Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
One may ask whether rings with completely normal spectrum can be characterized in a way
similar to rings with normal spectrum. A first approach is to talk about the maximal ideals in all
quotient rings As of A, where s A . In this way one obtains an axiomatization, but it is not first
order since the equality of two elements in As can only be expressed using quantification over
natural numbers. This is exactly the same problem as to recognize whether an element belongs to
the radical of some principle ideal, which requires in general that one can talk about some
unspecified power of an element – once again quantification over natural numbers is needed.
Here is a characterization that uses quantification over natural numbers.
Proposition
The ring A has completely normal prime spectrum if and only if the following statement holds:
 a,b A :
 y, z A:a
2

 y  a  b  & b 2  z  a  b 
 ( c, d A s,t A u, v, w A k  :

c  s  a & d  t  b & c  d  0 & a  b   u  c  v  d  w  a  b )
k
Proof
First assume that the prime spectrum is completely normal, pick two elements a,b A such that
a 2 ,b 2 a  b . The subspace D a  b   D a  D b  is normal, i.e., Aa  b is a ring with
ab
Aab . The previous Proposition shows that there are
normal prime spectrum, and
1
c
d
c
d
c
d
 a
 b
such that:
 0 and
i   ,
j   ,
i ,
j Aab
i 
a  b   1  a  b   1  a  b  a  b j
a  b  a  b 
 c
d
a b
Aab  
,
,
i
j
 . It remains to translate these conditions into the ring A:
 a  b  a  b  1 




There are some p  and some s A such that a  b  c  s  a  b
p
i p
There are some q  and some t A such that a  b  d  t  a  b
j q
q
a .
b.
There is some h  such that a  b  c  d  0 .
There are e, f , g,l  and u,v,w A such that
h
1
u

c

v

d
a  b  a  b  a  b  a  b 
e
i
f
j

w
a  b 
g

ab
,
1
a  bei  f  j gl  u  a  bf  j gl  c  v  a  bei  gl  d  w  a  bei  f  j l  a  b .
Once the numbers p, q, h and l have been found, they can be replaced by any larger numbers.
p
Therefore one may assume that l  p  q and h  p  q . Replacing c by a  b   c , s by
page 1 of 9
Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
s  a  b 
i p
06.02.2016
, d by a  b   d , t by t  a  b 
j q
q
, u by u  a  b 
f  jg
, v by v  a  b 
ei  g
and w
by w  a  b 
one gets all the desired equalities.
For the converse, pick two incomparable prime ideals p and q. There are elements a  p \ q
and b q \ p . Setting a  a  a  b  and b  b  a  b  one obtains representations
ei  f  gl
a  b  a  b  , a 2  a 2  a  b  , b 2  b 2  a  b . Now
2
c,d A s,t A u,v,w A k  :
c  s  a & d  t  b & c  d  0 & a  b   u  c  v  d  w  a  b
k

It follows that a  b  p  q , c  p \ q , d q \ p and Gen  p  Gen q   D d  D c    .
Thus the spectrum is completely normal. Ω
An example shows that the rings with completely normal prime spectrum do not have a first
order axiomatization. Recall that an elementary class of structures is always closed under the
formation of ultraproducts (cf. [ChaKe], p. 173, Theorem 4.1.12).
Example
n
Let K be any field and, for each 1  n  , let An  K X,Y  X,Y  . The spectrum of each ring
An has only one element, hence it is completely normal. The direct product is denoted by A. Let
U be a free ultrafilter on  , and let B  A U . It will be shown that the prime spectrum of B is
not completely normal.
b B
Some
notation:
If
and
if
is
a
representative
then
a A
 a   n  1  n & a n   0 . Note that b  0 if and only if  a  U .

The first step is to prove that Spec B  has only one minimal prime ideal. So, let b1 ,b2  B be
two elements whose product is nilpotent. It must be shown that one of them is nilpotent.
k
Suppose that b1  b2   0 .





