Noether-Lasker Theorem I'm on the final part of my project, where I have to prove the Noether-Lasker Theorem (or copy out the following proof and "fill in the gaps"). I'm looking for an explanation of what's going on at a macro-level. I think I could follow the proof, but I don't understand how it proves what it says it proves. I've already proved that every ideal is the intersection of finitely many primary ideals, and somehow the following proves that every ideal is the reduced intersection whose radicals are unique. I feel like if I understood the "macro logic" then I'd be away. Here's the link:
https://dl.dropboxusercontent.com/u/17606191/noether.gif
Thanks for any replies!
 A: After proving an ideal can be expressed at least one way as an intersection of primary ideals, the next logical step is to make that expression as "tight" as possible by eliminating redundancy.
The same idea applies to generating sets of vector spaces. After you find one generating set, then you can remove redundant elements to get down to a minimal generating set, and that set is a basis. Linear independence is basically the quality of a set of vectors to generate their span with no redundancy.
Anyhow, it's obvious if your decomposition of an ideal $I$ has the same ideal in it twice, then you would just remove one of those copies because you would still get $I$, but you'd be writing one less ideal in your decomposition.
Actually you can do better than that. If there are primary ideals $Q,Q'$ in the decomposition with the same radical $P$, then $Q\cap Q'=Q''$ is also primary. So why not delete $Q\cap Q'$ and replace it with $Q''$? You'd be writing out one less ideal, and you'd still be getting $I$.
Once you get down to "minimal" (or "optimal") decompositions, then you can start asking about what properties these decompositions share. All the bases of a vector space, for example, must have the same cardinality. One thing that the minimal primary decompositions share is that they all share the same associated primes.
A: If you're looking for another way to see that the radicals of the primary ideals in a primary decomposition are unique, here is a basic fact which should make the connection to associated primes clearer (this is Lemma 4.4 in Atiyah-MacDonald):
$\newcommand{q}{\mathfrak{q}}$
$\newcommand{p}{\mathfrak{p}}$
Lemma: Let $\q$ be $\p$-primary. For any $x \in A$, we have:
i) If $x \in \q$, then $(\q : x) = 1$
ii) If $x \not \in \q$, then $(\q : x)$ is $\p$-primary (so $r(\q : x) = \p$)
iii) If $x \not \in \p$, then $(\q : x) = \q$
This lemma follows directly from the definitions of radical and primaryness: if you have not done it, I would advise working it out. Once you have this though, and existence of primary decompositions, uniqueness follows. For any ideal $I$, write $I = \cap_i \q_i$ as a reduced intersection of primary ideals, and write $\p_i = r(\q_i)$ for their radicals. On the other hand, the associated primes of $I$, $\text{Ass}(R/I)$, are the prime ideals appearing in the set $\{(I : x) \mid x \in A\}$, and by definition depends only on $I$. Then for any $x$,
$$(I : x) = (\cap_i \q_i : x) = \cap_i (q_i : x) = \cap_{x \not \in \q_i} (q_i : x)$$
Thus $(I : x)$ is prime iff $(I : x) = (\q_i : x)$ is prime for some $x \not \in \q_i$ (since a prime ideal that is a finite intersection of ideals must be one of them), but then $(\q_i : x) = r(\q_i : x) = \p_i$, so every associated prime is one of the $\p_i$'s. Conversely, since the decomposition is reduced, for each $i$ there exists $x_i \not \in \q_i$, $x_i \in \cap_{j \ne i} \q_j$, so $(I : x_i) = (\q_i : x_i)$ and $r(I : x_i) = \p_i$ is prime, so each $\p_i$ is an associated prime.
(Note: the approach I have chosen here differs slightly from that of Atiyah-MacDonald: they define $\text{Ass}(R/I)$ as the prime ideals appearing in the set $\{r(I : x) \mid x \in A\}$. In a Noetherian ring though, it turns out a prime ideal of the form $r(I : x)$ is also of the form $(I : y)$, so the two notions coincidence in the Noetherian case).
