variety over the ¬eld k, its (geometric) dimension is the
transcendence degree over k of the function ¬eld (the fraction ¬eld of the coordinate
ring). We now show that the geometric dimension coincides with the algebraic (Krull)
dimension. We abbreviate transcendence degree by tr deg.
5.6. PROPERTIES OF AFFINE K-ALGEBRAS 15

5.6.7 Theorem
If R is an a¬ne k-algebra, then dim R = tr degk Frac R.
Proof. By Noether™s normalization lemma, there are elements x1 , . . . , xn ∈ R, alge-
braically independent over k, such that R is integral over k[x1 , . . . , xn ]. Since an integral
extension cannot increase dimension (see Problem 4), dim R = dim k[x1 , . . . , xn ] = n by
(5.6.5). Let F = Frac R and L = Frac k[x1 , . . . , xn ]. Then F is an algebraic extension
of L, and since an algebraic extension cannot increase transcendence degree, we therefore
have tr degk F = tr degk L = n = dim R. ™
It follows from the de¬nitions that if P is a prime ideal of R, then ht P + coht P ¤
dim R. If R is an a¬ne k-algebra, there is equality.

5.6.8 Theorem
If P is a prime ideal of the a¬ne k-algebra R, then ht P + coht P = dim R.
Proof. By Noether™s normalization lemma, R is integral over a polynomial algebra. We
can assume that R = k[X1 , . . . , Xn ] with ht P = h. (See Problems 4,5 and 6. An integral
extension preserves dimension and coheight, and does not increase height. So if height
plus coheight equals dimension in the larger ring, the same must be true in the smaller
ring.) By the strong form (5.5.2) of Noether™s normalization lemma, along with (5.6.6),
there are elements y1 , . . . , yn algebraically independent over k such that R is integral
over k[y1 , . . . , yn ] and Q = P © k[y1 , . . . , yn ] = (y1 , . . . , yh ) Since k[y1 , . . . , yn ]/Q ∼
=
k[yh+1 , . . . , yn ], it follows from (5.3.1) and (5.6.5) that coht Q = dim k[yh+1 , . . . , yn ] =
n ’ h. But coht Q = coht P (Problem 5), so ht P + coht P = h + (n ’ h) = n = dim R. ™
Chapter 6

Depth

6.1 Systems of Parameters
We prepare for the study of regular local rings, which play an important role in algebraic
geometry.


6.1.1 De¬nition
Let R be a Noetherian local ring with maximal ideal M, and let M be a ¬nitely generated
R-module of dimension n. A system of parameters for M is a set {a1 , . . . , an } of elements
of M such that M/(a1 , . . . , an )M has ¬nite length. The ¬niteness of the Chevalley
dimension (see (5.3.2) and (5.3.3) guarantees the existence of such a system.


6.1.2 Example
Let R be a Noetherian local ring of dimension d. Then any set {a1 , . . . , ad } that generates
an ideal of de¬nition is a system of parameters for R, by (5.4.1). In particular, if R =
k[[X1 , . . . , Xn ]] is a formal power series ring over a ¬eld, then X1 , . . . , Xn form a system
of parameters, since they generate the maximal ideal.


6.1.3 Proposition
Let M be ¬nitely generated and of dimension n over the Noetherian local ring R, and let
a1 , . . . , at be arbitrary elements of the maximal ideal M. Then dim(M/(a1 , . . . , at )M ) ≥
n ’ t, with equality if and only if the ai can be extended to a system of parameters for
M.
Proof. Let a be any element of M, and let N = M/aM . Choose b1 , . . . , br ∈ M such that
N/(b1 , . . . , br )N has ¬nite length, with r as small as possible. Then M/(a, b1 , . . . , br )M
also has ¬nite length, because

(M/aM )/(b1 , . . . , br )(M/aM ) ∼ M/(a, b1 , . . . , br )M.
=

1
2 CHAPTER 6. DEPTH

It follows that the Chevalley dimension of M is at most r + 1, in other words,

δ(M/aM ) ≥ δ(M ) ’ 1.

The proof will be by induction on t, and we have just taken care of t = 1 as well as the
key step in the induction, namely

dim(M/(a1 , . . . , at )M ) = dim(N/a1 N )

where N = M/(a2 , . . . , at )M . By the t = 1 case and the induction hypothesis,

dim(N/a1 N ) ≥ dim N ’ 1 ≥ dim M ’ (t ’ 1) ’ 1 = dim M ’ t

as asserted. If dim(M/(a1 , . . . , at )M ) = n ’ t with n = dim M , then we can choose a
system of parameters at+1 , . . . , an for N = M/(a1 , . . . , at )M . Then

N/(at+1 , . . . , an )N ∼ M/(a1 , . . . , at , at+1 , . . . , an )M
=

has ¬nite length. Thus a1 , . . . , an form a system of parameters for M . Conversely, if
a1 , . . . , at can be extended to a system of parameters a1 , . . . , an for M , de¬ne N =
M/(a1 , . . . , at )M . Then N/(at+1 , . . . , an )N ∼ M/(a1 , . . . , an )M has ¬nite length, hence
=
dim N ¤ n ’ t. But dim N ≥ n ’ t by the main assertion, and the proof is complete. ™


6.2 Regular Sequences
We introduce sequences that are guaranteed to be extendable to a system of parameters.


6.2.1 De¬nition
Let M be an R-module. The sequence a1 , . . . , at of nonzero elements of R is an M -
sequence, also called a regular sequence for M or an M -regular sequence, if (a1 , . . . , at )M =
M and for each i = 1, . . . , t, ai is not a zero-divisor of M/(a1 , . . . , ai’1 )M .


6.2.2 Comments and Examples
We interpret the case i = 1 as saying that a1 is not a zero-divisor of M , that is, if x ∈ M
and a1 x = 0, then x = 0. Since (a1 , . . . , at )M = M , M = 0 and the ai are nonunits.
It follows from the de¬nition that the elements a1 , . . . , at form an M -sequence if
and only if for all i = 1, . . . , t, a1 , . . . , ai is an M -sequence and ai+1 , . . . , at is an
M/(a1 , . . . , ai )M -sequence.
1. If R = k[X1 , . . . , Xn ] with k a ¬eld, then X1 , . . . , Xn is an R-sequence.
2. (A tricky point) A permutation of a regular sequence need not be regular. For example,
let R = k[X, Y, Z], where k is a ¬eld. Then X, Y (1 ’ X), Z(1 ’ X) is an R-sequence, but
Y (1 ’ X), Z(1 ’ X), X is not, because the image of Z(1 ’ X)Y is zero in R/(Y (1 ’ X)).
6.2. REGULAR SEQUENCES 3

6.2.3 Theorem
Let M be a ¬nitely generated module over the Noetherian local ring R. If a1 , . . . , at is
an M -sequence, then {a1 , . . . , at } can be extended to a system of parameters for M .
Proof. We argue by induction on t. Since a1 is not a zero-divisor of M , we have
dim M/a1 M = dim M ’ 1 by (5.4.7). (Remember that the ai are nonunits (see (6.2.2))
and therefore belong to the maximal ideal of R.) By (6.1.3), a1 is part of a system of
parameters for M . If t > 1, the induction hypothesis says that a1 , . . . , at’1 is part of a
system of parameters for M . By (6.1.3), dim M/(a1 , . . . , at’1 )M = n ’ (t ’ 1), where
n = dim M . Since at is not a zero-divisor of N = M/(a1 , . . . , at’1 )M , we have, as in the
t = 1 case, dim N/at N = dim N ’ 1. But, as in the proof of (6.1.3),

N/at N ∼ M/(a1 , . . . , at )M,
=

hence

dim M/(a1 , . . . , at )M = dim N/at N = dim N ’ 1 = n ’ (t ’ 1) ’ 1 = n ’ t.

