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Exercise 11.7 from Gathmann's commutative algebra notes reads:

Let $I = Q_1 \cap \ldots \cap Q_n$ be a primary composition of an ideal $I$ in a Noetherian ring $R$. Show that $$\dim R/I = \max \{\dim R/P : \text{$P$ is an isolated prime ideal of $I$}\}.$$ What is the geometric interpretation of this statement?

Here $P$ being isolated means that it's a minimal in the set of prime ideals associated to $I$. The dimension is the Krull dimension.

"$\ge$": A chain of prime ideals in $R/P$ corresponds to a chain of prime ideals in $R$ containing $P$. But $P = \sqrt{Q_i}$ (the radical) for some $i$, and so $P \supset Q_i \supset I$. So we get a corresponding chain of prime ideals in $R/I$.

I'm stuck on the other direction. Any hints would be much appreciated!

kipf
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1 Answers1

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OK, the part I was missing was this corollary from Gathmann's notes:

Corollary 8.30. Let $I$ be an ideal in a Noetherian ring $R$. Then the isolated prime ideals of $I$ are exactly the minimal prime ideals over $I$.

I think this works now:

"$\le$": Let $\bar{P_0} \subsetneq \ldots \subsetneq \bar{P_n}$ be a maximal chain in $R/I$. This corresponds to a chain of prime ideals in $R$, all containing $I$. By maximality, $P_0$ is minimal over $I$. So $P_0 = P$ for some isolated prime ideal $P$, by the corollary. Now apply the canonical projection $\pi : R \to R/P$ to each $P_i$: we get a chain $0 \subsetneq \ldots \subsetneq \pi(P_n)$ of prime ideals in $R/P$, showing that $\dim R/I \le \dim R/P$. Here we used that the image of a prime ideal under a surjection is prime, see e.g. here.

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