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There is a proof I don't get in Lang's Algebra:

(page 423) Theorem 3.3. Let $M$ be a Noetherian $A$-module. Let $N$ be a submodule. Then N admits primary decomposition.

Proof. We consider the set of submodules that do not admit a primary decomposition. If this set is not empty, then it has a maximal element because M is Noetherian. Let N be this maximal element. Then N is not primary and there exists $a \in A$ such that $a_{M / N}$ (This is $x \mapsto ax$ for $x \in M/N$) is neither injective nor nilpotent. The increasing sequence of modules $$\ker a_{M / N} \subset \ker a^2_{M / N} \subset \ker a^3_{M / N} \subset \cdots$$ stops, say at $a^r_{M / N}$. Let $\phi : M/N \to M/N$ be the endomorphism $\phi = a^r_{M / N}$. Then $\ker \phi = \ker \phi^2$. Hence $0 = \ker \phi \cap \text{im} \phi$ in $M/N$ and neither the kernel nor the image of $\phi$ is $0$. Taking the inverse image in M we see that N is the intersection of two submodules of M, unequal to N. We conclude, from the maximality of N, that each one of these submodules admits a primary decomposition, and therefore N admits one also, contradiction.

The sentence in bold is what I don't understand. The inverse image of what? Is there the canonical homomorphism involved somehow? And how would an inverse image show such a thing at all?

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  • $\begingroup$ I changed "homeomorphism" to "homomorphism" because "homeomorphism" is a term associated with topology, which is not being used here. $\endgroup$
    – rschwieb
    Commented Oct 3, 2012 at 12:28
  • $\begingroup$ I feel silly now. I always thought they were the same thing. They do have the same etymology, don't they? $\endgroup$ Commented Oct 3, 2012 at 15:02
  • $\begingroup$ Very close etymologies: dictionary.com seems to indicate that the Greek "homoios" means "similar" and "homos" means "the same". $\endgroup$
    – rschwieb
    Commented Oct 3, 2012 at 15:31

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I'm sure it is a typo, but it's Noetherian first of all.

Second, to answer your question: You have that $$0 = \ker( \phi)\cap \mathrm{im}(\phi)$$ in M/N. The claim that neither the image or the kernel is 0 translates to the fact that, if you consider the canonical homomorphism, $\psi: M \rightarrow M/N$ that $\psi^{-1}(\ker( \phi)) \neq N$. This is since 0 in $M/N$ is represented by the coset $N$.

Now, you do know however that since $0 = \ker( \phi)\cap \mathrm{im}(\phi)$, that taking the inverse image of $\ker( \phi)$ and $\mathrm{im}(\phi)$ under the canonical morphism, that their intersection is equal to N.

The key fact to note here is that the inverse image of 0 in $M/N$ under the quotient map is equal to N. This should make it all clear, I hope.

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