By (6.1.3), a1 , . . . , at extend to a system of parameters for M . ™

6.2.4 Corollary
If R is a Noetherian local ring, then every R-sequence can be extended to a system of
parameters for R.
Proof. Take M = R in (6.2.3). ™

6.2.5 De¬nition
Let M be a nonzero ¬nitely generated module over the Noetherian local ring R. The
depth of M over R, written depthR M or simply depth M , is the maximum length of an
M -sequence. We will see in the next chapter that any two maximal M -sequences have
the same length.

6.2.6 Theorem
Let M be a nonzero ¬nitely generated module over the Noetherian local ring R. Then
depth M ¤ dim M .
Proof. Since dim M is the number of elements in a system of parameters, the result follows
from (6.2.3). ™

6.2.7 Proposition
Let M be a ¬nitely generated module over the Noetherian ring R, and let a1 , . . . , an be an
M -sequence with all ai belonging to the Jacobson radical J(R). Then any permutation
of the ai is also an M -sequence.
Proof. It su¬ces to consider the transposition that interchanges a1 and a2 . First let
us show that a1 is not a zero-divisor of M/a2 M . Suppose a1 x = 0, where x belongs
4 CHAPTER 6. DEPTH

to M/a2 M . Then a1 x belongs to a2 M , so we may write a1 x = a2 y with y ∈ M . By
hypothesis, a2 is not a zero-divisor of M/a1 M , so y belongs to a1 M . Therefore y = a1 z
for some z ∈ M . Then a1 x = a2 y = a2 a1 z. By hypothesis, a1 is not a zero-divisor of M ,
so x = a2 z, and consequently x = 0.
To complete the proof, we must show that a2 is not a zero-divisor of M . If N is the
submodule of M annihilated by a2 , we will show that N = a1 N . Since a1 ∈ J(R), we can
invoke NAK (0.3.3) to conclude that N = 0, as desired. It su¬ces to show that N ⊆ a1 N ,
so let x ∈ N . By de¬nition of N we have a2 x = 0. Since a2 is not a zero-divisor of
M/a1 M , x must belong to a1 M , say x = a1 y with y ∈ M . Thus a2 x = a2 a1 y = 0. But a1
is not a zero-divisor of M , hence a2 y = 0 and therefore y ∈ N . But x = a1 y, so x ∈ a1 N ,
and we are ¬nished. ™

6.2.8 Corollary
Let M be a ¬nitely generated module over the Noetherian local ring R. Then any per-
mutation of an M -sequence is also an M -sequence.
Proof. By (6.2.2), the members of the sequence are nonunits, hence they belong to the
maximal ideal, which coincides with the Jacobson radical. ™

6.2.9 De¬nitions and Comments
Let M be a nonzero ¬nitely generated module over a Noetherian local ring R. If the
depth of M coincides with its dimension, we call M a Cohen-Macaulay module. We say
that R is a Cohen-Macaulay ring if it is a Cohen-Macaulay module over itself. To study
these rings and modules, we need some results from homological algebra. The required
tools will be developed in Chapter 7.
Chapter 7

Homological Methods

We now begin to apply homological algebra to commutative ring theory. We assume as
background some exposure to derived functors and basic properties of Ext and Tor. In
addition, we will use standard properties of projective and injective modules. Everything
we need is covered in TBGY, Chapter 10 and the supplement.


7.1 Homological Dimension: Projective and Global
Our goal is to construct a theory of dimension of a module M based on possible lengths
of projective and injective resolutions of M .

7.1.1 De¬nitions and Comments
A projective resolution 0 ’ Xn ’ · · · ’ X0 ’ M ’ 0 of the R-module M is said to
be of length n. The largest such n is called the projective dimension of M , denoted by
pdR M . (If M has no ¬nite projective resolution, we set pdR M = ∞.)

7.1.2 Lemma
The projective dimension of M is 0 if and only if M is projective.
Proof. If M is projective, then 0 ’ X0 = M ’ M ’ 0 is a projective resolution, where
the map from M to M is the identity. Conversely, if 0 ’ X0 ’ M ’ 0 is a projective
resolution, then M ∼ X0 , hence M is projective. ™
=

7.1.3 Lemma
If R is a PID, then for every R-module M , pdR M ¤ 1. If M is an abelian group whose
torsion subgroup is nontrivial, then pdR M = 1.
Proof. There is an exact sequence 0 ’ X1 ’ X0 ’ M ’ 0 with X0 free and X1 , a
submodule of a free module over a PID, also free. Thus pdR M ¤ 1. If pdR M = 0, then
by (7.1.2), M is projective, hence free because R is a PID. Since a free module has zero
torsion, the second assertion follows. ™

1
2 CHAPTER 7. HOMOLOGICAL METHODS

7.1.4 De¬nition
The global dimension of a ring R, denoted by gldim R, is the least upper bound of pdR M
as M ranges over all R-modules.

7.1.5 Remarks
If R is a ¬eld, then every R-module is free, so gldim R = 0. By (7.1.3), a PID has
global dimension at most 1. Since an abelian group with nonzero torsion has projective
dimension 1, gldim Z = 1.
We will need the following result from homological algebra; for a proof, see TBGY,
subsection S5.7.

7.1.6 Proposition
If M is an R-module, the following conditions are equivalent.
(i) M is projective;
(ii) Extn (M, N ) = 0 for all n ≥ 1 and all R-modules N ;
R
(iii) Ext1 (M, N ) = 0 for all R-modules N .
R

We can now characterize projective dimension in terms of the Ext functor.

7.1.7 Theorem
If M is an R-module and n is a positive integer, the following conditions are equivalent.
1. pdR M ¤ n.
2. Exti (M, N ) = 0 for all i > n and every R-module N .
R
3. Extn+1 (M, N ) = 0 for every R-module N .
R
4. If 0 ’ Kn’1 ’ Xn’1 ’ · · · ’ X0 ’ M ’ 0 is an exact sequence with all Xi
projective, then Kn’1 is projective.

