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.

  • 2
    $\begingroup$ The radicals of the ideals participating in a decomposition need not be different. $\endgroup$ Jan 19, 2010 at 17:34
  • $\begingroup$ The intersection of P-primary ideals is P-primary, so instead of using a bunch of ideals with the same radical, you can just use the intersection. This would contradict minimality. $\endgroup$
    – Joel Dodge
    Jan 20, 2010 at 2:22
  • $\begingroup$ Thanks for the comments. My question is not explicitly about primary decomposition, it's about decomposition using irreducible ideals. I found some answers on the internet which said 7.19 was true because the analogous statement was true for primary decomposition. I think 8.6 shows this argument is incorrect. It shows you may need to include multiple ideals with the same radical, which as Joel says, you do not need to do in a minimal primary decomposition. $\endgroup$
    – CJD
    Jan 20, 2010 at 11:10

4 Answers 4


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.)

  • $\begingroup$ Excellent! Best 100 reputation points I've ever spent. If you don't mind, I'm going to add something below which I think finishes the proof. If you see an easier way to finish the proof, I'd be happy to hear it! Lastly, I'm going to wait to accept your answer at least a day to see if anyone has any objection. (Accepted bounty answers last forever, I believe I read in the FAQ.) $\endgroup$
    – CJD
    Feb 15, 2010 at 17:44
  • $\begingroup$ Dear CJD, Take your time. I remember seeing this question the first time around and not seeing how to solve it straight away, and so was happy to be reminded of it this second time around. $\endgroup$
    – Emerton
    Feb 15, 2010 at 20:07
  • $\begingroup$ Today I tried to solve this problem. Now I've read this discussion, but I still don't understand, why example 8.6 does not contradict the statement of Atiyah MacDonald's 7.19? Furthermore, how do you construct I_{j}, what is the choice of c_j corresponding to b_i? $\endgroup$
    – user38485
    Aug 11, 2013 at 14:05
  • $\begingroup$ @Efim: Dear Efim, I don't have a copy of A&M, so I can' answer your question about the relationship b/w 8.6 and 7.19. As for the $I_j$, I don't really understand your question. The ideal $I_j$ is defined as in my answer, i.e. we fix an $i$, and then for each $j$ the ideal $I_j$ is defined by a certain formula in terms of the corresponding ideal $\mathfrak c_j$. So there is no particular "choice" of $\mathfrak c_j$; for each $\mathfrak c_j$ there is a corresponding $I_j$ (while the index $i$ is fixed throughout the discussion). Regards, $\endgroup$
    – Emerton
    Aug 12, 2013 at 11:54

This is meant to finish the proof left by Professor Emerton. Thanks to him for explaining the part I was stuck on. If anyone sees an easier way to finish the proof, I'd be happy to hear it!

As stated, the claim r = s is clear. So it remains to show that $\sqrt{\mathfrak{b}_i} = \sqrt{\mathfrak{c}_i}$, after possibly reordering. First, note that we can get from a minimal irreducible decomposition to a minimal primary decomposition by intersecting all the ideals with the same radical. So we know at least that the set of radicals which occur is uniquely determined, though not necessarily how many times they occur.

Our proof will be by induction on the maximal length of a chain of prime ideals in $\text{Ass}(\mathfrak{a})$. Consider first the base case, in which every prime ideal in the set is minimal. Consider any $\mathfrak{p}$ in this set. The primary ideal to which it corresponds is uniquely determined by Theorem 4.10 in Atiyah-MacDonald: the crucial thing is that this is a minimal ideal in the set $\text{Ass}(\mathfrak{a})$. Thus, replacing $\mathfrak{a}$ by this primary ideal, which we have written as an intersection of irreducible ideals $\mathfrak{b}_i$ and irreducible ideals $\mathfrak{c}_j$, our comment that $r=s$ above shows that the number of irreducible ideals $\mathfrak{b}_i$ with radical $\mathfrak{p}$ equals the number of irreducible ideals $\mathfrak{c}_j$ with radical $\mathfrak{p}$.

Now assume the result for chains of maximal length $r-1$. Consider a chain of maximal length $r$. By Theorem 4.10 again, we know that the intersection of the primary ideals which correspond to it is uniquely determined. Ditto for the intersection of the primary ideals corresponding to the first $r-1$ primes in the chain. Using our $r=s$ claim in both cases and subtracting, we finish the proof.


