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Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module or Noetherian $A$-module. In my commutative algebra class, I was given the following theorems.

The first was about the existence of irreducible decomposition of submodules of $M$ :

Every submodule $N$ of $M$ has an irreducible decomposition $N = \bigcap_{i=1}^r N_i$ such that $N_i$ is irreducible.

The irreducible decomposition was called a naive decomposition by the teacher since it did not work well on vector spaces.

Then, I was given the primary decomposition by the following 3 theorems :

1) An irreducible submodule of $M$ is primary.

2) If $N = \bigcap_{i=1}^r N_i$ with $N_i$ being $\mathfrak{p}_i$-primary is an irredundant decomposition, then, $\mathbf{Ass}_A(M/N) = \{\mathfrak{p}_1,\dots,\mathfrak{p}_r\}$.

3) For any shortest primary, irredundant decomposition $N = \bigcap_{i=1}^r N_i$ with $N_i$ being $\mathfrak{p}_i$-primary, if $\mathfrak{p}_i$ is minimal in $\mathbf{Ass}_A(M/N)$, then, $N_i = \phi^{-1}(N_{\mathfrak{p}_i})$ where $\phi : M \rightarrow M_{\mathfrak{p}_i}$ is the canonical morphism.

In addtion, 1) implies the existence of irredundant, shortest primary decomposition.

So, my question is what exactly is the relationship between primary decomposition and irreducible decomposition ?

It seemed from 1) that such two decompositions are the same.

Can anyone help me ? Thank you.

For your information, the following are some related definitions.

A proper submodule $N$ of $M$ is irreducible if $N = N_1 \cap N_2 \implies N = N_1 \lor N = N_2$ for any submodules $N_1,N_2$ of $M$.

$N = \bigcap_{i=1}^r N_i$ is an irreducible decomposition of $N$ if every $N_i$ is irreducible.

A submodule $N$ of $M$ is $\mathfrak{p}$-primary if $\mathfrak{p}$ is the only associated prime ideal of $M/N$ over $A$, that is, $\mathbf{Ass}_A(M/N) = \{\mathfrak{p}\}$.

The decomposition $N = \bigcap_{i=1}^r N_i$ is a primary decomposition if every $N_i$ is primary.

The decomposition $N = \bigcap_{i=1}^r N_i$ is shortest primary if $N_i$ are $\mathfrak{p}_i$-primary and $\forall i \neq j, \mathfrak{p}_i \neq \mathfrak{p}_j$.

The decomposition $N = \bigcap_{i=1}^r N_i$ is irredundant if the intersection with any $N_i$ omitted does not equal $N$.

Again, thank you for your help. This question may be a little too long, sorry to take your time.

Yu Ning
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1 Answers1

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In the context of algebraic geometry, a decomposition $I=\bigcap I_i$ corresponds to a decomposition of $V=\bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.

For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2\subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)\cap(x,y^2).$$

Note both are $(x,y)$-primary, so it is not a shortest decomposition.

Matt Keller
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