arXiv:1309.6694v1 [] 25 Sep 2013

THE THEORY OF ORDINAL LENGTH
HANS SCHOUTENS

A BSTRACT. We generalize the notion of length to an ordinal-valued invariant defined on
the class of finitely generated modules over a Noetherian ring. A key property of this
invariant is its semi-additivity on short exact sequences. We show how to calculate this
combinatorial invariant by means of the fundamental cycle of the module, thus linking the
lattice of submodules to homological properties of the module. Using this, we equip each
module with its canonical topology.

1. I NTRODUCTION
The length of an Artinian, finitely generated module M is defined as the longest chain
of submodules in M . Since we have the descending chain condition, such a chain is finite,
and hence can be viewed as a finite ordinal (recall that an ordinal is a linearly ordered
set with the descending chain condition). Hence we can immediately generalize this by
transfinite induction to arbitrary Artinian modules, getting an ordinal-valued length function. To remain in the more familiar category of finitely generated modules, observe that
at least over a complete Noetherian local ring, the latter is anti-equivalent with the class
of Artinian modules via Matlis duality. We could have used this perspective (which we
will discuss in a future paper), but a moment’s reflection directs us to a simpler solution:
just reverse the order. Indeed, if we view the class of all submodules of a Noetherian
module M , the Grassmanian GrassR pM q, as a partially ordered set by reverse inclusion,
then GrassR pM q admits the descending chain condition, and hence any subchain is wellordered, that is to say, an ordinal. We then simply define lenpM q as the supremum of all
such chains/ordinals in GrassR pM q. Viewed as a module over itself, this yields the length
lenpRq of a Noetherian ring R. In this paper, however, we will actually define length
through a foundation rank, and then show that it coincides with the above notion.
The key property of ordinary length is its additivity on short exact sequences. An ex2
ample like 0 Ñ Z ÝÝÑ Z Ñ Z{2Z Ñ 0 immediately shows this can no longer hold in the
general transfinite case. Moreover, even the formulation of additivity becomes problematic since ordinal sum, denoted in this paper as i, is not commutative. There does exist a
different, commutative sum, ‘, called in this paper the shuffle sum (see Appendix 12; for a
brief introduction to ordinals, see §2.1). As our first main result shows, both additions play
a role:
Theorem (Semi-additivity, Theorem 3.1). If 0 Ñ M Ñ N Ñ Q Ñ 0 is an exact sequence
of Noetherian modules, then lenpQq i lenpN q ď lenpM q ď lenpQq ‘ lenpN q.
To appreciate the power of this result, notice that we instantaneously recover Vasconcelos’ observation that a surjective endomorphism on a module is also injective (see Corollary 10.1 below). In fact, we can prove the following generalization, which essentially says
that endomorphisms cannot ‘expand’:
Date: November 8, 2017.
1

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HANS SCHOUTENS

Theorem (Non-expansion, Theorem 10.3). Let f be an endomorphism on a Noetherian
module M and let N be a submodule such that N Ď f pN q. Then f pN q “ N and the
restriction f |N is an automorphism of N .
We also get a new proof and a generalization of a result by Miyata [13]: any exact
sequence of the form M Ñ M ‘ N Ñ N Ñ 0 must be split exact (this was proved
independently in [21]). As a last application, let us call a a non-unit x P R a parameter if
dimpR{xRq ă dim R (in the local case this is equivalent with dim R{xR “ dim R ´ 1,
but in general, the dimension can drop more than one).
Theorem (Parameter Criterion, Theorem 10.9). Let R be a d-dimensional ring. An element f P R is a parameter if and only if as a module, AnnR pf q has dimension at most
d ´ 1.
Although defined as a combinatorial invariant, length turns out to also encode some
homological properties of a module. To formulate this, we must assume that the ring
R, or at least the module M , has finite Krull dimension, an assumption we henceforth
make. In intersection theory, one associates
to a module M its fundamental cycle cycpM q
ř
as the formal sum (Chow cycle) op pM qrps, where the coefficient op pM q is the local
multiplicity of M at p (defined as the length of the zero-th local cohomology of Mp ; see
§4). Length turns out to be an ordinal variant of this fundamental cycle:
Theorem (Cohomological Rank, Theorem 4.1). The length of a finite-dimensional Noetherian R-module M is equal to the shuffle sum
à
lenpRq “
op pM qω dim pR{pq .
pPAsspMq

It follows that the degree of the length of M is equal to the dimension of M . In particular, a ring R is a domain if and only if lenpRq “ ω d , where d is the dimension of R. In
fact, the latter two results can also be proven by the theory of deviations and generalized
Krull dimension initiated by Gabriel and Renschler ([5, 6]); see for instance, [12, Proposition 6.1.10]. However, our current theory is entirely distinct from this theory, as it applies
only to Noetherian modules: a module has ordinal length if and only if it is Noetherian. On
the other hand, ordinal length is a much finer invariant than Krull dimension. So, although
ordinals have been used in the past by algebraists ([6, 7, 15, 20]), it seems that our theory
is their first foray into commutative algebra (perhaps with the exception of their short appearance, very much in the spirit of the current paper, in [1]). For the reader less adept at
this concept from logic, the paper starts with a section on ordinals, and in an appendix, I
explain shuffle sums.
The last two sections contain applications of the theory that do not mention length itself: acyclicity (§10), and degradation (§11). These are merely meant as an illustration
of the power of the theory. However, since any two domains of the same dimension have
the same length (Theorem 3.4), as do and any two projective modules of the same rank
(Proposition 4.6), length is not very sensitive to singularities nor to local/global phenomena. However, one can build ordinal-valued invariants from it that do measure these, such
as the filtration rank of a module. This, and other ordinal invariants, will be discussed
in a series of future papers on applications of ordinal length to local cohomology, Koszul
homology, Cohen-Macaulay singularities, prime filtrations, ordinal Hilbert and Poincare
series, algebraic entropy, endomorphisms, . . . . The preprints [17] and [18], on the other
hand, study modules whose ordinal lengths have special properties.

THE THEORY OF ORDINAL LENGTH

2. N OTATION AND

3

GENERALITIES ON ORDERED SETS

An ordered set P (also called a partially ordered set or poset), is a set together with a
reflexive, antisymmetric and transitive binary relation ďP , called the ordering of P , and
almost always written as ď, without a subscript. Almost always, our posets will have
endpoints, that is to say, a (unique) least element K and greatest element J. A partial order
is total if for any two elements a, b P P either a ď b or b ď a. A subset C Ď P is called
a chain, if its induced order is total. If a ď b, then we may express this by saying that a is
below b; if a ă b (meaning that a ď b and a ‰ b), we also say that a is strictly below b.
More generally, for subsets A, B Ď P , we say A is below B, and write A ď B, to mean
that a ď b for all a P A and all b P B.
The initial closed interval determined by a P P is by definition the set of b P P with
b ď a and will be denoted pP, as. Dually, the terminal closed interval of a, denoted ra, P q,
is the collection of all b P P with a ď b.
2.1. Ordinals. A partial ordering is called a partial well-order if it has the descending
chain condition, that is to say, any descending chain must eventually be constant. A total
order is a well-order if and only if every non-empty subset has a minimal element.
Recall that an ordinal a is an equivalence class, up to an order-preserving isomorphism,
of a total well-order (we will use a special font to distinguish ordinals from ordinary numbers). We say that a ď b if a is isomorphic to an initial segment of b. This well-orders
the class of all ordinals, and in particular, any set of ordinals has a minimum and a supremum. The finite ordinals are just the natural numbers (where we identify n with the order
0 ă 1 ă ¨ ¨ ¨ ă n ´ 1); the first infinite ordinal is the order-type of pN, ăq and is denoted ω.
Ordinal sum is defined by concatenation: a i b is the ordinal obtained by putting a before
b (see §2.6). As this depends on the order, this sum is not commutative: 1 i ω ‰ ω i 1
since the former is just ω. (Ordinal sum is usually denoted simply by a ` b, but this might
be misleading for algebraists, as this is not a commutative operation, and so our notation
reflects that the right hand side is ‘dominant’: smaller terms on the left are ‘gobbled up’.)
With a few minor exceptions, we will not use ordinal multiplication explicitly, and so ω n
will just mean the order-type of Nn with its lexicographical ordering, and aω n , for a P N,
will mean the a-fold sum of ω n with itself (in textbooks, this would normally be denoted
ω n a, following Cantor’s original notation, but as this is quite awkward for the algebraically
inclined, we keep the more natural ‘scalar’ multiplication notation).
The supremum of all ω n for n P N is denoted ω ω . All ordinals considered in this paper
will be less than ω ω . They therefore admit a unique Cantor normal form
(1)

a “ ad ω d i ad´1 ω d´1 i . . . i a1 ω i a0

with ai P N. We call the ai the Cantor coefficients of a, and denote them by oi paq :“
ai . We call the least i (respectively, the largest i) such that oi paq ‰ 0 the order ordpaq
(respectively, the degree deg a) of a; the sum of all oi paq is called its valence valpaq.
An ordinal a is called a successor ordinal if it has an immediate predecessor. This is
equivalent with ordpaq “ 0. The second addition, the shuffle sum ‘, can be defined
using Cantor normal forms as follows (see Appendix 12 for details): for each i we have
oi pa ‘ bq “ oi paq`oi pbq. Thus, for instance, pω 3 iωq‘pω 2 i2ωq “ ω 3 iω 2 i3ω, which
we will therefore also denote as ω 3 ‘ω 2 ‘3ω. Note, however, that pω 3 iωqipω 2 i2ωq “
ω 3 ‘ ω 2 ‘ 2ω, where the first ω gets ‘gobbled up’ by the ω 2 to its right. An alternative
way of viewing ordinals is as those surreal numbers (a la Conway, see, for instance [10])
born last on any given day; the addition in the field of surreals then corresponds to the
shuffle sum ‘. Taking the latter point of view also endows the ordinals with a commutative

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HANS SCHOUTENS

multiplication, but the only instance we need is multiplication with some power ω n or some
scalar n P N. We have the obvious rule ω n ¨ ω m “ ω m`n and by ‘linearity’, we extend
this for an arbitrary ordinal a with Cantor normal form (1) to
(2)

ω n ¨ a :“ ad ω d`n i ad´1 ω d`n´1 i . . . i a1 ω n`1 i a0 ω n .

