Prime ideals of $R[x]$ that intersect $R$ in $P$

Question

Let $R$ be a noetherian ring and $P$ a prime ideal of height $h$. Show that the prime ideals $Q\subset R[x]$ that intersect $R$ in $P$ are of the following two kinds, with height as shown:

1. $Q=PR[x]$; in this case $\operatorname{ht}(Q)=h$

2. $Q \supsetneq PR[x]$; in this case $\operatorname{ht}(Q)=h+1$, and there is $f(x)\in Q$ with leading coefficient not in $P$ such that $$Q=\{g\in R[x]:ag\in PR[x]+(f) \text{ for some } a\in R-P\}$$

Just to make sure I understand the notation, is $PR[x]$ the set of all polynomials in $x$ with coefficients in $P$? (Equivalently, the set of all products $p\cdot f(x),p\in P, f(x)\in R$?)

The first task is to show that $Q$ is one of the two forms. The second task is to show the statements about heights. (Maybe I post the second part as a separate question?)

Since $\operatorname{ht}(P)=h$, there is a sequence of prime ideals of $R$ with $$P_0\subsetneq P_1\subsetneq\dots\subsetneq P_h=P$$

Let $Q\subset R[x]$ be an ideal with $Q\cap R=P$. If $PR[x]$ is what I wrote above, then $PR[x]$ is one such $Q$. It remains to show that if $Q\ne PR[x]$, then $Q=\{g\in R[x]:ag\in PR[x]+(f) \text{ for some } a\in R-P\}.$ I can't see why this holds (and the same goes for the statements about heights).

Answer 1

We first do the following case. Let $R$ be a domain, and $Q$ a non-zero prime ideal of $R[x]$ such that $Q\cap R=0$. Take $0\neq f\in Q$ of minimal degree. We want to show that $$Q=\{g\in R[x] : ag\in (f)\textrm{ for some }0\neq a\in R\}.$$ Let $g\in Q$. Then by considering leading terms we can find non-zero $a,b\in R$ and $d\geq0$ such that $ag-bx^df$ has strictly smaller degree than $g$. Since this element also lies in $Q$, we see by induction on degree that the inclusion $\subseteq$ holds. Conversely, if $g$ lies in the right hand side, then $ag\in(f)\subseteq Q$ and $a\not\in Q$, so $Q$ prime implies $g\in Q$. This gives the reverse inclusion. Note that this does not require $R$ to be Noetherian.

In general, let $P\lhd R$ be prime. Then $P[x]\lhd R[x]$ is the ideal consisting of those polynomials, all of whose coefficients lie in $P$. Clearly $R[x]/P[x]\cong(R/P)[x]$, which is a domain, so $P[x]$ is prime. Also, $P[x]\cap R=P$. Thus $P[x]$ is a prime of $R[x]$ lying over $P$.

Now let $Q\lhd R[x]$ be any other prime lying over $P$. Then $P\subset Q$ implies $P[x]\subset Q$, and passing to $(R/P)[x]$ and using the result above shows that $$ Q = \{g \in R[x] : ag\in P[x]+(f) \textrm{ for some }a\in R-P\} $$ for some $f\in Q$ with leading coefficient not in $P$.

We have thus described all primes of $R[x]$ lying over $P$, and this does not require $R$ to be Noetherian.

Answer 2

Again, let $R$ be a domain, and $Q\subseteq Q'$ non-zero primes of $R[x]$ such that $Q'\cap R=0$. We can write $Q=\{g:ag\in(f)\}$ and $Q'=\{g:ag\in(f')\}$ as above, where $f\in Q$ has minimal degree, and $f'\in Q'$ has minimal degree. Then $f\in Q$, so $af=bf'\neq0$ for some $a\in R$ and $b\in R[x]$. Since $\deg f'>0$, we must have $\deg b<\deg f$, so $bf'\in Q$ and $b\not\in Q$, which implies $f'\in Q$. Hence $Q=Q'$.

Next let $P'\lhd R$ be a height one prime, and assume $R$ is Noetherian. Take $0\neq a\in P'$, and consider a prime ideal $Q\lhd R[x]$ with $a\in Q\subseteq P'[x]$. Then $a\in Q\cap R\subseteq P'$, so $Q\cap R=P'$, and hence $Q=P'[x]$. Thus $P'[x]$ is a minimal prime over $a$ in $R[x]$. By Krull's Hauptidealsatz it follows that $P'[x]$ has height one.

In general, let $Q\subset Q'$ be distinct primes in $R[x]$. The first result shows that if $P=Q\cap R=Q'\cap R$, then $Q=P[x]$, and the second result shows that if $P\subset P'$ with $\mathrm{height}(P')=\mathrm{height}(P)+1$, then there is no prime strictly between $P[x]$ and $P'[x]$.

Thus, by induction on height, we can show that if $\mathrm{height}(P)=h$, then $\mathrm{height}(P[x])=h$ and $\mathrm{height}(Q)=h+1$ for any other prime $Q$ lying over $P$. This is where we need the Noetherian hypothesis.