# THE FINITENESS OF / WHEN R[x]/I IS FLAT

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TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 171, September
1972
THE FINITENESS OF / WHEN R[x]/I IS FLAT
BY
JACK OHM(7 AND DAVID E. RUSH
ABSTRACT.
Let J? be a commutative
ring with identity,
let X be an indeterminate,
and let / be an ideal of the polynomial
ring /?[X]. Let min / denote the
set of elements
of I of minimal degree and assume henceforth
that min / contains
a regular
element.
Then /?[Xj//
is a flat /?-module
implies
7 is a finitely
generated ideal.
Under the additional
hypothesis
that R is quasi-local
integrally
closed,
the stronger
conclusion
that / is principal
holds.
(An example
shows that
the first statement
is no longer valid when min / does not contain
a regular
element.)
Let c(7) denote
the content
ideal of 7, i.e. c(7) is the ideal of R generated
by the coefficients
of the elements
of /. A corollary
to the above theorem
asserts
that 7?[Xj// is a flat R-module
if and only if / is an invertible
ideal of R[X] and
c(/) b R. Moreover,
if R is quasi-local
integrally
closed,
then the following
are
equivalent:
(i) /?[X|/7
is a flat T&icirc;-module;
(ii) R[X]/I
is a torsion
free R-module
and c(/) = R; (iii) / is principal
and c(/) = R.
Let zf denote the equivalence
class
of X in R[X]/I,
and let (1, f, •••, &pound; &gt;
denote the T&icirc;-module generated
by 1, &pound;, •••, &pound; . The following
statements
are
also equivalent:
(i) (1, &lt;f, ---, &lt;?') is flat forall
t &gt; 0; (ii) (1, &lt;f, ••-, &lt;?') is flat
for some t &gt; 0 for which
1, &pound;, ,-••, &iquest;s are linearly
dependent
over R; (iii) / =
(ft,'&quot;,
fn&gt;, fi e mm /&gt; and c(/) = R; (iv) c(min /) = R. Moreover,
if R is integrally closed,
these are equivalent
to R[X\/1
being a flat 7?-module.
A certain
symmetry
enters
in when &lt;f is regular
in R[^ \, and in this case
(i)—(iv)
are also
equivalent
to the assertion
that R[&iquest;&iexcl;] and 7?[l/f]
are flat Ti-modules.
The main results
ties
of these
reader
might
of this paper
sections
already
benefit
by first
occur
Additional
technical
difficulties
ring and
7 is subject
element.
&sect;4
condition
on min /. In particular,
rings
then
modules
that
is easily
R[X]/I
Presented
answered
arise
to the Society,
Research
supported
R is an integral
to this
The
implies
class
that
this
as to the class
generated
R with
the property
generated
is exactly
of rings
by the editors
of
is R-flat,
for finitely
I is finitely
R
a regular
when one removes
/ of R[X], if R[X]/I
question
and the
to the case
min / contain
a conjecture
March 27, 1971; received
Polynomial
ideal,
finitely
domain,
case.
that
happens
there
The corresponding
R-module
3- Many of the difficul-
when one proceeds
of what
we make
AMS 1970 subject classifications.
(7
and
to the restriction
as follows:
flat
Key words and phrases.
module,
content,
invertible
the ring
that for every ideal
generated.
is a finite
in &sect;&sect;2
his attention
to a discussion
R with the property
/ is finitely
when
confining
is an arbitrary
is devoted
are found
May 10,1971.
Primary 13A15, 13B25, 13C10; Secondary 13B20.
ring, flat module,
generated
ideal,
torsion-free
module,
principal
ideal.
projective
in part by NSF Grant GP-8972.
Copyright
&copy; 1972, American
377
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Mathematical
Society
378
JACK OHM AND D. E. RUSH
the class
cludes
of rings
section
in more than
R
denotes
R[X].
this
fact
with
maximal
the integral
closure
If R C R
(RAS,
If R and
allow
When
ring with
A non-zero-divisor
will always
\
denotes
of R.
cp: R —&gt; R
with
defining
us to avoid
/ an ideal
be as-
to signify
quotient
ideal
that
R is quasi-local
ring of R, and
that
(R')p
set-complement.)
mention
denotes
We use
is a ring homomorphism,
explicit
R denotes
R is integrally
of R, then
homomorphism
For any subsets
emphasize
cp. The
then
A
that
the element
regard
R'[X]
phism
cpx obtained
for an ideal
also
that
zf of R
as an R[X]-algebra,
generates
we shall
by applying
R
have
R
is isomorphic
as an R -algebra
as if
AA =
R' = R[&lt;f]
Whenever
in mind the defining
of elements
/ of R[X], /R'[X] = the ideal of R'[X] generated
R'[X]
may
nota-
of R , we define
as an R-algebra.
cp to the coefficients
fot
R
following
of cp and to treat
A of R and
the
](R)
&iexcl;2 cp(a,)a'.\ a. &pound; A, a'. &pound; A'\; similarly, A' n R is defined to be 0-1U').
signifies
of
of the polynomial
we sometimes
R = R, we say
ideals
and
of R.
and
is a domain,
P is a prime
(Here
ate rings
an overring
question.
mean commutative
homomorphisms
the total
T(R)).
and
an R-algebra
conventions
it were
of R (in
S = R\P.
R
of a polynomial
work on this
zero-divisors.
R. R, m is used
denotes
of the maximal
be considered
the case
Nagata's
indeterminate,
discussion
than
222. T(R)
ate rings
where
the intersection
tional
X a single
D rather
ideal
briefly
Ring
in-
into the identity.
When the ring under
by using
closed.
ring
a ring,
element.
(and hence
rings).
in particular,
a ring without
a regular
are projective
Ring will always
will mean
to map the identity
ring
discusses
and,
and terminology.
and domain
R-modules
and noetherian
of the paper
a ring will be called
sumed
flat
rings,
one variable
0. Notation
identity,
finite
quasi-local
The final
ring
for which
domains,
[September
we
homomor-
of R[X].
Then
by &lt;px(/). Note
to R' &reg;„ R[X]
via the map
r &reg; f(X) -* r ■f\X).
By &quot;module&quot;
called
we shall
torsion-free
if
of R. We sometimes
etc.)
to emphasize
generators
the rank
Let
whenever
say
that
mean
a unitary
module.
An R-module
rm = 0, r &pound; R, m / 0 &pound; M, then
that
M is
R-torsion-free
we are regarding
of M, we write
M=
M as an
(222,,•••,
of M. If S is a multiplicative
R&ccedil;-module
clear)
always
r is a zero-divisor
(or R-flat,
R-projective,
R-module.
If 222., • • • , 222 are
222). When
system
(m.s.)
M is R-free,
of R, then
R^ &reg;R M; when M is an ideal of R, we shall identify
H be a nonempty
denotes
More generally,
the ideal
subset
of R[Xl.
of R generated
if U is a subring
generated
by the coefficients
a 1X + ...
+ a n X&quot; with
c „(H)
of R, c^H)
a ft 4 0, then
(or merely
by the coefficients
of the elements
denotes
of H. Let
a fa is called
M is
rkR M denotes
zM„ denotes
the
if the
JJ-submodule
/ /= 0 &pound; R{X].
&deg;
the
M~ with MRS.
of the elements
the leading
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c(H)
R-free,
coefficient
R is
of H.
of R
If / = aQ +
of '/, and
19721
THE FINITENESSOF / WHEN R[x]/I IS FLAT
we write
deg,,
subscript
j = n. When there
X. Similarly,
More generally,
is no possibility
of confusion,
if an = a. = -..=a._.
if H is a nonempty,
= 0 and
nonzero
379
subset
we shall
a ./
omit the
0 we write subd /= i.
of R[X],
we define
deg 7/ =
min {deg f\ f / 0 &pound; H\ and subd H = min [subd f\ f / 0 &pound; H\. For any ideal / of
R[X], we define min / to be {f / 0 &pound; l\ deg / &lt; deg g for all g ^ 0 e /!. Note that
min / U [Oi is then
an &szlig;-module
where
coefficient
a . = leading
We shall
call
an ideal
and that,
for any
f.,
A of R locally
trivial
ly principal
P
(resp.,
locally
of R[X],
IR[X]pi
Our primary
lar, C denotes
it is clear
1.1 Content.
and
content
of R[X]
finitely
P of R,
local-
if for every prime
generated).
will be Zariski-Samuel
containment.
sense;
ideal
Also,
[ZS].
(A : B)r
the subscript
C will
In particu-
denotes
be dropped
and flatness.
If l = 1,a.b.&pound;L,
and when
has
makes
on content
c &iexcl;&iexcl;(I), is defined
(/-module;
&lt; strict
this
at primes
(resp.,
prime
/ of R[X] will be called
Let L be a free R-module with basis
of R.
denoted
= a f.,
from the context.
1. Preliminaries
a subring
generated)
for terminology
containment
{c &pound; C\ cB C A !, whenever
when
finitely
is principal
reference
&pound; min /, &ouml;./,
if for every
either AR p = 0 or AR p = Rp. Similarly, an ideal
ideal
/,
/..
a . &pound; R, the
(/-content
(/-module
X a . U. c
to be the
U = R, c
the following
(I) is independent
B = {b \.
., and let U he
of 7 with respect
(I) is a finitely
of the choice
to B,
generated
of basis
B.
The
properties:
(a) For any I &pound; L and any ideal A of fi, I &pound; AL &laquo;\$^ cy(0 c A.
(b) c u(rl) = 7 - cAl)
for r &pound; R, I &pound; L.
If H is any nonempty
tained
subset
in R and generated
indeterminates,
this
basis.
a free
R-module
(c) Content
[No].
large
With one exception,
content
formula
content
formula
We also
R
that
For any
in the proof
that
is that
some
is an &szlig;-algebra
and
con-
then
of power-products
function
defined
multiplicative
with
C
of the
respect
to
property:
/, g &pound; C, c^/g^^g)&quot;
of 2.17,
our further
an element
need
(/-module
= c&ouml;(f)c
v(g)n + l
72.
confine
r / 0 &pound; R such
to be the
of indeterminates),
consisting
a (/-content
has the following
formula
R = (/; so we shall
R (in any number
with basis
C has
function
for all sufficiently
exists
ring over
and hence
This
c ..(H)
by {c (h)\ h &pound; H\.
If C is a polynomial
is, in particular,
of L, we define
deal
to this
g &pound; C is a zero-divisor
r • c
cR(gh)
we only
discussion
(g) = 0. Another
= cR(h)
properties
L is a free
if c
2?-module
the case
that
It follows
of C if and only
immediate
from the
if there
consequence
of the
(g) = R.
of the content
R' &reg;„ L is a free -R'-module with basis
with
case.
formula
under
localization.
with basis
{b.\.
., then
L =
il &reg; &egrave;.}.&pound;. [L, p. 418, Proposition
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If
8].
380
JACK OHMAND D. E. RUSH
For any' I &pound; L, I = 2 r zb z'., r.efi;
z
c Al)R
= cd'(1
submodule
cular,
and hence
'
&reg; /). More generally,
of L' generated
[September
1 &reg; / = 2 r z.(1 &reg; b z.). Therefore,
if K is a submodule
of L and K is the
by \l &reg; l\ I &pound; K\, then c AK) • R = cR'(K
if S is a multiplicative
system
of R, then
cR(K)Rc
= c „ (KA
). In parti(= c AK)&lt;.
by notation).
We also
Let
need,
in the proofs
G be a multiplicative
of 1.5,
system
1.6, and
2.17,
of the polynomial
the following
ring
observation.
C with the property
that
g &pound; G =^-cR(g) = R, and suppose in CQ, f /g x = (h/g 2)(l/g ?), f, h, I &pound; C, g{ &pound; G.
Then cR(f) = cR(hl); and if cR(h) = R, then cR(f) = cR(l). For, there exist
g, g
&pound; G such
formula.
free
that
(The
R-module
initions.
gf = g hi, and then
awkwardness
here
and hence
does
One can get around
not have
such
free
R-algebra
(b), (c) above.
[OR]
for a detailed
1.2 Flatness
R-modules
with
criteria.
Let
L free.
Then
a content
problems
of a not necessarily
See
the asssertion
follows
is due to the fact that
which
function
by working
has
study
Cr
in the sense
function
satisfying
setting
(a),
matters.)
0 —►K —&gt; L —&raquo; M —&gt; 0 be an exact
the following
a
of our def-
in the more general
a content
of such
from the content
is not necessarily
sequence
of
are equivalent:
(a) M is flat.
(b) For every
ideal
A of R and every
r &pound; R, (A : r)M Z) AzM : r (where
AM : r
= \m &pound; M \ rm &pound; AM\).
(c) For every finitely
(d)
For every
generated
ideal
A of R, K DAL
C AK.
x &pound; K, x &pound; c(x)K.
(a) &lt;=&pound;►
(c) follows from [B(a), p. 33, Corollary],
(a) &lt;#&gt; (b) from [B(a), p. 65, Exercise
(a) &laquo;#\$►(d) is easy
[B(a),
22]; and
p. 65, Exercise
23].
Our basic
reference
for flat-
ness will be [B(a)].
We shall
an exact
then
also
sequence
0 —&raquo; R
R'-modules
Corollary
use
the following
of R-modules,
&reg;R K —&gt; R'
[B(a), p. 30, Proposition
1.3 Corollary.
Let
By localizing
local with maximal ideal
following
and let
&reg;„ L —&raquo; R'
R
0 —&gt; K —&gt;L —►/Vl—&gt;0 be
be an R-algebra.
If M is
M —►0 is an exact
sequence
4] and R' &reg;r M is R'-flat
0 —&gt;K —&gt;L —&raquo;M —►0 be an exact
R-flat,
of
[B(a), p. 34,
sequence
of R-modules
Then c(K) is locally trivial.
at
P, it suffices
to prove
the corollary
P. If c(K) / R, then K C PL; hence
when
R is quasi-
c(K) = 0 by the
lemma:
1.4 Lemma [B(a), p. 66, Exercise
Jacobson
&reg;r
Let
2].
with L free and M flat.