Similarly, we define n ¨ a as the n-fold shuffle sum a ‘ . . . ‘ a, that is to say,
(3)

n ¨ a :“ nad ω d i nad´1 ω d´1 i . . . i na1 ω 1 i na0 .
Given any e ě 0, we will write

(4)

aěe :“ od paqω d ‘ . . . ‘ oe paqω e

and

aďe :“ oe paqω e ‘ . . . ‘ o0 paq,

and a similar meaning for aąe and aăe . In particular, a “ aąe i aďe is the decomposition
in all terms of degree respectively bigger than e and at most e.
2.2. The length of a partial well-order. Let P be a partial well-order. We define the
height rank hP p¨q on P by transfinite induction as follows: at successor stages, we say that
hP paq ě r ‘1, if there exists b ă a with hP pbq ě r, and at limit stages, that hP paq ě r,
if there exists for each a ă r some ba ď a with hP pba q ě a. We then say that hP paq “ r
if hP paq ě r but not hP paq ě r ‘1. In particular, hP pKq “ 0. For a subset A Ď P , we
set hP pAq equal to the supremum of all hP paq with a P A. Finally, we define the (ordinal)
length of P as lenpP q :“ hP pP q “ supthP paq|a P P u. If P has a maximal element J,
then lenpP q “ hP pJq.
2.3. Example (Ordinals). Note of caution: the length of an ordinal a can be different from
the ordinal itself. Indeed, if a is a successor ordinal, with predecessor a1 , then, as a chain,
it is given by 0 ă 1 ă ¨ ¨ ¨ ă a1 , with maximal element Ja “ a1 . An easy induction shows
that ha pbq “ b, and so lenpaq “ ha pJa q “ ha pa1 q “ a1 . On the other hand, if a is a limit
ordinal, then it has no maximal element, and lenpaq is the supremum of all ha pbq “ b
with b ă a, that is to say, lenpaq “ a. We may summarize this into a single formula
(5)

lenpaq “ sup r0, aq

Let us say that a partial well-order P admits a composition series, if there exists a chain
C in P with lenpP q “ lenpCq. Not every partial well-order has a composition series as
the following example shows:
2.4. Example. Let P 1 be the disjoint union of all finite ordinals (meaning that there are
no order relations among the different disjuncts) and let P be obtained from P 1 by adding
a single element J above all elements in P 1 . Hence hP pJq “ ω, but any chain in P has
finite length. It is true that lenpP q is equal to the supremum of all lenpCq with C a chain
in P . However, if instead we let Q be obtained from P 1 by adding two elements a and J
above each element in P 1 with a ă J, then lenpQq “ hQ pJq “ ω ‘ 1 but the supremum
of all chain lengths is just ω, so even the supremum of all chain lengths is less than the
actual length.
2.5. Lemma. Let P be a partial well-order and let A, B Ď P be subsets. If A ď B, then
(6)

hA pAq i hB pBq ď hP pBq

Proof. Let a :“ hA pAq “ lenpAq. Since hP pBq is the supremum of all hP pbq with b P B,
it suffices to show that
(7)

a i hB pbq ď hP pbq.

THE THEORY OF ORDINAL LENGTH

5

We will prove (7) by induction on b :“ hB pbq. Assume first that b “ 0. Let p :“ hP pAq.
Since a is the supremum of all hA paq for a P A, and since hA paq ď hP paq, we get a ď p.
Since A ď b, we have p ď hP pbq, and hence we are done in this case.
Next, assume b is a successor ordinal with predecessor by b1 . By definition, there exists
1
b P B below b such that hB pb1 q ě b1 . By induction, we get hP pb1 q ě a i b1 . This in turn
shows that hP pbq is at least aib. Finally, assume b is a limit ordinal. Hence for each c ă b,
there exists bc P B below b such that hB pbc q “ c. By induction, hP pbc q ě a i c ď a i b
and hence also hP pbq ě a i b.

2.6. Sum Orders. By the sum P ` Q of two partially ordered sets P and Q (which,
after taking an isomorphic copy, we may assume to be disjoint), we mean the partial order
induced on their union P \ Q by declaring any element in P to lie below any element in
Q. In fact, if a and b are ordinals, then their ordinal sum a i b is just a \ b. We may
represent elements in the disjoint union P \ Q as pairs pi, aq with i “ 0 if a P P and i “ 1
if a P Q. The ordering P ` Q is then the lexicographical ordering on such pairs, that is to
say, pi, aq ď pj, bq if i ă j or if i “ j and a ď b.
2.7. Proposition. If P and Q are partial well-orders, then so is P ` Q. If P has moreover
a maximum, then lenpP ` Qq “ lenpP q i lenpQq.
Proof. We leave it as an exercise to show that P ` Q is a partial well-order. Let p :“
hP pJP q “ lenpP q. For a pair pi, aq in P ` Q, let npi, aq be equal to hP paq if i “ 0 and
to p i hQ paq if i “ 1. The assertion will follow once we showed that npi, aq “ hpi, aq, for
all pi, aq P P ` Q, where we wrote hpi, aq for hP `Q pi, aq. We use transfinite induction.
If i “ 0, that is to say, if a P P , then the claim is easy to check, since no element from Q
lies below a. So we may assume i “ 1 and a P Q. Let a :“ hQ paq and suppose first that
a “ 0. Since any element of P lies below a, in any case p ď hpi, aq. If this were strict,
then there would be an element pj, bq below pi, aq of height rank p. Lest hp0, JP q would
be bigger than p, we must have j “ 1 whence b P Q. Since b ďQ a, we get hQ paq ě 1,
contradiction. This concludes the case a “ 0, so assume a ą 0. We leave the limit case to
the reader and assume moreover that a is a successor ordinal with predecessor a1 . Hence
there exists some b P Q below a with oQ pbq “ a1 . By induction, hp1, bq “ np1, bq “ pia1 ,
and hence hp1, aq ě p i a. By a similar argument as above, one then easily shows that this
must in fact be an equality, as we wanted to show.

Increasing functions. Let f : P Ñ Q be an increasing (=order-preserving) map between
ordered sets. We say that f is strictly increasing, if a ă b then f paq ă f pbq. For instance,
an increasing, injective map is strictly increasing.
2.8. Theorem. Let f : P Ñ Q be a strictly increasing map between partial well-orders. If
P has a minimum KP , then
hQ pf pKP qq i hP paq ď hQ pf paqq.
for all a P P .
Proof. From the context, it will be clear in which ordered set we calculate the rank and
hence we will drop the superscripts. Let c :“ hpf pKqq. We induct on a :“ hpaq, where
the case a “ 0 holds trivially. We leave the limit case to the reader and assume that a is a
successor ordinal with predecessor a1 . By definition, there exists b ă a with hpbq “ a1 . By
induction, the height rank of f pbq is at least c i a1 . By assumption, f pbq ă f paq, showing
that f paq has height rank at least c i a.


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HANS SCHOUTENS

Even in the absence of a minimum, the inequality still holds, upon replacing the first
ordinal in the formula by the minimum of the ranks of all f paq for a P P . In particular,
height rank always increases.
3. S EMI - ADDITIVITY
Let R be a ring and M a Noetherian R-module. The Grassmanian of M (over R) is by
definition the collection GrassR pM q of all submodules of M , ordered by reverse inclusion.
The height rank of GrasspM q will be called the length lenR pM q of M as an R-module.
This is well-defined, since GrassR pM q is a well-partial order. Thus, for N Ď M , we have
hpN q ě a ‘ 1, if there exists N 1 properly containing N with hpN 1 q ě a; and lenpM q is
then given as hp0M q, where 0M denotes the zero submodule of M . Since the initial closed
interval pGrasspM q, N s is isomorphic to GrassR pM {N q, we get
(8)

hpN q “ hGrassR pMq pN q “ hGrassR pM{N q p0M{N q “ lenR pM {N q.

Similarly, rN, GrassR pM qq consists of all submodules of M contained in N , whence is
equal to GrassR pN q. Note that if I is an ideal in the annihilator of M , then GrassR pM q “
GrassR{I pM q, so that in order to calculate the length of M , it makes no difference whether
we view it as an R-module or as an R{I-module. We call the length of R, denoted lenpRq,
its length when viewed as a module over itself. Hence, the length of R{I as an R-module
is the same as that of R{I viewed as a ring. We define the order, ordR pM q, and valence,
valpM q, as the respective order and valence of lenR pM q.
3.1. Theorem (Semi-additivity). If 0 Ñ N Ñ M Ñ Q Ñ 0 is an exact sequence of
Noetherian R-modules, then
(9)

lenR pQq i lenR pN q ď lenR pM q ď lenR pQq ‘ lenR pN q

Moreover, if the sequence is split, then the last inequality is an equality.
Proof. The last assertion follows from the first, Theorem 12.2, and the fact that then
GrassR pN q ˆ GrassR pQq Ď GrassR pM q.
To prove the lower estimate, let A be the initial closed interval pGrassR pM q, N s and
let B be the terminal closed interval rN, GrassR pM qq. By our discussion above, A “
GrassR pM {N q “ GrasspQq, since M {N – Q, with maximum, viewed in GrassR pQq,
equal to 0Q . By the same discussion, B “ GrassR pN q with maximum 0N . Since A ď B,
we may apply Lemma 2.5 to get an inequality
hGrassR pQq p0Q q i hGrassR pN q p0N q ď hGrassR pMq p0M q,
from which the assertion follows.
To prove the upper bound, let f : GrassR pM q Ñ GrassR pN q ˆ GrassR pQq be the
map sending a submodule H Ď M to the pair pH X N, πpHqq, where π denotes the
morphism M Ñ Q. It is not hard to see that this is an increasing function. Although it is in
general not injective, I claim that f is strictly increasing, so that we can apply Theorem 2.8.
Together with Theorem 12.2, this gives us the desired inequality. So remains to verify the
claim: suppose H ă H 1 but f pHq “ f pH 1 q. Hence H 1 Ł H, but H X N “ H 1 X N and
πpHq “ πpH 1 q. Applying the last equality to an element h P HzH 1 , we get πphq P πpH 1 q,
whence h P H 1 ` N . Hence, there exists h1 P H 1 such that h ´ h1 lies in H X N whence
in H 1 X N . This in turn would mean h P H 1 , contradicting our assumption on h.


THE THEORY OF ORDINAL LENGTH

7

3.2. Corollary. Let R be a Noetherian ring. If x is an R-regular element and I Ď R an
arbitrary ideal, then
lenpR{xRq i lenpR{Iq ď lenpR{xIq.
Proof. Apply Theorem 3.1 to the exact sequence
x

0 Ñ R{I ÝÝÑ R{xI Ñ R{xR Ñ 0.

Applying Corollary 3.2 to the zero ideal and observing that a i b “ b if and only if
deg a ă deg b, we get:
3.3. Corollary. If x is an R-regular element, then the degree of lenpR{xRq is strictly less
than the degree of lenpRq.

3.4. Theorem. Let M be a finitely generated module over a Noetherian ring R. Then the
degree of lenR pM q is equal to the dimension of M . In particular, R is a d-dimensional
domain if and only if lenpRq “ ω d .
Proof. Let m be the length of M and d its dimension. We start with proving the inequality
ω d ď m.

(10)

We will do this first for M “ R, by induction on d, where the case d “ 1 is clear, since
R does not have finite length. Hence we may assume d ą 1. Taking the residue modulo
a d-dimensional prime ideal (which only can lower length), we may assume that R is a
domain. Let p be a pd ´ 1q-dimensional prime ideal and let x be a non-zero element
in p. By Corollary 3.3, the degree of lenpR{xRq is at most deg m ´ 1. By induction,
ω d´1 ď lenpR{pq ď lenpR{xRq, whence d ´ 1 ď deg m ´ 1, proving (10).
For M an arbitrary R-module, let p be a d-dimensional associated prime of M , so that
R{p is isomorphic to a submodule of M , whence lenpR{pq ď m by Theorem 3.1. As we
already proved that ω d ď lenpR{pq, we obtain (10).
Next we show that
m ă ω d`1 ,

(11)

again by induction on d. Assume first that M “ R is a domain. Since R{I has then
dimension at most d ´ 1 for any non-zero ideal I, we get lenpR{Iq ă ω d by our induction
hypothesis. By (8), this means that any non-zero ideal has height rank less than ω d , and
hence R itself has length at most ω d . Together with (10), this already proves one direction
in the second assertion. For the general case, we do a second induction, this time on m.
Let p be again a d-dimensional associated prime of M , and consider an exact sequence
0 Ñ R{p Ñ M Ñ M̄ Ñ 0. By Theorem 3.1, we get m ď lenpM̄ q ‘ lenpR{pq. By what
we just proved, lenpR{pq “ ω d , and hence by induction m ď lenpM̄ q ‘ ω d ă ω d`1 . The
first assertion is now immediate from (10) and (11).
Conversely, if R has length ω d , then for any non-zero ideal I, the length of R{I is
strictly less than ω d , whence its dimension is strictly less than d by what we just proved.
This shows that R must be a domain.

