Primality in R[x, y^{-1}] and coprime height bound #
If x is transcendental over R and r is prime in R with r not dividing y, then r remains prime in R[x, y^{-1}]. As a consequence, if y_1 and y_2 are coprime in R, no height-one prime of T contracting to a nonzero ideal of R can contain both y_1 and y_2.
If r is prime in NSubring R and r ∤ y, then y ∉ P for any P ∈ Ass(T/(r·T)). Uses the NSubring height bound: ht(P∩R) ≤ 1 forces P∩R = (r), so y ∈ P → r | y.
y is a non-zero-divisor on T/(r·T) when r is prime in NSubring R and r ∤ y.
A prime element r of R that doesn't divide y is "prime" in adjoinLocSetY R x y when x is transcendental over R: if a₁ * b₁ = (r:T) * c in T with a₁, b₁, c ∈ adjoinLocSetY, then r divides a₁ or b₁ within adjoinLocSetY.
The proof has three key ingredients:
- Polynomial factorization: from a₁b₁ = rc and transcendence, derive C(r)|f₁ or C(r)|f₂ in R[X] (using C(r) prime and C(r) ∤ C(y)^n).
- NSubring height bound: y is a non-zero-divisor in T/(r·T) because y ∉ P for every P ∈ Ass(T/rT). This uses: P∩R has height ≤ 1, r ∈ P∩R with ht((r))=1, so P∩R = (r), and y ∉ (r) since r ∤ y.
- Divisibility transfer: from a₁y^m = re and y regular mod r, conclude r|a₁ in T. Then d := a₁/r ∈ adjoinLocSetY with witness (h₁, m₁).
Coprime height bound #
Key lemma: if y₁, y₂ ∈ R are coprime (no prime divides both), then no height-1 prime P of T (that contracts to a nonzero ideal of R) can contain both y₁ and y₂. This is what makes the intersection approach work.
If y₁, y₂ are coprime in R and P∩R has height ≤ 1 and is nonzero, then y₁ ∉ P or y₂ ∉ P.