Proof. To show that (1) implies (2), observe that by hypothesis, there is a projective
resolution 0 ’ Xn ’ · · · ’ X0 ’ M ’ 0. Use this resolution to compute Ext, and
conclude that (2) holds. Since (3) is a special case of (2), we have (2) implies (3). If (4)
holds, construct a projective resolution of M in the usual way, but pause at Xn’1 and
terminate the sequence with 0 ’ Kn’1 ’ Xn’1 . By hypothesis, Kn’1 is projective, and
this gives (4) implies (1). The main e¬ort goes into proving that (3) implies (4). We
break the exact sequence given in (4) into short exact sequences. The procedure is a bit
di¬erent from the decomposition of (5.2.3). Here we are proceeding from right to left,
and our ¬rst short exact sequence is
G K0 G X0
GM G0
i0
0
where K0 is the kernel of . The induced long exact sequence is

· · · ’ Extn (X0 , N ) ’ Extn (K0 , N ) ’ Extn+1 (M, N ) ’ Extn+1 (X0 , N ) ’ · · ·
R R R R
7.2. INJECTIVE DIMENSION 3

Now if every third term in an exact sequence is 0, then the maps in the middle are
both injective and surjective, hence isomorphisms. This is precisely what we have here,
because X0 is projective and (7.1.6) applies. Thus Extn+1 (M, N ) ∼ Extn (K0 , N ), so as
= R
R
we slide from right to left through the exact sequence, the upper index decreases by 1.
This technique is referred to as dimension shifting.
Now the second short exact sequence is
G K1 i1 G X1 d1 G K0 G 0.
0
We can replace X0 by K0 because im d1 = ker = K0 . The associated long exact sequence
is
· · · ’ Extn (X1 , N ) ’ Extn (K1 , N ) ’ Extn+1 (K0 , N ) ’ Extn+1 (X1 , N ) ’ · · ·
R R R R

and dimension shifting gives Extn (K0 , N ) ∼ Extn’1 (K1 , N ). Iterating this procedure, we
=
R R
∼ Ext1 (Kn’1 , N ), hence by the hypothesis of (3), Ext1 (Kn’1 , N ) = 0.
n+1
get ExtR (M, N ) = R R
By (7.1.6), Kn’1 is projective. ™

7.1.8 Corollary
gldim R ¤ n if and only if Extn+1 (M, N ) = 0 for all R-modules M and N .
R
Proof. By the de¬nition (7.1.4) of global dimension, gldim R ¤ n i¬ pdR M for all M i¬
(by (1) implies (3) of (7.1.7)) Extn+1 (M, N ) = 0 for all M and N . ™
R



7.2 Injective Dimension
As you might expect, projective dimension has a dual notion. To develop it, we will need
the analog of (7.1.6) for injective modules. A proof is given in TBGY, subsection S5.8.

7.2.1 Proposition
If N is an R-module, the following conditions are equivalent.
(i) N is injective;
(ii) Extn (M, N ) = 0 for all n ≥ 1 and all R-modules M ;
R
(iii) Ext1 (M, N ) = 0 for all R-modules M .
R
We are going to dualize (7.1.7), and the technique of dimension shifting is again useful.

7.2.2 Proposition
Let 0 ’ M ’ E ’ M ’ 0 be an exact sequence, with E injective. Then for all
n ≥ 1 and all R-modules M , we have Extn+1 (M, M ) ∼ Extn (M, M ). Thus as we slide
= R
R
through the exact sequence from left to right, the index of Ext drops by 1.
Proof. The given short exact sequence induces the following long exact sequence:
· · · ’ Extn (M, E) ’ Extn (M, M ) ’ Extn+1 (M, M ) ’ Extn+1 (M, E) ’ · · ·
R R R R

By (7.2.1), the outer terms are 0 for n ≥ 1, hence as in the proof of (7.1.7), the map in
the middle is an isomorphism. ™
4 CHAPTER 7. HOMOLOGICAL METHODS

7.2.3 De¬nitions and Comments
An injective resolution 0 ’ N ’ X0 ’ · · · ’ · · · Xn ’ 0 of the R-module N is said
to be of length n. The largest such n is called the injective dimension of M , denoted by
idR M . (If N has no ¬nite injective resolution, we set idR M = ∞.) Just as in (7.1.2),
idR N = 0 if and only if N is injective.


7.2.4 Proposition
If N is an R-module and n is a positive integer, the following conditions are equivalent.

1. idR N ¤ n.
2. Exti (M, N ) = 0 for all i > n and every R-module M .
R
3. Extn+1 (M, N ) = 0 for every R-module M .
R
4. If 0 ’ N ’ X0 ’ · · · ’ Xn’1 ’ Cn’1 ’ 0 is an exact sequence with all Xi injective,
then Cn’1 is injective.

Proof. If (1) is satis¬ed, we have an exact sequence 0 ’ N ’ X0 ’ · · · ’ Xn ’ 0, with
the Xi injective. Use this sequence to compute Ext, and conclude that (2) holds. If we
have (2), then we have the special case (3). If (4) holds, construct an injective resolution
of N , but pause at step n ’ 1 and terminate the sequence by Xn’1 ’ Cn’1 ’ 0. By
hypothesis, Cn’1 is injective, proving that (4) implies (1). To prove that (3) implies (4),
we decompose the exact sequence of (4) into short exact sequences. The process is similar
to that of (5.2.3), but with emphasis on kernels rather than cokernels. The decomposition
is given below.

0 ’ N ’ X0 ’ K0 ’ 0, 0 ’ K0 ’ X1 ’ K1 ’ 0, . . . ,


0 ’ Kn’2 ’ Xn’1 ’ Cn’1 ’ 0

We now apply the dimension shifting result (7.2.2) to each short exact sequence. If the
index of Ext starts at n + 1, it drops by 1 as we go through each of the n sequences, and
it ends at 1. More precisely,

Extn+1 (M, N ) ∼ Ext1 (M, Cn’1 )
= R
R

for any M . The left side is 0 by hypothesis, so the right side is also 0. By (7.2.1), Cn’1
is injective. ™


7.2.5 Corollary
The global dimension of R is the least upper bound of idR N over all R-modules N .
Proof. By the de¬nition (7.1.4) of global dimension, gldim R ¤ n i¬ pdR M ¤ n for all
M . Equivalently, by (7.1.7), Extn+1 (M, N ) = 0 for all M and N . By (7.2.4), this happens
R
i¬ idR N ¤ n for all N . ™
7.3. TOR AND DIMENSION 5

7.3 Tor and Dimension
We have observed the interaction between homological dimension and the Ext functor,
and this suggests that it would be pro¬table to bring in the Tor functor as well. We will
need the following result, which is proved in TBGY, subsection S5.6.

7.3.1 Proposition
If M is an R-module, the following conditions are equivalent.
(i) M is ¬‚at.
(ii) TorR (M, N ) = 0 for all n ≥ 1 and all R-modules N .
n
(iii) TorR (M, N ) = 0 for all R-modules N .
1

In addition, if R is a Noetherian local ring and M is ¬nitely generated over R, then M is
free if and only if M is projective, if and only if M is ¬‚at. See Problems 3-6 for all the
details.