4. L ENGTH

AS A COHOMOLOGICAL RANK

In this section, R wil always be a Noetherian ring of finite Krull dimension, and all
R-modules will be finitely generated. We denote the collection of all associated primes of
M (= prime ideals of the form Annpaq with a P M ) by AsspM q; it is always a finite set.

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HANS SCHOUTENS

We will make frequent use, for a short exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0, of the
following two inclusions (see, for instance, [4, Lemma 3.6])
(12)

AsspN q Ď AsspM q;

(13)

AsspM q Ď AsspN q Y AsspQq

Let M be a finitely generated R-module, and a Ď R an ideal. Recall that the a-torsion
or (zero-th) local cohomology of M at a, denoted Γa pM q, is given as the intersection of
all AnnM pan q, that is to say, as all elements in M that are killed by some power of a. This
is a left exact functor and its higher derived functors will be denoted Hja pM q. Following
common practice, we will identify Γa pM q with its zero-th derived functor and henceforth
denote it H0a pM q. If pR, mq is local, then H0m pM q has finite length, denoted om pM q and
called the local multiplicity of M .
Note that H0p pM qp – H0pRp pMp q. Since localization is exact, we get
(14)

Hip pM qp – HipRp pMp q

for all i. We will write op pM q for opRp pMp q and call it the local multiplicity of p on M ;1
it is non-zero if and only if p P AsspM q.
We now define the cohomological rank of a module M as
à
cohrkpM q :“
op pM qω dimpR{pq .
p

It is instructive to view this from the point of view of Chow cycles. Let ApRq be the Chow
ring of R, defined as the free Abelian group on SpecpRq.
ř An element D of ApRq will be
called a cycle, and will be represented as a finite sum ai rpi s, where rps is the symbol
denoting the free generator corresponding to the prime ideal p. The sum of all ai is called
the degree degpDq of the cycle D. We define
a partial order on ApRq by the rule that
ř
D ĺ E, if ai ď bi , for all i, where E “ bi rpi s. In particular, denoting the zero cycle
simply by 0, we call a cycle D effective , if 0 ĺ D, and we let A` pRq be the semi-group
of effective cycles. This allowsř
us to define a map from effective cycles to ordinals by
sending the effective cycle D “ i ai rpi s to the ordinal
à
ai ω dimpR{pi q .
opDq :“
i

Clearly, if D and E are effective, then opD ` Eq “ opDq ‘ opEq. Moreover, if D ĺ E,
then opDq ĺ opEq, so that we get a map pA` pRq, `, ĺq Ñ pORD, ‘, ĺq of partially
ordered semi-groups.
To any R-module M , we can assign its fundamental cycle, by the rule
ÿ
(15)
cycR pM q :“
op pM qrps
p

In particular, opcycpM qq “ cohrkpM q. Our main result now links this cohomological
invariant to our combinatorial length invariant :
4.1. Theorem. For any finitely generated module M over a finite-dimensional Noetherian
ring R, we have lenpM q “ cohrkpM q.
1See, for instance, [4, p. 102]; not to be confused with the multiplicity of a module at a primary ideal given
in terms of its Hilbert function.

THE THEORY OF ORDINAL LENGTH

9

Before we give the proof, we derive two lemmas. It is important to notice that the first
of these is not true at the level of cycles.
4.2. Lemma. If M Ñ Q is a proper surjective morphism of R-modules, then
cohrkpQq ă cohrkpM q.
Proof. Let N be the (non-zero) kernel of M Ñ Q, and let d be its dimension. If p P
AsspM q but not in the support of N , then Mp – Qp , so that they have the same local
cohomology. This holds in particular for any p P AsspM q with dimpR{pq ą d, showing
that cohrkpQq and cohrkpM q can only start differing at a coefficient of ω i for i ď d. So
let p P AsspM q X SupppN q have dimension d. In general, local cohomology is only left
exact, but by Lemma 4.3 below, we have in fact an exact sequence (16). Since op pN q ‰
0, we must therefore have op pQq ă op pM q. It now easily follows that cohrkpQq ă
cohrkpM q.

4.3. Lemma. Given an exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0, if p is a minimal prime
of N , then
(16)

0 Ñ H0pRp pNp q Ñ H0pRp pMp q Ñ H0pRp pQp q Ñ 0

is exact.
Proof. Upon localizing, using (14), we may assume pR, pq is local and N has finite length.
We only need to prove exactness at the final map. By assumption, N is annihilated by
some power pn . Let a P M be such that its image ā in Q lies in H0p pQq, that is to say,
pm ā “ 0 in Q, for some m. Therefore, pm a P N , whence pm`n a “ 0 in M , showing that
a P H0p pM q.

4.4. Corollary. Given an exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0, if M has no embedded
primes, then we have an equality of cycles
(17)

cycR pM q ` D “ cycR pN q ` cycR pQq

where D is an effective cycle supported on AsspQqz AsspM q.
Proof. Let D be the cycle given by (17), so that D has support in AsspM q Y AsspQq by
(13). We need to show that D is effective and supported on AsspQqz AsspM q. Since any
associated prime p of M is minimal, D is not supported in p by Lemma 4.3. On the other
hand, any associated prime of Q not in AsspM q appears with a positive coefficient in D,
showing that the latter is effective.

Proof of Theorem 4.1. Let us first prove lenpM q ď cohrkpM q by transfinite induction
on cohrkpM q, where the case cohrkpM q “ 0 corresponds to M “ 0. Let N be any
non-zero submodule of M . By Lemma 4.2, we have cohrkpM {N q ă cohrkpM q, and
hence our induction hypothesis applied to M {N yields lenpM {N q ď cohrkpM {N q ă
cohrkpM q. Since lenpM {N q “ hpN q by (8), continuity yields that lenpM q “ hp0M q
can be at most cohrkpM q, as we needed to show.
To prove the converse inequality, we induct on the length of M . Choose an associated
prime p of M of minimal dimension, say, dimpR{pq “ e. By assumption, there exists
m P M such that AnnR pmq “ p. Let H be the submodule of M generated by m. Since
H – R{p, we get cohrkpHq “ ω e , and so by what we already proved, lenpR{pq ď ω e .
By Theorem 3.4, this then is an equality. So we may assume that Q :“ M {H is non-zero.
By Lemma 4.3, we get op pM q “ op pQq ´ 1. By semi-additivity, we have an inequality
lenpQq i lenpHq ď lenpM q

10

HANS SCHOUTENS

and therefore, by induction
(18)

cohrkpQq i ω e ď lenpM q

Let g be any associated prime of M different from p. By minimality of dimension, g cannot
contain p. In particular, Mg – Qg, whence og pM q “ og pQq. Let b :“ cohrkpQqěe (see
(4)). Putting together what we proved so far, we can find an ordinal a with ordpaq ě e
(stemming from primes associated to Q but not to M ), such that b ‘ω e “ cohrkpM q ‘ a.
Since cohrkpQq i ω e “ b ‘ω e , we get, from (18) and the first part, inequalities
cohrkpM q ‘ a ď lenpM q ď cohrkpM q
which forces a “ 0 and all inequalities to be equalities.



4.5. Corollary. The order of a module is the smallest dimension of an associated prime,
and its valence is the degree of its fundamental cycle.

By [3, Proposition 1.2.13], over a local ring, we have
(19)

depthpM q ď ordpM q.

This inequality can be strict: for example, a two-dimensional domain which is not CohenMacaulay, has depth one but order two by Theorem 3.4. As an illustration of the use of
Theorem 4.1, let us calculate the length of some special modules.
4.6. Proposition. Suppose P is a finitely generated projective module over a d-dimensional
Noetherian ring R. If P has rank n, then lenpP q “ n ¨ lenpRq.
Proof. For any prime ideal p of R, we have Pp – Rpn , and hence op pP q “ nop pRq. The

result now follows from Theorem 4.1 and (3).
Note that this is false if P is a projective module without rank, like the example P “
Z{2Z over the ring R “ Z{6Z already shows. By Theorem 4.1, any d-dimensional local Cohen-Macaulay ring R has length eω d , where e “ od pRq is its generic length (see
Corollary 4.12 below), since any associated prime is d-dimensional (see, for instance, [11,
Theorem 17.4]). In a future paper, we will show that e is at most the multiplicity of R.
4.7. Proposition. If a Cohen-Macaulay local ring R admits a canonical module ωR , then
lenpRq “ lenR pωR q.
Proof. Since ωR is a maximal Cohen-Macaulay module, it has the same associated primes
p as R. Given such a p, since Rp is Artinian, and since canonical modules localize, pωR qp
is the injective hull of the residue field of Rp . By [3, Proposition 3.2.12(e)], the length of
this injective hull is equal to that of Rp , showing that op pRq “ op pωR q, for every prime
ideal p of R. The result now follows from Theorem 4.1.

Our next result gives a constraint on the possible length of a submodule, which is exploited in [17] to study binary modules. Let us say that a is weaker than b, denoted a ĺ b,
if oi paq ď oi pbq, for all i. Clearly, a ĺ b implies a ď b, but the converse fails in general
(e.g., ω is smaller than ω 2 but not weaker than it).
4.8. Theorem. If N Ď M , then lenpN q ĺ lenpM q. Conversely, if n ĺ lenpM q, then
there exists a submodule N Ď M of length n.
Proof. The first assertion is immediate from Theorem 4.1, inclusion (12), and the fact that
(16) is always left exact. For the second assertion, let m :“ lenpM q. We induct on the
(finite collection) of ordinals n weaker than m to show that there exists a submodule of
that length. The case n “ 0 being trivial, we may assume n ‰ 0. Let i be the order of

THE THEORY OF ORDINAL LENGTH

11

n and write n “ q ‘ ω i for some q ĺ n. Since then q ĺ m, there exists a submodule of
length q by induction. Let H Ď M be maximal among all submodules of length q. By
Theorem 4.1, there exists an i-dimensional associated prime p of M , such that H0pRp pHp q
is strictly contained in H0pRp pMp q. Hence we can find x P M outside H such that px Ď H.
Let N :“ H ` Rx and let x̄ be the image of x modulo H, so that N {H “ Rx̄. Since
px̄ “ 0, the length of Rx̄ is at most ω i . By semi-additivity applied to the inclusion H Ď N ,
we have an inequality lenpN q ď q ‘ lenpRx̄q, and hence lenpN q ď n. Maximality of
H yields q ă lenpN q. On the other hand, since lenpN q ĺ m by our first assertion,
minimality of i then forces lenpN q “ n, as we needed to show.

