Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
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
    
Ah, I thought you were assuming finite dimension. –  Dylan Moreland Jul 20 '12 at 13:50
add comment

4 Answers 4

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.

share|improve this answer
    
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
1  
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
show 1 more comment

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

share|improve this answer
add comment

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

share|improve this answer
2  
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 9:58
    
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 2:59
add comment
up vote 1 down vote accepted

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 $$ I(C) = (x^4-y^3) + R[z]\cap\left( \left(z-\frac{x^2}{y}\right) R_{(x^4-y^3)}[z] \right). $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.