# Irreducible Components of the Prime Spectrum of a Quotient Ring and Primary Decomposition

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!

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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 $\mathfrak{a}.$ –  user17794 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.) –  user17794 Oct 4 '12 at 22:08
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

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 $2^n\mathbb{Z}_{(2)}\times\mathbb{Q}$, $n\in\mathbb{N}^*$.

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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