4.9. Remark. In fact, if N Ď M , then cycpN q ĺ cycpM q, so that the fundamental cycle
map is a morphism GrasspM q˝ Ñ ApRq of partially ordered sets, where GrasspM q˝ is
the opposite order given by inclusion. On the other hand, by (8) and Theorem 4.1, the
map GrasspM q Ñ ApRq given by N ÞÑ cycpM {N q factors through the length map
GrasspM q Ñ ORD, but there is no natural ordering on ApRq for which this becomes a map
of ordered sets.
We may improve the lower semi-additivity by replacing ď by ĺ:
4.10. Corollary. If 0 Ñ N Ñ M Ñ Q Ñ 0 is exact, then lenpQq i lenpN q ĺ lenpM q.
Proof. This is really just a fact about ordinals. Let m, n, q be the respective lengths of M ,
N , Q. By semi-additivity, q i n ď m ď q ‘ n, whereas Theorem 4.8 gives n ĺ m, and we
now show that these inequalities imply that q i n ĺ m. Write m “ n ‘ a and let d be the
dimension of n. For an arbitrary ordinal b, we have a unique decomposition b “ běd ibăd
as described in (4). By assumption, něd “ aω d with a “ od pnq, and hence the semiadditivity inequalities at degree d and higher become qěd ‘aω d ď aěd ‘aω d ď qěd ‘aω d ,
showing that qěd “ aěd . By definition of ordinal sum, q i n “ qěd ‘ n and so
pq i nq ‘ aăd “ qěd ‘ n ‘ aăd “ aěd ‘ aăd ‘ n “ a ‘ n “ m
proving that q i n ĺ m.



To give a more detailed version of semi-additivity, let us write oi pM q :“ oi plenpM qq
for the i-th Cantor coefficient of lenpM q. By Theorem 4.1, each oi pM q is equal to the
sum of all op pM q, where p runs over all i-dimensional associated primes of M .
4.11. Proposition. Let 0 Ñ N Ñ M Ñ Q Ñ 0 be an exact sequence of modules and let
r, d, s be the respective dimensions of N , M , and Q.
(4.11.a) or pM q “ or pN q ` or pQq,
(4.11.b) If r ă d, then d “ s and oi pM q “ oi pQq, for all r ă i ď d,
(4.11.c) If s ă d, then r “ d and oi pN q “ oi pM q, for all s ă i ď d.
Àd
Àr
Proof. Write
of N , M , and Q, respectively as n :“ i“0 ni ω i , m :“ i“0 mi ω i ,
Às the length
and q :“ i“0 qi ω i , with nr , md , qs ‰ 0. In particular
qin “

d
à

qi ω i ‘ pqr ` nr qω r ‘

i“r`1

à

ni ω i .

iăr

By semi-additivity, we have
(20)

q i n ď m ď q ‘ n,

and the first two relations follow now easily by comparing Cantor coefficients.

12

HANS SCHOUTENS

So assume s ă d, so that q i n “ n. We prove by downward induction on t ą s that
nt “ mt . The case t “ d is covered by (4.11.a). So assume we have already proven this
for all i ą t. In particular, each of the three ordinals in (20) have the same part of degree
t ` 1 and higher, and so we may subtract it from each of them. The resulting inequality
has become
nt ω t ` năt ď mt ω t ` măt ď nt ω t ` v
with v of degree at most t´ 1 (note that by assumption degpqq ď t´ 1). Hence, the leading
coefficients must be equal, that is to say, nt “ mt , and so we are done by induction.

Since dim N ď dim R, we immediately get:
4.12. Corollary (Top additivity). If R has dimension d, then od p¨q (called the generic
length) is additive on exact sequences.

I conclude this section with an example illustrating how length captures the nilpotent
structure of a ring (the proof will be given in a future paper on the length of monomial
algebras).
4.13. Example. Let R be the quotient of the polynomial ring krx, y, zs modulo the monomial ideal
I :“ px5 yz, x4 y 3 z, x3 y 2 z 2 , x2 y 3 z 2 , x2 y 2 z 3 , x4 yz 4 , x6 z 4 , x5 z 5 q.
Without proof we state that the minimal primes are pxq and pzq with respective local multiplicities 2 and 1; the one-dimensional primes are px, yq, px, zq, py, zq, with respective
multiplicities 5, 1, and 3; and the maximal ideal px, y, zq has local multiplicity 7. Therefore, by Theorem 4.1, the length is
lenpRq “ 3ω 2 ` 9ω ` 7.
5. BASE

CHANGE

The behavior of length under base change is intricate, and so we will only discuss some
basic facts. Recall that a homomorphism R Ñ S is called cyclically pure, if I “ ISXR for
every ideal I Ď R. Faithfully flat maps are examples of cyclically pure homomorphisms.
5.1. Lemma. If R Ñ S is cyclically pure, then lenpRq ď lenpSq.
Proof. Consider the canonical map f : GrasspRq Ñ GrasspSq given by sending an ideal
I Ď R to its extension IS. Being cyclically pure means that f is injective, and the inequality is now immediate from Theorem 2.8.

Even under extensions of scalars can this be a strict inequality: let R be the domain
Rrx, ys{px2 ` y 2 q and note that R bR C is not a domain and hence must have length
different from lenpRq “ ω by Theorem 3.4. Using Theorem 4.1, one easily calculates
that lenpR bR Cq “ 2ω. A similar phenomenon occurs for completion: let R be the
p is not a
analytically reduced domain krx, ys{px2 ` y 2 ` y 3 q, so that its completion R
domain and hence has length different from lenpRq “ ω (again using Theorem 4.1, one
p “ 2ω). Another factor to take into account, is that the rings might
calculates that lenpRq
have different dimension. Apart from the latter, length behaves well under polynomial
extensions:
5.2. Corollary. Let R be a finite-dimensional Noetherian ring and t an n-tuple of variables. Then lenpRrtsq “ ω n ¨ lenpRq.

THE THEORY OF ORDINAL LENGTH

13

Proof. By an inductive argument, it suffices to treat the case that t is a single variable.
The associated primes of Rrts are precisely the extensions pRrts with p P AsspRq. Moreover, oRrts ppRrtsq “ oR ppq. Since Rrts{pRrts – pR{pqrts, its dimension is equal to

dim pR{pq ` 1, and the result now follows from Theorem 4.1.
We can now identify one useful class of extensions that preserve length: recall that a
Nagata extension of a local ring pR, mq is a localization of some polynomial extension
Rrts with respect to the prime ideal mRrts, and will be denoted Rptq, where t is some
tuple of variables. The extension R Ñ Rptq is a scalar extension in the terminology of
[19], that is to say, faithfully flat and unramified. In particular, both rings have the same
dimension, and so an argument similar but easier as the above gives:
5.3. Corollary. Nagata extensions do not change the length, that is to say, lenpRq “
lenpRptqq, for any Noetherian local ring R and any tuple of variables t.

6. E QUILATERAL SEQUENCES
Let us call an exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0 equilateral, if the length
of M is equal to lenpN q ‘ lenpQq (the upperbound in (9)); if this is moreover also
equal to lenpQq i lenpN q (the lowerbound in (9)), then we call it strongly equilateral. Similarly, we say that 0 Ñ N Ñ M Ñ Q Ñ 0 is equilateral up to degree r, if
lenpN qěr ‘ lenpQqěr “ lenpM qěr , that is to say, if oi pN q ` oi pQq “ oi pM q, for all
i ě r. Immediately from Proposition 4.11, we get:
6.1. Corollary. An exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0 is equilateral up to degree r,
where r is the minimum of dim N and dim Q ` 1.

By Theorem 3.1, split exact sequences are equilateral. More generally, suppose we can
embed N ‘ Q into M , then, if 0 Ñ N Ñ M Ñ Q Ñ 0 is exact, it is equilateral: semiadditivity gives lenpM q ď lenpN q ‘ lenpQq whereas N ‘ Q ãÑ M gives lenpQq ‘
lenpN q ď lenpM q.
6.2. Proposition. An exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0 such that dim Q ă ordpN q
is equilateral.
Proof. Let n, m, and q be the lengths of N , M , and Q respectively. By Theorem 4.8, we
have n ĺ m and q ĺ m, whence n _ q ĺ m, where n _ q denotes the ĺ-supremum of n
and q. However, since the degree of q is less than the order of n, we have n _ q “ n ‘ q.
Semi-additivity, on the other hand, yields the converse inequality m ď n ‘ q.

Note that such a sequence will never be strongly equilateral. In fact, being strongly
equilateral is really a property of ordinals: a i b “ a ‘ b if and only if the degree of b is
at most the order of a, and hence
6.3. Corollary. An exact sequence of finitely generated R-modules
0 Ñ N Ñ M Ñ Q Ñ 0,
is strongly equilateral if and only if dim N ď ordpQq.



Immediately from semi-additivity, the left exactness of local cohomology and Theorem 4.1, we get:
f

6.4. Proposition. A short exact sequence 0 Ñ N Ñ M ÝÝÑ Q Ñ 0 is equilateral if
and only if f is a cohomological epimorphism, meaning that fp : H0p pM q Ñ H0p pQq is
surjective, for all prime ideals p.


14

HANS SCHOUTENS

In this paper,2 we will call a module M unmixed of dimension d if all its associated
primes have dimension d. Theorem 4.1 then shows that M is unmixed if and only if
ordpM q “ dim M , in which case lenpM q “ od pM qω d . Corollary 6.3 yields:
6.5. Corollary. Let R be a d-dimensional Noetherian ring and 0 Ñ N Ñ M Ñ Q Ñ 0
a short exact sequence. If Q is unmixed of dimension d, then this sequence is strongly
equilateral.

Recall the dimension filtration d0 pM q Ď d1 pM q Ď ¨ ¨ ¨ Ď dd pM q “ M of a ddimensional finitely generated R-module M defined by Schenzel in [16], where di pM q
is the submodule of all elements of dimension at most i, and where we define the dimension of an element x P M as the dimension of the module it generates, that is to say,
dimpR{ AnnR pxqq. Equivalently, di pM q is the largest submodule of M of dimension at
most i.
6.6. Proposition. Given a d-dimensional module M , the exact sequence
0 Ñ di pM q Ñ M Ñ M {di pM q Ñ 0
is strongly equilateral, for each i. In particular, lenpdi pM qq “ lenpM qďi , for all i.
Proof. It follows from [16, Corollary 2.3] that the associated primes of M {di pM q are
precisely the associated primes of M of dimension strictly larger than i. By Corollary 4.5,
this means that lenpM {di pM qq has order at least i ` 1. Since lenpdi pM qq has degree at
most i by Theorem 3.4, the result follows from Corollary 6.3.

