Seidenberg's theorems about Krull dimension of polynomial rings ...

Seidenberg’s theorems about Krull dimension of polynomial rings
Darij Grinberg
written back in 2009 if not earlier; translated into English in 30 minutes in 2011.
Rough writing and mistakes guaranteed.
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In the following, ”ring” always means ”commutative ring with 1”.
By dim R, we always denote the Krull dimension of a ring R.
If we have a ring R and speak of R [X] or R [X 1 , X 2 , ..., X n ] without having defined
the variables X resp. X 1 , X 2 , ..., X n beforehand, it is to be understood that X resp.
X 1 , X 2 , ..., X n mean free variables (for polynomial rings).
The free coefficient of a polynomial always means its coefficient before X 0 (or, if
we work with several variables, the coefficient before X 0 1X 0 2...X 0 n).
A ring is called nontrivial if its 0 is non-equal to its 1.
Not only the zero element of a ring, but also the zero ideal of a ring will be denoted
by 0.
The following theorems are results by A. Seidenberg, quoted after
http://planetmath.org/?op=getobj&from=objects&id=7196
Theorem 1. Let R be a nontrivial ring. We are going to show two ways
to obtain an ideal of R [X] from an ideal of R.
a) For every ideal a ⊆ R, the set
aR [X] = {P ∈ R [X] | all coefficients of P lie in a}
is an ideal of R [X]. If a ⊆ R is a prime ideal, then aR [X] ⊆ R [X] is a
prime ideal, too.
b) For every ideal a ⊆ R, the set
aR [X] + XR [X] = {P ∈ R [X] | the free coefficient of P lies in a}
is an ideal of R [X]. If a ⊆ R is a prime ideal, then aR [X]+XR [X] ⊆ R [X]
is a prime ideal, too.
Using a) and b), we can show the following property of polynomial rings:
c) We have dim (R [X]) ≥ dim R + 1.
Proof of Theorem 1 (sketched). a) The only nontrivial part here is to show that
if a ⊆ R is a prime ideal, then aR [X] ⊆ R [X] is a prime ideal. The easiest way to
show this is by noticing that (R [X]) (aR [X]) ∼ = (Ra) [X], and that (Ra) [X] is
an integral domain if Ra is one. For an alternative proof, consider two polynomials
P, Q ∈ R [X] \ (aR [X]); each of these has at least one coefficient not lying in a; now
take the highest such coefficients and use these to show that P Q /∈ aR [X] . Proceeding
in this way one should not forget proving that aR [X] is not the whole R [X].
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) The only nontrivial part here is to show that if a ⊆ R is a prime ideal, then
aR [X] + XR [X] ⊆ R [X] is a prime ideal. This can be shown by proving that
(R [X]) (aR [X] + XR [X]) ∼ = Ra (in fact, the map R [X] → Ra defined by
n∑
a i X i ↦→ (a 0 ) a
is surjective, and its kernel is aR [X] + XR [X]), or again by hand.
i=0
c) Each chain p 0 p 1 ... p n−1 p n of prime ideals of R yields (due to a) and
b)) a chain p 0 R [X] p 1 R [X] ... p n−1 R [X] p n R [X] p n R [X] + XR [X] of
prime ideals of R [X]. This latter chain is one element longer than the former. This
yields Theorem 1 c).
Theorem 2. Let R be a ring. Theorem 1 a) and b) gave two ways to
construct ideals of R [X] from ideals of R; the converse direction is simpler:
a) For every ideal a ⊆ R [X], the set a ∩ R is an ideal of R. If a ⊆ R [X] is
a prime ideal, then a ∩ R ⊆ R is a prime ideal.
Next let us show how ideals of R [X] can be lifted to ideals of (Quot R) [X]
when R is an integral domain:
b) Let R be an integral domain, and let Quot R be the quotient field of R.