Proof.
observations:
radical
23-d]. Under the hypothesis
of R and K C JL, then K = 0.
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of 1.3, if ] =
19721
THE FINITENESS OF / WHEN R[X]/l IS FLAT
Proof.
For any
flatness.
x &pound; K, x &pound; c(x)L
Therefore
x &pound; c(x)JL,
c(x) = 0, which implies
Note
assertion
c(Kp),
that
when
that
either
if Kp/0
Let
Proof.
the conclusion
of C.
l]
ring over
) and
assume
(c(x)
■ I)
R is quasi-local
If c(I) = 0, then
to prove
s &pound; C\ M'.
C\m'
Therefore
with maximal
formula,
by the content
exist
formula
that
l)Mt.
The following
Theorem
result
finitely
primes
finitely
of R[X]
Proof.
Let
we may assume
(R/P)[X]
Then
c(x) = c(g).
by first
of R[X]
such
Thus
x/l
M, we may
Therefore
that
with maximal
R = c(l) C c(f)
c(l)M =
at
this.
is immediate.
&pound; C\ M'
and
is principal
localizing
M in proving
assume
= (f/l)(g/s),
g &pound; C,
s .x = fgs 2. But
ideal
implies
= (f/l)(g/s)
with Corollary
Let I bean
c(l)
If /,,'
&pound; (c(x) ■ I)M&lt;. Note that
= (fg/s)
&pound;
ideal of R[X] suchthat
at primes
is locally
a converse
ring in one variable
and
ideal
to
/
is
of C.
of R[X],(2)
then
R[X]/I is R-flat.
I is locally
principal
If I
at
trivial.
R is quasi-local
is a principal
by
= R. It follows
P' be a prime of R[X], and let P = P' Cl R. By localizing
that
s . &pound;
M. Therefore
c(f)
1.3 will provide
C is a polynomial
at primes
generated
and
of indeter-
at primes
Q.E.D.
that
generated
1.6 Proposition.
is locally
s ., s
together
1.5 in the case
locally
ideal
R is quasi-local
c(x) = c(fg).
(c(g)- DM- =(c(x)-
principal
for some / &pound; I. For any x &pound; I, (x/l)
there
=^- c(s .) = R since
the content
R (in any number
the following:
■ I ) i; and thus,
/ = 0 and the assertion
c(l) = R. IM&gt; = (//1)CM-
=
M' of C, and let M = M Cl R. By 1.2(d)
it suffices
&lt; = (c(x)
c(K)p
is a flat R-module.
and e(l)M = 0 or RM, then for any x &pound; I, x/1
c(l
to the
R, since
c(K) = R.
If I is locally
then C/l
Fix a maximal ideal
p. 112, Corollary
from the
lemma,
of 1.3 is equivalent
for arbitrary
P of R, then
C be a polynomial
I be an ideal
trivial,
following
By Nakayama's
Q.E.D.
for every prime
and let
the equality
c(x) = c(x)j.
K = 0 or c(K) = R. Also,
and c(l) is locally
[B(a),
x = 0.
R is a domain
1.5 Theorem.
minates),
Cl K = c(x)K,
and hence
381
domain;
with
maximal
and hence
ideal
there
P.
exists
at P,
R[X]/PR[X]
=
/ &pound; I such
that
/ =
fR[X] + (PR[X] n /). But R[X]/l is K-flat implies PR[X] n / = PR[X] ■I. Therefore
I pi =fR[X]pi
+ PR[X]pi
• I pi.
Since
by Nakayama's
lemma that /„' = fR[X]pi.
then c(l) = R.
Q.E.D.
/„'
is finitely
generated,
Moreover, by Corollary
we conclude
1.3, if I/O,
(2) Heinzer
and Ohm, The finiteness
of I when R[x]/&iexcl;
is R-flat.
II, Proc. Amer.
Math. Soc. (to appear),
have recently
proved that R[x]/l
is /&icirc;-flat
implies
/ is locally
finitely generated
at primes of &auml;[x1.
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382
JACK OHM AND D. E. RUSH
1.7
zero
The condition
ideal
degree.
leading
dition
is
we have
min / contains
and let
proof
R{X]/I
follows.
First
closed
with
prove
assumption
That
2.13,
There
(ii)
that
R{X]/I
these
is
that
R is inte-
closed
c(l) = R, rather
R-flat.
This
assump-
the quasi-local
R uses
than
is explained
are equivalent
statements
be discussed
thoroughly
lemma.
Let
I be an ideal
of R[X].
exists
There
will
the following
/ = /R[X],
(iii)
that
asserts
field
an integrally
free and
as
integrally
residue
next remove
The
proceeds
the assumption
concerns
need
Lemma.
(i)
f &pound; min / such
the leading
exists
the
by
for an inte-
that
coefficient
g &pound; min / such
Then
c(f)
the following
For the
are equivalent:
= R.
of f is regular
that
in &sect;3-
c(g) = aR,
in R, and
a regular
c(l) = R.
in R, and
R{X]/l
R-torsion-free.
Moreover,
c(g)
hypothesis
2.7);
R-torsion
element.
ideal
R is quasi-local
the infinite
remove
which
is
for / to be principal
we only
2.1
R{X]/I
a regular
indeterminate,
R.
Conditions
present,
(Theorem
and finally,
c(l) = R.
generated
that
then remove
of the proof
that
which
grally closed
in the case
if the
so if in ad-
X be a single
min / contains
I is a finitely
field;
2.16);
part
stronger
Corollary
that
an indeterminate
the hypothesis
apparently
such
let
= 0 for
of / is regular.
trivial;
that
for a non-
if and only
coefficient
is locally
it follows
R be a ring,
implies
residue
(Corollary
closed.
Let
that
of / of least
/ is regular
c(I)
then
Recall
elements
if and only if rc(f)
if and only if some
in 1.3 that
the theorem
infinite
is regular
element,
of R[X]
R-flat
element.
if / &pound; min /, then
seen
generated.
is
tion by adjoining
grally
/ &pound; R[X]
a regular
/ be an ideal
that
a regular
the set of nonzero
of / is regular
R-flat,
2. I is finitely
is
formula,
r = 0. In particular,
coefficient
If R{x]/I
min / contain
min / denotes
By the content
r &pound; R implies
only
that
/ of R[X],
[September
when
is a principal
Proof,
these
d
then
R{X]/l
is
R-flat
and,
for any
g &pound; min /,
ideal.
(i) =#►(ii):
of /. Then
hold,
is regular
Let
/ be as in (i), and let
in R. For
d be the leading
bd = 0 for some
b &pound; R implies
coefficient
bf = 0 or
deg bf &lt; deg /. Since bf &pound; I and / 6 min /, it follows that bf = 0. But c(f) = R,
so b = 0.
Now let g &pound; L By the division algorithm
tot some
integer
c(h) = dnc(g)
7z and some
by the content
h &pound; R[X].
formula,
dng
and hence in R[Xl, so then g &pound; c(g)fR[X].
(ii)
=^ (iii):
R{X]/I
is
free [B(a), p. 29, Proposition
R-flat
[ZS(a), p. 30, Theorem 9], d&quot;g = hf
Therefore,
dng &pound; c(h)
&pound; d&quot;c(g)fR{X].
Therefore
by Theorem
But
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
and since
d is regular
in R
/ = /R[X].
1.5 and hence
3l.
■ /R[X];
is also
R-torsion-
19721
free
THE FINITENESS OF / WHEN R[X]/I IS FLAT
(iii)
=^
and
g &pound; I implies
(i):
c(g) = aR
implies
c(g) = aR implies
c(f) = R.
We have
seen
R-flat.
(ii)
since
already
also
g = af fot some
/ &pound; I. Therefore,
implies
/ &pound; R[Xl.
g &pound; min / implies
in the course
of the proof
for any g &pound; min /, g = af,
383
R[X]/I
is R-torsion-
/ &pound; min /. Also,
that
(ii)
implies
a &pound; R; and then
ac(f)
R[X]/I
=
is
c(g) = aR
c(f) = R.
2.2
cipal)
Remark.
If D is a Bezout
domain
(i.e.
and D[Xl/7 is D-flat, then 2.1(iii)
principal.
This
discussed
in &sect;5.
then
D[X]/I
Theorem
observation
is
D-flat
2.15.
includes
The case
domain
/ is finitely
/ need
generated
case
are prin-
of Nagata's
D is disposed
generated,
not be principal
ideals
and hence, by 2.1(ii),
the one variable
of a Pr&uuml;fer
implies
However,
finitely
is satisfied,
which
in this
/ is
theorem
of almost
as easily;
will follow
from our
case,
as can be seen
from
Example 2.14(3).
We impose
regular
element.
is extended
that
on / in much of what
T(R)
2.3
This
condition
to a torsion-free
denotes
Let
insures
overring
the total
Lemma.
follows
the condition
that
that
deg min / does
not decrease
of R, as the following
quotient
lemma
a
when
shows.
/
Recall
ring of R.
I be an ideal
of R[X],
let
a regular
R'
be a torsion-free
and let
I = IR [X].
If min / contains
regular
element
deg min / = deg min / . // R C R' C T(R)
and
min / contain
element,
then
R-algebra,
min /
contains
and there
a
exists
f &pound; min / such that c (f)R' = R', then f is regular in R[X] and /' = fR'[X].
Proof.
the leading
For the first
coefficient
ficient
algorithm
= hg fot some
of g is regular
the image
h &pound; R [X].
in R and
c „(f)
contains
cients
of / is zero
assertion,
a regular
element.
Note
contain
by Lemma 2.1.
that
it can easily
a regular
element.
contains
a nonnilpotent
putation
shows
is
is
R-torsion-free.
2.4 Lemma.
R
of min /. Then
r. &pound; R ,
9], ang . &pound; gR[X]
&gt; deg g since
It then
C T(R)
that
/ is regular
for some
the leading
also
follows
coefthat
/ &pound; min /'.
and
c Af)
the annihilator
in R[X].
But
c „'(f)
• R
= R
implies
of the coeffi-
Therefore,
= c „(f)
deg / =
• R' = R',
so
Q.E.D.
happen
that
R[Xl//
For an example
But then
element
of min /'.
that
It follows
zero-divisor
/ / R[X].
deg g
R-torsion-free.
element
that
deg min / = deg min /', and hence
p. 30, Theorem
Thus
R
observe
and hence
g is a regular
in R. If g '&pound; I , g =%r.g.,
[ZS(a),
of g in R [X] is a regular
For the second
/' =/R'[X]
suppose
a of g is regular
g . &pound; I. By the division
n; so a&quot;g
assertion,
By Theorem
(with
a, and let
1.5,
is
and yet
/ / 0), let
/ = (1 + aX).
R[X]//
/ Cl R = min / U [0|
R-flat
is
R-flat
consists
Let I be an ideal of R[Xl, and let R
min / does
R be any ring which
Then
a simple
and hence
com-
a fortiori
only of zero-divisors.
be a ring such that
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
not
384
JACK OHMAND D. E. RUSH
fiCfi'C
[September
T(R). If R[X]/I is R-torsion-free, then IR'{X] n R[X] = /.
Proof. Let '/ &pound; IR'[X] n R[X]. Then '/= 2 (a iAs i&quot;
)f.i with 'iii
f. &pound; I, a., s . &pound; R,'
and
s . regular
But since
for every
R{X]/l
2.5
is
Lemma.
i. Then
sf = 2 a. f. &pound; I tot some regular
R-torsion-free,
Let
then
b &pound; R, and let
element
s of R.
/ &pound; I.
cp denote
the
R-automorphism
of R[X]
defined by &lt;p(X)= X + b. Then c(f) = c(rp(/)) for each f &pound; R[Xl.
Proof.
Let anX&quot; + • • • + aQ = f(X) &pound; R[X]. Then
cpifiX)) = aniX + b)n + an_xiX
= a\Y\
+ b)n~l
)xn-ib')+a
+ ia n b&quot; + a n— 1 h&quot;-1
+ ■■■+ aQ
,(y[
)X&quot;-1~i*!l
+ ■■■+aA.u
2.6 Lemma. Lez: R' be an R-algebra,
= h(0) = 1. // the coefficients
g and
h are integral
over
+ ■■■+aAX + b) + an
and let g, h &pound; R'[X] be such that g(0)
of gh are integral
over
R, then
the coefficients
of
R.
Proof. Let g(X) = a Xn + • • • + a XX+ 1, and let h(X) = bjA71 + ■■■ + b XX+ 1.
Let g
= a n + a n- ,X
+ • • • + a ,Xn~
' + X&quot; and similarly J h = b mm—1
+ b
&deg;
1
1
b ,Xm~
cients
+ Xm.
integral
over
The proof
that
Then
of g h
ate
g h
integral
has
over
the same
R.
coefficients
But then
R by [B(c), p. 17, Proposition
of the following
R/222 is finite
requires
2.7
Let
theorem
Lemmas
2.9
as
2.10
of g
the coeffi-
and
h
ate
Q.E.D.
will be broken
and
gh, and hence
the coefficients
ll].
.X + • • • +
into two parts.
and will be given
The case
after
Lemma
2.10.
Theorem.
that fiCfi'C
such
that
T(R) and R is integrally
R[Xl/7
g &pound; min /, then
Proof
R, m be a quasi-local
is
R-torsion-free
I is principal
(in the case
that
and
ring,
and let
R
be a ring such
closed in R'. If I is an ideal of R[X]
c(l) = R, and if c Ag)R'
and is generated
R/222 is infinite).
by a regular
Let
element
/ e / be such
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
= R' for some
that
of min /.
c(f)
= R.
1972]
THE FINITENESS OF / WHEN R[X]/I IS FLAT
Since
R/ttz
/(a)
there
is a unit of R.
X + a.
Then
same
further
some
a &pound; R such
cp he the
h and
for each
exists
h &pound; R[X]
that
assume
= 1. By Lemma
/(0)
(g/g(0))
• (g(0)
over
and hence
Thus,
g/g(0)
f(a)
cp(f)(0)
2.3,
/R'[Xl
by Lemma
= /(a),
and the
we may, after
replacing
= gR [X].