7.3.2 Proposition
Let R be Noetherian local ring with maximal ideal M and residue ¬eld k. Let M be
a ¬nitely generated R-module. Then M is free (⇐’ projective ⇐’ ¬‚at) if and only if
TorR (M, k) = 0.
1
Proof. The “only if” part follows from (7.3.1). To prove the “if” part, let {x1 , . . . , xn }
be a minimal set of generators for M . Take a free module F with basis {e1 , . . . , en }
and de¬ne an R-module homomorphism f : F ’ M via f (ei ) = xi , i = 1, . . . , n. If K
is the kernel of f , we have the short exact sequence 0 ’ K ’ F ’ M ’ 0. Since
TorR (M, k) = 0, we can truncate the associated long exact sequence:
1

0 = TorR (M, k) ’ K —R k ’ F —R k ’ M —R k ’ 0
1

where the map f : F —R k ’ M —R k is induced by f . Now f is surjective by construction,
and is injective by minimality of the generating set [see (0.3.4) and the base change device
below]. Thus K —R k = ker f = 0. But (TBGY, subsection S7.1)

K —R k = K —R (R/M) ∼ K/MK
=

so K = MK. By NAK, K = 0. Therefore f is an isomorphism of F and M , hence M is
free. ™

7.3.3 Theorem
Let R be a Noetherian local ring with maximal ideal M and residue ¬eld k. If M is a
¬nitely generated R-module, the following conditions are equivalent.
(i) pdR M ¤ n.
(ii) TorR (M, N ) = 0 for all i > n and every R-module N .
i
6 CHAPTER 7. HOMOLOGICAL METHODS

(iii) TorR (M, N ) = 0 for every R-module N .
n+1

(iv) TorR (M, k) = 0.
n+1

Proof. If (i) holds, then M has a projective resolution of length n, and if we use this
resolution to compute Tor, we get (ii). There is no di¬culty with (ii) =’ (iii) =’ (iv),
so it remains to prove (iv) =’ (i). Let 0 ’ Kn’1 ’ Xn’1 ’ · · · ’ X0 ’ M ’ 0
be an exact sequence with all Xi projective. By (7.1.7), it su¬ces to show that Kn’1
is projective. Now we apply dimension shifting as in the proof of (7.1.7). For example,
the short exact sequence 0 ’ K1 ’ X1 ’ K0 ’ 0 [see(7.1.7)] induces the long exact
sequence · · · ’ TorR (X1 , k) ’ TorR (K0 , k) ’ TorR (K1 , k) ’ TorR (X1 , k) ’ · · ·
n n n’1 n’1
and as before, the outer terms are 0, which implies that the map in the middle is an
isomorphism. Iterating, we have TorR (Kn’1 , k) ∼ TorR (M, k) = 0 by hypothesis. By
=
1 n+1
(7.3.2), Kn’1 is projective. ™

7.3.4 Corollary
Let R be a Noetherian local ring with maximal ideal M and residue ¬eld k. For any
positive integer n, the following conditions are equivalent.

(1) gldim R ¤ n.
(2) TorR (M, N ) = 0 for all ¬nitely generated R-modules M and N .
n+1

(3) TorR (k, k) = 0.
n+1


Proof. If (1) holds, then pdR M ¤ n for all M , and (2) follows from (7.3.3). Since (3) is
a special case of (2), it remains to prove that (3) implies (1). Assuming (3), (7.3.3) gives
TorR (k, N ) = TorR (N, k) = 0 for all R-modules N . Again by (7.3.3), the projective
n+1 n+1
dimension of any R-module N is at most n, hence gldim R ¤ n. ™


7.4 Application
As promised in (6.2.5), we will prove that under a mild hypothesis, all maximal M -
sequences have the same length.

7.4.1 Lemma
Let M and N be R-modules, and let a1 , . . . , an be an M -sequence. If an annihilates N ,
then the only R-homomorphism h from N to M = M/(a1 , . . . , an’1 )M is the zero map.
Proof. If x is any element of N , then an h(x) = h(an x) = h(0) = 0. Since an is not a
zero-divisor of M , the result follows. ™

7.4.2 Proposition
Strengthen the hypothesis of (7.4.1) so that each ai , i = 1, . . . , n annihilates N . Then
Extn (N, M ) ∼ homR (N, M/(a1 , . . . , an )M ).
=
R
7.4. APPLICATION 7

Proof. The short exact sequence 0 ’ M ’ M ’ M/a1 M ’ 0, with the map
from M to M given by multiplication by a1 , induces the following long exact sequence:
G Extn’1 (N, M/a1 M ) δ G Extn (N, M ) a1 G Extn (N, M )
Extn’1 (N, M ) R R
R R
where the label a1 indicates multiplication by a1 . In fact this map is zero, because a1
annihilates N ; hence δ is surjective. By induction hypothesis, Extn’1 (N, M ) is isomor-
R
phic to homR (N, M/(a1 , . . . , an’1 )M = 0 by (7.4.1). (The result is vacuously true for
n = 1.) Thus δ is injective, hence an isomorphism. Consequently, if M = M/a1 M ,
we have Extn’1 (N, M ) ∼ Extn (N, M ). Again using the induction hypothesis, we have
= R
R
ExtR (N, M ) ∼ homR (N, M /(a2 , . . . , an )M = homR (N, M/(a1 , . . . , an )M ). ™
n’1
=
We prove a technical lemma to prepare for the main theorem.


7.4.3 Lemma
Let M0 be an R-module, and I an ideal of R. Then homR (R/I, M0 ) = 0 if and only if
there is a nonzero element of M0 annihilated by I. Equivalently, by (1.3.1), I is contained
in some associated prime of M0 . (If there are only ¬nitely many associated primes, for
example if R is Noetherian [see (1.3.9)], then by (0.1.1), another equivalent condition is
that I is contained in the union of the associated primes of M0 .)
Proof. If there is a nonzero homomorphism from R/I to M0 , it will map 1+I to a nonzero
element x ∈ M0 . If a ∈ I, then a + I is mapped to ax. But a + I = 0 + I since a ∈ I, so
ax must be 0. Conversely, if x is a nonzero element of M0 annihilated by I, then we can
construct a nonzero homomorphism by mapping 1 + I to x, and in general, r + I to rx.
We must check that the map is well de¬ned, but this follows because I annihilates x. ™


7.4.4 Theorem
Let M be a ¬nitely generated module over the Noetherian ring R, and I an ideal of R
such that IM = M . Then any two maximal M -sequences in I have the same length,
namely the smallest nonnegative integer n such that Extn (R/I, M ) = 0.
R
Proof. In (7.4.2), take N = R/I and let {a1 , . . . , an } be a set of generators of I. Then

Extn (R/I, M ) ∼ homR (R/I, M0 )
=
R

where M0 = M/(a1 , . . . , an )M . By (7.4.3), Extn (R/I, M ) = 0 if and only if I is not
R
contained in the union of all associated primes of M0 . In view of (1.3.6), this says that
if a1 , . . . , an is an M -sequence in I, it can be extended to some an+1 ∈ I as long as
Extn (R/I, M ) = 0. This is precisely the statement of the theorem. ™
R



7.4.5 Remarks
Under the hypothesis of (7.4.4), we call the maximum length of an M -sequence in I the
grade of I on M . If R is a Noetherian local ring with maximal ideal M, then by (6.2.2),
the elements ai of an M -sequence are nonunits, hence belong to M. Thus the depth of
M , as de¬ned in (6.2.5), coincides with the grade of M on M .
8 CHAPTER 7. HOMOLOGICAL METHODS

Again let M be ¬nitely generated over the Noetherian local ring R. By (7.4.4), the
depth of M is 0 if and only if homR (R/M, M ) = 0. By (7.4.3) and the maximality of M,
this happens i¬ M is an associated prime of M . Note also that by (6.1.1) and (6.2.3), if
a1 , . . . , ar is an M -sequence of maximal length, then the module M/(a1 , . . . , ar )M has
¬nite length.
Chapter 8

Regular Local Rings

In algebraic geometry, the local ring of an a¬ne algebraic variety V at a point P is the set
O(P, V ) of rational functions on V that are de¬ned at P . Then P will be a nonsingular
point of V if and only if O(P, V ) is a regular local ring.


