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.

Recently I encountered a problem (the first exercise from chapter four of Atiyah & McDonald's Introduction to Commutative Algebra) stating that if $\mathfrak{a}$ is a decomposable ideal of $A$ (a commutative ring with unity), then the prime spectrum of $A/ \mathfrak{a}$ has finitely many irreducible components. This follows easily from the recognition that the maximal irreducible subspaces of $\textrm{ Spec } (A / \mathfrak{a})$ are precisely the "zero loci" of the minimal prime ideals of $A /\mathfrak{a}$.

I'm curious about the converse - the proof isn't easily reversed since the notion of minimal ideals of $\mathfrak{a} $ doesn't make sense before we know what we are trying to prove. My intuition says that it is false based on the general premise that images are badly-behaved (and the fact that it isn't part of the exercise.) However, I've had some difficulty constructing a counterexample, so that the main purpose of this post is to ask for a reasonable procedure or heuristic for doing so (or, of course, proof that my intuition is false.)

If it helps, if $\textrm{Spec }(A / \mathfrak{a})$ is irreducible then the nilradical $\mathcal{R}_{A /\mathfrak{a}}$ is prime so that $r(\mathfrak{a}) = \rho^{-1} ( \mathcal{R}_{A / \mathfrak{a}} ) = \displaystyle\cap_{i=1}^n \rho^{-1} (p_i),$ where $p_i$ are the minimal prime ideals of $A /\mathfrak{a}$ and $\rho $ is the associated projection, is also prime. Thanks!

share|improve this question
    
If I understand well, you are looking for an ideal $\mathfrak{a}$ that has finitely many minimal primes but $\mathfrak{a}$ doesn't have a finite primary decomposition. Right? –  user26857 Oct 4 '12 at 21:59
    
Well, "minimal primes" of $\mathfrak{a}$ doesn't make sense (in the context of my book) unless $\mathfrak{a}$ has a primary decomposition, so the property $\mathfrak{a}$ should have is that there are finitely many minimal elements in the set of primes containing $\mathfrak{a}.$ –  Tim Duff Oct 4 '12 at 22:03
    
I dont see any reason for "minimal primes" not having sense in the absence of a primary decomposition. Anyway, if you want what I said before, please let me know. –  user26857 Oct 4 '12 at 22:06
    
The definition given in the book - "the minimal elements of the set $\{ \mathfrak{p_1} , ..., \mathfrak{p_n } \} $ are minimal prime ideals belonging to $\mathfrak{a}.$" This is in reference to the first uniqueness theorem (AM p. 52.) –  Tim Duff Oct 4 '12 at 22:08
2  
Let's forget about AM for the moment. Irreducible components of the spectrum correspond to minimal ideals and this has nothing to do with some special definition given in a book. Let's make it clear: a prime ideal $P$ is minimal over an ideal $I$ iff there is no other prime $Q$ strictly contained in $P$ which contains $I$. Is this okay for you? –  user26857 Oct 4 '12 at 22:10
add comment

1 Answer

up vote 8 down vote accepted
+500

In order to find an ideal which doesn't have a primary decomposition, the following construction is useful. Let $R$ be a commutative ring and $M$ an $R$-module. On the set $A=R\times M$ one defines the following two algebraic operations:

$$(a,x)+(b,y)=(a+b,x+y)$$

$$(a,x)(b,y)=(ab,ay+bx).$$

With these two operations $A$ becomes a commutative ring with $(1,0)$ as unit element. ($A$ is called the idealization of the $R$-module $M$ or the trivial extension of $R$ by $M$). Let's list some important properties of this ring:

  1. $\{0\}\times M$ is an ideal of $A$ isomorphic to $M$ (as $R$-modules) and there is a ono-to-one correspondence between the ideals of $R$ and the ideals of $A$ containing $\{0\}\times M$.

  2. $A$ is a Noetherian ring if and only if $R$ is Noetherian and $M$ is finitely generated.

  3. All prime (maximal) ideals of $A$ have the form $P\times M$, where $P$ is a prime (maximal) ideal of $R$.

  4. If $R$ is an integral domain and $M$ is divisible, then all the ideals of $A$ have the form $I\times M$ with $I$ ideal of $R$, or $\{0\}\times N$ with $N$ submodule of $M$.

Now I suggest to consider $R=\mathbb{Z}_{(2)}$ (the localization of $\mathbb{Z}$ at the prime ideal $2\mathbb{Z}$), $M=\mathbb{Q}$, $A=R\times M$ (as before) and $\mathfrak{a}=\{0\}\times H$ with $H$ a proper $\mathbb{Z}_{(2)}$-submodule of $\mathbb{Q}$. There are only two prime ideals of $A$ containing $\mathfrak{a}$ (and one of them is minimal over $\mathfrak{a}$), and $\mathfrak{a}$ has no finite primary decomposition because the primary ideals of $A$ have the following form: $\{(0,0)\}$, $\{0\}\times\mathbb{Q}$ and $2^n\mathbb{Z}_{(2)}\times\mathbb{Q}$, $n\in\mathbb{N}^*$.

share|improve this answer
3  
The easy way to remember this ring is "the set of matrices of the form $\begin{bmatrix}a&m\\0&a\end{bmatrix}$ with $a\in R$ and $m\in M$, added and multiplied with matrix operations." –  rschwieb Oct 5 '12 at 0:08
    
Dear navigetor23, this is a very impressive answer! Are there cases other than 4. where you can describe all the ideals of $A$? –  Georges Elencwajg Oct 5 '12 at 8:57
    
Dear Georges Elencwajg, I don't know other cases when one can completely describe the ideals of $A$, but this doesn't means that these do not exist. –  user26857 Oct 5 '12 at 14:26
    
Thanks for answering, navigetor23. And, once again, congratulations for this great answer. –  Georges Elencwajg Oct 6 '12 at 6:15
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.