Spectrum of $R[x]$

Question

The spectrum of $\Bbb Z[x]$ is well known : a prime ideal of $\Bbb Z[x]$ is or $(Q, p)$, with $Q \in \Bbb Z[x]$ zero or irreducible modulo $p$, and $p$ prime or zero.

If I'm not mistaken, we have a similar result for $R[x]$, when $R$ is a domain with finite Krull dimension : under this hypothesis, a prime ideal of $R[x]$ is $(Q) + \mathfrak p$, with $\mathfrak p$ a prime ideal of $R$ and $Q$ zero of irreducible modulo $\mathfrak p$.

Some years ago I saw a paper dealing with this question of the spectrum of $R[x]$ in great details, but I'm unable to find it again. Could you help me ?

Eh, sorry for even mentioning it. It seems like this should fall out of the various proofs that $\dim R[x] = \dim R + 1$, but I can't remember how that goes at the moment. I hope someone provides a good reference! — Dylan Moreland, Jul 20 '12 at 13:26
@DylanMoreland: Yes, but what about non finite dimensional ring ? — Lierre, Jul 20 '12 at 13:31

score 11 · Answer 1 · edited Apr 13 '17 at 12:20

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The conjecture that every prime ideal $\mathfrak P\subset R[z]$ is of the form $\mathfrak P =\mathfrak p+(Q)$ is not true.

Let $k$ be a field and $R=k[x,y]$, the polynomial ring over $k$ with two indeterminates.
Consider the curve in $C\subset \mathbb A^3_k$ given parametrically by $(t^3,t^4,t^5) \quad (t\in A^1_k)$.
Its ideal is the prime ideal $I(C)=\mathfrak P=(y^2-xz,x^3-yz,z^2-x^2y)\subset k[x,y,z]=R[z]$.

The important point is that this ideal cannot be generated by two polynomials, i.e. $C$ is not an ideal-theoretic complete intersection: see a proof here.

So it is certainly not true that $\mathfrak P$ is of the required form $ \mathfrak P=\mathfrak p+(Q)$ (with $\mathfrak p$ prime in $R=k[x,y]$), because we would have $\mathfrak p=(f(x,y))$ and thus $\mathfrak P=(f,Q)$, which would falsely imply that $C$ is a complete intersection.

edited Apr 13 '17 at 12:20

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answered Jul 20 '12 at 15:20

Georges Elencwajg

150,790

What conjecture do you presume? – Bill Dubuque Jul 20 '12 at 15:31
@Bill: The conjecture that every ideal $\mathfrak P\subset R[z]$ is of the form $\mathfrak P=\mathfrak p+(Q)$ where $\mathfrak p$ is prime in $R$ and $Q\in R[z]$. By the way, does your answer settle this question or did you have something else in mind? – Georges Elencwajg Jul 20 '12 at 15:44
I presumed that he was trying to recall the theorem that I cited. – Bill Dubuque Jul 20 '12 at 15:52
@Bill: Ah, thanks for your answer. And since my formulation in terms of conjecture wasn't clear, I'll edit my answer to make it more explicit. – Georges Elencwajg Jul 20 '12 at 16:01
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Yes, I see now, reading more closely. Hopefully between the two he'll have his answer and then some. – Bill Dubuque Jul 20 '12 at 16:03
@BillDubuque & Georges — So I was mistaken, thank you for your answers. I have to think about it. – Lierre Jul 21 '12 at 08:54

score 5 · Answer 2 · answered Jul 20 '12 at 14:45

5

By factoring out primes then localizing, it reduces to a simple case, e.g. see the exposition below from Kaplansky's Commutative Rings enter image description here

answered Jul 20 '12 at 14:45

Bill Dubuque

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score 1 · Answer 3 · answered Jul 21 '12 at 01:31

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Alternatively, you could look at pages 22-23 of Miles Reid's "Undergraduate Commutative Algebra" (LMS Student Texts 29).

answered Jul 21 '12 at 01:31

Chris Leary

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Dear Chris: alternatively to what? Reid describes the spectrum of $\mathbb Z[x]$ (and of $k[x,y]$), which is well known to Lierre, as he writes himself in his very first sentence. Lierre is asking about $Spec(R[x])$ where $R$ has Krull dimension $\gt 1$. – Georges Elencwajg Jul 21 '12 at 09:58
Reid has the same setup as Kaplansky's Theorem 36, $R$ a domain and $K$ its field of fractions. "Alternatively" was probably a bad choice of words. Just suggesting another reference for the basic case, really. – Chris Leary Jul 23 '12 at 02:59

score 1 · Accepted Answer · answered Jul 22 '12 at 23:12

I still haven't found again the paper I was looking for, but the answers of Bill and Georges show me that I was mistaken and also show me how to fix it.

Finally, it has nothing to with the Krull dimension.

So let $R$ be a ring (commutative, unital). The prime ideals of $R[x]$ are precisely : $$\mathfrak p R[x] + R[x]\cap(Q R_{\mathfrak p}[x])$$ with $\mathfrak p$ a prime ideal of $R$ and $Q$ a polynomial of $R[x]$ which is zero irreducible in $k(\mathfrak p)[x]$, where $k(\mathfrak p)$ is the residual field of $\mathfrak p$.

More over, we have $$ \mathfrak p_1 R[x] + R[x]\cap(Q_1 R_{\mathfrak p_1}[x]) \varsubsetneq \mathfrak p_2 R[x] + R[x]\cap(Q_2 R_{\mathfrak p_2}[x])$$ if and only if one of the following holds :

$\mathfrak p_1 = \mathfrak p_2$ and $Q_1 = 0$
$\mathfrak p_1 \varsubsetneq \mathfrak p_2$ and $Q_2$ divides $Q_1$ in $k(\mathfrak p_2)$.

In the example of Georges, we have $$ I(C) = (x^4-y^3) + R[z]\cap\left( \left(z-\frac{x^2}{y}\right) R_{(x^4-y^3)}[z] \right). $$

Spectrum of $R[x]$

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