One picks representatives a1 , a2  A . Then  a1  a2  U . For each n  a1  a2 
k
k
 one
a1 n  a2 n  0 in An . Let Fi,n be a polynomial representing ai n  . Let v be the order
k
k
valuation of K X,Y . Then F  X,Y   0  X,Y  An if and only if v F   n . It follows
n
   
v F  n .

n

 
k
k
2k
 v F2,n
 n for all n  a1  a2  . For each such n it follows that v F1,n
 n or
that v F1,n
k
2k
2,n


 n  1  n & F
 a1  a2   n  1  n & F1,n  F2,n  X,Y 

k
2k
1,n
k
n
X,Y 
n

 n  1  n & F
2k
2,n
X,Y 
n
 U
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
shows that the union of the two sets belongs to the ultrafilter, hence one of them belongs to U .
Thus b12k  0 or b22k  0 , and the proof of the claim is finished.
The canonical images of X and Y in B are denoted by x and y. The homomorphism
f : K X,Y  B , X  x , Y  y , is injective: If f P   0 then the set
n  P X,Y  

n
belongs to U , and P belongs to infinitely many ideals X,Y  , which implies that P  0 .
n
If P K X,Y  \ X,Y  , say P    P1 ,  K  , P1 X,Y , then P  X,Y  An is
n
invertible with inverse  1   2  P1   3  P12 K   n  P1n1 . Now pick any element


b B \ x, y  and a representative a A . Then n  a n  X,Y   X,Y  does not belong

to U . Hence the complement belongs to U . For each n in this set a n  K X,Y \ X,Y  ,
n
hence a n   X,Y  An has an inverse c n  X,Y  An . For all other components of A,
c n  can be chosen arbitrarily, e.g., c n   0 . If d is the image of c in B then b  d  1 . In fact, B
is local with maximal ideal x, y . One only needs to show that x, y  is a proper ideal. If not,
then B  x, y  , i.e., there are b1 ,b2  B such that 1  b1  x  b2  y . If one picks representatives
n
n
a1 , a2  A for b1 and b2 then it follows that the set
n  1  a n X  a nY X,Y  

n
1
2
belongs to U . It would follow that 1 a1 n X  a2 nY  X,Y   X,Y  , a contradiction.
This shows that x, y  is a proper ideal of B.
If q  K X,Y  is a non-trivial prime ideal that is contained in X,Y  then it is either equal to
n
X,Y  , or it is generated by an irreducible polynomial P. In both cases q is the restriction of a
prime ideal of B:
If q  X,Y  then q is the restriction of the maximal ideal x, y   B .
 f P   B .
Now let q  P  and consider
Suppose that
f Q  
 f P  ,
say
f Q  b  f P  . Choosing a representative a A this means that
k
n  Q  an P X,Y  U

n
k
In the factor ring K X,Y  P  this implies that Q  P  belongs to infinitely many powers of
k
the
ideal
Q  P 
k
X  P,Y  P.
Krull's
Intersection
Theorem
it
follows
that
 0  P  , i.e., Q k P . By irreducibility of P one concludes Q P . These
arguments show that
P   I
By
f 1
 f P P.
Writing
 f P   I   r one sees that
r V  f P  that does not contain both
rV f P
r . There must be some prime ideal
x and y. Thus, P   f 1 r   X,Y . Since there is no prime ideal that is properly in between
P  and X,Y  one obtains the desired equality P   f 1 r .
rV  f P 
f
1
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
The functorial map Spec B  Gen X,Y  Spec K X,Y  is surjective. In particular,
there are non-minimal prime ideals in B that are not comparable. Hence the prime spectrum of B
is not completely normal.
The class of rings with completely normal prime spectrum is not closed under the formation
of ultra products, hence they do not form an elementary class. Ω
This Example shows much more:
 The class of rings with Boolean spectrum is not elementary. For, the rings An have Boolean
spectrum, but the spectrum of the ultraproduct B is not zero-dimensional.
On the other hand, the class of reduced rings with Boolean spectrum is well-known to be
elementary – these are the von Neumann regular rings.
 For the same reason the class of rings whose spectrum has only one point is not elementary.
On the other hand, the class of reduced rings whose spectrum is a singleton is well-known to
be elementary – these are the fields.
 The class of rings with finite spectrum is not elementary. The rings An have finite spectrum,
but the spectrum of the ultraproduct B is infinite. In fact this can be seen much more easily
than through the above example: If the class of reduced rings with finite spectrum is not
elementary then the class of arbitrary rings with finite spectrum cannot be elementary either.
An obvious construction is the formation of a free ultraproduct of the family of rings
K n n (with componentwise operations).