8.1 Basic De¬nitions and Examples
8.1.1 De¬nitions and Comments
Let (R, M, k) be a Noetherian local ring. (The notation means that the maximal ideal is
M and the residue ¬eld is k = R/M.) If d is the dimension of R, then by the dimension
theorem [see (5.4.1)], every generating set of M has at least d elements. If M does in fact
have a generating set S of d elements, we say that R is regular and that S is a regular
system of parameters. (Check the de¬nition (6.1.1) to verify that S is indeed a system of
parameters.)

8.1.2 Examples
1. If R has dimension 0, then R is regular i¬ {0} is a maximal ideal, in other words, i¬ R
is a ¬eld.
2. If R has dimension 1, then by (3.3.11), condition (3), R is regular i¬ R is a discrete
valuation ring. Note that (3.3.11) assumes that R is an integral domain, but this is not
a problem because we will prove shortly that every regular local ring is a domain.
3. Let R = K[[X1 , . . . , Xd ]], where K is a ¬eld. By (5.4.9), dim R = d, hence R is regular
and {X1 , . . . , Xd } is a regular system of parameters.
4. Let K be a ¬eld whose characteristic is not 2 or 3, and let R = K[X, Y ]/(X 3 ’ Y 2 ),
localized at the maximal ideal M = {X ’ 1, Y ’ 1}. (The overbars indicate calculations
mod (X 3 ’ Y 2 ).) It appears that {X ’ 1, Y ’ 1} is a minimal generating set for M,
but this is not the case (see Problem 1). In fact M is principal, hence dim R = 1 and
R is regular. (See Example 2 above, and note that R is a domain because X 3 ’ Y 2 is
irreducible, so (X 3 ’ Y 2 ) is a prime ideal.)

1
2 CHAPTER 8. REGULAR LOCAL RINGS

5. Let R be as in Example 4, except that we localize at M = (X, Y ) and drop the
restriction on the characteristic of K. Now it takes two elements to generate M, but
dim R = 1 (Problem 2). Thus R is not regular.
Here is a convenient way to express regularity.

8.1.3 Proposition
Let (R, M, k) be a Noetherian local ring. Then R is regular if and only if the dimension
of R coincides with dimk M/M2 , the dimension of M/M2 as a vector space over k. (See
(3.3.11), condition (6), for a prior appearance of this vector space.)
Proof. Let d be the dimension of R. If R is regular and a1 , . . . , ad generate M, then the
ai + M2 span M/M2 , so dimk M/M2 ¤ d. But the opposite inequality always holds
(even if R is not regular), by (5.4.2). Conversely, if {a1 + M2 , . . . , ad + M2 } is a basis for
M/M2 , then the ai generate M. (Apply (0.3.4) with J = M = M.) Thus R is regular.


8.1.4 Theorem
A regular local ring is an integral domain.
Proof. The proof of (8.1.3) shows that the associated graded ring of R, with the M-
adic ¬ltration [see (4.1.2)], is isomorphic to the polynomial ring k[X1 , . . . , Xd ], and is
therefore a domain. The isomorphism identi¬es ai with Xi , i = 1, . . . , d. By the Krull
intersection theorem, ©∞ Mn = 0. (Apply (4.3.4) with M = R and I = M.) Now let
n=0
a and b be nonzero elements of R, and choose m and n such that a ∈ Mm \ Mm+1 and
b ∈ Mn \ Mn+1 . Let a be the image of a in Mm /Mm+1 and let b be the image of b in
Mn /Mn+1 . Then a and b are nonzero, hence a b = 0 (because the associated graded ring
is a domain). But a b = ab, the image of ab in Mm+n+1 , and it follows that ab cannot
be 0. ™
We now examine when a sequence can be extended to a regular system of parameters.

8.1.5 Proposition
Let (R, M, k) be a regular local ring of dimension d, and let a1 , . . . , at ∈ M, where
1 ¤ t ¤ d. The following conditions are equivalent.
(1) a1 , . . . , at can be extended to a regular system of parameters for R.
(2) a1 , . . . , at are linearly independent over k, where ai = ai mod M2 .
(3) R/(a1 , . . . , at ) is a regular local ring of dimension d ’ t.
Proof. The proof of (8.1.3) shows that (1) and (2) are equivalent. Speci¬cally, the ai
extend to a regular system of parameters i¬ the ai extend to a k-basis of M/M2 . To
prove that (1) implies (3), assume that a1 , . . . , at , at+1 , . . . , ad is a regular system of
parameters for R. By (6.1.3), the dimension of R = R/(a1 , . . . , at ) is d ’ t. But the d ’ t
elements ai , i = t + 1, . . . , d, generate M = M/(a1 , . . . , at ), hence R is regular.
Now assume (3), and let at+1 , . . . , ad be elements of M whose images in M form a
regular system of parameters for R. If x ∈ M, then modulo I = (a1 , . . . , at ), we have
8.1. BASIC DEFINITIONS AND EXAMPLES 3

d d
x ’ t+1 ci ai = 0 for some ci ∈ R. In other words, x ’ t+1 ci ai ∈ I. It follows
that a1 , . . . , at , at+1 , . . . , ad generate M. Thus R is regular (which we already know by
hypothesis) and a1 , . . . , at extend to a regular system of parameters for R. ™

8.1.6 Theorem
Let (R, M, k) be a Noetherian local ring. Then R is regular if and only if M can be
generated by an R-sequence. The length of any such R-sequence is the dimension of R.
Proof. Assume that R is regular, with a regular system of parameters a1 , . . . , ad . If
1 ¤ t ¤ d, then by (8.1.5), R = R/(a1 , . . . , at ) is regular and has dimension d ’ t. The
maximal ideal M of R can be generated by at+1 , . . . , ad , so these elements form a regular
system of parameters for R. By (8.1.4), at+1 is not a zero-divisor of R, in other words,
at+1 is not a zero-divisor of R/(a1 , . . . , at ). By induction, a1 , . . . , ad is an R-sequence.
(To start the induction, set t = 0 and take (a1 , . . . , at ) to be the zero ideal.)
Now assume that M is generated by the R-sequence a1 , . . . , ad . By repeated appli-
caion of (5.4.7), we have dim R/M = dim R ’ d. But R/M is the residue ¬eld k, which
has dimension 0. It follows that dim R = d, so R is regular. ™