Let us canonically identify R with a subring of Quot R. Each element of
(Quot R) [X] has the form P for some P ∈ R [X] and some r ∈ R \ {0} .
r
b 1 ) For every ideal a ⊆ R [X], the set
{ }
P
a ((Quot R) [X]) =
r | P ∈ a; r ∈ R \ {0}
is an ideal of (Quot R) [X] .
b 2 ) If a ⊆ R [X] is a prime ideal such that a ∩ R = 0, then a ((Quot R) [X])
is a prime ideal as well.
b 3 ) If a is a prime ideal of R [X] and b is an ideal of R [X] with a b and
a ∩ R = 0, then a ((Quot R) [X]) b ((Quot R) [X]) .
c) Let R be an integral domain. If a and b are two prime ideals of R [X]
with 0 a ⊆ b and b ∩ R = 0, then a = b.
d) Let R be an arbitrary ring again. Let p, q and r be three prime ideals
of R [X] with p q r. Then, p ∩ R = q ∩ R = r ∩ R cannot hold. (In
other words, at least one of the two inclusions in p ∩ R ⊆ q ∩ R ⊆ r ∩ R is
a strict inclusion.)
e) Let R be an arbitrary ring. Then, dim (R [X]) ≤ 2 dim R + 1.
Proof of Theorem 2 (sketched).
a) is completely trivial.
b 1 ) Proving a ((Quot R) [X]) =
{ P
r | P ∈ a; r ∈ R \ {0} }
is easy (just notice
that every element of (Quot R) [X] has the form P for some P ∈ R [X] and some
r
r ∈ R \ {0}). That it is an ideal is more than obvious.
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2 ) Let P r
and Q with P, Q ∈ R [X] and r, s ∈ R \ {0} be two elements of
s
(Quot R) [X] which satisfy P r · Q
s ∈ a ((Quot R) [X]). Then, P
r · Q
s = T for some
{ t }
P
T ∈ a and some t ∈ R \ {0} (due to a ((Quot R) [X]) =
r | P ∈ a; r ∈ R \ {0} ).
Thus, tP Q = rsT. Since T ∈ a, this leads to tP Q ∈ a, and thus t ∈ a or P ∈ a or
Q ∈ a. But t ∈ a is impossible, since t ∈ R \ {0} and R ∩ a = 0. Thus, P ∈ a or
Q ∈ a. Therefore, P r ∈ a ((Quot R) [X]) or Q ∈ a ((Quot R) [X]) . It now remains to
s
prove that a ((Quot R) [X]) is not the whole ring (Quot R) [X]. But this is easy again
(assume the contrary; then, 1 ∈ a ((Quot R) [X]) ; thus there exist some P ∈ a and
some r ∈ R \ {0} such that 1 = P , so that P = r, so that r ∈ a, which is impossible
r
because of r ∈ R \ {0} and R ∩ a = 0).
b 3 ) From a ⊆ b we immediately obtain a ((Quot R) [X]) ⊆ b ((Quot R) [X]). The
only thing we must show is therefore a ((Quot R) [X]) ≠ b ((Quot R) [X]) . For the sake
of contradiction, assume that a ((Quot R) [X]) = b ((Quot R) [X]) . Since a b, there
exists some element P ∈ b such that P /∈ a. Now, P ∈ b yields P ∈ b ((Quot R) [X])
(we identify P with P ), so that P ∈ a ((Quot R) [X]) . Therefore there exists an S ∈ a
1
and an r ∈ R \ {0} such that P = S . Hence, rP = S. Since S ∈ a, this yields rP ∈ a.
r
Since a is a prime ideal, this entails r ∈ a or P ∈ a. But r ∈ a is impossible (since
r ∈ R \ {0} and a ∩ R = 0), and P ∈ a is impossible as well (remember P /∈ a). This
is the contradiction we wanted.
c) Assume that a ≠ b. Then, 0 a ⊆ b becomes 0 a b.
Let Quot R be the quotient field of R. Since a ∩ R = 0 (because b ∩ R = 0 and
a ⊆ b) and b ∩ R = 0, Theorem 2 b 2 ) yields that the ideals a ((Quot R) [X]) and
b ((Quot R) [X]) are prime ideals of (Quot R) [X] . Moreover, 0 a b yields (according
to Theorem 2 b 3 )) that 0 ((Quot R) [X]) a ((Quot R) [X]) b ((Quot R) [X])
(since 0 and a are prime ideals, a and b are ideals, and we have 0∩R = 0 and a∩R = 0).