 The class of rings with totally ordered spectrum is not elementary.
The same statement is also true about reduced rings. However, this requires another
construction, which will be exhibited below. Such an example will show at the same time
that the reduced rings with completely normal prime spectrum do not form an elementary
class.
 The class of rings with only one minimal prime ideal is not elementary. This needs some
further explanation, to be given below.
On the other hand, the reduced rings with only one minimal prime ideal are the integral
domains, and they clearly form an elementary class.
 The class of rings A with Nil A   J A  is not elementary: All the rings An clearly have this
property. But the ultra product is a local ring of higher dimension, hence the nilradical and
the Jacobson radical are distinct. (This can be shown with a less elaborate example: It
suffices to consider polynomial rings with only one variable.)
Example
A further elaboration of the example above provides a proof that reduced rings with linearly
ordered prime spectrum do not form an elementary class. The main idea is the following:
Suppose that each ring An (defined as in the previous example) is a factor ring of a ring C n ;
let  n : Cn  An be the canonical surjection. Then A is a factor ring of the direct product
C   Cn ,     n : C  A . If U is a free ultra filter on  again then the composition
n
n


 A o : C  A  A U factors as  Uo U :C  C U  A U . Thus, Spec A U is considered
as a closed subspace of Spec C U.
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
If each ring C n is a domain with totally ordered spectrum then it follws that the class of
reduced rings with totally ordered spectrum is not closed under the formation of ultraproducts,
hence is not an elementary class. It remains to produce such domains.


The order function v for the polynomial ring K X,Y  (where v P   sup n P X,Y  )
n
defines a discrete valuation v : K X,Y     . The valuation ring is denoted by V, its

maximal ideal by M. The ring C n is defined to be K X,Y  M n . (This is the " D  M construction" as in [Gilmer].) It is a domain, being a subring of V. Besides the zero ideal it has
only one prime ideal, namely X,Y   M n :
As X,Y   M n  Cn  M it is clear that this is a prime ideal. If c Cn does not belong to
this ideal then it can be written in the form c    c1  c2 with  K  , c1 X,Y  and
c2 M n . It is claimed that c Cn . Of course one may assume that   1 . Then
n1