/ by /(0),
in R',
c „(g/g(0))
/ = gh tot
we have
/ =
of g/g(0)
it follows
being
we may
In particular,
in R'[X]
2.6, the coefficients
closed
by cp(X) =
2.5)
dividing
&pound; R[X] Cl IR [X] = /, the equality
&pound; min / and
defined
(by Lemma
is a unit of R'; and hence
R is integrally
g/g(0)
/ 0 (mod m); and hence
of R[X]
content
is a unit of R. After
&quot; ^)- Therefore,
R. Since
But then
g(0)
the same
and since
/(0)
that
R-automorphism
cf&gt;(b) have
&lt;/&gt;(/), assume
h &pound; R [X].
tegral
Let
since
degree
/ with
2.4.
is infinite
385
that
a consequence
= R, so the result
are in-
g/g(0)
&pound; R[X];
of Lemma
follows
from
Lemma 2.1.
2.8
Definition
infinite,
we pass
of R(Y).
In removing
to a certain
overring
from 2.7
R(Y)
the restriction
of R which
that
is defined
R/m
be
as follows
[Na3, p. 18]. Let Y be an indeterminate,
and let S = {f &pound; R[Y]| c(f) = R\. S is a
multiplicative
system
of regular
is customarily
denoted
(i) An ideal
ideal
of R[Y]
R(Y).
consisting
We list
ztz of R(Y)
elements,
a few properties
is maximal
and the ring
of this
RfY],.
ring.
if and only if there
exists
a maximal
m of R such that m = 77zR(V) [Na3, p. 18].
(ii) For each ideal A of R, R(Y)/AR(Y) ^ (R/A)(Y) [Na
(iii)
R(Y) is R-flat
[Na3, p. 64, Exercise
p. 18].
2]; and hence by (i) R(Y) isa
faithfully flat R-module.
(iv) R(Y)[X] is a faithfully
2.9
Lemma.
Let
flat R[X]-module
R, m be a quasi-local
[B(a), p. 48, Proposition
ring,
and let
I be an ideal
If IR(Y)[X] = fR(Y)[X] for some regular element f &pound; min(/R(Y)[X]),
principal
and is generated
Proof. Since
element
R(Y)[X] is a faithfully
extends
and contracts
to show
that
there
by a regular
to itself
exists
[B(a),
g &pound; I such
flat R[Xl-module,
deg.,
We may assume
and hence
degx /&gt; degx /.. But / &pound; min(/R(Y)[X])
/ &lt; degx
/&iquest;. Thus
• • • + &ouml;(Y'.
Hence
degx
atYl)f;
/. = degx
/. Since
tion that
2.14(3)
and since
a . g mR(Y)
will
Therefore
/ and
show
that
2.9
=
+ f( Yl, j. &pound; I;
and p. &pound; 7R(Y)[X], so
the leading
/ is regular
for some
it suffices
IR(Y)[X]
coefficient
of / is
we have /. = a ./ with a. &pound; R(Y). Thus
;*. Then
is the unique maximal ideal of R(Y). Therefore
Example
every ideal of R[X]
/ &pound; &iexcl;R[Y , X]. Then / = /n + f {Y + ...
regular in R(Y) and IR(Y)[X] = /R(Y)[X],
/ = (a0 + ajY+--.+
9].
g = degx
gR(Y)[X].
degx
of R[X],
then I is
of min /.
p. 51, Proposition
that
5].
in R(Y)[X],
l = ar] + a]Y
a . is a unit of R(Y), since
f.R(Y)[X] =/R(Y)[X].
is no longer
valid
R is quasi-local.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
without
+
77zR(Y)
Q.E.D.
the assump-
386
JACK OHMAND D. E. RUSH
2.10
Lemma.
Let
I be an ideal
of R[X]
such
element of R[X]. // R{X]/I is R-torsion-free
that
[September
min / contains
then R(Y){X]/IR(Y){X]
a regular
is Ro-
tors ion-free.
Proof.
aAsy
c(s)
Suppose
f = f'/s2,
= R\.
af &pound; 1R(Y){X], a regular in R(Y) and / &pound; R(Y){X]. Then a =
where a' is regular in R{Y], f
There
exists
show R{Y, X]/IR{Y,
Let
s &pound; S such
sa f'
&pound; R{Y, X], s . &pound; S = \s &pound; R{Y]\
&pound; IR{Y, X], and thus
it suffices
to
X] is R[Y]-torsion-free.
T be the total
min /. Since
that
quotient
the leading
ring
coefficient
of R, and let
g be a regular
of g is regular
element
in R, we have
of
c Ag)
= T and
IT{X] = gT{X] by Lemma 2.3. Hence by Theorem 1.5, T{X]/IT{X] is T-flat.
Therefore, T{Y, X]/IT{Y, X] ( =S T[X]//T[X] &reg;_ T{Y]) is T[Y]-flat by [B(a),
p. 34, Corollary
2l and a fortiori
T[Y]-torsion-free.
lar, let / &pound; R{Y, X], and suppose
T[y];
so / &pound; IT{Y, X].
rf &pound; IR{Y, X].
c R\x](rf)
When
Let
contains
is
A contains
a regular
have
r • cd(&quot;v-i(/)
R-torsion-free,
=
and hence
element,
then
that,
of 2.10
element
remains
valid
statement
can be
to
of omitting
this
assump-
(0 : c Aa)) R = 0; and
af as a polynomial
r Af) C / for some
that
Thus,
ideal
a domain
or a noetherian
in
Y
22 &gt; 1. If
cDry~|(/)
/ &pound; IR{Y, X].
generated
without
without
interest
so that
we can conclude
(e.g.
valid
of some
Now regard
and hence
for any finitely
a regular
(a related
of R[Y],
c Aa)nc
R-torsion-free),
remains
perhaps
the possibility
/ &pound; R{Y, X].
to obtain
2.10
element
It is thus
element
By [Kaj,
example
shows
p. 63, Exercise
(i) R is its own total
pA
has
zero
quotient
that
(p.,
such
that
25,722. = 0, p ym2 = 0 and
this
*- '
if R is a
A of R, (0 : A)R = 0
the assumption
assumption
exists
ring,
annihilator,
that
(iii)
/ = gR[X].
a is a regular
element
of R{Y, X]; and by (i) R[Xl/7
cannot
a ring
(ii)
there
there
222
2 &lt;&iquest; p XR. Let
and let
Then
By (ii),
that
7], there
such
722jX.
is
that
ring),
then
min / contains
element.
The following
erties:
r of R such
Y, we then
of Lemma
to consider
formula
with the property
the conclusion
eral.
in order
a regular
R{X]/I
implies
element
R{X]/I
a regular
18.12]).
a be a regular
the content
(assuming
in
the conclusion
af &pound; IR{Y, X] fot some
c Aa)
a regular
^ ' slnce
min / contains
for a moment
and apply
exists
Q.E.D.
that
from 2.10.
suppose
ring
c R\x~\if)
in [Na ,, p. 63, Theorem
digress
a &pound; R{Y] be regu-
af &pound; IR{Y, X] C IT{Y, X]. Then a is regular in
there
/ as a polynomial
R is noetherian,
the assumption
tion
Regarding
C /. Therefore,
/ &pound; IR{Y, X].
found
Thus,
Now let
element
is trivially
exist
exist
m2&iquest;
pxR. Thus,
nonzero
nonzero
in gen-
p.,
R-torsion-free.
X].
p2&pound;
R
g = p l + p 2X,
g is a regular
Let /=
However,
/ &iquest;
R{Y, X]/IR{Y, X] is not R[Y]-torsion-free.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
prop-
272,, 222 &pound; R
a = p , + p 2Y, let
of R{Y] and
af = p ,2222 + p 2m xYX = (2222 + 222xY)g &pound; IR{y,
gR{Y, X] since
be omitted
R with the following
272 +
1972]
THE FINITENESS OF / WHEN R.[X]/I IS FLAT
We shall
Proof
now complete
of Theorem
the previous
proof.
the proof
2.7
in the general
Otherwise
R(Y), which is quasi-local
R(Y)[X]/IR(Y)[X]
R(Y).
By Lemma
regular
element
R(Y),
contained
closed
in R
of Theorem
we must
with infinite
2.9
it suffices
in the total
and such
is infinite
case
ring
c&szlig;(y)(g)l?&quot;
apply
to the ring
field by 2.8. By Lemma 2.10,
/R(y)[X]
For this
we merely
by passing
and c(l) = R implies
to show
quotient
that
If R/m
to this
residue
is R(Y)-torsion-free;
of min (/R(Y)[xl).
2.7.
case.
reduce
387
cR(y)(/R(Y)[X])
is principal
it suffices
of R(Y),
generated
by a
to find an overring
such
that
= R&quot;• We claim
R(Y)
=
R
of
is integrally
R = R [Ylr
is such a ring,
where S = {f &pound; R[Y]\ cR(f) = R\. Since cR(g)R' = R', it follows that cR(y)(g)R'[Y]s
= R'[Y]S.
Also, R[Y] is integrally
and hence
16].
R(Y) = RMs
Finally,
R[Y]
and
is integrally
any element
cR(g)
total quotient
=sR,
with
corollary
closed
of R'[Y]„
to Theorem
2.7
Let
R, m be a quasi-local
such
that
of R[X]
and
R'[Y]S
is the main result
2.11
torsion-free
p. 22, Proposition
in the form f/g
in R; and hence
and integrally
be an ideal
[B(c),
13],
with
/, g &pound;
is contained
in the
Q.E.D.
R is quasi-local
Corollary.
in R'[V]S
can be written
s regular
ring of R[Y].
The next
closed in R'[Y] [B(c), p. 19, Proposition
closed
(in its total
a regular
I is principal
for the case
that
ring).
integrally
min / contains
c(l) = R, then
in &sect;2
quotient
closed
ring,
element.
and is generated
and let
If R[X]/I
is
by a regular
I
R-
ele-
ment of min /.
Proof.
Take
min /, then
T(R)
to be the
the leading
coefficient
R
of Theorem
2.7.
of g is regular
If g is a regular
in R and hence
element
is a unit
of
of
T(R). Q.E.D.
Another
consequence
relationship
between
following
lemma
2.12
Then
torsion-free
Tp
there
is
T = T(R),
and
R[X]/l
the corollary
of R[X]
such
2.13,
which
being
R-torsion-free.
in a global
that
illuminates
the
The
form.
min / contains
a regular
(if and) only if Rp[X]/lRp[X]
is Rp-
P of R.
let
a &pound; R, s, s'
and hence
exists
therefore
R-flat
is R-torsion-free
prime
is Corollary
P be a prime
&pound; r\p,
f &pound; R[X],
of R, and suppose
and
a/s
(a/s)(f/s
is regular
) &pound;
in R p. Then
&pound; &iexcl;Rp[X] C 7Tp[X]. But T[X]//T[X] is T-flat by Lemma 2.3 and Lem-
Rp-flat,
element
being
2.7
us to give
I be an ideal
R[x]/7
Let
where
(a/s)(f/s')
ma 2.1,
Let
for every
Proof.
}Rp[X]
R[X]/I
will enable
Lemma.
element.
of Theorem
Tp[X]/ITp[X]
so a/s
s&quot; &pound; R\P
r &pound; R such
//l
that
is regular
such
that
is
Tp-flat
in Tp,
s&quot;f &pound; IT[X],
rs&quot;f &pound; I. But since
and a fortiori
and hence
f/s
and hence
R[X]/I
is
T „-torsion-free.
&pound; ITp[X].
there
exists
R-torsion-free,
&pound; IR p[X].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
Also
Therefore
a regular
then
s&quot;f &pound; I;
388
JACK OHMAND D. E. RUSH
2.13
element
Corollary.
Let
and consider
I be an ideal
the following
(i) For every prime ideal
ment of min(/R„[X]);
(ii)
For every
of R{X] such
[September
that
min / contains
a regular
properties:
P of R,
}R p{X]
is principal
generated
by an ele-
and c(l) = R.
prime
ideal
P of R, IR p{X] is principal;
and
c(l) = R.
closed,
then
(iii) R{X]/I is R-flat.
(iv) R{X]/I
is R-torsion-free,
Then
(ii)
(i) =^
=^&raquo; (iii)
and c(l) = R.
=&gt; (iv); and if R is integrally
(iv) =&pound;&raquo; (i).
Proof,
(i) =#►(ii): Trivial.
(ii) =\$&gt; (iii): Apply Theorem 1.5.
(iii)
=^&gt; (iv):
c(l) = R by Corollary
1.3 and the fact that
c(I)
contains
a regu-
lar element.
(iv) =^
(i):
Let
P be a prime
ideal
of R and let
g be a regular
min /. Then Rp C T(R) „ C T(R A, and R p is integrally
[B(c),
p. 22, Proposition
16].
Also,
in Rp{X] is in min(/Rp[X]);
cR (g/l)T(R)p.
Since
Rp[X]//Rp[X]
lows from Theorem 2.7.
Note
Lemma
that
To see
this
ting of zero-divisors
that
min / contains
it suffices
and such
to exhibit
that
torsion-free.
existence
R 0 be a noetherian
prime ideals
exists
ting
R = RAA,
of a ring
ring of Krull
and
P,
A of R.
whose
we get a ring
= P AA.
a regular
2.14
integral
tains
To see
prime
proceeding
can happen
further,
if one alters
Examples.
closure
a regular
this,
g/l
of g
R with
such
dimension
a prime
&gt; 3. Then
is needed
a prime
element
but Rp{X]/PRp{X]
associated
R whose
Thus
be an associated
Before
element
R having
P2
primes
in
P consis-
of Rp.
For
would not be R p-
can be seen
as follows.
there
a chain
are exactly
associated
consists
primes
exists
R.
we shall
the hypothesis
(1) Let
Then
element
choose
of (0) in R [B(b),
exists
and such
that
and let
R[X] —&gt;R[&lt;f]. R[rf] is R-torsion-free
2.13(iv)
of
P,.
Let-
P 2 C P,;
P x
but
_be an
P2
5].
to illustrate
what
2.13.