Another way to formulate this result is as the following formula for calculating length
(21)

lenpM q “

d
à

lenpdi pM q{di´1 pM qq,

i“0

and each non-zero di pM q{di´1 pM q is unmixed of dimension i and of length ai ω i , where
ai is its generic length.
6.7. Example. Let R be the coordinate ring of a plane with an embedded line inside three
dimensional space over k given by the equations x2 “ xy “ 0 in the three variables x, y, z.
Using Theorem 4.1, one easily calculates that lenpRq “ ω 2 ‘ ω, where the associated
primes are p “ pxq and q “ px, yq. The ideals of length ω are exactly those contained in
p. The ideals of length ω 2 ‘ ω (the open ideals in the terminology from §8), are precisely
those that contain a non-zero multiple of x and a non-zero multiple of y (this follows, for
instance, from [17, Proposition 3.10]). Finally, the remaining (non-zero) ideals of length
ω 2 , are those contained in q but disjoint from p (note that if I is not contained in q, then
IRp “ Rp and IRq “ Rq , so that I must be open).
In a future paper, we will use equilateral sequences to define a (new) Grothendieck
group on the category of finitely generated R-modules, namely, let Geq pRq be the free
Abelian group on isomorphism classes rM s modulo the relations rM s ´ rN s ´ rQs, for
every equilateral exact sequence 0 Ñ N Ñ M Ñ Q Ñ 0. Sending a class of a module
M in Geq pRq to its length, is a surjective semi-group homomorphism, when we view the
ordinals as a semi-group with respect to the shuffle sum ‘, and hence in particular, Geq pRq
is non-trivial.
2Our terminology differs slightly from the literature, where, at least for local rings, the condition is formulated
over the completion.

THE THEORY OF ORDINAL LENGTH

7. C OMPOSITIONS

15

SERIES

We will call a strictly descending chain C of submodules of M simply a chain in M .
When discussing composition series, we may assume that a chain has a first and last element, so that we can represent it as C :“ pMa |a ď rq. Since C – r ‘1, formula (5) then
yields lenpCq “ r. The factor module Ma {Ma‘1 will be called the a-th cokernel of C. Our
goal is to show that each module M admits a composition series C, that is to say, a chain
C such that lenpM q “ lenpCq. In other words, a partial well-order as in Example 2.4 can
never be a Grassmanian.
7.1. Theorem. Every finitely generated module over a Noetherian ring of finite Krull dimension admits a composition series.
Proof. We induct on the length m :“ lenpM q. For m ă ω, this is just the classical JordanHolder theorem for modules of finite length (note that the length of a finite chain is defined
to be one less than the number of members in the chain, explaining why we have to take the
length of the ordinal, not the ordinal itself). So assume m ě ω. If m has valence at least two,
we can write it as m̄ ‘ ω e , where e is the order of m. Let N be a submodule of height rank m̄,
so that M̄ :“ M {N has length m̄ by (8). If n :“ lenpN q, then by semi-additivity, we get
m̄ i n ď m̄ ‘ ω e ď m̄ ‘ n. The latter implies that ω e ď n, and by the former inequality, it
¯ “ lenpM̄ q.
cannot be bigger either. By induction we can find a chain C¯ in M̄ with lenpCq
¯
Taking the pre-image of each module in C in M gives a chain C in M , of length m̄, in which
each term contains N . Induction also gives a chain D in N with lenpDq “ lenpN q “ ω e .
The union C Y D is therefore equal to the chain C ` D, which by Proposition 2.7 has length
lenpCq i lenpDq “ m̄ ‘ ω e “ m, showing that it is a composition series in M .
So remains the case that m “ ω d . We induct this time on d, where the case d “ 1
is classical: if there were no infinite chains, then M is both Artinian and Noetherian,
whence of finite length ([2, Proposition 6.8]). Put M0 :“ M and choose a submodule
M1 of height rank ω d´1 . Applying the induction hypothesis to the quotient M {M1 , which
has length ω d´1 , we can find, as above, a chain C1 of submodules containing M1 , of
length ω d´1 . By semi-additivity, the length of M1 is again ω d . Choose a submodule M2
of M1 of height rank ω d´1 in GrasspM1 q, and as before, find a ω d´1 -chain C2 in this
Grassmanian of submodules containing M2 . Continuing in this manner, we get an ω-chain
M0 Ń M1 Ń M2 Ń . . . and ω d´1 -chains Cn from Mn down to Mn`1 . The union of all
these chains is therefore a chain of length ω d , as we needed to construct.

7.2. Remark. We could ask for additional properties of a composition series (which would
hold automatically for finite ones), such as being maximal, etc. By Zorn’s lemma, any
composition series can be refined to a maximal one, but, as Example 7.3 below shows,
being maximal is not a sufficient condition for being a composition series. A more subtle
question is the existence of a composition series pMa |a ď lenpM qq with the property that
hpMa q “ a for all a. In a future paper, I will show that they exist for excellent, reduced
Jacobson rings and for monomial algebras.
7.3. Example. Let R :“ krx, y, zs{zpx, yq. Its associated primes are p “ pzq and q “
px, yq, and both localizations are fields, so that lenpRq “ ω 2 ‘ω by Theorem 4.1. Consider
the ω-chain ai :“ px, y, z i q with intersection equal to aω “ px, yq, and continue now with a
(continuous) ω 2 -chain api`1qω‘j :“ pxj`1 y i , y i`1 q, with intersection equal to aω2 “ p0q.
Hence the total chain has length ω 2 and is maximal, but it is too short to be a composition
series.

16

HANS SCHOUTENS

8. O PEN

SUBMODULES

By semi-additivity, lenpN q is at most lenpM q, and, in fact, lenpN q ĺ lenpM q by
Theorem 4.8. If M has finite length and N is a proper submodule, then obviously its length
must be strictly less, but in the non-Artinian case, nothing excludes this from being an
equality. So, we call N Ď M open, if lenpN q “ lenpM q. Immediately from Theorem 4.1
and Theorem 4.8, we get
8.1. Corollary. A submodule N Ď M is open if and only if H0pRp pNp q “ H0pRp pMp q for
all p, if and only if cycR pN q “ cycR pM q, if and only if valpN q “ valpM q.

8.2. Remark. In particular, the long exact sequence of local cohomology yields that N Ď
M is open if and only if the canonical morphism H0pRp pMp {Np q Ñ H1pRp pNp q is injective,
for every (associated) prime p (of M ).
8.3. Proposition. Given an exact sequence
0 Ñ N Ñ M Ñ Q Ñ 0,
if dimpQq ă ordpM q, then N is open. In particular, any non-zero ideal in a domain is
open, and more generally, any ideal in an unmixed ring containing a parameter is open.
ř
ř
ř
Proof. Let n “ ai ω i , m “ bi ω i , and q “ ci ω i be the respective lengths of N , M
and Q. By semi-additivity, m ď q ‘ n, whence bi ď ai ` ci , for all i. By assumption,
ci “ 0 whenever bi ‰ 0, so that in fact bi ď ai , for all i, that is to say, m ĺ n. Since the
other inequality always holds, N is open. To prove the last assertion, let x be a parameter
in a d-dimensional unmixed ring R, so that R{xR has dimension at most d ´ 1. Since
ordpRq “ d by Theorem 4.1, our first assertion shows that the ideal pxq, and hence any
ideal containing x, is open.

Let us call a submodule N Ď M equilateral, if 0 Ñ N Ñ M Ñ M {N Ñ 0 is
equilateral, that is to say, by (8), if lenpM q “ lenpN q ‘ hpN q. Hence a direct summand
is equilateral by semi-additivity. By Corollary 6.3, any submodule N such that dim N ď
ordpM {N q, is equilateral, but the converse need not hold.
8.4. Proposition. A maximal (proper) submodule is either equilateral or open. In particular, if M has positive order, then any maximal submodule is open.
Proof. Let N Ł M be maximal, so that Q :“ M {N is simple, of length one. Let n and
m be the respective lengths of N and M . By semi-additivity, we have n ď m ď n ‘ 1. If
the former inequality holds, the submodule is open, and if the latter holds, it is equilateral.
The latter case is excluded when m is a limit ordinal, that is to say, when M has positive
order.

In the ring case, we can even prove:
8.5. Proposition. If pR, mq is a non-Artinian local ring, then m is open.
Proof. Let m and r be the respective lengths of m and R. We induct on r. In view of
Proposition 8.4, to rule out that m is equilateral, we may assume that it is an associated
prime. Choose x P R with Annpxq “ m and put R̄ :“ R{xR. Since xR has length
one, Corollary 6.3 applied to the exact sequences 0 Ñ xR Ñ R Ñ R̄ Ñ 0 and 0 Ñ
xR Ñ m Ñ mR̄ Ñ 0 yields r “ lenpR̄q ‘ 1 and m “ lenpmR̄q ‘ 1. By induction,
lenpmR̄q “ lenpR̄q, and hence m “ r.


THE THEORY OF ORDINAL LENGTH

17

8.6. Remark. As for primary ideals n, they will not be open in general if R has depth zero.
More precisely, suppose lenpRq “ r ‘n with r a limit ordinal and n P N. If n H0m pRq “ 0
(which will be the case if lenpR{nq ě n), then lenpnq “ r. Indeed, the case n “ 0 is
trivial, and we may always reduce to this since n is a module over R{ H0R pmq, and the latter
has length r by Proposition 8.3.
8.7. Corollary. If N Ď M is open, then SupppM {N q is nowhere dense in SupppM q. The
converse holds if M has no embedded primes.
Proof. Let Q :“ M {N . The condition on the supports means that no minimal prime p of
M lies in the support of Q. However, for such p, we have Mp “ H0pRp pMp q. Hence Np “
Mp by Corollary 8.1, showing that Qp “ 0. Suppose next that M has no embedded primes
and SupppQq is nowhere dense in SupppM q. Let D :“ cycpN q ` cycpQq ´ cycpM q as
given by Corollary 4.4, and let p be an associated prime of Q. Since SupppQq is nowhere
dense, p is not an associated prime of M . Since the cycle cycpM q ´ cycpN q has support
in AsspM q and is equal to cycpQq ´ D, it must be the zero cycle, since cycpQq and D

have support disjoint from AsspM q. Hence N Ď M is open by Corollary 8.1.
It is useful to reformulate the above equivalence for modules without embedded primes:
N Ď M is open if and only if AsspM q X AsspM {N q “ H. Corollary 8.7 together with
Proposition 8.3 yields:
8.8. Corollary. If N Ď M is open then dimpM {N q ă dimpM q, and the converse holds
if M is unmixed.

8.9. Proposition. Let M be a d-dimensional module. The following are equivalent for a
submodule N Ď M :
(8.9.a) N is open in M ;
(8.9.b) dim M {N ă d and dd´1 pN q is open in dd´1 pM q;
(8.9.c) dim di pM q{di pN q ă i, for all i.
Proof. Since di pN q “ di pM q X N , assumption (8.9.a) implies the other two by Corollary 8.8 and Theorem 9.1 below. Assume (8.9.b) holds. By the first of these two conditions
and top additivity applied to 0 Ñ N Ñ M Ñ M {N Ñ 0, we get od pM q “ od pN q.
By Proposition 6.6, the second condition yields lenpN qăd “ lenpM qăd , and hence
lenpN q “ lenpM q, that is to say, N is open. Finally, assuming (8.9.c), let us prove
by induction on i that di pN q is open in di pM q, where the case i “ d gives the desired
result. If i “ 0, the assumption states that d0 pN q and d0 pM q are actually equal, and so
we may assume i ą 0. Applying the first equivalence to the inclusion di pN q Ď di pM q
and using that by induction di´1 pN q is open in di´1 pM q, we see that the former is open
too.

