An ideal $\mathfrak{a}$ is called irreducible if $\mathfrak{a} = \mathfrak{b} \cap \mathfrak{c}$ implies $\mathfrak{a} = \mathfrak{b}$ or $\mathfrak{a} = \mathfrak{c}$. Atiyah-MacDonald Lemma 7.11 says that in a Noetherian ring, every ideal is a finite intersection of irreducible ideals. Exercise 7.19 is about the uniqueness of such a decomposition. 


7.19. Let $\mathfrak{a}$ be an ideal in a noetherian ring. Let 
$$\mathfrak{a} = \cap\_{i=1}^r \mathfrak{b}\_i = \cap\_{j=1}^s \mathfrak{c}\_j$$ be two minimal decompositions of $\mathfrak{a}$ as an intersection of irreducible ideals. [I assume minimal means that none of the ideals can be omitted from the intersection.] Prove that $r = s$ and that (possibly after reindexing) $\sqrt{\mathfrak{b}\_i} = \sqrt{\mathfrak{c}\_i}$ for all $i$.


Comments: It's true that every irreducible ideal in a Noetherian ring is primary (Lemma 7.12), but I don't think our result follows from the analogous statement about primary decomposition. For example, here is Example 8.6 from Hassett's $\textit{Introduction to Algebraic Geometry}$. 


8.6 Consider $I = (x^2, xy, y^2) \subset k[x,y]$. We have $$I = (y, x^2) \cap (y^2, x) = (y+x, x^2) \cap (x, (y+x)^2),$$ and all these ideals (other than $I$) are irreducible. 


If my interpretation of "minimal" is correct, then this is a minimal decomposition using irreducible ideals, but it is not a minimal primary decomposition, because the radicals are not distinct: they all equal $(x,y)$. 


There is a hint in the textbook: Show that for each $i = 1, \ldots, r$, there exists $j$ such that $$\mathfrak{a} = \mathfrak{b}\_1 \cap \cdots \cap \mathfrak{b}\_{i-1} \cap \mathfrak{c}\_j \cap \mathfrak{b}\_{i+1} \cap \cdots \cap \mathfrak{b}\_r.$$ I was not able to prove the hint.


I promise this exercise is not from my homework. 


**Update.** There doesn't seem to be much interest in my exercise. I've looked at various solution sets on the internet, and I believe they all make the mistake of assuming that a minimal irreducible decomposition is a minimal primary decomposition. Does anyone know of a reference which discusses irreducible ideals? Some google searches have produced Hassett's book that I mention above and not much else. 



Atiyah-MacDonald, exercise 7.19 - "decomposition using irreducible ideals"
Here is a solution to the hint:


First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.


Next, the condition that $\mathfrak b\_1 \cap \cdots \cap \mathfrak b\_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b\_1 \times \cdot \times
A/\mathfrak b\_r$ (the product ot the natural quotient maps) is injective, while the 
condition that $\mathfrak b\_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b\_i$, then $I \cap J \neq 0$ also.


Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I\_j := \mathfrak b\_1 \cap \cdots \cap \mathfrak b\_{i-1} \cap \mathfrak c\_j\cap \mathfrak b\_{i+1}
\cap \cdots \cap \mathfrak b\_r,$ for each $j =1,\ldots,s$.


Then $I\_1\cap \cdots \cap I\_s = 0$ (since it is contained in the intersection
of the $\mathfrak c\_j$, which already vanishes).


Now we recall that $A$ embeds into the product of the $A/\mathfrak b\_{i'}$.
Note that $I\_j$ is contained in $\mathfrak b\_{i'}$ for $i'\neq i$. Thus,
if we let $J\_j$ denote the image of $I\_j$ in $A/\mathfrak b\_i$, then
we see that the image of $I\_j$ under the embedding
$A \hookrightarrow A/\mathfrak b\_1\times\cdots\times A/\mathfrak b\_i \times \cdots \times A/\mathfrak b\_r$ 
is equal to $0 \times \cdots \times J\_j \times\cdots \times 0$.
Thus the intersection of the images of the $I\_j$, which is the image
of the intersection of the $I\_j$ (since we looking at images under an embedding)
 is equal to $0\times \cdots \times (\bigcap J\_j) \times \cdots \times 0.$
Thus, since the intersection of the $I\_j$ is equal to $0$, we see
that $\bigcap J\_j = 0.$ But $\mathfrak b\_i$ is irreducible, and so
one of the $J\_j = 0$. Equivalently, the corresponding $I\_j = 0.$


This proves the hint.


(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)