R is different
/ of R[X]
is satisfied
/ be the kernel
since
and
P~R~&iquest;
2 P 2 would
8] and hence
examples
of Corollary
an ideal
since
p. 134, Proposition
some
R be any ring for which
there
&lt;f &pound; R\R
give
P.
of (0) are exactly
of zero-divisors
P 2,R-t;P 2 contains
a regular
element
of R--,
&amp;
P 2 , for otherwise
associated
prime of (0) in Rp
[B(b), p. 137, Proposition
would
image
T(R)p =
P Q &lt; P x &lt; P 2 &lt; P} of R Q, and by [ZS(a), p. 230, Theorem 21] there
an ideal
= P j/A
a regular
a ring
PR p contains
would be R-torsion-free
The
the canonical
= T(R) then
is R p-tors ion-free by Lemma 2.12, (i) fol-
then R[X]/PR[Xl
Let
2.3
cR(g)T(R)
of
by
Q.E.D.
the assumption
2.12.
by Lemma
and since
element
closed in T(R)p
such
but
of the
that
from its
min / con-
2.13(iii)
is not.
R-homomorphism
R[&pound;] C T(R), and c(l) = R since
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
&lt;f
1972]
THE FINITENESS OF / WHEN R[X]/l IS FLAT
is integral
regular
over
R. Moreover,
element
since
&ccedil;=
of min /. However,
a/b,
R[&ccedil;J
389
a, b &pound; R, b regular,
is not
R-flat
since
bX — a is a
R &lt; R[&lt;f] C R [A, p.
803, Corollary 2].
(2) [Na2, p. 446, Example
which
is not integrally
2].
closed,
fied but not 2.13(a).
Let
This
is an example
and an ideal
k he a field
of a quasi-local
/ of D[X]
and
such
that
t he an indeterminate,
k[t2, z*3] 2 , , and let / be the kernel of the D-homomorphism
r
D[l/t\
/ i
does
is the quotient
field
not have principal
(3) This
local,
(Corollary
exists
domain
an ideal
the quotient
osition
shows
closed.)
(iii)
Let
that
D[X] —&raquo; D[b/a],
of D, D[b/a]
implies
class
domain
domain
D, which
but
group
is not quasi-
R is quasi-local
is not Bezout
Since
of the
7]).
D[b/a]
and hence
1.3-
&pound; min / and
/ is not princi-
[C, p. 222, Theorem
if / is the kernel
c(l) = D by Corollary
when
which
is not principal.
D =
by 2.1.
is satisfied
/ is principal
is D-torsion-free
Therefore,
then
closed
y
but z&deg;X - t
be principal
2.13(iii)
D be a Pr&uuml;fer
with nontrivial
D-flat.
/ cannot
such
(a, b) of D which
field
3] is
so
let
D,
is satis-
D[X] —&gt;D[l/t].
a
is D-flat;
of an integrally
/ of D[Xl
2.13
and integrally
Dedekind
content,
is an example
and an ideal
pal.
of D and hence
domain
2.13(iii)
(e.g.
any
Then
there
is contained
by [B(a),
in
p. 29, Prop-
D-homomorphism
Moreover,
/ is not principal.
For,
if / = /D[X], then c(l) = D implies c(f) = D. But aX - b &pound; I implies aX - b = df,
d &pound; D; and then
by equating
(a, b).
[Na2,
(Nagata
Note
that
in Lemma
Example
2.9-
For
D(Y)[X]/ID(Y)[X]
contents,
p. 446,
3 also
D[X]/l
(a, b) = (d), a contradiction
Example
shows
is D-flat
is D(y)-flat;
l] gives
that
a similar
the quasi-local
implies
to the choice
of
example.)
assumption
(by tensoring;
see
and D(Y) is the Kronecker
is needed
1.2)
that
function ring of D
[G, p. 384, Theorem 27.4], which is Bezout [Kr, p. 559, Satz 14], so ID(Y)[X] is
principal
by 2.2.
The next
2.13(i)
theorem
gives
some
and is the tool needed
2.15
element,
Theorem.
and assume
(i) For every
element
Let
conditions
which
are equivalent
to pass
from the local
to the global
I be an ideal
of R[X] such
that
c(l) = R.
prime
further
ideal
Then
the following
P of R, &iexcl;R Ax]
to
case.
min / contains
a regular
are equivalent:
is principal
generated
by an
of min(/Rp[X]).
(ii) c(min /) = R.
(iii) / = (f y • • •, fn), f. &pound; min /.
Proof,
prime
(i) =^ (ii):
P of R, it follows
that /p = fRp[X].
Since
that
by Lemma
for each
2.3, deg min / = deg min(/p)
prime
P of R there
exists
for each
/ &pound; min / such
Then c(f) p = R p, and hence c(f) t P. It follows that
c(min 7) = R.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
390
JACK OHM AND D. E. RUSH
(ii) ==&gt; (iii):
R. Therefore
By (ii)
there
for any prime
/. in R p{X] is in min(/p),
shows
that
such
/,,•••,/
it follows
that
c(f x, • • •, / ) =
i; and since
the image
prime
P of R, and hence
of
This
/ = if x, • • •, / )
Theorem l].
=^
that
&pound; min / such
j&pound; P for some
from Lemma 2.1 that Ip =/.Rp[X].
I p = (f y • • • , f ) p fot each
[B(a), p. Ill,
(iii)
exist
P of R, c(/.)
[September
(i):
c(/.)
Since
\t P.
c(/,,
• • •, / ) = R, for any prime
But since
the image
P of R there
of /. in R p[X]
exists
is in min(/p)
/.
(by Lemma
2.3), it follows from 2.1 that lp = /.Rp[X].
2.16 Corollary.
such
that
ated
by finitely
Proof.
R-flat,
ated.
give
the assumption
Note also
eses
2.16,
of Corollary
additional
condition
dimension
zero,
lar element
principal
that
then
rem l], and hence
2.15(iii).
not
by Lemma
2.1,
R to R, where
R denotes
2.17
Let
2.14(3)
that
shows
c(g)
R{X]/I
gener-
a regular
that under
as follows:
Let
2.1, it follows
is principal
ele-
the hypoth-
if one imposes
of R be a noetherian
and Lemma
But then
that
/ is not finitely
however,
spectrum
that
is satisfied,
R such
min / contains
to be principal
2.13
P of R.
that
be principal;
2.15(i)
closed
but such
and 2.16
that
Q.E.D.
of an integrally
the maximal
The final part of the proof
Lemma.
is just
/ may be seen
prime
c(l) = R.
shows
that Example
of min /. By Corollary
for every
I is gener-
1.3 and the assumption
is satisfied
/ need
then
2.13
in 2.15
ment is essential.
R-flat,
from Corollary
an example
c(l) = R, and 2.15(i)
Thus,
is
Corollary
conclusion
we shall
If R{X]/I
of min / and
element.
the remaining
closed, and let I be an ideal of R[X]
element.
c(/) = R follows
a regular
In 2.21
is
a regular
many elements
That
min / contains
and then
Let R be integrally
min / contains
the
space
of
g be a regu-
that
c(g)p
by [S, p. 318,
is
Theo-
/ is principal.
of the main theorem
the integral
closure
R, m be a quasi-local
of &sect;2
consists
of R in its total
ring,
let
in going
quotient
I be an ideal
from
ring
T(R).
of R{X], let
g &pound; I, and let P' be a prime ideal of R[X] lying over m. If c(l) = R and
IR{X]p&quot; =gR{X]p&quot;
for every prime P&quot; of R[X] lying over m, then IR{X]p&gt; =
gR{X]p&gt;.
Proof.
it suffices
First we show /R[X]p- = gR[X]pz, and by [B(a), p. Ill,
to check
the integralness
tracts
to P'
this
locally
of R[X]p&lt;
in R[X].
over
Therefore
R[X]p'
R[X]p&quot;,
in R[X] and hence
to 222in R. Since
we have
that
ideals
P
/R[X]p&quot;
ideal
of R[X]p'
is a prime
l]
of R[X] p&gt;. It follows
any maximal
the localizations
are all of the form
hypothesis,
where
at the maximal
Theorem
of R[X]
= gR{X]p&quot;
of R[X]p&lt;
from
con-
at its maximal
which
contracts
tot all such
ideals
to P'
P&quot; by
&iexcl;R{X]pi = gR{X]p&lt;.
Now we show /R[X]pz = gR[X]p-.
Since IR{X] p&lt; = gk~{X]p&lt;, for any / &pound; I
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1972]
THE FINITENESS OF / WHEN R[x]/&iexcl; IS FLAT
there
exist
h &pound; R[X]
R = 772implies
there
exists
yields
and
s
s
&pound; R[X\\P'
cR(s&iquest;) = R. Therefore,
/.
&pound; / such
that
c Af
such
cR(j)
that
s f = s 2gh.
= cR(gh)
by the lying-over
Moreover,
C c~^(g)- Since
) = R; so the above
R = c (/ ) C c~j&iuml;(g)- Hence
391
reasoning
theorem,
c(/) = R,
applied
cR(g)
to /.
= R. Therefore,
cR(/) = cR(gh) = cR(/&gt;), so cR(h) CR and h &pound; R[X]. Thus, / &pound; gR[X]p-.
2.18
Theorem.
element.
Then
Let
I be an ideal
the following
of R[X] such
that
P' Cl
min / contains
Q.E.D.
a regular
are equivalent:
(i) R[X]/IJs R-flat.
(ii) R[X]//R[X] is R-flat.
(iii)
I pi
Proof,
is principal
(i) =&pound;&gt;(ii):
for every
prime
Since R[X]/I
P
of R[X],
and
c(l) = R.
is R-flat, the sequence
0-&gt; 7 &reg;R R -♦ R[X]
&reg;R R -+ (R[X]/l) &reg;R R -&gt; 0 is exact; and hence (R[X]/7) &reg;R R S R[X]//R[X]
(as R-algebras). But (R[Xj//) &reg;R R is R-flat by [B(a), p. 34, Corollary 2l.
(ii) =&pound;&gt; (iii): Let P' he a prime ideal of R[x], and let P = P' n R. By localizing
at
P, we may assume
(R/P)[X]
is a principal
R is quasi-local
ideal
domain,
with maximal
so there
exists
ideal
P.
g &pound; I such
Then
that
/ = gR[X]
+
(PR[X] n /). Therefore, IR~[X]= g~R[X]+ (PR~[X]Cl 1R~[X]).Since R[X]//R[X] is
R-flat, PR[X] n /R[X] = PR[X] ■ /R[X] [B(a), p. 33, Corollary]. Therefore, 1R~[X]=
gR[X] + PR[X]
so
7R[X]
lemma,
• IR[X].
is a finitely
IR[X]p&quot;
But
min(/R[X])
generated
= gR[X]p&laquo;
ideal
for every
contains
a regular
by Corollary
2.16.
prime
P&quot; of R[X]
apply Lemma 2.17 to get IR[X]pi = gR[X]p'
R[X]//R[X]
is R-flat implies
element
Thus,
lying
provided
by Lemma
2.3,
by Nakayama's
over
P.
We can now
we first note that
c(IR[X]) = R by Corollary
1.3 and hence that
c(l) = R.
(iii) =#&raquo;(i): Apply Theorem 1.5.
Note
prime
that
(ii) =\$&gt; (iii)
P of R there
R[X]
lying
primes
over
exists
P.
Thus,
P of R which
ideals
of the above
g &pound; I such
if (ii)
[Kaj,
argument
that
is valid
are contractions
of R by Kaplansky
Q.E.D.
actually
proves
Ipi = gR[X]pi
and
that
fot every
a is the cardinality
of maximal
p. 17, Theorem
ideals
27]),
of R[X]
then
for each
prime
P
of
of the set
(called
of
the
G-
/ can be generated
by
a. elements.
2.19. Main theorem of &sect;2.
regular
element.
If R[X]/l
Let I be an ideal of R.[x] such that min / contains
is a flat
R-module,
then
I is a finitely
a
generated
idea I.
Proof.
flat;
ma 2.3-
exists
As observed
and min / contains
Therefore,
/R[X]
a finitely generated
previously,
a regular
is finitely
R[X]/l
element
is R-flat implies
implies
generated
min(/R[X])
by Corollary
ideal A of R[X] such that AC/
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R[X]//R[X]
does
2.16.
also
Hence
is Rby Lemthere
and AR[X] = /R[X].
392
JACK OHMAND D. E. RUSH
Claim.
A = I. It suffices
localizing
Let
at
(R/P){X]
to show A p&lt; = Ip&lt; tot every prime
P = P fl R, we may assume
P be a prime
ideal
of R lying
R is quasi-local
over
P, let
and let &lt;p be the homomorphism
(or applies
[B(a), p. 48, Proposition
(R/P){X]
extends
to itself
P' of R[X].
with maximal
since
R/P
in (R/P)[X].
—&raquo;
easily
every
Therefore,
P.
R[X]
One checks
is a field,
By
ideal
&lt;p be the homomorphism
R[X] -♦ (R/P)[X].
5]) that
and contracts
c/&gt;(/R[X]) implies
[September
ideal
of
cp~(AR[X]) =
&lt;p(A) = &lt;p(/). Thus, / = A + (PR{X] n /). Since
R[X]//
is R-flat,
PR{X] n / = PR{X] ■ I [B(a), p. 33, Corollary]; so I = A + PR{X] ■ I. Then I p&gt; =
A p&lt; + PR{X]pi
conclude
and since
by Nakayama's
Note
eses
■ Ipi;
that
the last
/p'
is finitely
lemma that
part
generated
Ip&lt; = A p&gt;.
of the above
proof
by Theorem
portion
of the proof
could
also
we
Q.E.D.
actually
shows
of 2.19) for any ideal A of R[X] such that AC/,
A = /. This
2.18,
be done
that (under the hypoth-
AR{X] = IR{X] implies
using
(ii) &lt;\$&gt; (i) of 2.18
and
Lemma 2.4.
2.20
Corollary.
lar element.
Let
Then
I be an ideal
R{X]/I
is
R-flat
of R[X]
such
if and only
that
min / contains
if I is an invertible
a regu-
ideal
and
c(l) = R.
Proof.
