Which of the following is required for the going down theorem to hold for an integral extension R ⊆ S?
AS is a finitely generated R-module
BR is integrally closed and S is a domain
CS is Noetherian
DR is a local ring
The going down theorem for integral extensions requires R to be an integrally closed domain and S to be a domain. Without these hypotheses, going down can fail. For example, Z ⊆ Z[i] satisfies going down because Z is integrally closed. But the integral extension Z[√5] over Z[2√5] can fail going down when the base is not integrally closed.
Question 2 True / False
If R ⊆ S is an integral extension and P is a prime of R, then there exists a prime Q of S with Q ∩ R = P.
TTrue
FFalse
Answer: True
This is the lying-over theorem. The proof goes through the localization: R_P ⊆ S_P is still integral, and the maximal ideal of R_P contracts from some prime of S_P. Pulling back gives Q lying over P. Lying-over is the starting point; going up and going down extend it to chains.
Question 3 Short Answer
Let k be a field and consider the integral extension k[x^2] ⊆ k[x]. The prime (x^2 - 1) of k[x^2] has the prime (x - 1) lying over it. What other prime of k[x] lies over (x^2 - 1)?
Think about your answer, then reveal below.
Model answer: (x + 1), since x^2 - 1 = (x - 1)(x + 1) and (x + 1) ∩ k[x^2] = (x^2 - 1).
In k[x], the ideal (x^2 - 1) of k[x^2] generates (x-1)(x+1), which lies in both (x-1) and (x+1). The incomparability theorem guarantees these two primes lying over (x^2-1) are not comparable by inclusion, consistent with both being maximal in k[x].
Question 4 True / False
The going up theorem holds for any integral extension of commutative rings (no extra hypotheses beyond integrality).
TTrue
FFalse
Answer: True
Going up is unconditional for integral extensions: given R ⊆ S integral and primes P ⊆ P' of R with Q lying over P, there exists Q' ⊇ Q lying over P'. This is the Cohen-Seidenberg going up theorem. Going DOWN is the one that requires extra hypotheses (integrally closed base, or flatness).
Question 5 Short Answer
Explain the geometric meaning of the going up theorem in terms of morphisms of varieties.
Think about your answer, then reveal below.
Model answer: An integral extension R ⊆ S corresponds to a finite surjective morphism f: Spec S → Spec R. Going up says the map f is closed: it maps closed sets to closed sets. Chains of subvarieties in the base can be lifted to the total space.
Algebraically, going up says prime chains lift through integral extensions. Geometrically, finite morphisms (those corresponding to integral extensions) are closed maps. The lying-over theorem says f is surjective. Going up gives closedness. Going down (when it holds) gives an 'open-like' property: the map preserves generization.