c   1  c1i  1  1
i 0
i
n1

n1
 c1n  c2   1  c1i 1  M n ;
i
i0
write it in the form 1  d . In the valuation ring this element is invertible: 1  1  d  z . The
equality can be rewritten as z  1  d  z 1  M n  Cn , and one concludes that c is invertible in
C n . It is now clear that C n is local with maximal ideal X,Y   M n .
Pick any element c  c1  c2 X, X   M n . Then the radical ideal generated by c is the
maximal ideal. For the proof it suffices to show that X,Y 
b V with v b   v c   n belongs to the principle ideal
c  . First note that any element
c  .
 b
For, v    n implies that
 c
b
b
M n  Cn . Hence, b   c c  . In particular, X vcn ,Y vcn c, and the claim is proved.
c
c
The spectrum of the ring C n consists of two prime ideals and is totally ordered. The factor
ring Cn M n is clearly isomorphic to An . 
The following considerations are mostly concerned with the minimal prime spectrum. Some
properties of minimal prime spectra have already been mentioned, e.g., minimal prime spectra
are always Hausdorff; they are compact if and only if they are pro-constructible. The notions of
potency and almost cleanliness have been mentioned briefly. Now follows a more profound
investigation. Frequently it will be assumed that the rings are reduced.
The first main question is concerned with compactness of the minimal prime spectrum. There
are various characterizations of rings with compact minimal spectrum:
Theorem
The following statements about the reduced ring A are equivalent:
(a) The minimal prime spectrum is compact.
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
(b) Every open subset of Spec A  that contains Specmin A  also contains the constructible
closure of Specmin A  .
(c) For all a A there is a finitely generated ideal J  Ann a  such that the ideal A  a  J is
dense.
(d) Every dense ideal contains a finitely generated dense ideal.
Proof
Denseness of an ideal I means that Ann I   0 . In a reduced ring this is equivalent to the
condition that I is not contained in any minimal prime ideal.
The equivalence of (a) and (c) can be found in [Glaz], Theorem 4.2.15.
(a)  (b) This is a trivial immplication since Specmin A  is its own constructible closure.
(a)  (d) The condition that I is dense can be expressed on the level of the spectrum:
Specmin A   U D a  . The constructible closure of Specmin A  , say C, is also contained in
U D a .
aI
The constructible closure is quasi-compact, hence there is a finite subcover
aI
C
U D a  , where E  I
is finite. Then E   I , and E  is dense as well.
aE
(a)  (d) Suppose that Specmin A   UU i is some open cover. It can be refined by a cover
iI
using basic open sets D s  , s S . Then the ideal generated by the set S is dense, and there is a
finitely generated dense ideal E   S . Each generator a E has a finite presentation
a  ca,1  sa,1 K  ca,ra  sa,ra as a linear combination of elements of S. Only finitely many
elements of S intervene in these representations; let F  S be this finite set. Then
Specmin A   U D a   U D s  , and the finite subcover has been found. 
aE
sF
Most of the following result has first been proved by Henriksen and Jerison (M. Henriksen, M.
Jerison, The space of minimal prime ideals of a commutative ring. Trans. AMS 115, 110 – 130
(1965)).
Theorem
In a reduced ring A the following statements are equivalent:
(a) Specmin A  is compact, and the set D a   Specmin A  a A is closed under the
formation of finite unions.
(b) a b : a  b  0 &  x : x  a  b  0  x  0.
(c) Tot A  , the total ring of quotients, is von Neumann regular.
(d) Every dense ideal contains a dense principle ideal.
Proof
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
(a)  (b) Pick a A and a finite subset J  Ann a  such that the ideal A  a  J  is dense.
There is some element b  A such that D b  Specmin A  U D  j  Specmin A. Then
j J
a  b belongs to every minmal prime ideal, hence is equal to 0. And a  b does not belong to any
minimal primme ideal, hence is a non-zero divisor.
(b)  (c) The spectrum of Tot A  is considered as a generically closed subspace of the
spectrum of A. It consists of the intersection of the sets D s  , where s varies in the set of nonzero divisors. It is zero-dimensional if and only if every non-minimal prime ideal p contains a
non-zero divisor. So, pick a non-minimal prime ideal p. It contains some minimal prime ideal q.
There is an element a  p \ q . According to condition (b) there is some element b  A with
a  b  0 and a  b a non-zero divisor. As a q , but a  b q , it follows that b q  p . This
implies a  b  p .
(c)  (a) The spectrum of Tot A  , again considered as a subspace of Spec A, contains all
minimal points. In fact, being zero-dimensional, it is exactly equal to Specmin A  . As the
spectrum of Tot A  is compact, the minimal prime spectrum is compact as well. For the rest of
the claim it suffices to show that D a  Specmin A D b  Specmin A is equal to