(==H
every
prime
ment,
/ is invertible
Corollary
P
/ is finitely
of R[X]
generated
by Theorem
by Theorem
by [B(a),
1.3 and the fact
c(I)
then
contains
element
of degree
for a moment
1. Then
give
a simple
lai.
By 1.2, (yR : x)R{x/y]
fore there
R[X]//
construction
exist
R[X]//
the special
= R[&lt;f]
= (yR[x/y]
since
4].
case
ele-
from
element.
at primes
that
of R[X]
min / contains
&lt;f &pound; T(R).
of /. Let
f=l+a.
'
for
a regular
c(l) = R follows
is R-flat by Theorem 1.5.
: x) and hence
that
Ip&gt; is principal
/ contains
That
principal
where
for the generators
a z. &pound; (yR
: x) such
w
and
a regular
/ is locally
[B(a), p. 148, Theorem 4]. Therefore
Let us consider
Thus,
p. 148, Theorem
that
(&lt;^=) If / is invertible,
2.18.
2.19,
In this
&lt;f = x/y,
by
Q.E.D.
a regular
case
we can
x, y &pound; R, y r&eacute;gu-
1 &pound; (yR : x)R{x/y].
+ a,X+...+a
0
1
There-
X&quot; &pound; I. Let
zz
b I. &pound; R be such that a.x
= Jyb l.. Then it follows that / = (/,
a.X
- bn,
• • •,' a zz X —b 72).
l
'
0
0'
For,
if P'
then
/ &pound; P'
is a prime
ideal
implies
a . i P'
/Rp[X] = (a.X - b^RpiX]
of R[X]
containing
fot some
(/, a X - b., • • • , a X - b ),
i. Therefore,
if P = P'nR,
by Lemma 2.1, and then a fortiori
then
/R[X]p&lt; =
(/, a0X-b^,...,anX-bn)R{X]pl.
We conclude
of our other
lar element.
results
Recall
this
section
that
a ring
is flat [B(a), p. 64, Exercises
2.21
Example.
with
an example
are not true without
This
R is said
which
shows
the hypothesis
that
to be absolutely
16—17; p. 168, Exercise
is an example
of a ring
that
2.19
flat
if every
a regu-
R-module
9; p. 173, Exercise
R and an ideal
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
and many
min / contains
/ of R[X]
17].
such
1972]
that
THE FINITENESS OF 7 WHEN R[X]/I IS FLAT
R[X]/I
Let
example,
but yet
/ is not a finitely
R to be a countable
sum ideal.)
R-flat
chosen
is
R-flat
take
the direct
is
is
A be any non-finitely-generated
because
to be
Then
every
direct
/ = AR[X]
R-module
R-projective
A slight
modification
with
is.
Added
of this
the additional
c(l) = R. In this
case
January
R[X]/7
of copies
note
the direct
that
(For
A to be
and
R[X]/I
if
sum ideal),
R.
and
generated,
reference,
with
flat ring
of a field
not finitely
For later
to the recent
platitude
et de projectivit&eacute;,
paper
is the
a domain,
&laquo;-variable
A is
then
3. / principal.
an ideal
AR[X]
being
in this
that
that
is dropped.
R[Xl//
Thus,
being
principal.
that
min / contains
when
R-flat,
R[zf]
l/f
is in T((p(R)[cf]),
when
to denote
no confusion
We remind
generators
3.1
&ccedil; denote
element.
2.1
closed.
what
ring
that
to /
/ is principal
in Corollary
and
2.11
Example
closure
2.14(2)
assumption
conditions
in addition
to conclude
solution
R and
are equivalent
However,
in order
a satisfying
R is
The problem
which
when the integral
on R[Xl/7
that
of the theo-
2.
and we have proved
class
to this
so
of X in R[X]/I.
is the canonical
the ring
reader
that
to
/ is
problem
(when
&lt;Jj(R)[f].
&lt;/6(R)[l/f]
If f is a regular
makes
Then R[X]/l = c/&gt;(R)[f ],
homomorphism.
R —&gt; cf&gt;(R), we can
that
R(a
•-.,
• . . , a . When the ring
) for this
Lemma.
in the case
sense;
Regarding
more
&lt;/&gt;(R)[rf] as
simply
element
and we again
use
the nota-
of &lt;/&gt;(R)[(f], then
write
R[l/f]
can arise.
the
a
(ctj, ...,a
a regular
is to determine
via the homomorphism
tion
de
element).
tp: R[X] —&gt; R[X]/l
R-algebra
our
Crit&egrave;res
of a quasi-local
on R[X]//
valid
provides
zf will denote the equivalence
where
the case
R is integrally
be imposed
3.6
a regular
has brought
generalization
R. We know by Lemma
is
the main problem
The Summary
2.19
and Ohm in footnote
is no longer
must
and
R-module.
and L. Gruson,
?z-variable
min / contains
R[X]/l
R-flat
the
for the moment
and of content
is true
flat
gives
element
Vasconcelos
of our Theorem
contains
is to find conditions
the converse
a cyclic
with
work of M. Raynaud
to Heinzer
such
R implies
the converse
shows
3.4.2
section
principal
of content
is even
a regular
Invent. Math. 13(1971), 1—89- Corollary 3-4.7 of that
Consider
/ of R[X]
treated
R/A
/ = AR[X] + XR[x],
/ contains
A conversation
generalization
attributed
namely
that
important
and Corollary
rem previously
example,
property
=
28, 1972.
attention
let
product
is also
(as is the case
ideal.
of an absolutely
R[X] projective.
an example
an
generated
ideal
393
Let
R is
a ) denotes
understood,
the
R-module
we shall
also
having
merely
use
of R[X],
and
R-module.
R, m be a quasi-local
the equivalence
class
+ • • • + a X&quot; &pound; / with aQ, • • • , a ._
ring,
let
of X in R[X]/l.
I be an ideal
If there
exists
f = a . + a .X
&pound; m and a . = 1 (/ &gt; 0), then ( 1, f, - • • , f &quot;+ A
= (1, f, •&quot; - , f7&quot; \ fl+1 +\ - - - , e+i &gt; for all i &gt; 0.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
394
JACK OHM AND D. E. RUSH
Proof.
Let
[September
E = (1, {;, • ■ • , &lt;f&quot; *); and for e &pound; E, let
e denote
its image
in
E/272E.ThenJ' + Vi?'+_! + ' ' ' Aa J&quot; &quot; &deg; i^lies
?' &pound; ^
&lt;f'*V
Therefore E = (1, &pound;, • • • ,
' ' .f&quot;'1)
tf _z, &lt;f
lemma.
!, • • • , cf&quot;
&pound; (&pound;'+&raquo;♦', ■•-, t&quot;+i).
*\, and hence
the
lemma
follows
l.' &quot; ' ' ?!&gt;&laquo; ?'+1 &pound;
from Nakayama's
Q.E.D.
Before
known
,_:• • , ^
proceeding
facts
which
be a finitely
generating
generated
set
generating
for
sets
module).
to the proof
will be used
R-module.
M contains
contain
Moreover,
of Theorem
there.
Let
It follows
a minimal
the same
3.2, we shall
from Nakayama's
generating
number
mention
R, 222be a quasi-local
and let
lemma
that
M
any
set and that any two minimal
of elements
if zM is free of rk 72, then
a few well-
ring,
( = dim M/mM as an
any minimal
generating
set
R/m-
is free
[B(a), p. 109, Corollary].
Theorem
3.2
shows
that
( 1, tf, • • - , &lt;f ) is equivalent
equivalent
regular
false,
conditions
element,
as we have
as Example
2.14(2)
be integrally
closed,
that
a finitely
free
[B(a),
3.2
for
&quot;free&quot;
Theorem.
&pound;222rilet
to the equivalent
conditions
R{X]/I
already
by Corollary
module
over
3-e],
Let
is
2.13
of all the modules
of Lemma
R-flat
in Lemma
If one imposes
and
These
2.1; but the converse
the converse
a quasi-local
2.1.
min / contains
the additional
is flat
&quot;flat&quot;
that
Recall
R
also
if and only
could
a
is
condition
is valid.
ring
so that the word
in Theorem
rf denote
( = R[&lt;f])
noted
shows.
then
generated
R, flatness
that
p. 167, Exercise
substituted
R[X],
imply
for a quasi-local
if it is
equally
well
be
3.2.
R, m be a quasi-local
ring,
the equivalence
of X in R{X]/I.
class
let
I be a nonzero
Then
ideal
of
the following
are equivalent:
(i) (1, zf,• • • , tf ' ) is a free R-module for all t &gt; 0.
(ii) (1, ff, • • • , &lt;f') is a free R-module f or some t &gt; 0 for which 1, &lt;f,• • • , (*
are linearly
(iii)
c(l) - R.
dependent
/ = /R[X]
(Note:
Proof,
some
there
exists
elements
for some
this
(iii):
First
/ &pound; I such
of / having
over,
ther
that
f(a)
consider
that
content
the case
as a linear
c(f)
leading
coefficient
assertions
that
= R. Choose
is regular,
of Lemma
term 1. It follows
the
R-automorphism
we have
the residue
field
of
/ to have
72 = deg /.
of / are in 222and
a = 0. Thus,
that
combination
R, and let
= u &eacute; m; and by replacing
by applying
assume
f &pound; R{X] whose
is one of the equivalent
ef7 can be written
all of the coefficients
such
R.
and
2.1.)
(i) =&pound;&gt; (ii): Trivial.
(ii) =^
Since
over
minimal
Claim,
R/222 is infinite,
/ by f/u
of R[X]
reduced
R/222 is infinite.
1, zf, • • • , f7, • • • , &ccedil;
degree
there
exists
we may assume
given
among
f &pound; min /. Since
not
a &pound; R
u - 1. More-
by X —►X + a, we may fur-
to the case
that
/ has
constant
that rf is a unit of R[f ]. By Lemma 3.1, (1, &lt;f,• • •, &iquest;jn l) =
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1972]
THE FINITENESS OF / WHEN R[X]/&iexcl; IS FLAT
(fm,---,
f&quot;+!)
for all i&gt; 0. In particular, for i = t-n,
zfz
(1, f, • • • , f &quot;~ 1) is free;
that
(1, ff, • • •, &lt;f&quot;_1)
pendent
over
R, for otherwise
quasi-local),
and hence,
since
is free. Therefore,
contradicting
some
(1, f, •••, f*+7 =
f is a unit of R[f ], it follows
1, f, • • • , f&quot;
proper
the choice
395
subset
forms
of /. Thus
must be linearly
a basis
(because
/ &pound; min /. Therefore
indeR is
(iii)
holds
by Lemma 2.1.
It remains
device
used
by R(Y),
have
to remove
in &sect;2
mR(Y).
the restriction
of adjoining
There
a commutative
that
R/m
be infinite.
an indeterminate
are then
a few things
Y, thus
that
must
We do this
replacing
by the
the ring
be checked.
R, 772
For one,
we
diagram
0 — / &reg;R R(Y) -, R[X] &reg;r R(Y) — (R[X]/I) &reg;R R(Y) -&gt; 0
]
1R(Y)[X]
0—
The top row is exact
isomorphisms.
implies
since
Let
is
there
exists
and hence,
Also
the
!
— R(Y)[X]//R(Y)[X]-&gt;0.
R-flat,
and the vertical
maps
R-module
(1, &lt;f, • • • , f).
Then
and since
R(Y)
we have
a regular
element
by Lemma
=^ (i):
2.9,
Let
R-flat,
argument
/ = a. + ajX
are
R(Y)-algebra
E is
R-free
an injection
for the ring R(Y) and
Therefore
by the above argument,
such
7R(Y)[X]
/ &pound; min (IR(Y)[X])
7 is principal
generated
that
by a regular
= /R(Y)[X],
element
of min /.
as above.
+ ...+a
flQ, • • ■, a,_ j &pound; ??z and
a, = 1. Since
each
t = n + i fot some
t &lt; n; so assume
is
&reg;R R(Y). Thus, (ii) is satisfied
(1 &reg; 1, &lt;f &reg; 1, • • • , (f &reg; 1)').
c(l) = R by the same
(iii)
is
R(Y)-free;
E &reg;R R(Y)—. (R[X]/7)
the R(Y)-module
R(Y)
E denote
E &reg;R R(Y)
i
R(Y)[X]
-&gt;
X&quot;.
/ &pound; min /,
Since
c(f ) = R, we may assume
(1, &lt;f, • • • , &lt;f')
i &gt; 0. By Lemma
is certainly
3.1,
(1, f, •••,
free for
f&quot;+z)
= &lt;1,f, - • • , f '&quot; &raquo;, f/+1+&iquest;,• • • , f &quot;+&iquest;&gt;.Clatm. I, &pound; ..., f'-\
f/+1+&iquest;,• • • , f &quot; +&iquest;
are linearly
,rf ~
independent
b,Z+ ,l+Z^.&Ccedil;l+l+i+...
over
R.
Suppose
&egrave;n + &egrave; ,rf + • • • + &egrave;,_
+ b72 +2.&lt;fn+&iquest;=0,
b.&pound;R,
=
J
+
and let g&deg; = b n0 + b ,X
+ • • • + b72+2.Xn+i.
1
Then g &pound; 1; and since
/ = /R[X],
g = hf fot some z? &pound; R[X]. Let 7&gt;= c0 + CjX +
•••
of h must
be
+ c. X1 (the degree
lar element
of R).
Equating
z since
coefficients
the leading
coefficient
of / is a regu-
of X , X + , • • • , X +1 in g = hf, we get
a set of 7 + 1 equations:
0 = c„a,+
c,a,
u /
any
c .a,
z I—j
0=c0a/ +1 +c1&laquo;r+e^/_1
+ ... +&lt;7&laquo;,_i+i
0= c a,
+ c.a,.
