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

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
4