If $P$ is a prime ideal of $R[x;\delta]$ such as $P\cap R=\{0\}$, is $P(Q[x;\delta])$ also prime?

Question

I'm currently learning about Ore extensions in McConnell's book (Noncommutative Noetherian Rings) and Marubayashi's book (Prime Divisors and Noncommutative Valuation Theory). On the second book, I learn until part 2.3.17, where he talk about a function from $Spec_0(R[x;\delta])$ to $Spec(Q[x;\delta])$. It maps a prime ideal $P$ of $R[x;\delta]$ such that $P\cap R=\{0\}$ to $P(Q[x;\delta])$.

Let $R$ be a HNP (hereditary Noetherian prime) ring, $Q$ its ring of quotient (or ring of fractions, or localization ring), and $R[x;\delta], Q[x;\delta]$ be their differential polynomial ring (or skew polynomial ring of derivation type, or Ore extension of derivation type). If $P$ is a prime ideal of $R[x;\delta]$ such that $P\cap R=\{0\}$, then $P(Q[x;\delta])$ is an ideal of $Q[x;\delta]$. How to prove that $P(Q[x;\delta])$ is prime ideal? Thanks in advance.

Answer 1

Let $S$ be $R \setminus \{0\}$, $A= R[x;\delta], B = Q[x;\delta]=S^{-1}A$. Recall that there is an isomorphism $$(*) \ \ \{ I \text{ ideal of } A: I \cap S = \emptyset \} \to \{ J \subsetneq S^{-1}A \} $$ given by $I \to S^{-1}I := \{ \frac{i}{s}: i \in I, s \in S\}, J \to J \cap A $. It is easy to show:

1. injectivity: if $S^{-1}I = S^{-1}J$, then $I = S^{-1}I \cap A = S^{-1}J \cap A = J $.
2. surjectivity: given $J \subset S^{-1}A$, let $I=J \cap A$. Surely $S^{-1}I \subset J$. Conversely, given $j \in J$ by definition we can write it as $j=a/s$ with $a \in A, s \in S$. Thus $a=sj \in J \cap A = I$ and we are done.

The isomorphism (*) correctly restricts to prime ideals. Infact the contraction of a prime is always a prime, and given $P \in Spec(A)$ we have $S^{-1}(A/P) \simeq S^{-1}A/S^{-1}(P)$ thus $S^{-1}(P)$ is a prime (localization of a domain is a domain).

Remark. For $A \to S^{-1}A$, the function $I \to S^{-1}I$ is actually the extension of ideals. Infact $S^{-1}I$ is an ideal of $S^{-1}A$ which contains $I$, and given $I \subset J$ where $J$ is a $S^{-1}A$ ideal, $S^{-1}I \subset J$.

Answer 2

By simplicity, call $A = R[x;\delta ], B=Q[x;\delta]$. You have to pass from the fact that

(#) *There exist an ideal $P' \subset B$ such that $P' \cap A = P$. *

Why does this conclude?

1. (#) implies $PB \cap A=P$. Infact $\supset$ follows from $P \subset PB, A$, and $\subset$ from $$ P \subset P' \text{ B ideal } \Rightarrow PB \subset P' \Rightarrow PB \cap A \subset P' \cap A = P$$.
2. Consider the morphism $A \to B/PB$ which compose the inclusion $A \to B$ and the projection. The ker is exactly $PB \cap A = P$, thus $A/P\simeq B/PB$. The LHS is a domain because $P$ is prime, so the RHS is a domain. This makes $PB$ a prime.

Let's prove (#) in this case. The "abstract but clear" proof, if you have a little scheme theory, is the following. Let's introduce the fibered product. Suppose you have $ f: X\to Y, f': X'\to Y$, which we call connection maps. The fibered product is the following:

$$X \times_Y X' = \{(x,x') \in X \times X' \text{such that} f(x) =f'(x') \} $$

The simple theorem i'm going to cite is $$ Spec(R \otimes_S R') \simeq X \times_Y X' $$ Where $S \subset R, R'$ and $X=Spec (R), Y=Spec(S) $, and the connection maps are given by the intersection with $S$. The isomorphism is given by $Q \to (Q \cap R, Q \cap R') $ (why this is an element of the fibered product?)

In your case, $B = Q \otimes_R A$, and the theorem will give us exactly (#). Notice that $Spec Q = 0$ because it is a field. Then, using the definition of fibered product we get

$$Spec B \simeq \{(P,0) : P \cap R = 0 \} $$ Recall that the isomorphism is given by $P' \to (P' \cap A, P' \cap Q) $. In other words, we showed (#).