In some cases, we have to check less:
8.10. Proposition. Let I be an ideal in a d-dimensional ring R. If dim R{I ă d and I
contains an ideal J such that R{J is unmixed of dimension d, then I is open.
Proof. Let lenpRq “ eω d ‘a with a of degree at most d´1. By assumption, lenpR{Jq “
bω d , for some non-zero b. Since R{I has dimension less than d, top additivity (Corollary 4.12) applied to 0 Ñ I Ñ R Ñ R{I Ñ 0 yields od pIq “ od pRq “ e. Hence
lenpIq “ eω d ‘ b, for some ordinal b of degree at most d ´ 1. On the other hand,

18

HANS SCHOUTENS

by Corollary 6.3, the exact sequence 0 Ñ J Ñ R Ñ R{J Ñ 0 is equilateral, so that
lenpRq “ lenpJq ‘ bω d , and hence lenpJq “ pe ´ bqω d ‘ a. By Theorem 4.8, we have
lenpJq “ pe ´ bqω d ‘ a ĺ lenpIq “ eω d ‘ b ĺ eω d ‘ a “ lenpRq.
Therefore, a “ b, showing that I is open.



8.11. Remark. The module version is proven exactly in the same way: given a module M
and submodules K Ď N such that M {K is unmixed of dimension d :“ dim M and M {N
has dimension strictly less than d, then N is open. Note that M {K is unmixed if and only if
dd´1 pM q Ď K, and so, an equivalent criterion is that dim M {N ă d and dd´1 pM q Ď N ,
then N is open. The converse may fail: in R “ krx, y, zs{px2, xyq, the ideal pxz, yq is
open by [17, Proposition 3.10], but it does not contain d1 pRq “ pxq.
8.12. Corollary. In an equidimensional ring R, a prime ideal is open if and only if it is
non-minimal. If, moreover, dim R ą 0, then R admits a composition series consisting
entirely of height one ideals (and the zero ideal).
Proof. Let d “ dim R, let p be a non-minimal prime ideal, and let g be a minimal prime
ideal contained in p. Hence lenpR{gq “ ω d and dim R{p ă d, so that p is open by
Proposition 8.10. On the other hand, g is not open by Corollary 8.8, proving the converse.
Let p be a height one prime. By what we just proved, lenppq “ lenpRq, and hence,
there exists a composition series inside p by Theorem 7.1, the non-zero members of which
therefore have height one.

We conclude with some applications of open submodules to morphisms (more will be
proven in §11 below):
8.13. Corollary. If dimpM q ă ordpN q, then HomR pM, N q “ 0.
Proof. Let d ă e be the respective dimension of M and order of N , and let x P M . Since
x has dimension at most d, so does f pxq. Hence f pxq must be zero, since de´1 pN q “ 0
by Proposition 6.6.

8.14. Theorem. If N is an open submodule of an unmixed module M , then two endomorphisms of M are equal if and only if their restriction to N are equal.
Proof. Since N is open, M {N has dimension strictly less than the dimension, whence the
order, of M by Corollary 8.8. Hence HompM {N, M q “ 0 by Corollary 8.13. From the
canonical exact sequence
0 Ñ HompM {N, N q Ñ EndpM q Ñ HompN, M q
we get an inclusion EndpM q Ñ HompN, M q, from which our claim follows.
9. T HE CANONICAL



TOPOLOGY

As the name indicates, there is an underlying topology. To prove this, we need:
9.1. Theorem. The inverse image of an open submodule under a morphism is again open.
In particular, if U Ď M is open, and N Ď M is arbitrary, then N X U is open in N .
Proof. We start with proving the second assertion. Let m :“ lenpM q, n :“ lenpN q and
a :“ lenpN X U q. We have an exact sequence
0 Ñ N X U Ñ N ‘ U Ñ N ` U Ñ 0.

THE THEORY OF ORDINAL LENGTH

19

Since N ` U is again open, semi-additivity yields lenpN ‘ U q ď a ‘ m. Since the former
is equal to n ‘ m, we get n ď a, showing that N X U is open in N . To prove the first
assertion, let f : M Ñ N be an arbitrary homomorphism and let V Ď N be an open
submodule. Let G Ď M ‘ N be the graph of f and let p : G Ñ M be the projection onto
the first coordinate. Since M ‘ V is open in M ‘ N by semi-additivity, pM ‘ V q X G is
open in G, by what we just proved. Since p is an isomorphism, the image of pM ‘ V q X G
under p is therefore open in M . But this image is just f ´1 pV q, and so we are done.

We can now define a topology on M by letting the collection of open submodules be
a basis of open neighborhoods of 0 P M . This is indeed a basis since the intersection
of finitely many opens is again open by Theorem 9.1. An arbitrary open in this topology
is then a (possibly infinite) union of cosets x ` U with U Ď M open and x P M . If a
submodule N is a union of cosets xi ` Ui of open submodules Ui Ď M , then one such
coset, x ` U say, must contain 0, so that U Ď N , and hence the inequalities lenpU q ď
lenpN q ď lenpM q are all equalities, showing that N is indeed open in the previous sense.
We call this the canonical topology on M , and Theorem 9.1 shows that any homomorphism is continuous in the canonical topology. Moreover, multiplication on any ring is
continuous: given a1 , a2 P R and an open ideal I such that a1 a2 P I, let Ji :“ ai R ` I.
Hence Ji is an open neighborhood of ai and J1 ¨ J2 Ď a1 a2 R ` I “ I. If M has dimension
zero, then the canonical topology is trivial, since M is then the only open submodule. The
complement of an open module N is the union of all cosets a`N with a R N , and hence is
also open. In particular, an open module is also closed, and the quotient topology on M {N
is discrete, whence in general different from the canonical topology. The zero module is
closed if and only if the intersection of all open submodules is 0, that is to say, if and only
if the canonical topology is Haussdorf.
9.2. Corollary. A module is non-Artinian if and only if its canonical topology is non-trivial.
Proof. One direction is immediate since an Artinian module has no proper open submodules. For the converse, we show, by induction on lenpM q, that M has a non-trivial, open
submodule. Assume first that m is a limit ordinal. Choose a submodule N of M of height
rank 1. By (8), this means lenpM {N q “ 1, and so N is open by Proposition 8.3. Next, assume m “ n ‘ 1. Let H be a submodule of height rank n, so that by (8) again, M̄ :“ M {H
has length n. By induction, we can find a proper open submodule of M̄ , that is to say, we
can find N Ł M containing H such that lenpN {Hq “ n. Semi-additivity applied to the
inclusion H Ď N yields ni lenpHq ď lenpN q. Since lenpHq ‰ 0 and lenpN q ď n‘ 1,
we get equality, that is to say, lenpN q “ n ‘ 1, whence N is open.

9.3. Proposition. The closure of a submodule N Ď M is equal to N ` d0 pM q. In particular, a submodule is closed if and only if o0 pN q “ o0 pM q.
Proof. Let H :“ d0 pM q, and let W be an open submodule containing N . By Theorem 9.1,
the intersection H X W is open in H, and since H has finite length (so that its topology is
trivial), we must have H “ W X H, proving that H lies in W . As this holds for all opens
W containing N , the closure of N contains H.
Using Corollary 6.3, one easily shows that the quotient topology on M {H is equal to
the canonical topology. Therefore, to calculate the closure, we may divide out H, assume
that M has positive order, and we then need to show that N is closed. Let x P M be any
element not in N and let m be a maximal ideal containing pN : xq. Since M {mk M has
finite length, each mk M is open by Proposition 8.3, whence so is each N ` mk M . If x is
contained in each of these, then it is contained in their intersection W :“ XpN ` mk M q.

20

HANS SCHOUTENS

By Krull’s Intersection theorem, there exists a P m such that p1 ` aqW Ď N . In particular,
1 ` a P pN : xq Ď m, contradiction. So x lies outside some open N ` mk M , and hence
does not belong to the closure of N . The last assertion is now also clear by Theorem 4.1,
since o0 pM q is the length of d0 pM q.

9.4. Corollary. A module has a separated canonical topology if and only if its order is
positive if and only if any submodule is closed.

9.5. Example. If pR, mq is local and M has positive depth, then the canonical topology
refines the m-adic topology on M : indeed mk M is then open by Proposition 8.3. We
already showed that this is no longer true in rings of positive depth in Remark 8.6. Also
note that the canonical topology on a local domain is strictly finer than its adic topology,
since all non-zero ideals are open.
The ring krrx, yss{px2 , xyq, of length ω ‘ 1, is not Haussdorf as the closure of the zero
ideal is the ideal pxq by Proposition 9.3. It is the only closed, non-open ideal, since the
closure of any ideal must contain x whence is open when different from pxq. In particular,
whereas the canonical topology is not Haussdorf, the adic one is.
Recall that a submodule N Ď M is called essential (or large), if it intersects any nonzero submodule non-trivially.
9.6. Corollary. An open submodule is essential. In particular, 0 is a limit point of any
non-zero submodule.
Proof. Let N Ď M be open and H Ď M arbitrary. Suppose H X N “ 0, so that H ‘ N
embeds as a submodule of M . In particular, lenpHq ‘ lenpN q ď lenpM q by semiadditivity, forcing lenpHq, whence H, to be zero.
To prove the second assertion, let N be non-zero and let U be an open containing 0.
Since U is the union of cosets of open submodules and contains 0, it must contain at least
one open submodule W . Since W is essential by our first assertion, W X N ‰ 0.

The converse is false: in an Artinian local ring, the socle is essential, but it is clearly not
open. A less trivial example is given by the ideal p :“ px, yq in the ring R “ krx, y, zs{p2,
which is essential but not open. Indeed, cycpRq “ 3rps and hence R has length 3ω by
Theorem 4.1. Since R{p is a one-dimensional domain, its length is ω by Theorem 6.3, and
hence lenppq “ 2ω by Corollary 6.3. To see that p is essential, we can use the following
proposition with S “ krx, ys{p2 (so that R “ Srzs).
9.7. Proposition. Let pS, pq be a local ring and S Ñ R a flat extension. Then pR is an
essential ideal of R.
Proof. If pR were not essential, we could find a non-zero x P R such that pR X xR “ 0.
In particular, xp “ 0. By flatness, x P AnnS ppqR Ď pR, contradiction.

9.8. Theorem. If M has no embedded primes, then the induced topology on a submodule
N Ď M is the same as the canonical topology on N .
Proof. In view of Theorem 9.1, we only have to show that if W Ď N is open, then there
exists U Ď M open such that W “ U X N . Let U be maximal such that U X N “ W .
Suppose U is not open, so that there exists an associated prime p of M with H0pRp pUp q Ł
H0pRp pMp q by Corollary 8.1. In particular, we can find x P M zU with spx Ď U , for
some s R p. By maximality, pU ` Rxq X N must contain an element n not in W . Write
n “ u ` rx with u P U and r P R. In particular, spn “ spu lies in U X N “ W . In other

THE THEORY OF ORDINAL LENGTH

21

words, we showed that H0pRp pNp {Wp q ‰ 0. In particular, Np ‰ 0, so that p is a minimal
prime of N . By Lemma 4.3, we have op pW q ă op pW q ` op pW {N q “ op pN q, so that
lenpW q ă lenpN q by Theorem 4.1, contradiction.