. + c a,
0 Z+2
(We define
,+■■■+
1 /— 1
a . having
subscript
. .+•••
1 l+i—1
less
than
2 z
zero
or greater
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
than
tz in these
396
[September
JACK OHM AND D. E. RUSH
equations
to be zero.)
unknowns
cQ, • • •, c ., the coefficient
elements
If we regard
of 272above
fore by Cramer's
these
the diagonal,
rule
equations
matrix
and hence
[L, p. 330], cQ = c
as
has
z' + 1 equations
in the
a, = 1 down the diagonal
z+ 1
and
its determinant
is a unit of R. There-
= • • • = c . = 0. Thus
h = 0, and hence
g = 0. Q.E.D.
If one imposes
(= R[X]//),
then
sense
to speak
then
be replaced
the mild restriction
l/cf
is
in the
of the ring
3.2(i)
implies
limits
by symmetric
R[&lt;f] and
and
R[l/&lt;f]
shall
next
demonstrate
then
following
a suitably
3.3
R'
(ii)
AR
Proof.
exists
mal ideal
P.
czR'. Since
Therefore
a prime
the equivalent
let
modules
conditions
when
in the case
that
Consider
this
of Theorem
r.1 &pound; R such
and that
homomorphism,
R' = R{b/a\
We
element.
R is a domain,
since
of R.
ideal
Then
of R which
there
exists
an
Then
that
R{a/b]
has
that
R.
lemma.
&ccedil;=
the lemma
when
satisfies
(iii)
be satisfied
since
+ ■■• + xjt(b/a),
properties.
Let
taking
use
/
X -&raquo; l/&lt;f.
^ R'.
f. &pound;
lemma,
(a, b)1
■• • , b*). Therefore
It follows
by lying
that
over,
b/a
is a
PR{a/b]
^
Q.E.D.
&lt;p: R[X] —&gt; R{X]/I
be the canon-
qS(X) and that we are denoting
for the kernel
=
PR
i &gt; 1. There-
b)1; so by Nakayama's
But then,
maxi-
(a, b)R
x. &pound; P,
b)1 tot some
a2-2b2,
A = (a, b)
with
if we had
&lt;p(R)[&lt;f] by merely R[g]. If f is regular in R[f], then &lt;p(R)[l/&pound;]
and we shall
set for A
R is quasi-local
+ ••• + r.(b/a)1.
z
the desired
technical
and recall
over
P O R = P, and
in a generating
a2 &pound; P(a,
a1' &pound;(ai~lb,
1 = r,(b/a)
1
is integral
a rather
R
1 = x xf x(b/a)
, • • • , b') + P(a,
that
that
a &pound; A.
to prove
of R, (ii) would
b1), and hence
a/b
such
element
of elements
we conclude
a2~2b
and the ring
rp(R)[l/ff]
3.2.
a regular
generated
ideal
P, we may assume
ideal
= R'.
■-.,
of R
it suffices
at
the ring
PR'
P
R -regular
element,
equation,
a2~2b2,
exist
ideal
By localizing
(a, b)1 = (a2~lb,
T(R[&lt;f]);
are then
properties.
A be a finitely
P be a prime
that
conversely
min / contains
ring has the required
on the number
a regular
We next need
ical
of whether
for some
assume
= t&laquo;*'&quot;**,
R{a/b]
these
and let
P is a maximal
From
unit of R
since
can
R-modules
It follows
arises
is the case
makes
3.2
the finite
thus
is well known
By induction
contains
there
since
R-flat
R be a ring,
= aR
zz regular.
fore,
l/rf
it thus
of Theorem
is R-torsion-free,
there
(iii)
R[X].
involving
and
(ii)
of R[&lt;f]
R such that
(i)
with
(i) and
are both
this
element,
ring of R[tf],
The question
valuation
Let
element
\ • &quot; ' &gt; 1} are isomorphic.
implies
that
lemma
a regular
R-algebra
R-flat
chosen
Lemma.
contains
which
are
quotient
conditions
R[l/tf]
&lt;f be a regular
Conditions
(l/&pound;)'~
of flat modules.
R[zf]
The
total
R[l/&lt;f].
(1, &pound;• •♦, &lt;f'} and ((!/&pound;)',
direct
that
of the homomorphism
the ring
is a subring of
R[X] —&raquo;
Then a QXn + a ,X&quot;- 1 + . • • + an, a . &pound; R, is in /* if
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1972]
THE FINITENESS OF / WHEN R[X]/l IS FLAT
and only if aQ + aX
subset
+ -'-+a
H of R[X],
degrees)
deg 77 (subd
of the nonzero
3.4
Lemma.
element,
Xn &pound; I. Recall
Let
H) denotes
elements
also
that,
397
for any nonempty
the minimum
of the degrees
which
I be an ideal
reduces
(sub-
of 77.
of R[X]
such
that
min / contains
a regular
let P be a prime ideal of R, and let if/: R[X] —►
(R/P)[X]
morphism
nonzero
coefficients
mod P.
//
zf is regular
be the homo-
in R[f]
and
R[&lt;f]
z's
R-flat, then deg / - deg xf/(l)&lt; subd xfAf).
Proof.
Let
/ be a regular
R-algebra
R
such
cR(f)R
that
which
is
element
of min 7. By Lemma
R-torsion-free,
= aR
for some
which
has
R -regular
a prime
element
3.3, there
P
lying
a &pound; cR(f).
exists
over
an
P, and
Also,
R'[X]//R'[X] ^(R[X]/7)
&reg;r R' is R'-flat by [B(a), p. 34, Corollary 2], and / • 1
&pound; min(/R
2.3
[X]) by Lemma
of R'[X]).
/jR
Therefore
coefficient
2.3, deg / = deg(/R'[X]).
that
P
over
of R
lying
of /.
will denote
fl &pound; R'[X]
is regular.
g = b0+b.X
g • 1 = hf
+ -''
Then
P, the homomorphism
IR'[X] =
that,
by Lemma
a. &pound; R', a^ 4 0, let /* =
+ • • • + b m &pound; I*,I the
+ b Xm &pound; I. Therefore
g* • 1=7)
the identity
suchthat
Note also
. For any7 g*
= b&ntilde;Xm
+ &egrave;.X&quot;2-1
72
&deg;
0
1
polynomial
h &pound; R [X] such
a subscript
If fl = aQ + a yX + • ■• + aXn,
+ ••• +a
corresponding
1 without
by Lemma 2.1 there exists
[X] and the leading
anX&quot;
+ a,X&quot;-1
01
(where
/ .. Since
xfj &quot;extends&quot;
there
there
exists
exists
a prime
to a homomorphism
if/
of R'[X]; and it follows that subd xfj'(I*R'[X]) &gt; subd xfj'(f*&iquest; = n - deg xfj'(f,).
Similarly, deg xfj'(&iexcl;RAX]) &gt; deg &lt;/&gt;'(/j) since IR'[X] = /jR'[X]. Therefore,
subd
xfj (I R [X]) &gt; deg / - deg xfj (IR [X]).
i/-'(/V[X])
tify
R[X],
with an appropriate
Theorem.
and let
element
conclusion
follows
since
= &lt;/&gt;(/*)(R7P')[X] and xfj'(&iexcl;R'[X]) = xfj(&iexcl;)(R'/P')[X] provided we iden-
(R/P)[X]
3.5
The desired
Let
f denote
of R[f],
then
subring
of (rVp')[X].
R, m be a quasi-local
ring,
the equivalence
of X in R[X]/I.
the following
class
is equivalent
let
I be a proper
ideal
of
If f is a regular
to conditions
(i)—(iii)
of Theo-
rem 3.2:
(iv)
min / contains
Proof.
a regular
element,
(=^0
(iii)
of Theorem
while
(i) implies
that
R[f]
direct
limits
(&lt;#=)
so a
of the modules
Let
principal
ideal.
a . of generators
z7 is as large
this
/ =
'01
is a regular
leads
a_+a,X
element
Since
and
3.2 implies
R[l/rf]
are
in (i) [B(a),
+ ...
+ a 72X&quot;,
for c(f)
from among
and
R[rf]
and
min / contains
R-flat,
since
are
R-flat.
a regular
these
element,
modules
are
2].
a 72 &iacute; 0,' he a regular
&deg;
2.1, it suffices
we can choose
the
R[l/f]
p. 28, Proposition
of R. By Lemma
R is quasi-local,
as possible
and
element
to show
a minimal
a ., and we do this
i'. &lt; &iquest;- &lt; • • • &lt; z,. We shall
to a contradiction.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
set
in such
assume
of min /,
c(f)
is a
a . , •••,
a way that
t &gt; 1 and show
398
JACK OHMAND D. E. RUSH
There
choose
[September
exist
b,. &pound; R such that a, = 2l. , b,.a . , 1 = 0, 1, • • • , n; and we can
z;
z
,=ll]ij'
= b.
= •. • = b . = I. Moreover , since
z. was chosen as large as
b.
21 1
possible,
222
&iquest;z.
itt
'
1
b
&pound; 722for /&gt; z'.. Then
/- ZVi. + IV,]* +■•■+
2X&laquo;,-.)*&quot;
z= l
= &laquo;f/fcoi
By applying
+ *„X
+ • ■• + &iquest;7nlX&quot;) + . • • + fl;^0i
the flatness
of R[&lt;f]
c . &pound; (a . , • • • , a .) : a . Cm
z
z2'
If
such
ij
(1)
via criterion
a .
by applying
1.2(b),
we conclude
that
there
exist
that
lH + *&raquo;i+Similarly,
+ bXtX + ■■■+ bntX&quot;).
+ *,lf'*Efif'
the flatness
of R[l/&lt;f],
there exist
d. &pound; (a . ,••• ,
) : a . C m such that
zz-1
2t
(2)
b0td/on
+ bxtd/ct)n-1
+.. • + *wf = &pound;
zi.d/f )&laquo;-.
1=0
If i/z denotes
the homomorphism
mod 222,we have
from
R[X] —►(R/t2z)[X]
(1) that
deg &lt;//(/) &lt; i,
since
which
b.
reduces
coefficients
= 1 and
b.
&pound; m tot
I &gt; i..
Similarly,
it follows from (2) and the fact that b . = 1 that subd xp(l ) &lt; n - i
(where
is the kernel
deg
/
ip(l) &gt; 22- i
a contradiction
3.6
R{X]/I.
inequality
to the assumption
We shall
contains
of the homomorphism
the first
now globalize
Summary.
Let
a regular
element,
Then
that
t &gt; 1.
and summarize
R be a ring,
and let
the following
(i) (1, I, • •• , &pound;')
R[X] —►R{l/&iquest;;]).
following
let
from Lemma
n - i &gt; n -
Therefore
i &lt; z,,
Q.E.D.
the preceding
/ be an ideal
&lt;f denote
Then
3.4.
local
results.
of R[X]
the equivalence
such
class
that
min /
of X in
are equivalent:
is a flat R-module for all t &gt; 0.
(ii) (1, &lt;f,• • • , rf') is a flat R-module for some t &gt; 0 for which 1, zf, • • • , &lt;f'
are linearly
(iii)
dependent
over
R.
c(l) = R; and for every
the leading
coefficient
prime
P of R, there
of / is regular
in R p and
exists
/ &pound; Rp[X]
such
that
IR p[X] = /Rp[X].
(iv) / = (fx, • • •, fn), f. &pound; min /, and c(l) = R.
(v) c(min /) = R.
(vi)
every
R[zf]
regular
is
R-torsion-free
and
c(g)
is invertible
for some
g &pound; min /).
If &lt;f is regular
in R[f],
then
(i)—(vi)
are equivalent
(vii) R[f] and R[l/&lt;f] are R-flat.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
to
g &pound; min / (or for
1972]
THE FINITENESS OF 7 WHEN R[X]/&iexcl; IS FLAT
If R is integrally
closed,
then
(i)—(vi)
are also
399
equivalent
to
(viii) R[f] is R-flat.
Proof,
localizes
(i) &laquo;^ (ii) &lt;#&gt; (iii):
Apply Theorem 3.2 and the fact that flatness
well [B(a), p. 116, Proposition
15].
(iii) &lt;\$&gt; (iv) &lt;#&pound;►
(v): Apply Theorem 2.1.5(vi) &lt;^&gt; (iii):
Suppose
is Rp-torsion-free
by Lemma
if it is finitely
[B(a),
generated,
p. 148, Theorem
and is generated
Therefore
cR
(g/l)
4].
follows
2.12.
Recall
Then
that
a regular
Therefore
c(g)
element.
from Lemma
g &pound; min /, g/l
is principal
therefore
is satisfied.
contains
by a regular
(iii)
any regular
(vi)
by Lemma
an ideal
element,
2.1.
is invertible
implies
g/l
if and only
Conversely,
principal
c(g)p
&pound; min(Ip)
is principal
by Lemma
suppose
fot any prime
2.1, and hence
P of R, R[f]p
and is locally
is invertible
Moreover,
&pound; min (/p)
for any prime
(iii)
is valid.
P of R, by Lemma
c(g)
is locally
2.3For
2.3-
principal
Then
and
invertible.
(vii) &lt;^- (iii): Apply Theorem 3.5.
(viii) ^
(iii): Apply Corollary 2.13-
The simplest
case
deg min / = 1, i.e.
to which
the above
min / contains
a regular
Q.E.D.
results
apply
element
f is in the total quotient ring of R. Then rf = a/b,
is isomorphic
as an R-module
(1, f, • • • , f')
Theorem
grally
(viii)
domain
We have
shown
min / contains
pal.
We conclude
3.7
&sect;3
Theorem.
Let
Proof.
Let
that
field
that
[B(a), p. 148,
if D is an inte-
of D, then
D[f]
as a very special
if R is quasi-local
then
generalizations
a regular
element,
is flat
case
of our
R[X]/I
is
of this
ring,
integer
is
I bean
R-flat,
such
If t = n this
that
X&quot; be a regular
c(f)
is immediate,
= (a , a
element
,,•••,
closed
/ is princi-
ideal
then
of R[X]
IR[X]
is
72
)r..a.,'j
&iquest;^
j
z = 0, 1, ■■■ , t — 1,
i=t
and
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
of min /, and let
a ). We shall
so we may assume
r z;.. &pound; R such that
a.=
1
implies
result.
and let
&iexcl;f R[X]/I
integrally
R-flat
](R) C R, then i is principal.
/ = a0 + fljX + ...+a
the largest
= a(R.
