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If $M$ is Noetherian, any submodule $N \neq M$ can be written as a finite intersection of intersection-indecomposable ones.

I just don't know how to begin with the proof of this proposition.

Very kind of you for reading or answering.

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@ShinyaSakai: Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening. – Arturo Magidin Mar 14 '11 at 20:40
Maybe "indecomposable" is supposed to be "lattice irreducible"? – Jack Schmidt Mar 14 '11 at 20:53
@Arturo Magidin San: Thank you very much for your warning. I don't know how to reedit my question, but I will not make this kind of mistake any more. @Jack Schmidt San: This is a lemma on page 435 in the second edition of Nanthan Jacobson's Basic Algebra. I think there maybe something wrong in the proof on this book, so I tried to prove it on my own but failed. – ShinyaSakai Mar 14 '11 at 21:03
@ShinyaSakai: I believe you meant primary, not "indecomposable". At least, that's what Theorem 7.21 in my copy of Jacobson says (page 439, as it happens). – Arturo Magidin Mar 14 '11 at 21:10
@Arturo: indecomposable on page 438 for me (defined as meet-irreducible). – Jack Schmidt Mar 14 '11 at 21:13

2 Answers 2

up vote 3 down vote accepted

The three equivalent definitions of "noetherian module" (in the presence of the Axiom of Choice) are:

  1. $M$ is noetherian if and only if every submodule of $M$ is finitely generated.

  2. $M$ is noetherian if and only if every ascending chain of submodules of $M$ stabilizes: $N_1\subseteq N_2\subseteq\cdots$ implies that there exists $m$ such that for all $k\geq m$, $N_m=N_k$.

  3. $M$ is noetherian if and only if every nonempty collection of submodules of $M$ has maximal elements.

So look at the collection of all proper submodules that are not finite intersections of intersection-indecomposable submodules. If this collection is empty, you are done. If it is not empty, then by (3) above it has a maximal element $N$. Note that $N$ cannot be intersection-indecomposable, nor can $N$ be maximal in $M$ (if $N$ is maximal in $M$, then it is intersection indecomposable). So there must exist two proper submodules $N_1$ and $N_2$ of $M$ such that $N=N_1\cap N_2$, and $N\neq N_1$, $N\neq N_2$. Since $N$ is maximal among submodules that are not finite intersections of intersection-indecomposable submodules, then neither $N_1$ nor $N_2$ belong to the collection. So $N_1$ is a finite intersection of intersection-indecomposable submodules, as is $N_2$, $N_1 = \cap_{j=1}^k M_j$, $N_2 = \cap_{j=k+1}^{\ell}M_j$. Then $$N = N_1\cap N_2 = \cap_{j=1}^{\ell} M_j$$ is a finite intersection of intersection-indecomposable submodules, a contradiction. The contradiction arises from assuming that the collection of submodules that do not satisfy the condition is nonempty, so the collection is empty and we are done.

Note. Just before stating the lemma, Jacobson explicitly defines "indecomposable" as follows: "We shall call a submodule $N$ of $M$ (intersection) indecomposable if we cannot write $N$ as $N_1\cap N_2$ where $N_i\neq N$ for $i=1,2$." This is the definition I am using above, rather than the more common "not expressible as a direct sum of two proper submodules" (or even "subdirectly irreducible").

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Yes, I think we have read the same book~(maybe in different edition...) What I don't understand is: why can we say that $N=N_1\cap N_2$ when we only know that $N$ is indecoposable? I think the definition of "indecomposable" for $N$ is that it does not equal to the direct sum of any two of its proper submodules. Or, do you mean that if $N$ is not a maximal submodule of $M$, then it must be the intersection of two submodules of $M$, both of which contain it properly? – ShinyaSakai Mar 15 '11 at 6:48
Jacobson uses "indecomposable" weirdly. He means a module that is not an intersection. He does not mean a module that is not a direct sum (though, that is what I would mean). – Jack Schmidt Mar 15 '11 at 7:17
I see the definition of "indecomposability" of modules on his book Basic Algebra II, in section 3.4 "the krull-schmidt theorem": A module $M$ is decomposable if $M=M_1 \oplus M_2$ where $M_i \neq 0$. Otherwise, $M$ is called indecomposable. – ShinyaSakai Mar 15 '11 at 9:46
@ShinyaSakai: the paragraph just before Lemma 1 reads: "We now assume that $M$ is noetherian and we shall show that every submodule $N$ of $M$ can be expressed as a finite intersection of primary submodules We shall call a submodule $N$ of $M$ (intersection) indecomposable if we cannot write $N$ as $N_1\cap N_2$ where $N_i\neq N$ for $i=1,2$. We first prove Lemma 1. If $M$ is noetherian, any submodule $N\neq M$ can be written as a finite intersectio of indecomposable ones." – Arturo Magidin Mar 15 '11 at 13:43
@ShinyaSakai: This is the only reference to "indecomposable submodule" in the index to Jacobson, so that is the definition he uses. He even puts "intersection" in parenthesis before its first use. So that's what he means by "indecomposable", not the (more usual) notion not expressible as a direct sum of nontrivial submodules. – Arturo Magidin Mar 15 '11 at 13:44

This is obviously false; perhaps you meant something else? Consider a vector space of dimension 3 as the module M. It is Noetherian, but its directly indecomposable submodules are all dimension 1. If N has dimension 2, then N is not the intersection of one-dimensional submodules.

I suspect this is just a translation problem. There is the Lasker–Noether theorem that expresses each proper submodule of a Noetherian module as an intersection of meet-irreducible (or primary) submodules.

See page 84 and page 102 of Lam's Lectures on Modules and Rings for my understanding of it, or the proper chapter and exercises in Atiyah-MacDonald for a more commutative view.

The part you asked about I think just follows from the definition of Noetherian and should be true in any lattice with appropriate finiteness condition.

In section 7.13 of Jacobson's Basic Algebra II (pagination differs a little, page 438 for me), Jacobson uses (intersection) indecomposable to mean what I have called meet-irreducible. His proof is by Noetherian induction and holds in any Noetherian lattice.

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I agree with you. But this is a lemma on page 435 in the second edition of Nanthan Jacobson's Basic Algebra. This lemma is used in the proof of the decomposition for ideals in noetherian rings as finite intersection of primary ideals. – ShinyaSakai Mar 14 '11 at 21:05
What part of the proof do you have trouble with? This is just called induction and is very standard. – Jack Schmidt Mar 14 '11 at 21:12
I am wondering if we are reading Nanthan Jacobson's book in different editions. My book is in edition two, published by "W. H. freeman and company". This is lemma 1 on page 435 in section 7.12 of Basic Algebra II. I have doubt in it because I think of the example that if $M=N_1 \oplus N_2 \oplus N_3$, and all of $N_1,N_2,N_3$ are irreducible, $N=N_1 \oplus N_2$ cannot be written as a finite intersection of indecomposable submodules of $M$. – ShinyaSakai Mar 15 '11 at 6:58
If M=N1⊕N2⊕N3, then N1 is no longer intersection-indecomposable since N1 = (N1⊕N2⊕0) ∩ (N1⊕0⊕N3). Informally, intersection-indecomposable modules are sort of "big", while direct-sum-indecomposable modules are "small". Jacobson uses "indecomposable" to mean intersection-indecomposable (which I think is strange). – Jack Schmidt Mar 15 '11 at 7:16
Thank you very much. Now I know how I was mistaken... – ShinyaSakai Mar 15 '11 at 21:35

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