Since 0 ((Quot R) [X]) = 0, this becomes 0 a ((Quot R) [X]) b ((Quot R) [X]) .
We thus have two prime ideals a ((Quot R) [X]) and b ((Quot R) [X]) in the ring
(Quot R) [X] satisfying the inclusion 0 a ((Quot R) [X]) b ((Quot R) [X]). But
(Quot R) [X] is a principal ideal domain (since Quot R is a field), and thus any prime
ideal of Quot R distinct from 0 is a maximal ideal. This yields that no prime ideal
of Quot R distinct from 0 is contained in another prime ideal of Quot R. This contradicts
0 a ((Quot R) [X]) b ((Quot R) [X]) (since both a ((Quot R) [X]) and
b ((Quot R) [X]) are prime ideals of (Quot R) [X]). Contradiction! This proves Theorem
2 c).
d) Assume that p ∩ R = q ∩ R = r ∩ R. Set p ∩ R = q ∩ R = r ∩ R = n. Then,
n ⊆ R is a prime ideal (due to Theorem 2 a), since p ⊆ R [X] is a prime ideal). The
ring Rn thus is an integral domain.
The canonical surjection ( R → Rn induces a canonical surjection ν : R [X] →
n
)
∑ ∑
(Rn) [X] , given by ν a i X i = n (a i ) n
X i . It is easy to see that Ker ν = nR [X],
i=0
i=0
and thus the images ν (p) , ν (q) and ν (r) of the prime ideals p, q and r of R [X] under
this surjection ν are prime ideals of (Rn) [X] (this follows from the trivial fact that if
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f : A → B is a surjective ring homomorphism between two rings A and B, and p ⊆ A
is a prime ideal with Ker f ⊆ p, then f (p) ⊆ B is a prime ideal as well). Furthermore,
p q r yields ν (p) ν (q) ν (r) (here we use the following fact: if f : A → B is a
surjective ring homomorphism between two rings A and B, and if a b are two ideals
of A such that Ker f ⊆ a, then f (a) f (b)).
Since ν (p) ν (q), we have 0 ν (q) (because 0 ⊆ ν (p)). Thus, 0 ν (q) ν (r) .
Now we will show ν (r) ∩ (Rn) = 0. In fact, every element of Rn ( has the form
n
)
∑
a n for some a ∈ R. If such an element a n lies in ν (r), then a n = ν a i X i for
i=0
n∑
∑
some polynomial a i X i ∈ r, so that a n = n (a i ) n
X i . By comparing coefficients, we
i=0
therefore conclude that a n = (a 0 ) n
and (a i ) n
= 0 for all i > 0. Thus, a − a 0 ∈ n and
a i ∈ n for all i > 0. Since n = r ∩ R, this yields a i ∈ r for all i > 0 (since a i ∈ n for
∑
all i > 0). Combined with n a i X i ∈ r, this results in a 0 X 0 ∈ r, so that a 0 ∈ r. Since
i=0
a 0 ∈ R, this yields a 0 ∈ r ∩ R = n, so that a n = 0. We thus have shown that if an
element a n of Rn lies in ν (r), then a n = 0. In other words, ν (r) ∩ (Rn) = 0.
We summarize: The ring Rn is an integral domain. Furthermore, ν (q) and ν (r)
are two prime ideals of (Rn) [X] satisfying 0 ν (q) ν (r) and ν (r) ∩ (Rn) =
0. Thus, Theorem 2 c) yields ν (q) = ν (r) , contradicting to ν (q) ν (r). This
contradiction proves Theorem 2 d).
e) Every chain p 0 p 1 ... p n−1 p n of prime ideals of R [X] gives rise to a
chain p 0 ∩ R ⊆ p 1 ∩ R ⊆ ... ⊆ p n−1 ∩ R ⊆ p n ∩ R of prime ideals of R (according to
Theorem 2 a)). According to Theorem 2 d), no two subsequent ⊆ signs in this chains
can simultaneously be = signs. Thus, if we remove repeated terms from this chain, it
will be shortened by a factor of no more than 2. This easily yields Theorem 2 e).
i=0
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