See also Bourbaki, Algebra, Chapter II, exercise 17 to §2, for an alternative proof.


$\def\a{\mathfrak{a}} \def\b{\mathfrak{b}} \def\c{\mathfrak{c}} \def\q{\mathfrak{q}} \def\g{\mathfrak{g}} \def\p{\mathfrak{p}} \def\ass{\operatorname{Ass}}$I will give a detailed argument to show equality of multiplicities of the associated ideals. The following text is a development of the ideas in CJD's answer.

From the hint (proven by Prof. Emerton) plus some argument (cf. [Y, Exercise 7.19]) one deduces

($*$) If $\bigcap_{i=1}^r \b_i=\bigcap_{j=1}^s \c_j$ are minimal irreducible decompositions, then $r=s$.

Let $\{\p_1,\dots,\p_n\}$ be the primes belonging to $\a$. We will denote this set as $\ass\a$ (although we won't ever need this, the notation means that the $\p_i$'s are the primes associated to the $A$-module $A/\a$; cf. [AM, Exercise 7.18]). We might relabel the $\b_i$'s and the $\c_j$'s as $\b_{ij}$, $\c_{ik}$, where $r(\b_{ij})=\p_i=r(\c_{ik})$, $1\leq j\leq r_i$, $1\leq k\leq s_i$. Then $\q_i=\bigcap_{j=1}^{r_i}\b_{ij},\; \g_i=\bigcap_{k=1}^{s_i}\c_{ik}$ are $\p_i$-primary ideals [AM, 4.3] and $\a=\bigcap_{i=1}^n\q_i=\bigcap_{i=1}^n\g_i$ are minimal primary decompositions. We want to show that $r_i=s_i$. Let's proceed by induction in $\ell=$ maximal length of a chain in $\ass\a$. If $\ell=1$ that means every prime belonging to $\a$ is minimal. Thus every subset of $\ass\a$ is isolated (a subset of some poset is isolated if it is downward-closed) and $\q_i=\g_i$ by [AM, 4.10]. That is, we have an equality $\bigcap_{j=1}^{r_i}\b_{ij}=\bigcap_{k=1}^{s_i}\c_{ik}$ of minimal irreducible decompositions, whence $r_i=s_i$, by ($*$). Now suppose $\ell\geq 2$ and assume the result true for any ideal of $A$ whose maximal length of chains in the poset of its belonging primes is less than $\ell$. Let $T=\{\p_{i_1},\dots,\p_{i_m}\}$ be the primes sitting at the top of the chains of length $\ell$ in $\ass\a$. The sets $\Sigma=\ass\a\setminus T$ and $\Sigma_e=\Sigma\cup\{\p_{i_e}\}$ are isolated in $\ass\a$. Hence, by [AM, 4.10], $$ \bigcap_{\p_i\in\Sigma}\q_i=\bigcap_{\p_i\in\Sigma}\g_i,\quad \bigcap_{\p_i\in\Sigma_e}\q_i=\bigcap_{\p_i\in\Sigma_e}\g_i. $$ From the induction hypothesis applied to the former ideal equality, we deduce $r_i=s_i$ for all $1\leq i\leq n$ such that $\p_i\not\in T$. From ($*$) applied to the latter ideal equality, we get $r_{i_e}+\sum_{\p_i\not\in T}r_i=s_{i_e}+\sum_{\p_i\not\in T}s_i$, whence $r_{i_e}=s_{i_e}$.

Lastly, I would like to note that the last sentence in the statement of [AM, Exercise 7.19] (omitted by OP) is “state and prove and analogous result for modules.” The generalization can be done for a Noetherian module (the base ring doesn't need to be Noetherian). This is written in [Y, Exercise 7.19], except that Yu's argument doesn't account the proof of $r_i=s_i$. The above argument is directly generalized to the module case, but instead now one invokes the corresponding generalized results coming from [AM, Exercises 4.22, 4.23] (you may read them in [Y]).


AM. Atiyah, Macdonald, Introduction to Commutative algebra.

Y. B. Yu, Supplement and Solution Manual for Introduction to Commutative algebra. Also saved in the Wayback Machine.


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