8.1.7 Corollary
A regular local ring is Cohen-Macaulay.
Proof. By (8.1.6), the maximal ideal M of the regular local ring R can be generated by an
R-sequence a1 , . . . , ad , with (necessarily) d = dim R. By de¬nition of depth [see(6.2.5)],
d ¤ depth R. But by (6.2.6), depth R ¤ dim R. Since dim R = d, it follows that
depthR = dim R. ™
1

List of Symbols
J(R) Jacobson radical. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .0-2
»a multiplication by a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1
rM (N ) radical of annihilator of M/N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1
AP(M ) associated primes of M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-3
z(M ) zero-divisors of M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1-4
MS localization of M by S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-6
Supp M support of M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8
V (I) set of prime ideals containing I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8
N (R) nilradical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-12
lR (M ) length of the R-module M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-13

Rc integral closure of R in a larger ring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2-3
RT localized ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-5

I radical of an ideal I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-8

V valuation ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-3
|x| absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-5
v discrete valuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-5

{Rn } ¬ltration of a ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1
{Mn } ¬ltration of a module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1
gr(R) associated graded ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-2
gr(M ) associated graded module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-2
lim← Mn inverse limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-4
ˆ
M completion of a module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-5

∆G di¬erence of G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1
(r)
analog of xr in the calculus of ¬nite di¬erences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1
k
n >> 0 for su¬ciently large n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-2
l length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-3
h(M, n) Hilbert polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-4
sI (M, n) Hilbert-Samuel polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-5
d(M ) degree of the Hilbert-Samuel polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-5
dim dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-6
ht height . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-7
coht coheight. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5-7
δ(M ) Chevalley dimension of the module M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-7
tr deg transcendence degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-14

pdR M projective dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-1
gldim R global dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-2
Ext Ext functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-2
idR N injective dimension. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7-4
Tor Tor functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-5
I-depth maximum length of an M -sequence in I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-7
2

(R, M, k) local ring with maximal ideal M and residue ¬eld k . . . . . . . . . . . . . . . . . . . 8-1
1

Index
absolute value, 3-5
archimedean, 3-5
nonarchimedean, 3-5
additivity of length, 5-3
a¬ne k-algebra, 5-12
Artin-Rees lemma, 4-3
Artinian ring, 1-11
associated graded rings and modules, 4-2
associated primes, 1-3
calculus of ¬nite di¬erences, 5-1
Chevalley dimension, 5-7
coheight, 5-7
Cohen-Macaulay rings and modules, 6-4, 8-3
coherent sequences and homomorphisms, 4-4
completion of a module, 4-5
contraction, 2-9
decomposing a long exact sequence into short exact sequences, see (5.2.3), (7.1.4)
Dedekind domain, 5-7
degree, 4-1
depth, 6-3
determinant trick, 0-3, 2-1
dimension, 5-1, 5-6¬.
homological, 7-1¬.
dimension shifting, 7-3, 7-4, 7-6
dimension theorem, 5-8
consequences of, 5-9¬.
discrete valuation, 3-5
discrete valuation ring (DVR), 3-5, 3-6
embedded component, 1-6
equivalent ¬ltrations, 4-7
extension theorems, 3-1, 3-3
faithful module, 2-1
¬ltered ring and module, 4-1
¬ltration, 4-1
¬rst uniqueness theorem, 1-6
¬‚at module, 7-5
formal power series, 5-11
global dimension, 7-2
going down, 2-7, 2-9
going up, 2-6
grade, 7-7
graded module, 4-1
graded piece, 4-1,
graded ring, 4-1,
2

Hausdor¬ topology, 4-5
height, 5-7
Hensel™s lemma, 4-10
Hilbert polynomial, 5-4
Hilbert-Samuel polynomial, 5-5
homogeneous elements, 4-1
homological dimension, 7-1¬.
I-adic completion, 4-8
I-adic ¬ltration, 4-1
I-adic topology, 4-4
I-¬ltration, 4-3
I-stable ¬ltration, 4-3
ideal of de¬nition, 5-4
injective dimension, 7-4
integral closure, 2-3
integral elements, extensions, 2-1
integrally closed, 2-3
inverse limit, 4-4
inverse system, 4-4
irreducible submodule, 1-2
isolated component, 1-6
Jacobson radical, 0-2, 1-12
Krull dimension, 5-6
Krull intersection theorem, 4-8
Krull™s principal ideal theorem, 5-10
generalization, 5-10
Laurent series, 3-2
length of a module, 1-13
local ring, 0-2, 0-4
lying over, 2-4, 2-5
M -sequence, 6-2 minimal component, 1-6
Nakayama™s lemma (NAK), 0-3
Noether™s normalization lemma, 5-11
nilradical, 1-12
polynomial-like function, 5-2
primary decomposition, 1-2
existence of, 1-3
reduced, 1-2
uniqueness of, 1-6, 1-8
primary ideal, 1-1
primary submodule, 1-1
prime element, 3-6
P -primary submodule, 1-1
prime avoidance lemma, 0-1
projective dimension, 7-1
regular local ring, 8-1
3

regular sequence, 6-2
regular system of parameters, 8-1
residue ¬eld, 0-4
second uniqueness theorem, 1-8
stabilizing a module, 2-2
support of a module, 1-8
system of parameters, 6-1
transcendence degree, 5-14
transitivity of integral extensions, 2-2
UFD (unique factorization domain), 2-3
uniformizer, 3-6
valuation ring, 3-2
Zariski topology, 1-8
zero-divisors, 1-4
Exercises

Chapter 1
1. What are the primary ideals of Z?
2. Let R = k[x, y] where k is a ¬eld. Show that Q = (x, y 2 ) is P -primary, and identify
P.
3. Continuing Problem 2, show that Q is not a power of a prime ideal.
4. Let R = k[x, y, z]/I where I = (xy ’ z 2 ). Let x = x + I, y = y + I, z = z + I. If
2
P = (x, z), show that P is a power of a prime ideal and its radical is prime, but it
is not primary.
5. Let R = k[x, y] where k is a ¬eld, and let P1 = (x), P2 = (x, y), Q = (x2 , y), I =
(x2 , xy). Show that I = P1 © P2 and I = P1 © Q are both primary decompositions of
2

I.
6. Let M and N be ¬nitely generated modules over a local ring R. Show that M —R N = 0
if and only if either M or N is 0.
7. Continuing Problem 6, show that the result fails to hold if R is not local.
8. Let S be a multiplicative subset of R, and MS = S ’1 M . Use base change formulas
in the tensor product to show that (M —R N )S ∼ MS —RS NS as RS -modules.
=
9. If M and N are ¬nitely generated R-modules, show that Supp(M —R N ) = Supp M ©
Supp N .

