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Let the below ideals be in a commutative Noetherian ring $R$.

Corollary 22. (3) There are prime ideals $P_1, \dots, P_n$ (not necc. distinct) $\supset I$ such that $P_1\cdots P_n \subset I$.

(Out of D&F)

Prove (3) of Corollary 22 directly by considering the coll. $\mathcal{S}$ of ideals that do not contain a finite product of prime ideals. [If $I$ is a maximal element in $\mathcal{S}$, show that since $I$ is not prime there are ideals $J, K$ properly containing $I$ (hence not in $\mathcal{S}$) with $JK \subset I$.]

I know:

  • $I$ is not prime $\implies \exists$ ideals $J,K$ such that $JK \subset I$ yet $J \not\subset I$ and $K \not\subset I$.
  • $I$ is not prime $\implies$ in particular not maximal $\implies$ $I$ properly contained in some maximal ideal $J$.
  • From examining proof to Proposition 20 the proof of this would go something like if $\mathcal{S}$ were not empty, then since $R$ is Noetherian, all chains in $\mathcal{S}$ are upper bounded and so $\mathcal{S}$ contains a maximal element $I$.

I can't piece it together from these facts alone, what am I missing?

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Assume that $\mathcal{S}$ is nonempty and let $I$ be a maximal element. Then $I$ is in $\mathcal{S}$ so it is not prime. Since $I$ is not prime choose $a,b \in R$ such that $ab \in I$ and $a,b \not\in I$. Then $I+(a)\neq R$ because if not we can write $1=x+sa$, thus $b=xb+sab \in I$. Similarly $I+(b) \neq R$. By maximality of $I$ there are prime ideals $\mathfrak{p}_1,...,\mathfrak{p}_k$ with $I+(a) \subset \mathfrak{p}_j$, $\prod \mathfrak{p}_j \subset I+(a)$ and $\mathfrak{q}_1,...\mathfrak{q}_l$ with $I+(b)\subset \mathfrak{q}_j$, $\prod \mathfrak{q}_j \subset I+(b)$. Then $\prod \mathfrak{p}_j \prod \mathfrak{q}_j \subset (I+(a))(I+(b))\subset I$. This is a contradiction. So $\mathcal{S}$ is empty.

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  • $\begingroup$ How can you choose $a, b \in R$ such that $ab \in I$? $\endgroup$ – StudySmarterNotHarder Dec 30 '15 at 23:33
  • $\begingroup$ I don't like this answer at all. $\endgroup$ – StudySmarterNotHarder Dec 30 '15 at 23:34
  • $\begingroup$ Well this is basically your first fact. You have $J\not\subset I $ and $K \not\subset I$ so there are $a\in J$ and $b\in K$ such that $a$ and $b$ are not in $I$. But $JK \subset I$, so $ab\in I$ $\endgroup$ – math635 Dec 30 '15 at 23:49
  • $\begingroup$ i extended the answer. What don't you like about it? $\endgroup$ – math635 Dec 30 '15 at 23:50
  • $\begingroup$ I like your first answer now, the only part I don't understand is how $JK \subset I$ and $J \supsetneq I, \ K \supsetneq I$ is a contradiction. $\endgroup$ – StudySmarterNotHarder Dec 31 '15 at 0:01
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by maximality of $I$ in $S$, in the ring $R/I$ every ideal $K \subset R/I$ is contained in each of a finite set of prime ideals $K_i$ such that some product formed from them (possibly with repetitions) is contained in $K$. we may call the $K_i$ a primal set for $K$ and the product contained within $K$ as a primal product for $K$

since $I$ is not prime the quotient $R/I$ has zero-divisors. let us say $PQ=0$ for the corresponding principal ideals.

let $P_i,Q_i$ be primal sets for $P$ and $Q$ respectively. since any $T'$ (the pre-image of a prime ideal $T \subset R/I$) is prime in $R$ we have that $P'_i,Q'_j$ are primal sets for $P',Q'$.

since $P'Q' \subset I$, $I \subset P'_i$ and $I \subset Q'_j$ the $P'_i$ and $Q'_j$ together form a primal set for $I$ (with the corresponding primal product being the product of the primal products for $P'$ and $Q'$ separately

this contradicts the assumption $I \subset S$ showing that $S$ is empty

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