I do not know whether this remains still true if there are embedded primes, but see [18,
Corollary 4.4].
10. A PPLICATION I: ACYCLICITY
For the remainder of this paper R is a d-dimensional Noetherian ring. Furthermore, M ,
N , . . . are finitely generated modules over R, of length m, n, etc. We start with reproving the
observation of Vasconcelos [22] that a surjective endomorphism on a Noetherian module
must be an isomorphism (the usual proof uses the determinant trick; see for instance [11,
Theorem 2.4]).
10.1. Corollary. Any surjective endomorphism is an isomorphism.
Proof. Let M Ñ M be a surjective endomorphism with kernel N , so that we have an
exact sequence 0 Ñ N Ñ M Ñ M Ñ 0, and therefore, by Theorem 3.1, an inequality
lenpM q i lenpN q ď lenpM q. By simple ordinal arithmetic, this implies lenpN q “ 0,
whence N “ 0.

10.2. Remark. Our argument in fact proves that any surjection between modules of the
same length must be an isomorphism, or more generally, if f : M Ñ N is an epimorphism and lenpM q ď lenpN q, then f is an isomorphism and lenpM q “ lenpN q. More
generally, endomorphisms are non-expansive in the following sense (note that by taking
N “ M , we recover Vasconcelos’ result):
10.3. Theorem. Let f be an endomorphism on a Noetherian module M and let N be
a submodule such that N Ď f pN q. Then f pN q “ N and the restriction f |N is an
automorphism of N .
Proof. Since N Ď f pN q, we have lenpN q ď lenpf pN qq. On the other hand, f induces a surjection N Ñ f pN q, showing that lenpf pN qq ď lenpN q. Hence lenpN q “
lenpf pN qq, implying by (8) that the surjection N Ñ f pN q is an isomorphism. Now,
from N Ď f pN q, we get f pN q Ď f 2 pN q, and applying the same argument to the submodule f pN q, shows that f is injective on f pN q. Repeating in this way, we see that f is
injective on each f n pN q. Suppose now that the inclusion N Ď f pN q is strict, so that we
can find a P N with f paq R N . Suppose for some n, we have f n paq P f n´1 pN q, say,
f n paq “ f n´1 pbq with b P N . Since f is injective on each f k pN q, we get b “ f paq, contradiction. Hence the chain N Ł f pN q Ł f 2 pN q Ł . . . is strictly ascending, contradicting
Noetherianity. In conclusion, N “ f pN q and the assertion follows.

10.4. Corollary. If N is a homomorphic image of M which contains a submodule isomorphic to M , then M – N .
Proof. Since M ãÑ N , semi-additivity yields lenpM q ď lenpN q. By Remark 10.2, the
epimorphism M ։ N must then be an isomorphism.

The following result generalizes Miyata’s result [13] as we do not need to assume that
the given sequence is left exact.
10.5. Theorem. An exact sequence M Ñ N Ñ C Ñ 0 is split exact if and only if the
(abstract) modules N and M ‘ C are isomorphic.

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HANS SCHOUTENS

Proof. One direction is just the definition of split exact. Let M̄ be the image of M and apply Theorem 3.1 to 0 Ñ M̄ Ñ N Ñ C Ñ 0 to get lenpN q ď lenpCq ‘ lenpM̄ q. On the
other hand, N – M ‘ C yields lenpN q “ lenpM q ‘ lenpCq, whence lenpM q ď
lenpM̄ q. Since M̄ is a homomorphic image of M , they must be isomorphic by Remark 10.2. Hence, we showed M Ñ N is injective. At this point we could invoke [13],
but we can as easily give a direct proof of splitness as follows. Given a finitely generated
R-module H, since M b H – pN b Hq ‘ pC b Hq, the same argument applied to the
tensored exact sequence
M b H Ñ N b H Ñ C b H Ñ 0,
gives the injectivity of the first arrow. We therefore showed that M Ñ N is pure, whence
split by [11, Theorem 7.14].

10.6. Remark. Independently, Striuli proved the same result in [21]. However, as she only
proves exactness in the local case, she still needs to invoke Miyata’s original result, whereas
our proof stands alone.
10.7. Theorem. Let X be a non-singular variety over an algebraically closed field k.
Then a closed subscheme Y Ď X with ideal of definition I is non-singular if and only if
ΩX{k b OY is locally isomorphic to I{I 2 ‘ ΩY {k .
Proof. Since X is non-singular, its module of differentials ΩX{k is locally free ([8, Theorem 8.15]), whence so is ΩX{k b OY , and therefore so is its direct summand ΩY {k .
Moreover, by Theorem 10.5, the conormal sequence
I{I 2 Ñ ΩX{k b OY Ñ ΩY {k Ñ 0
is then split exact, and the result now follows from [8, Theorem 8.17].



10.8. Theorem. Let A be a finitely generated R-algebra, I Ď A an ideal, and Ā :“ A{I.
The closed immersion Spec Ā Ď SpecpA{I 2 q is a retract over R if and only if we have an
(abstract) isomorphism of A-modules
L
(22)
ΩA{R IΩA{R – ΩĀ{R ‘ I{I 2 .
Proof. One direction is easy, and if (22) holds, then the conormal sequence
L
I{I 2 Ñ ΩA{R IΩA{R Ñ ΩĀ{R Ñ 0
is split exact by Theorem 10.5, so that the result follows from [4, Proposition 16.12].



Given a module M and an element f P R, we say that f is a parameter on M , if
M {f M is non-zero but has dimension strictly less than M . For a local ring pR, mq, an
element f P m is a parameter if and only if dim R{f R “ dim R ´ 1, but this does not
necessarily hold in non-local rings, like krrxssrys with f “ xy ´ 1.
10.9. Theorem. An element f P R is a parameter on M if and only if the dimension of
AnnM pf q as a module is less than that of M .
Recall that AnnM pf q is the submodule of all y P M such that f y “ 0.
Proof. Let d :“ dim M . Replacing R by R{ AnnpM q, we may assume d “ dim R. Put
M̄ :“ M {f M and consider the exact sequence 0 Ñ f M Ñ M Ñ M̄ Ñ 0. It follows
from top additivity (Corollary 4.12) that dim M̄ ă d, if and only if M and f M have the
same generic length, that is to say,
(23)

od pM q “ od pf M q.

THE THEORY OF ORDINAL LENGTH

23

Applying top-additivity instead to the exact sequence 0 Ñ AnnM pf q Ñ M Ñ f M Ñ 0,
we see that (23) in turn is equivalent with od pAnnM pf qq “ 0, meaning that AnnM pf q has
dimension at most d ´ 1.

11. A PPLICATION II: D EGRADATION
By degradation, we mean the effect that source and target of a morphism have on its
kernel. We start with a general observation about kernels: given two R-modules M and N ,
let us denote the subset of GrassR pM q consisting of all kerpf q, where f P HomR pM, N q
runs over all morphisms, by kerR pM, N q.
11.1. Theorem. As a subset of GrassR pM q, the ordered set kerR pM, N q has finite length.
Proof. Let f : M Ñ N be a morphism, and let q be the length of its image. By Theorem 4.8, we have q ĺ lenpN q. In particular, there are at most 2valpN q possibilities for q.
I claim that if g : M Ñ N is a second morphism and kerpgq Ł kerpf q, then lenpImpgqq
is strictly bigger than q. From this claim it then follows that any chain in kerR pM, N q
has length at most 2valpN q . To prove the claim, we have hpkerpgqq “ lenpN { kerpgqq “
lenpImpgqq, by (8). By assumption, kerpgq is strictly contained in kerpf q, and hence it
has strictly bigger height rank, showing the claim.

11.2. Corollary. If N is univalent (i.e., valpN q “ 1), then there are no inclusion relations
among the kernels of non-zero morphisms M Ñ N . In particular, for each K Ď M , the
set HK of morphisms M Ñ N with kernel equal to K together with the zero morphism, is
a submodule of HomR pM, N q.
Proof. By the proof of Theorem 11.1, there are only two possibilities for the height rank of
a kernel, one of which is zero (given by the zero morphism). Assume f, g P HK . Since K
then lies in the kernel of rf ` sg, for any r, s P R, the latter kernel is either K or M . 
For each v, let Grassv pM q be the subset of the Grassmanian GrasspM q consisting
of all submodules for which M {N has valence at most v. The same argument shows
that each Grassv pM q has finite length: indeed, for N P Grassv pM q, we have hpN q ĺ
vω d ‘ vω d´1 ‘ . . . ‘ v, where d is the dimension of M , and therefore, we only have
finitely many possibilities for hpN q. Note that the union of the Grassv pM q is GrasspM q,
so that the Grassmanian can be written as a union of suborders of finite length. We can
now list some examples of degradation:
11.3. Corollary. If M and N have no associated primes in common, then any morphism
between them has open kernel.
Proof. Let f : M Ñ N be a morphism and let K Ď M and Q Ď N be its respective kernel
and image, so that we have an exact sequence 0 Ñ K Ñ M Ñ Q Ñ 0. For any associated
prime p of M , we have H0pRp pNp q “ 0 whence also H0pRp pQp q “ 0. By left exactness of

local cohomology, H0pRp pKp q “ H0pRp pMp q, and hence K is open by Corollary 8.1.
11.4. Corollary. Let M be a module without embedded primes. A submodule U Ď M is
open if and only if every homomorphism f : M Ñ M {U has open kernel.
Proof. Sufficiency follows from the discussion following Corollary 8.7 and Corollary 11.3.
As for necessity, this is immediate when applied to the canonical projection M Ñ M {U
whose kernel is precisely U .


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HANS SCHOUTENS

Since endomorphism rings are in general non-commutative, the set of nilpotent elements is not necessarily an ideal, but we can always find a subcollection which is:
11.5. Proposition. Given a module M , let OpM q be the subset of all endomorphisms
whose kernel is open. Then OpM q is a two-sided ideal of EndpM q consisting entirely of
nilpotent endomorphisms.
Proof. Suppose a, b P O :“ OpM q and c P EndpM q, then the kernel of a ` b contains
the open kerpaq X kerpbq and hence is open itself. Likewise, the kernel of ca contains the
open kerpaq, and the kernel of ac is the inverse image of kerpaq under c, which is open by
continuity (Theorem 9.1). This shows that O is a two-sided ideal.
Finally, some power an satisfies kerpan q “ kerpan`1 q by Noetherianity, which implies
kerpan qXImpan q “ 0. Since kerpan q is a fortiori open, whence essential by Corollary 9.6,
the submodule Impan q must be zero, showing that a is nilpotent.

Note that the second part of the proof actually gives:
11.6. Corollary. If an endomorphism has open kernel, then it is nilpotent.



11.7. Theorem. Suppose M is torsion-free over Z. If M and N have no associated primes
in common, then there exists k P N, such that for any choice of morphisms fi : M Ñ N
and gi : N Ñ M , with i “ 1, . . . , k, the composition gk fk ¨ ¨ ¨ g1 f1 “ 0.
Proof. By Proposition 11.5, the ideal O :“ OpM q of endomorphisms with open kernel
contains only nilpotent elements. Hence, by the Nagata-Higman Theorem ([9, 14]), the
ideal O is nilpotent, that is to say, Ok “ 0, for some k. Since the kernel of each fi is
open by Corollary 11.3, so is the kernel of each gi fi , showing that gi fi P O, and the claim
follows.