In Problems 10-13, we consider uniqueness in the structure theorem (1.6.7) for Ar-
tinian rings.
r
10. Let R = 1 Ri , where the Ri are Artinian local rings, and let πi be the projection of
R on Ri . Show that each Ri has a unique prime ideal Pi , which is nilpotent. Then
’1
show that Mi = πi (Pi ) is a maximal ideal of R.

11. Let Ii = ker πi , i = 1, . . . , r. Show that I i = Mi , so by (1.1.2), Ii is Mi -primary.
12. Show that ©r Ii is a reduced primary decomposition of the zero ideal.
1
13. Show that in (1.6.7), the Ri are unique up to isomorphism.
14. Let M be ¬nitely generated over the Noetherian ring R, and let P be a prime ideal
in the support of M . Show that lRP (MP ) < ∞ if and only if P is a minimal element
of AP(M ).

1
2

Chapter 2
1. Let R = Z and S = Z[i], the Gaussian integers. Give an example of two prime ideals
of S lying above the same prime ideal of R. (By (2.2.1), there cannot be an inclusion
relation between the prime ideals of S.)
2. Let R = k[X, Y ]/I, where k is a ¬eld and I is the prime ideal (X 2 ’ Y 3 ). Write the
coset X + I simply as x, and Y + I as y. Show that ± = x/y is integral over R, but
± ∈ R. Thus R is not integrally closed.
/
3. Suppose we have a diagram of R-modules
f g
M ’’’ M ’’’ M
’’ ’’
with im f ⊆ ker g. Show that the following conditions are equivalent.
(a) The given sequence is exact.
(b) The sequence
fP gP
M P ’ ’ ’ MP ’ ’ ’ MP
’’ ’’
is exact for every prime ideal P .
(c) The localized sequence of (b) is exact for every maximal ideal P .
4. Let f : M ’ N be an R-module homomorphism. Show that f is injective [resp.
surjective] if and only if fP is injective [resp. surjective] for every prime, equivalently
for every maximal, ideal P .
5. Let R be an integral domain with fraction ¬eld K. We may regard all localized rings
RP as subsets of K. Let M be the intersection of all RP for maximal ideals P . If S is
any multiplicative subset of R, show that
S ’1 M ⊆ S ’1 RP
P ∈max R

where max R is the set of maximal ideals of R.
Continuing Problem 5, if Q is any maximal ideal of R, show that MQ ⊆ RQ .
6.
7. Continuing Problem 6, show that the intersection of all RP , P prime, coincides with
the intersection of all RP , P maximal, and in fact both intersections coincide with R.
8. If R is an integral domain, show that the following conditions are equivalent:
(a) R is integrally closed;
(b) RP is integrally closed for every prime ideal P ;
(c) RQ is integrally closed for every maximal ideal Q.
9. Let P be a prime ideal of R. Show that the ¬elds RP /P RP and Frac R/P are isomor-
phic. Each is referred to as the residue ¬eld at P .


Chapter 3
Let R and S be local subrings of the ¬eld K, with maximal ideals MR and MS respec-
tively. We say that S dominates R, and write (R, MR ) ¤ (S, MS ), if R is a subring of S
and R © MS = MR .
3

1. If V is a valuation ring of K, show that (V, MV ) is maximal with respect to the partial
ordering induced by domination.
Conversely, we will show in Problems 2 and 3 that if (V, MV ) is maximal, then V is
a valuation ring. Let k be the residue ¬eld V /MV , and let C be an algebraic closure
of k. We de¬ne a homomorphism h : V ’ C, by following the canonical map from V
to k by the inclusion map of k into C. By (3.1.4), it su¬ces to show that (V, h) is a
maximal extension. As in (3.1.1), if (R1 , h1 ) is an extension of (V, h), we may assume
R1 local and h1 (R1 ) a sub¬eld of C. Then ker h1 is the unique maximal ideal MR1 .
2. Show that (R1 , MR1 ) dominates (V, MV ).
3. Complete the proof by showing that (V, h) is a maximal extension.
4. Show that every local subring of a ¬eld K is dominated by at least one valuation ring
of K.


Chapter 4
1. Let R be the formal power series ring k[[X1 , . . . , Xn ]], where k is a ¬eld. Put the
I-adic ¬ltration on R, where I is the unique maximal ideal (X1 , . . . , Xn ). Show that
the associated graded ring of R is the polynomial ring k[X1 , . . . , Xn ].
2. Let M and N be ¬ltered modules over the ¬ltered ring R. The R-homomorphism
f : M ’ N is said to be a homomorphism of ¬ltered modules if f (Mn ) ⊆ Nn for
all n ≥ 0. For each n, f induces a homomorphism f n : Mn /Mn+1 ’ Nn /Nn+1 via
f n (x + Mn+1 ) = f (x) + Nn+1 . We write grn (f ) instead of f n . The grn (f ) extend to
a homomorphism of graded gr(R)-modules, call it gr(f ) : gr(M ) ’ gr(N ). We write

gr(f ) = grn (f ).
n≥0

For the remainder of this problem and in Problems 3 and 4, we assume that gr(f ) is
injective. Show that Mn © f ’1 (Nn+1 ) ⊆ Mn+1 for all n ≥ 0.
3. Continuing Problem 2, show that f ’1 (Nn ) ⊆ Mn for all n ≥ 0.
4. Continuing Problem 3, show that if in addition we have ©∞ Mn = 0, then f is
n=0
injective.
5. Show that in (4.2.10), the two ¬ltrations {I n N } and {N © I n M } are equivalent.
6. If we reverse the arrows in the de¬nition of an inverse system [see (4.2.1)], so that
maps go from Mn to Mn+1 , we get a direct system. The direct limit of such a system
is the disjoint union Mn , with sequences x and y identi¬ed if they agree su¬ciently
far out in the ordering. In other words, θn (xn ) = θn (yn ) for all su¬ciently large n.
In (4.2.6) we proved that the inverse limit functor is left exact, and exact under an
additional assumption. Show that the direct limit functor is always exact. Thus if
G Mn G Mn
fn gn
is exact for all n, and
Mn

M = lim Mn
’’
4

is the direct limit of the Mn (similarly for M and M ), then the sequence
GM GM
f g
is exact. (The maps f and g are induced by the fn and
M
gn .)
ˆ ˆ
7. Let M be an R-module, and let M [resp. R] be the I-adic completion of M [resp.
ˆ ˆ
R]. Note that M is an R-module via {an } {xn } = {an xn }. De¬ne an R-module
ˆ ˆ
homomorphism hM : R —R M ’ M by (r, m) ’ r m. If M is ¬nitely generated over
R, show that hM is surjective.
8. In Problem 7, if in addition R is Noetherian, show that hM is an isomorphism. Thus
if R is complete (R ∼ R), then M is complete (M ∼ M ).
=ˆ =ˆ
ˆ
9. Show that the completion of M is always complete, that is, M ∼ M .
ˆ= ˆ
ˆ ˆ
10. Let R be the I-adic completion of the Noetherian ring R. Show that R is a ¬‚at
R-module.
11. If M is complete with respect to the ¬ltration {Mn }, show that the topology induced
on M by {Mn } must be Hausdor¬.
ˆ
In Problems 12-16, R is the I-adic completion of the ring R. In Problems 12-14, R is
assumed Noetherian.
12. Show that I ∼ R —R I ∼ RI.
ˆ= ˆ =ˆ
13. Show that (I)n ∼ (I n ).
ˆ= ˆ
14. Show that I n /I n+1 ∼ (I)n /(I)n+1 .
=ˆ ˆ
ˆ ˆ
15. Show that I is contained in the Jacobson radical J(R).
ˆ
16. Let R be a local ring with maximal ideal M. If R is the M-adic completion of R,
ˆ ˆ
show that R is a local ring with maximal ideal M.