D c   Specmin A  for some c A . Since Tot A  is von Neumann regular every finitely
generated ideal is generated by an idempotent. In particular, there are idempotents e, f Tot A
such that e  Tot A  a and  f   Tot A b . Thus,
D e  D a  Specmin A , D  f   D b  Specmin A
D e  D  f   D e  f  e  f .
By construction of Tot A  there is an element c and a non-zero divisor s such that
c
c
s
e  f  e  f  , or:  e  f  e  f  . This implies
s
1
1
 c
 s
D c   Specmin A   D    D e  f  e  f   D  
 1
 1
 D e  f  e  f   D a   Specmin A   D b   Specmin A 
(c)  (d) If I is a dense ideal then I  Tot A   Tot A , and there are finitely many elements
b
a1 ,K , ak I , s1,K , sk Tot A  such that 1   ai  si . Write si  i , bi  A , ti a non-zero
ti
t
divisor. If ti  denotes the product
then t   ai  bi  ti I , and this is a non-zero divisor.
ti
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Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
(d)  (c) It suffices to show that Tot A  is zero-dimensional. So, pick a non-minimal prime
p Specmin A   ), hence it contains a non-zero divisor,
As  Tot A  , and Spec Tot A Spec As   D s  does not contain p. 
ideal p. The ideal p is dense (as
say s. Then
The class of rings that satisfy the equivalent conditions of the Theorem is elementary because of
condition (b).
The last conditions in the two Theorems exhibit the difference between the two classes of
rings very clearly.
For a while it was an open question whether the two classes coincide, i.e., it is enough to
require compactness of Specmin A  in condition (a) of the last Theorem. However, Quentel
showed that there are rings with compact minimal prime spectrum that do not satisfy (a) (Y.
Quentel, Sur la compacité du spectre minimal d'un anneau. Bull. Soc. Math. France 99, 265 –
272 (1971); see also S. Glaz, p.117 ff). The example will be worked out here in detail with some
minor modifications.
Theorem
Let k  2 be a fixed integer and suppose that a ring A satisfies the following conditions:
(i) A is reduced.
(ii)  a :  x :1  x  a   b : b  0 & a  b  0  (non-zero divisors are invertible or 0)
(iii)  a,b : a  b &  x :1  x  a &  y :1  y  b (there are at least two distinct units)
(iv) a,b : x :1  x  a& y :1  y  b a  b  0  z :1  z  a  b (the set of non-zero
divisors is additively closed)
(v) a,b : a  0 & b  0 & a  b  0 (there are zero divisors)
(vi) If a is not a unit then Ann a  contains an ideal J  a1 ,K , ak  such that A  a  J is dense.
This is expressed as a first order statement by the formula
 a : b : a  b  0 
a1 ,K , ak : a  a1  0 & K & a  ak  0 &
 x : x  a  0 & x  a1  0 & K & x  ak  0   x  0
Then K  A   0 is a subfield of A, K  3 . The ring A is not a field. The total quotient ring
of A coincides with A. The minimal prime spectrum of A is compact, and the set
D a Specmin A a A is not closed under finite unions. If a is a zero divisor and if
b Ann a  then the ideal a,b  is never dense.
Proof
To prove the first claim, note that non-zero divisors are invertible by (ii). It must be shown that
the sum of two non-zero divisors is either a non-zero divisor, or is 0. This is exactly condition
(iv). By (iii), K has at least three elements.
The ring A is not a field, since there are zero divisors (condition (v)).
The total quotient ring Tot A  coincides with A since non-zero divisors are invertible.
page 8 of 9
Niels Schwartz and Marcus Tressl
Minimal and maximal prime spectra
06.02.2016
For compactness of Specmin A  , pick a zero divisor a A . By condition (vi) there is a
finitely generated ideal a1 ,K , ak   Ann a  such that A  a  a1 ,K , ak  is dense. This is
criterion (c) for compactness of the minimal prime spectrum.
Suppose that a is a zero divisor, that b Ann a  and that a,b  is dense. Then both a  b
and
are non-zero divisors, where  K \ 0,1. It follows that
a  b
1    b  a  b  a    b is not 0, hence is invertible (condition (ii)). This is a
contradiction since the element is clearly a zero-divisor.
Pick any zero divisor c. If the set D a   Specmin A  a A is closed under finite unions
then there is some d such that c  d  0 and c  d is a non-zero divisor. (Note that the Theorem
of Henriksen and Jerison can be applied since the minimal prime spectrum is already known to
be compact.) The ideal c,d  would be dense, and this has been shown to be impossible. 
The conditions expressed in the Theorem are first order statements, hence they define an
elementary class of rings, which is denoted by k , where k is the integer of condition (vi). It is
clear that 2  3  K is an ascending sequence of classes of rings. At the moment it is not
clear whether any one of these classes is nonempty, as well as whether the difference classes
k 1 \ k are nonempty. Quentel's examplle, to be presented below, shows that 2   , which
implies that all k are non-empty. An adaptation of his method will be used to prove that each
difference set is non-empty as well.
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