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Noether-Lasker Theorem: In a Noetherian ring $R$, every ideal $I$ is an intersection of finitely many primary ideals $P_1, P_2, ..., P_n$. My textbook (Grillet's Abstract Algebra) proves that the radical of each primary ideal is an associated prime ideal $I:a=\{r\in R\mid ra\in I\}$ of $I$ for some $a\notin I$.

But I don't know the motivation of the associated prime ideal. That is, I don't know why we need to express these primary ideals as ideal quotient.

An idea is that the ideal quotient is easy to compute. But I am not sure.

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2 Answers 2

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I dont have your textbook, but I try to answer your question. If this is not what you want, I'll delete it.

In short it is because of "uniqueness theorems".
[Atiyah-MacDonald, Theorem 4.5], 1st uniqueness theorem: Let $\mathfrak{a}$ be a decomposable ideal in the ring $A$ and $\mathfrak{a}=\bigcap_{1\leq i\leq n}\mathfrak{q}_i$ be a minimal primary decomposition of $\mathfrak{a}$. Let $\mathfrak{p}_i=\mathrm{rad}(\mathfrak{q}_i)$, ${1\leq i\leq n}$. Then the $\mathfrak{p}_i$ are precisely the prime ideals which occur in the set of ideals $\mathrm{rad}(\mathfrak{a}:x)$, $x\in A$, and $\color{red}{hence}$ are independent of the particular decomposition of $\mathfrak{a}$.

Then they prove 2nd uniqueness theorem as Theorem 4.10 (they make free use of the 1st uniqueness Theorem).

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  • $\begingroup$ Thanks. I will read the Atiyah's book. $\endgroup$
    – bfhaha
    Jan 14, 2017 at 15:53
  • $\begingroup$ you are welcome. $\endgroup$
    – user 1
    Jan 14, 2017 at 18:14
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The decomposition of an ideal $I$ as intersection of primary ideals is the extension to an arbitrary noetherian ring of the decomposition of an ideal in a P.I.D., or more generally a Dedekind domain, of an ideal as a product of powers of prime ideals, which itself is the generalisation of the decomposition of a positive integer as the product of powers of primes.

In $\mathbf Z$, a $p$-primary ideal is the ideal generated by a power of the prime number $p$.

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  • $\begingroup$ I'm not sure this answers the question. $\endgroup$
    – user26857
    Jan 13, 2017 at 15:31

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