2.13
R, m be a quasi-local
If in addition
c(f)R
may be regarded
element,
with some
principal.
that
3] has proved
is in the quotient
which
in Corollary
min / contains
exist
f = a/b
a regular
that
/ denote
, • • • , a ) = (a, b)1. Moreover,
(vi).
and
such
and
(a, b) is invertible,
=^
a, b &pound; R; and (1, tf, • • • , rf')
(b , ab
[R, p. 797, Proposition
where
1, or equivalently,
is flat if and only if the ideal (a, b)' is invertible
4]. Richman
closed
implies
to the ideal
is the case
of degree
prove
t &lt; n. Then
there
400
JACK OHM AND D. E. RUSH
[September
zz
/-&iquest;a/(ro/-
+ 'i,Jf+&quot;-
+ ri-iyx&lt;&quot;l+x/)-
7=2
If tf denotes
the equivalence
class
r0z + rlz^+---
of X in R[X]//R[X],
+ rz-lz^_1
= Hat +X, •••,&laquo;„):
the equality
being
and Criterion
a consequence
1.2(b).)
Thus
+?
6{at+V
then
/(&lt;f) = 0 implies
■■■,an)R{c;]:at
at&Iacute;R{cf]C &laquo;R[f ],
of the
R-flatness
of R[f].
(See
g = r Q( + r X[X + • • • + rt_XtXt~l
Theorem
2.18
+ X1 &pound; mR{X] +
IR{X].
Let
272be anyJ maximal
R'[X]//R'[X]
by Lemma
ideal
is R'-flatby
2.3-
Since
p. 22, Proposition
R
of R, and let
R = R—,
222'= 222R— Then
mm
[B(a), p. 116, Proposition 15], and / • 1 &pound;min(/R'[X])
is quasi-local
l6l and since
and integrally
closed
R' C T(R)— C T(R-),
IR [X] is generated
element
h' &pound; min(/R [X]) with cR'(h ) = R' by Theorem
content
formula
element
of the
Claim,
that
q = t. Let
coefficients
c(f)R
is principal
form a . 1, where
cp: R [X] —►(R /m'){X]
mod 222'. Then
cp(b')(R'/m'){X].
and hence
t &lt; q &lt; 22 and
m of R, c(f)R = cz^R. Therefore
c(f)R
= a R ; and since
+ • • • + a 72Xn and
For the second
d.l &pound; R such
zzZ. = a.a„
Z
assertion
that
a.l = d.1 mod J/,
from our previous
coefficients
mod /.
this
regular
of R .
which
reduces
cp(g • 1) &pound;
Z
that,
d. &pound; R. Let
since
for an arbitrary
maxi-
by Lemma 2.1, /R[X] = /R~[X], where
it will
suffice
to show
ideal
only&iquest;_ to show
i/z: &szlig;[X]
r.
that
a,Z+F'tz
,,■■•,
for any maximal
q' = t. It remains
_
Then
is true
z = 0, • • • , 22. That
observation
a. = d. mod / for some
reduces
by some
the identity
a.Z &pound; R.
of the theorem,
c(f)R—
&lt; q —
&lt; n, then
'
772 = a 9 R—
772 for t —
have
by an
from the
Since deg cp(g • 1) = t, deg &lt;p(h') &lt; t. But deg (/&gt;(*') &gt; q &gt; t, so
mal ideal
follows
is generated
1 denotes
be the homomorphism
q = t. Thus,
exist
2.7. It follows
g • 1 &pound; m'R [X] + IR [X] implies
therefore
/'01= a. + a,X
in T(R)— by [B(c),
—♦ (R//)[X]
a
there
ate
in
_
/
J
272of R, if
that
for i —
&lt; t we
be the map which
+ r. X + • • • + r
Xl~
+ X1 &pound; IR{X]
+ ]R{X] and IR{X] = /r[X],
we have z/z(rQi+ 2-X(X+ • • • + x') &pound; iff(f)(R/J)[X]. But
i/KVoz + ''i^X + • • • + X ) and
xpif) have
if,(r.t) = z/z(a.) for z = 0, • - • , t.
Note
alent
to the
rem 2.18.
seem;
that
the hypothesis
(apparently
Also,
for at least
in the rather
and are both monic,
R[X]//
hypothesis
J(r)
is
R-flat
that
in the above
R[X]//R[X]
C R is not as contrived
trivial
case
of a 1-dimensional
D, ?22,it can be seen that
&iexcl;(D) C D it (and only if), for every
D{X]/I is D-flat implies
/ is principal.
The proof
of this
remark
so
Q.E.D.
that
weaker)
the hypothesis
the same degree
(which
is inspired
by [E,,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
theorem
is R-flat,
as it might
quasi-local
ideal
p. 345])
is equiv-
by Theoat first
domain
/ of D[X],
requires
a few
19721
THE FINITENESS OF / WHEN R[X]/l IS FLAT
preliminary
observations.
the ideal
For elements
a, b in a ring
{r &pound; R \ ra &pound; bR \. A criterion
overring
R we denote
of [A, p. 801, Theorem
by (bR : a)
l] says
that
an
R' of R, R' C T(R), is R-flat iff (bR : a)R' = R' for every a/b &pound; R' ; and
in the case
reduces
of a simple
extension
to the following:
or equivalently,
R = R[x/y],
R[x/y]
is
R-flat
if and only if \/(yR
it is easily
and hence
Lemma
= R[x/y].
that
Now let
this
(x, y) is not principal
= R[x/y],
if x/y
R[Xl —&gt;R[x/y]
criterion
is a unit of
defined by
(x) 4 U, y) / (y);
and it follows
from
/ is not principal.
D, 772be a 1-dimensional
the above
the form
then
that
Also,
then x/y, y/x I R. Therefore,
if R is quasi-local,
2.1
seen
if and only if (yR : x)R[x/y]
: x)R[x/y]
T(R), if / is the kernel of the R-homomorphism
X —&raquo;x/y, and if y/x &pound; R\R,
prove
401
remark
1 + r (x/y)
we need
quasi-local
only find
+ ■ • • + r^x/y)&quot;
x/y
domain
&pound; T(D)
= 0, with
r.&pound;
such
that
satisfying
](D)
t- D. To
an equation
of
m ( = -\J(yR : x)), and such
that y/x &iquest; D; for then y/x &pound; d\d
and yJ(yR : x)D[x/y] = D[x/y].
If mD &gt; m
then
772Dj&pound; D and
&pound; D and
it D.
Then
(ca/b)&quot;
b/ac
has
there
exist
+ • • • + c&quot;r
1
the required
](D)\D
suchthat
required
properties.
properties.
(a/b)&quot;
3.8
Proposition.
the nil radical
regular
is another
for noetherian
rings
Let
&iexcl;f R/A
72 &laquo;
let
closed
case
there
x/y
p. 445,
and
=
a/b
&pound;
such
that
R[X]/l
is
'
exists
= b/a
has
2.13.
A similar
Corollary
A he an ideal
of R[X]
x/y
z
of Corollary
in [Na2,
R be quasi-local,
is integrally
ca/b
if mD = m, then
generalization
I be an ideal
that
r. &pound; D; and hence
_
n, and in this
can be found
of R, and let
element,
c &pound; m such
= 0 fot some
Similarly,
&pound; ttz for some
The next proposition
statement
a/b
+ cr ,(ca/b)n~
the
4.3(iv)].
of R contained
in
min / contains
R-flat,
then
a
I is
principal.
Proof. Let M = R[X]/L Since M is R-flat, 0 -&gt; I/AI -&raquo; (R/A)[X] -*M/AM
—&raquo;0 is exact and M/AM is R/A-flat.
a principal
2.19,
ideal
of (R/A)[X].
it follows
4. The
hypothesis
examples
from Nakayama's
case
that
that
of arbitrary
ever,
to what
rings
R the condition
difficulties
extent
of &sect;2
/ or R[X]//.
Thus,
they nonetheless
We remind
Moreover,
free
this
lemma that
a regular
are false
condition
&sect;3
by means
although
give
that,
&lt;##&gt; projective
of &sect;2
of various
of this
perspective
R-module
for R quasi-local;
and
and indeed
this
In studying
for any
ideal
by Theorem
/ is principal.
hypothesis.
can be weakened,
the results
an interesting
the reader
generated
element,
without
can be dropped.
and
2.13 and Lemma 2.3, I/AI is
/ is a finitely
/. Most of the theorems
min / contains
the theorems
By Corollary
Since
&sect;3
involve
we have
One can
and in particular,
these
questions,
finiteness
section
by
ask,
how-
for what
assumptions
the
on either
superficial,
problem.
M, free =&pound;&gt; projective
and if M is finitely
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
shown
we avoid
are somewhat
to the general
the
=&pound;►flat.
generated,
402
JACK OHMAND D. E. RUSH
then projective
local
rings,
&laquo;#&gt; flat
and noetherian
As usual,
shall
begin
4.1
for a large
R[X]
rings.
denotes
Let
We refer
remarks
A be an ideal
(i) A is generated
of rings
including
the reader
the polynomial
with a few simple
Lemma.
class
[September
domains,
to [B(a)]
quasi-semi-
for these
ring in one indeterminate
facts.
over
R. We
on idempotents.
of R, and consider
the following
statements:
by an idempotent.
(ii) A is locally trivial.
(iii) A =A2.
Then (i) =^ (ii) =^ (iii); and if A is finitely
general,
(iii)
valuation
=&gt; (ii).
Take
A to be the maximal
ideal
then (iii) =^ (i). (/t2
of a nondiscrete
rank
1
ring.)
Proof,
(i) =^ (ii)
a quasi-local
the same
ring,
follows
either
localizations
4.2
Lemma.
Let
that,
e = 0 or e - 1 = 0; (ii) ^&gt; (iii)
and (iii)
3]. (See also
R
R . If c(l) satisfies
Supp / is open
from the observation
are equal;
by [B(a), p. 83, Corollary
then
generated,
=4&gt; (i) when
for any idempotent
since
two ideals
A is finitely
[B(a), p. 172, Exercise
be a polynomial
ring over
(Recall
that
with
generated
15].)
R, and let
I be an ideal
4.1(ii), then Supp / is open; while if c(l) satisfies
and closed.
e of
Supp / denotes
the set
of
4.1(i),
of primes
P' of R' such that lp&gt; 4 0 [B(a), p. 132].)
Proof.
Let P' be a prime of R', and let P = P' C\ R. If c(I) satisfies
(ii),
I p. 4 0 &lt;\$&gt; lp 4 0 &lt;&amp;&gt;c(lp) = Rp &lt;^ c(l) t P &lt;#►c(I)R' \t P', the nontrivial
implications
regular
all following
element.
from the fact
If in addition
c(l) = eR
l-eeP^(l-e)fi'CP'.
asserts
open and closed
in the proof
4.3
(i)
free ideal
that if c(I)
= Rp
tot an idempotent
of R , then
is generated
set and hence
of the next
Theorem.
c(/p)
implies
Ip
e, then
contains
a
c(I) f_ P ■&lt;\$&raquo;
Q.E.D.
If / is a locally
ma 4.2
that
Let
/ is a projective
rk / = 0 or 1. For such an ideal,
by an idempotent,
rk / is locally
constant.
then
Lem-
rk / = 1 on an
We need
this
observation
theorem.
I be an ideal
of R[X].
Then
ideal of RlX], and Ai)
the following
is generated
are equivalent:
by an idempotent.
(ii) R{X]/I is R-flat, and I is a finitely generated ideal of R[xl.
(iii)
/ is a finitely
generated
flat
ideal
of R{X], and
c(l)
is generated
by an
idempotent.
Proof,
(i) ==&gt; (ii):
Since
/ is R[X]-projective,
it is locally
free at primes
R[X] by [Ka2, p. 374, Theorem 2]. Therefore by Theorem 1.5, R[X]//
By Lemma
4.2, rk / is locally
[V2, p. 431, Proposition
constant,
and hence
/ is finitely
1.4].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
of
is R-flat.
generated
by
1972]
THE FINITENESS OF 7 WHEN R[X]/&iexcl; IS FLAT
(ii)
=^ (i):
Lemma
By Proposition
4.1, c(I)
1.6, / is locally
is generated
4.2, rk / is locally
free;
by an idempotent.
constant,
and hence
403
and by Corollary
By the remark
by [B(a),
1.3
following
p. 138, Theorem
and
Lemma
l], / is projec-
tive.
(i) and (ii) =&pound;&raquo; (iii):
Projective
implies
(iii) =^ (ii): If / is R[X]-flat,
tion 15; and p. 167, Exercise
flat by [B(a), p. 28].
then it is locally
3-e]. Therefore,
free [B(a), p. 116, Proposi-
R[X]//
is R-flat by Theorem
1.5- Q.E.D.
Note
trarily
that
the above
many indeterminates
(i).
Also, Example
4.3
are essential.
4.4
R
proof
such
2.21
except
shows
Theorem.
Let
that
is a finite
R'//
would
R
carry
through
for a polynomial
for the appeal
that
the various
be a finitely
to Proposition
finiteness
generated
projective
then
in arbi-
1.6 in (ii) =^&raquo;
hypotheses
R-algebra.
R-module,
ring
of Theorem
If I is an ideal
I is a finitely
of
generated
ideal.
Proof.
Let
of x . in R //
polynomial
ated
l/L
R = R[x
satisfies
an R-integral
/.(X) &pound; R[X] suchthat
it splits.
a finitely
gether
with
Therefore,
generated
/j(*j),
&lt;4-5)
(2)
(3)
mains;
and,
rings.
generated
'
finitely
are finitely
exists
a monic
the ideal of R' gener-
R-module.
The sequence
and since
generated
implies
l/L
in / of the generators
form a generating
set for /.
has shown
0 —►
R'//
is
is
of l/L
to-
Q.E.D.
that the following
are
ideals
(4.5) include
in keeping
flat
R-modules
generated
are projective.
projective
R-modules
generated.
