Proving that $\dim(R[t]) \leq 2 \dim(R) + 1$ using localization

Question

I'm studying for an exam, this is part of an exam question.

Let $S = R[t]$ where $R$ is an integral domain. Let $Q < P$ be prime ideals of $S$ such that $Q \cap R = P \cap R = P_0$. Let $Y = R-P_0$ Show that $P_0S$ is a prime ideal of $S$ and that:

$$\frac{SY^{-1}}{P_0SY^{-1}} \cong F[t]$$

For a field $F$. Hence prove that $P_0S = Q$

For this part, I think that $P_0S = P_0[t]$ is prime since if you take two polynomials from outside $P_0[t]$, $f$ and $g$, they have some coefficients not in $P_0$. Taking the product, you will at some point get a sum involving a product of two coefficients, neither in $P_0$. $P_0$ is prime, so this coefficient won't be in $P_0$, thus the result $fg$ isn't in $P_0[t]$. I think this works.

I really have no idea what's going on with this quotient. I think it's the same thing as:

$$\frac{RY^{-1}[t]}{P_0Y^{-1}[t]}$$

Because $Y$ lives inside $R$, so it only affects the coefficients. $P_0Y^{-1}$ is maximal in the localization of $R$, $RY^{-1}$ so this is how the field comes about. The "hence" bit presumably follows from the solution to this bit, but I'm not really sure how.

Here are the other parts of the question:

With $S$ as before, prove that $\dim(S) \leq 2\dim(R) + 1$

Not sure how the previous parts really help with this.

Now suppose $S$ is a finitely generated $A$ algebra where $A$ is a subring and $\dim(A)$ is finite. Prove that $\dim(S)$ is finite.

Here I guess since $S = A[s_1, \ldots, s_n]$, it's the homomorphic image of a polynomial ring, and we can use the bound from before to bound $\dim(S)$. I guess the exact numbers matter less since I'm only asked to prove it's finite.

Answer 1

I'm going to start backwards and first show how the first part yields the bound on the Krull dimension of $R[t]$.

An immediate consequence is that

Any proper chain of prime ideals in $R[t]$ lying over a prime ideal in $R$ has length at most $2$.

Indeed if $Q_1, Q_2 \subsetneq P$ is such that $P_0 = Q_1 \cap R = Q_2 \cap R = P \cap R$ then we saw that $Q_1 = P_0[t] = Q_2$.

So start with a chain of maximal length in $R[t]$, $$P_1 \subsetneq P_2 \subsetneq \ldots \subsetneq P_k$$ in $R[t]$. Contraction gives us a chain $$Q_1 \subsetneq \ldots \subsetneq Q_m$$ in $R$ (after we remove redundancies), where each $Q_i = P_j \cap R$ for some $j \geq i$.

We know that each $P_i$ lies over exactly one or two $Q_j$s,

therefore $$dim(R[t]) + 1 = k \leq 2m \leq 2(dim(R) + 1)$$

Now for the justification....

Your argument that $P_0R[t] = P_0[t]$ is prime goes through fine, but I'll mention that a slick, traditional way to show this is by using the fact that $\frac{R}{M}$ is a domain iff $M$ is prime and the natural isomorphism $\frac{R[t]}{P_0[t]} \cong \left(\frac{R}{P_0}\right)[t]$.

For the main part of the question.... Start by noting that if $Q$ is $0$ then trivially $P_0 =0$ and $P_0R[t] = Q = 0$. So we can assume $0 \subsetneq Q \subsetneq P$.

Now let's break down the quotient, I'm going to change the notation to what is standard. $RY^{-1}$ is usually referred to as the localization at $P_0$, and written $R_{P_0}$.

Now the two things we need to know about localization are: (1) for $\mathfrak{p}$ a prime ideal of $R$, we have a bijection between ideals of $R_{\mathfrak{p}}$ and ideals of $R$ contained in $\mathfrak{p}$. In particular $R_{\mathfrak{p}}$ is local ring with maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$ (2) localization commutes with polynomial extension (check this by universal properties or direct computation)

Thus your quotient is

$$\frac{(R[t])_{P_0}}{(P_0 R[t])_{P_0}} \cong \frac{R_{P_0}[t]}{P_0R_{P_0}[t]} \cong \left(\frac{R_{P_0}}{P_0R_{P_0}} \right)[t] $$

and therefore taking $F = \frac{R_{P_0}}{P_0R_{P_0}}$ we have the claim.

Lets break this up as $$R[t] \xrightarrow{\iota} R_{P_0}[t] \xrightarrow{\pi} F[t] $$

So what are we even doing with this quotient? The core idea here is that $F[t]$ is a $PID$ so every non-zero prime ideal in $F[t]$ has height $1$. We can thus get information about $Q$ and $P$ by looking at their homomorphic images in $F[t]$ under the composition $\pi \cdot \iota$. In particular, the equality $Q=P_0R[t]$ would follow immediately if we can show that $Q \subset ker(\pi \cdot \iota)$, which would mean every polynomial in $Q$ takes coefficients in $P_0$.

Of course there are some details to work out, but the path is pretty clear. This amounts to a straightforward exercise, which i think is worth completing on your own. I'll outline details below in case that is helpful.

We start with the observation that for $\mathfrak{p}$ a prime ideal of $R[t]$ and $\mathfrak{p_0} = \mathfrak{p} \cap R$, $\mathfrak{p}R_{\mathfrak{p_0}}[t]$ is a prime ideal of $R_{\mathfrak{p_0}}[t]$ (check this for yourself). (For the special case $\mathfrak{p} \cap R = 0$, this just says $\mathfrak{p}$ extends to a prime ideal of $K[x]$ where $K$ is the quotient field of $R$.) Moreover if $\mathfrak{p} \subsetneq \mathfrak{q}$ for any ideal $\mathfrak{q} \subset R[t]$, then $\mathfrak{p}R_{\mathfrak{p_0}}[t] \subsetneq \mathfrak{q}R_{\mathfrak{p_0}}[t]$. Thus the chain of prime ideals $$0 \subsetneq Q \subsetneq P$$ in $R[t]$ extends to a chain of prime ideals $$0 \subsetneq QR_{P_0}[t] \subsetneq PR_{P_0}[t]$$ in $R_{P_0}[t]$. Letting $\pi$ be the projection $R_{P_0}[t] \rightarrow F[t]$, we look at the chain $$0 \subset \pi(QR_{P_0}[t]) \subsetneq \pi(PR_{P_0}[t])$$ (That this is a chain of prime ideals and that the second inclusion is strict are because $\pi$ is surjective and $QR_{P_0}[t]$ contains the kernel of $\pi$). Now since this chain is in a $PID$, we deduce $\pi(QR_{P_0}[t]) = 0$, and indeed $Q \subset ker(\pi \cdot \iota)$.