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

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
3