Chapter 5
1. In di¬erential calculus, the exponential function ex is its own derivative. What is the
analogous statement in the calculus of ¬nite di¬erences?
2. Let M be nonzero and ¬nitely generated over the local ring R with maximal ideal M.
Show that V (ann(M/MM )) = {M}.
3. If I is an arbitrary ideal and P a prime ideal of R, show that (R/I)P = 0 i¬ P ⊇ I.
In Problems 4-7, the ring S is integral over the subring R, J is an ideal of S, and
I = J © R. Establish the following.
4. dim R = dim S.
5. coht I = coht J.
6. ht J ¤ ht I.
7. If R and S are integral domains with R integrally closed, then ht J = ht I.
If P is a prime ideal of R, then by de¬nition of height, coheight and dimension, we have
ht P + coht P ¤ dim R. In Problems 8 and 9 we show that the inequality can be strict,
5

even if R is Noetherian. Let S = k[[X, Y, Z)]] be a formal power series ring over the ¬eld
k, and let R = S/I where I = (XY, XZ). De¬ne X = X + I, Y = Y + I, Z = Z + I.
8. Show that the dimension of R is 2.
9. Let P be the prime ideal (Y , Z) of R. Show that P has height 0 and coheight 1, so
that ht P + coht P < dim R.


Chapter 6
1. Let R be a Noetherian local ring with maximal ideal M, and suppose that the elements
a1 , . . . , at are part of a system of parameters for R. If the ideal P = (a1 , . . . , at ) is
prime and has height t, show that ht P + coht P = dim R.
2. let S = k[[X, Y, Z]] be a formal power series ring over the ¬eld k, and let R = S/I,
where I = (XY, XZ). Use an overbar to denote cosets mod I, for example, X =
X + I ∈ R. Show that {Z, X + Y } is a system of parameters, but Z is a zero-
divisor. On the other hand, members of a regular sequence (Section 6.2) cannot be
zero-divisors.


Chapter 7
1. Let Z4 = Z/4Z, a free Z4 -module. De¬ne f : Z4 ’ Z4 by 0 ’ 0, 1 ’ 2, 2 ’ 0, 3 ’ 2,
i.e., f (x) = 2x mod 4. Let M = 2Z4 ∼ Z2 (also a Z4 -module), and de¬ne g : Z4 ’ M
=
by 0 ’ 0, 1 ’ 1, 2 ’ 0, 3 ’ 1, i.e., g(x) = x mod 2. Show that
G Z4 f G Z4 f G Z4 G Z4 GM G 0 is a free, hence projective,
f g
···
resolution of M of in¬nite length.
2. Given an exact sequence
‚n+1
G An fn G Bn gn G Cn ‚n G An’1 fn’1 G Bn’1 gn’1 G Cn’1
G Cn+1 G ···
···
Show that if the maps fn are all isomorphisms, then Cn = 0 for all n.

Let R be a Noetherian local ring with maximal ideal M and residue ¬eld k = R/M. Let
M be a ¬nitely generated R-module, and de¬ne uM : M —R M ’ M via uM (a — x) =
ax, a ∈ M, x ∈ M . We are going to show in Problems 3,4 and 5 that if uM is injective,
then M is free. If M is generated by x1 , . . . , xn , let F be a free R-module with basis
e1 , . . . , en . De¬ne a homomorphism g : F ’ M via ei ’ xi , 1 ¤ i ¤ n. We have an
exact sequence 0 ’ K ’ F ’ M ’ 0, where f : K ’ F, g : F ’ M , and K = ker g.
The following diagram is commutative, with exact rows.
G M—F G M—M G0
M—K
uK uF uM
  
GK GF GM
0
Applying the snake lemma, we have an exact sequence
f— g—
G coker uK G coker uF G coker uM .
δ
ker uM
6

3. Show that coker uM ∼ k —R M , and similarly coker uF ∼ k —R F .
= =
4. Show that coker uK = 0.
5. Show that g is injective. Since g is surjective by de¬nition, it is an isomorphism, hence
M ∼ F , and M is free.
=
6. Let M be a ¬nitely generated module over the Noetherian local ring R. Show that M
is free if and only if M is projective, if and only if M is ¬‚at.
7. Show that in (7.2.1), M can be replaced by R/I, I an arbitrary ideal of R.
8. Show that the global dimension of a ring R is the least upper bound of pdR (R/I),
where I ranges over all ideals of R.
9. Let f : R ’ S be a ring homomorphism, and let M be an R-module. Prove that the
following conditions are equivalent.
(a) TorR (M, N ) = 0 for all S-modules N .
1
(b) TorR (M, S) = 0 and M —R S is a ¬‚at S-module.
1



Chapter 8
1. In (8.1.2), Example 4, show that X ’ 1 and Y ’ 1 are associates.
2. Justify the assertions made in Example 5 of (8.1.2).
Let (R, M, k) be a Noetherian local ring, and let grM (R) be the associated graded
ring with respect to the M-adic ¬ltration [see(4.1.2)]. We can de¬ne a homomorphism
of graded k-algebras • : k[X1 , . . . , Xr ] ’ grM (R) via •(Xi ) = ai + M2 , where the
ai generate M. (See Chapter 4, Problem 2 for terminology.) In Problems 3-5, we
are going to show that • is an isomorphism if and only if the Hilbert polynomial
h(n) = h(grM (R), n) has degree r ’ 1. Equivalently, the Hilbert-Samuel polynomial
sM (R, n) has degree r.
3. Assume that • is an isomorphism, and let An be the set of homogeneous polynomials
of degree n in k[X1 , . . . , Xr ]. Then An is isomorphic as a k-vector space to I =
(X1 , . . . , Xr ). Compute the Hilbert polynomial of grM (R) and show that it has degree
r ’ 1.
4. Now assume that • is not an isomorphism, so that its kernel B is nonzero. Then
B becomes a graded ring •n≥0 Bn with a grading inherited from the polynomial ring
A = k[X1 , . . . , Xr ]. We have an exact sequence

0 ’ Bn ’ An ’ Mn /Mn+1 ’ 0.

Show that
n+r’1
’ lk (Bn ).
h(n) =
r’1

5. Show that the polynomial-like functions on the right side of the above equation for
h(n) have the same degree and the same leading coe¬cient. It follows that the Hilbert
polynomial has degree less than r ’ 1, completing the proof.
7

6. Let (R, M, k) be a Noetherian local ring of dimension d. Show that R is regular if

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