11.8. Corollary. Suppose Q Ď R. Let e be the maximal dimension of a common associated
prime of M and N . Then there is some k, such that for any choice of morphisms fi : M Ñ
N and gi : N Ñ M , with i “ 1, . . . , k, the image of the composition gk fk ¨ ¨ ¨ g1 f1 has
dimension at most e.
Proof. We take the convention that e “ ´1 if there are no common associated primes
and we assign ´1 to the dimension of the zero module. Hence the assertion is now just
Theorem 11.7 in case e “ ´1. For e ě 0, let M 1 :“ M {de pM q and N 1 :“ N {de pN q.
Since M 1 and N 1 have no associated primes in common by maximality of e, we can find
some k as in Theorem 11.7. Choose k many morphisms fi and gi as in the hypothesis,
and let h be their composition. Since morphisms cannot increase dimension, they induce
morphisms between M 1 and N 1 , and hence, by choice of k, the endomorphism on M 1
induced by h is zero. It follows that hpM q Ď de pM q, as claimed.

11.9. Remark. The torsion restrictions above and below come from our application of the
Nagata-Higman Theorem, which requires some form of torsion-freeness (see [17, Remark
6.5] for a further discussion). One can weaken these assumptions: for instance, in Corollary 11.8, we only need that p X Z “ 0, for any associated prime p of M .
To extend Theorem 11.7 to several modules, let us say that an endomorphism f P
EndpM q reflects through a collection of modules N , if we can factor f as M Ñ N Ñ M ,
for each N P N . Of course, any endomorphism reflects through M itself. We can now
prove:

THE THEORY OF ORDINAL LENGTH

25

11.10. Theorem. Let N be a collection of R-modules such that no prime ideal is associated to every N P N . For any module M P N without Z-torsion, there exists k P N, so
that any product of k-many endomorphisms on M reflecting through N is zero.
Proof. As in the proof of Theorem 11.7, it suffices to show that any endomorphism f P
EndpM q reflecting to N has open kernel. Let K be its kernel and let p be an associated
prime of M . By assumption, there exists N P N such that p is not an associated prime of
N . By definition, there exists a factorization f “ hg with g : M Ñ N and h : N Ñ M .
Let H be the kernel of g. By the argument in the proof of Corollary 11.3 applied to g, we
get H0pRp pHp q “ H0pRp pMp q. Since H Ď K, this implies H0pRp pKp q “ H0pRp pMp q, and
since this holds for all associated primes p of M , we proved by Corollary 8.1 that K is
open.

If, instead, there are common associated primes, let e be the maximum of their dimensions. By the same argument as in the proof of Corollary 11.8, we may then conclude that
the image of any product of k-many endomorphisms reflecting through N has dimension
at most e.
12. A PPENDIX :

SHUFFLE SUMS

Recall that (standard) addition on ordinals is not commutative. We will give three different but equivalent ways of defining a different, commutative addition operation on the
¯ and ‘.
˜ The sum ‘ is also
class of ordinals, which we temporarily will denote as ‘, ‘
known as the natural (Hessenberg) sum and is often denoted #. Recall our convention for
scalar multiplication on the left (see §2.1). Every ordinal a—we no longer restrict to those
of finite degree—can be written as a sum
(24)

a “ an ω nn i . . . i a1 ω n1

where the ni (called the exponents) form a strictly ascending chain of ordinals, that is to
say, n1 ă ¨ ¨ ¨ ă nn , and the ai (called the coefficients) are non-negative integers. This
decomposition (in base ω) is unique if we moreover require that all coefficients ai are nonzero, called the Cantor normal form (in base ω) of a. If (24) is in Cantor normal form, then
we call the highest (respectively, lowest) occurring exponent, the degree (respectively, the
order) of a and we denote these respectively by degpaq :“ nn and ordpaq :“ n1 . Note
that a is a successor ordinal if and only if ordpaq “ 0.
Given a second ordinal b, we may assume that after possibly adding some more exponents, that it can also be written in the form (24), with coefficients bi ě 0 instead of the ai .
We now define
a ‘ b :“ pan ` bn qω nn i . . . i pa1 ` b1 qω n1 .
It follows that a ‘ b is equal to b ‘ a and is greater than or equal to both a i b and b i a.
For instance if a “ ω i 1 then a ‘ a “ 2ω i 2 whereas a i a “ 2ω i 1. In case both
ordinals are finite, a ‘ b “ a i b. It is easy to check that we have the following finite
distributivity property:
(25)

pa ‘ bq i 1 “ pa i 1q ‘ b “ a ‘ pb i 1q.

In fact, this follows from the more general property that pa ‘ bq i q “ pa i qq ‘ b “
a ‘ pb i qq for all q ă ω o`1 , where o is the minimum of ordpaq and ordpbq.
For the second definition, we use transfinite induction on the pairs pa, bq ordered lexi¯ 0 :“ a and 0 ‘
¯ b :“ b so that we may assume a, b ą 0. If
cographically.3 Define a ‘
3As alluded to above, the counterintuitive notation normally adopted for the resulting ordinal is ba.

26

HANS SCHOUTENS

¯ b as pa1 ‘
¯ bq i 1. Simia is a successor ordinal with predecessor a1 , then we define a ‘
1
¯ b as pa ‘
¯ b1 q i 1.
larly, if b is a successor ordinal with predecessor b , then we define a ‘
Note that by transfinite induction, both definitions agree when both a and b are successor
ordinals, so that we have no ambiguity in defining this sum operation when at least one of
the components is a successor ordinal. So remains the case that both are limit ordinals. If
¯ b be equal to the supremum of the d ‘
¯ b for all d ă a.
ordpaq ď ordpbq, then we let a ‘
¯ b be equal to the supremum of
In the remaining case, when ordpaq ą ordpbq, we let a ‘
¯ d for all d ă b. This concludes the definition of ‘.
¯
the a ‘
˜ b as the supremum of all sums a1 i b1 i . . . i an i bn , where
Finally, define a ‘
the supremum is taken over all n and all decompositions a “ a1 i . . . i an and b “
˜ b is the largest possible ordering
b1 i . . . i bn , with ai , bi ordinals. Loosely speaking, a ‘
one can obtain by shuffling pieces of a and b. Since we may take a1 “ 0 “ bn , one checks
˜ b“b‘
˜ a.
that a ‘
¯ b“a‘
˜ b.
12.1. Theorem. For all ordinals a, b we have a ‘ b “ a ‘
¯ b and c̃ :“ a ‘
˜ b. We first prove c “ c̄ by induction on
Proof. Let c :“ a ‘ b, c̄ :“ a ‘
pa, bq (in the lexicographical order). Since the case a “ 0 or b “ 0 is trivial, we may take
a, b ą 0. If a is a successor ordinal with predecessor a1 , then
¯ bq i 1 “ pa1 ‘ bq i 1 “ a ‘ b “ c,
c̄ “ pa1 ‘
where the first equality is by definition, the second by induction and the third by the finite
distributivity property (25). Replacing the role of a and b, the same argument can be used
to treat the case when b is a successor ordinal. So we may assume that both are limit
ordinals. There are again two cases, namely ordpaq ď ordpbq and ordpaq ą ordpbq. By
symmetry, the argument for the second case is similar as for the first, so we will only give
the details for the first case. Write a as a1 i ω o where o :“ ordpaq. By definition, c̄ is
¯ b with d ă a. A cofinal subset of such d are the ones of the form
the supremum of all d ‘
1
o
¯ b for 0 ă q ă ω o . By
a i q with 0 ă q ă ω , so that c̄ is the supremum of all pa1 i qq ‘
induction, c̄ is the supremum of all
(26)

pa1 i qq ‘ b “ pa1 ‘ bq i q,

where the equality holds because o ď ordpbq. Taking the supremum of the ordinals in (26)
for q ă ω o , we get that c̄ “ pa1 ‘ bq i ω o . Using the remark following (25) one checks
that this is just pa1 i ω o q ‘ b “ a ‘ b “ c.
The inequality c ď c̃ is clear using the shuffle of the terms in the Cantor normal
forms (24) for a and b. To finish the proof, we therefore need to show, by induction
on a, that
(27)

a1 i b1 i . . . i an i bn ď c̄,

˜ is commutative,
for all decompositions a “ a1 i . . . i an and b “ b1 i . . . i bn . Since ‘
we may assume ordpaq ď ordpbq and, moreover, that an ą 0. Suppose first that a
is a successor ordinal with predecessor a1 . In particular, an is also a successor ordinal,
¯ bq i 1. Using the decomposition
with predecessor, say, a1n . By definition, c̄ “ pa1 ‘
¯ b.
a1 “ a1 i . . . i an´1 i a1n and induction, we get that a1 i b1 i . . . i bn i a1n ď a1 ‘
Taking successors of both ordinals then yields (27). Hence suppose a is a limit ordinal. Let
q ă an and apply the induction to each d :“ a1 i . . . i an´1 i q, to get
¯ b.
a1 i b1 i . . . i bn´1 i q i bn ď d ‘
Taking suprema of both sides then yields inequality (27).



THE THEORY OF ORDINAL LENGTH

27

We will denote this new sum simply by ‘ and refer to it as the shuffle sum of two
ordinals, in view of its third equivalent form.
Product Orders. The product of two partially ordered sets P and Q is defined to be the
Cartesian product P ˆ Q ordered by the rule pa, bq ď pa1 , b1 q if and only if a ď a1 and
b ď b1 . The map pa, bq ÞÑ pb, aq is an order-preserving bijection between P ˆ Q and
Q ˆ P . It is easy to check that if both P and Q are partial well-orders, then so is P ˆ Q.
12.2. Theorem (Product Formula). Given partial well-orders P and Q, we have an equality lenpP ˆ Qq “ lenpP q ‘ lenpQq.
Proof. We prove the more general fact that
(28)

hpa, bq “ hpaq ‘ hpbq

for all a P P and b P Q, from which the assertion follows by taking suprema over all
elements in P and Q. Note that we have not written superscripts to denote on which
ordered set the height rank is calculated since this is clear from the context. To prove (28),
we may assume by transfinite induction that it holds for all pairs pa1 , b1 q strictly below
pa, bq. Put a :“ hpaq, b :“ hpbq and c :“ hpa, bq. Since hpa, bq “ hpb, aq, via the
isomorphism P ˆ Q – Q ˆ P , we may assume that ordpaq ď ordpbq whenever this
assumption is required (namely, when dealing with limit ordinals). We start with proving
the inequality a ‘ b ď c. If a “ b “ 0 then a and b are minimal elements in respectively P
and Q, whence so is pa, bq in P ˆ Q, that is to say, c “ 0. So we may assume, after perhaps
exchanging P with Q that a ą 0. Suppose a is a successor ordinal with predecessor a1 .
Hence there exists a1 ă a in P with hpa1 q “ a1 . By induction, hpa1 , bq “ a1 ‘ b and
hence c “ hpa, bq is at least pa1 ‘ bq ‘ 1 “ a ‘ b, where the last equality follows from
Theorem 12.1. If a is a limit ordinal, then there exists for each d ă a an element ad ă a
of height rank d. By induction hpad , bq “ d ‘ b and hence hpa, bq is at least a ‘ b by
Theorem 12.1. This concludes the proof that a ‘ b ď c.
For the converse inequality, assume first that c is a successor ordinal with predecessor
c1 . By definition, there exists pa1 , b1 q ă pa, bq in P ˆ Q of height rank c1 . By induction,
c1 “ hpa1 , b1 q “ hpa1 q ‘ hpb1 q, from which we get c ď a ‘ b. A similar argument can be
used to treat the limit case and the details are left to the reader.

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10016, USA