Projective
of R are finitely
noetherian
with the terminology
We can obtain
characterization
there
the image
of R-modules;
is finitely
2.1]
R-module,
R:
Locally
Rings satisfying
A(0)
is a finite
Pre-images
(x ) then
(1) Finitely
,
and hence
sequence
[V-, p. 432, Theorem
for a ring
is a finite
&pound; I. If L denotes
R /L
R-module.
•••,/
Vasconcelos
equivalent
R /L
—&gt; R // —&gt; 0 is an exact
R-projective,
R //
equation,
fix.)
by f Ax A, • • •, f (x ), then
—* R /L
also
.,
• ••, x ]. Since
as a corollary
of such rings,
but first
The following
property
rings,
generated.
quasi-semi-local
of [CP],
to Theorem
we need
we shall
4.4
rings, and docall
a rather
a well-known
these
rings
interesting
intermediate
char-
acterization.
4.6
Lemma.
of a ring
R is equivalent
to the proper-
ties of (4.5):
(4) Every
locally
trivial
ideal
Proof. (1) =\$&raquo; (4): Let A be locally
of R is finitely
generated.
(0) or R p. Then R/A is locally
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
free
404
JACK OHMAND D. E. RUSH
and hence
R-flat.
Therefore
by (1), R/A
0 —&raquo;A —&raquo;R —&raquo;R/A —&raquo;0 splits,
(4) ==&gt; (1):
locally
fore,
Let
(0) or Rp
by [V.,
4.7
Theorem.
flat
p. 506]
by [V.,
R-projective.
Thus
the sequence
and it follows that A is principal.
M be a finite
M is projective
is
[September
R-module.
and hence
The invariant
are finitely
p. 506, Proposition
The following
property
factors
generated
of A4 are
by (4).
There-
1.3].
of a ring
R is equivalent
to the proper-
ties of (4.5):
// / is an ideal of R[x]
R-module,
Proof.
then
such
that
I is a finitely
R{x]/l
generated
is a finite
ideal.
By Theorem 4.4, (1) =&pound;&gt; (5). It remains to show (5) =&gt; (4). Let A be
as in (4), and let / = AR[X] + XR[X]. / is finitely generated
so we must
show
(0) or (1),
R/A
generated.
that
by &quot;cyclic
is locally
the proof
flat&quot;
&quot;finite
such
theorem
that
implies
4.7
=
R/A;
R-flat.
if and only if A is,
and since
Therefore
A is locally
by (5), / is finitely
should
1.6
1.6
shows
free
at primes
tive
and
ring
R-flat
then
probably
and thus
that
allow
R have
of R[X].
Under
c(I) = R, the corollary
R[X]
flat&quot;
can be replaced
can be replaced
by a poly-
the real problem
is to re-
the property
/ is a finitely
is
&quot;yes&quot;
that if / is an ideal
generated
when
ideal?
The main
R is a domain.
The next
be that of a quasi-local
to ask whether,
generated
at primes
us to delete
make it a true converse
an affirmative
&quot;finite
Of course,
weakened
finitely
would
that
in 4.7. Thus,
A(0)
can be further
question
tion
shows
of (5) and that
that the answer
/ is locally
to this
position
answer
the stronger
to the next
R.
for a ring
of R[X].(3)
R,
also
hypothesis
1.5.
Proposi-
/ is then
R{X]/l
shows
of Propo-
Moreover,
imply that
that
proposition
R{X]/I
An affirmative
part of the hypothesis
to Theorem
would
ring
of
that
is
locally
R-projec-
/ is,
in fact,
generated.
4.8
ideal
is
shows
to investigate
finitely
is
many indeterminates.
R{X]/I
of &sect;2
answer
of Theorem
Does every
The problem
R-flat
R{X]/I
free and hence
flat&quot; by merely &quot;flat&quot;
Question.
R[X]
generated.
in the statement
ring in finitely
place
case
/ is finitely
Q.E.D.
Note
nomial
flat
Proposition.
of R[X].
Then
Let
R{X]/l
R, m be a quasi-local
is R-free
(if and)
ring,
and let
only if there
I be a nonzero
exists
a monic
f &pound;
R[X] such that I = /R[X].
Proof.
R{X]/I.
c(l) = 0 or R by Corollary
1.3; and since
Since M is free and R is quasi-local,rk
(3) Heinzer
and Ohm have now proved
ple involving
a quasi-local
R which shows
/ /= 0, c(l) = R. Let M =
M = rkp/m M/mM [L, p. 418,
this-see
footnote
2. They also give an examthat the above question
must be modified.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
is
1972]
THE FINITENESSOF / WHEN R[X]/I IS FLAT
Proposition
8l.
Moreover,
since
M is
R-flat,
—*VM/mM —♦ 0 of R/?72-modules is exact.
hence
xfj is not an isomorphism.
Then
by Theorem
ing coefficient
polynomial
3.2,
and hence
4.9 Corollary.
and
R[X]/I
Also,
that
M/mM
and generated
M finitely
is generated
0 —►I/ml
But c(l) = R implies
It follows
/ is principal
is regular.
the sequence
405
by a monic
is a finite
by some
generated
—►(R/?72)[X]
I/ml 4 0, and
R/m-module.
/ &pound; R[X]
implies
whose
/ contains
lead-
a monic
polynomial.
Let R be a ring, and let I be an ideal of R[X]. // c(l) = R
is R-projective,
then
R[X]/l
is R-finite
and
I is a finitely
generated
ideal.(i)
Proof.
M = R[X]/I
is
R-projective
of R. Since c(Ip) = R _, 7^0
form 1, f/l,
• • • , (f/1)',
so there
ly at
P for every
P.
with
prime
c(l)
The appropriate
(ii)
4.3, while
[Naj,
is a finitely
in the questions
gives
this
section
the
5.1
Lemma.
Let
prime
c(l) = R imM local-
7 is finitely
in this
shown
I is a projective
if R[X]/l
is
paper.
R-flat.
of R[X]
Moreover,
assertion
the second
let
X
X .,•••,
is
that
proof
(i)&laquo;#^&gt;
assertion.
he indeterminates,
X ]/I is R-flat.
if R is a valuation
theorem
than the theorem
first
ring,
aroused
then
our interest
of this
theorem
states,
and we shall
actually
devote
of his proof.
of Nakayama's
the intersection
M be an R-module,
ideal
generalizes
Nagata's
theorem:
is R-finite.
of the first
that
this
and elaboration
of R, i.e.,
is the following
lemma.
Recall
of the maximal
let A be an ideal
that
ideals
](R)
of R.
of R contained
in
}(R), and let M' be a submodule of M such that M = M' + AM. If M = 0M ,
where
geneous,
esis
{M .} is a family
then
P
of the
generate
and then
X ] such that R[X „•'••',
3] has
with a generalization
/-radical
4.9
and it was
more information
to an analysis
We begin
denotes
ideal;
treated
considerably
1, cf, • • • , f
then R[X]/I
R be a ring,
p. 164, Theorem
generated
if and only
Corollary
Let
Then
generalization
and let / be an ideal of R[X ,,•■•,
Nagata
ring.
is R-projective,
theorem.
that
of Vasconcelos
nonnoetherian
of our Theorem
5. Nagata's
[V,]
by an idempotent
if c(l) = R and R[X]/I
t such
M = (1, &lt;f, • • • , f'\,
R be a noetherian
generated
fot every
Q.E.D.
of the paper
Let
exists
Therefore,
by Theorem 4.4.
Theorem.
R p-ttee
where R[X]// = R[zf]. But the hypothesis
rk M i s bounded,
The content
Mp is
and hence, by 4.8, Mp has a free basis
plies
generated
implies
of finitely
generated
submodules
of M, and
M' is homo-
M= M .
(4) We are indebted
to W. Heinzer
for showing
from the original
version
of this corollary.
us how
to delete
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
an extraneous
hypoth-
/
406
JACK OHMAND D. E. RUSH
Proof.
lows
that
Let /VI.= M n M .. Since M = M + AM and AI is homogeneous,
M . = M'. + AM .. But then
hence M = Ai'.
(R//)-module
(where
Thus,
satisfies
lar,
if R/J
R satisfies
noetherian
and
R[Xj,...,
X ]//
/ = ](R)),
then
ideal
R-module
zV is
the properties
if R//
such
Nakayama
theorem
then
This
includes
homogeneous
that
N/JN
R-projective
of (4.5),
is noetherian.
This
/ is any
ideal
[Vj,
it fol-
lemma,
and
is a projective
p. 508, Theorem
R does
also;
observation
the statement
of R[X ,,•••,
is R-flat, then / is finitely
2.1].
and in particu-
will
be used
that
if R/J
X ] such
in the
is
that
generated.
Let F = &copy; F., z = 0, 1, 2, •'•'•', be a graded ring which is a
generated
erated
flat
theorem.
5.2 Theorem.
finitely
generated
(4.5)
of the next
geneous
M . = M. by the usual
Q.E.D.
If N is a finitely
proof
[September
algebra
over
of F such
F f&iacute;. If F Aj(F
that
F/M
is a flat
A is noetherian
F „-module,
and
then
M is a homo-
M is a finitely
gen-
ideal of F.
Proof. Let / = J&Iacute;F A. Then F/JF
and therefore
of F/JF.
clude
F/JF
Taking
that
is noetherian
pre-images
M contains
and
is a finitely generated
M/(JF
and using
a finitely
the fact
generated
algebra over FQ/J,
O M) is a finitely
that
generated
ideal
/VI is homogeneous,
homogeneous
ideal
M
we con-
of F such
that
M = M' + iJF n M). Since F/M is Fn-flat, JF n M = JM; and hence M = M' + //VI.
Let AL = F . DM. Since
F ./M. is F.-flat
ule by [B(b),
for each
AI. is a finite
Example
such
in &sect;4,
[B(a), p. 28, Proposition
p. 10, Corollary].
i &gt; 0, and so each
Hence each
some
2.21
shows
condition,
it seems
quite
The next
theorem
stronger
hypothesis
device
Thus,
shows
than
&copy;z&gt;0
(P/^A
Thus, each
2l. Also, each F . is a finite
as remarked
of the sequences
that
above,
F AM ■ is
in Theorem
to us,
that
this
F/M
of homogenizing
that
5.2.
F .-flat.
by introducing
Q.E.D.
A is noetherian,
as We have
in the case
assumption
can be done
being
F AJ(F
However,
at least
the homogeneous
FQ-mod-
F„-projective
0 —►Af . —&gt;F . —&gt;F ./Al. —&gt; 0 splits.
the assumption
conceivable
that
F/M =
FQ-module; so AI = M by Lemma 5.1-
is needed
in one indeterminate,
tomary
M is homogeneous,
of a polynomial
can be deleted
provided
one imposes
The proof
or
mentioned
of this
a somewhat
involves
a new indeterminate,
ring
from 5.2.
the cus-
and then
ap-
plying Theorem 5.2.
5.3 Theorem.
finitely
generated
be the canonical
flat
F „-module
Let F = &copy; F ., i' = 0, 1, 2, • • • , be a graded ring which is a
algebra
over
homomorphism.
for each
F
let
M be an ideal
If E AJ(F
k &gt; 0, then
of F, and let
A is noetherian
M is a finitely
generated
and
2-_&ntilde;
cp: F —►F/M
cp(F ■) is a
ideal.
Proof. Let Z be an indeterminate; let Fh = &copy; FhkZk, where F* = &pound;JLfl F .;
and let Eh = &copy; EhkZk, where' E* = 2k{=Q cp(F .). If &lt;pA:Fh -&raquo; Eh denotes the
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
19721
THE FINITENESS OF 7 WHEN R[X]/&iexcl; IS FLAT
graded
F.-algebra
kernel
M
that
M
of cp
der first
generated.
which
We shall
the case
modules
f denotes
regular
element,
are also
there
then
are indispensable).
no longer
of a flat
lows
5.3-
noetherian
and
rem 5.3 that
has pointed
module
(b) submodules
closure
R[X j, • • • , X ]. Any finite
is a ring satisfying
direct
(a) and
[E2, pp. 116—117,
is
modules
R-flat
noetherian
Note
state-
ring is
ring,
over a valuation
that
theorem
fol-
(a) R/](R)
is
it follows
from Theo-
generated
satisfies
direct
5 and Corollary]).
are neither
such
then
domains
R is a finite
on min /
is a valuation
since
ideal
(b); and
sum of Pr&uuml;fer
also
a
), k &gt; 0,
closed
/ is a finitely
sum of Pr&uuml;fer
) studied
the corresponding
is why Nagata's
are flat,
implies
f
(1, f, • • • , f
F
if R is any ring
(b), then
Theorem
which
of 5.3, when
the
min / contains
over an integrally
and this
then
and the condition
out to us that
is flat;
of flat
X ]//
and
the modules
Consi-
If / is an
(1, &lt;f, •••,
S . _0 cp(F .) are flat,
More generally,
R[Xj,--.,
domains
implies
the
Q.E.D.
cases.
of X in R[X]/I,
closed
5.2
under
generated.
special
the modules
the integral
the modules
ring any submodule
from Theorem
class
are just
in the notation
implies
M is finitely
of M
in one indeterminate.
ring in 2 indeterminates
However,
is flat
R[X]
then the
from Theorem
image
5-3 in some
if R is integrally
that
W. Heinzer
valid.
F/M
5.3
that
shown
ment for a polynomial
Pr&uuml;fer
ring
R[f ] is R-flat
(and we have
that
Theorem
the equivalence
X . _Q cp(F .) of Theorem
We have proved
(see
1, it follows
of a polynomial
and
of F ; and it follows
M is the homomorphic
by discussing
in &sect;3.
then
Z to
by &lt;/&gt;(2 ct.Z1) = 2 cp(a )Zl,
ideal
Since
takes
conclude
of R[X]
defined
is a homogeneous
is finitely
homomorphism
ideal
homomorphism
407
that
if R
domains
there
nor quasi-semi-local
of
exist
and which satisfy
(a) (see [EO, p. 348, Example 4.5]).
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DEPARTMENT OF MATHEMATICS, LOUISIANA STATE UNIVERSITY, BATON ROUGE, LOUISIANA 70803
Current
Riverside,
address
California
(David
E. Rush):
Department
of Mathematics,
92502
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
University
of California,