A doubt on Krull's Principal Ideal Theorem Proof

Question

Sorry if it is a dumb question but i'm studying the proof of Krull's PIT from this pdf and i don't understand why the author uses in his proof the ideals $P^{(n)}=P^nR_P\cap R$ instead of the simpler $P^n$. It seems to me that it should work the same. Moreover, in an integral domain aren't $P^{(n)}$ and $P^n$ the same ideal?

Answer 1

The ideals $P^{(n)}$ are $P$-primary (why?), while $P^n$ is not necessarily.

This also leads to an answer to the question if they coincide in an integral domain: from the Wiki page dedicated to the primary ideals we learn that there are prime ideals in integral domains, e.g. $R=K[X,Y,Z]/(XY-Z^2)$ and $P=(x,z)$ such that $P^2$ is not primary. In particular, $P^2\ne P^{(2)}$.

Answer 2

Your question is two-fold:

(1) Why is $P^{(n)}$ used in the proof and not $P^n$ ?

(2) Is $P^{(n)}=P^n$ in domains ?

Answers:

(2) No. Eisenbud (3.9.1 in his commutative alg. book) figured out a counter-example: Let $R=k[x_{ij}\mid 1 \le i,j\le 3]$ be a polynomial ring over a field and let $d = \det(x_{ij})\in R$ be the generic determinat. Then there is a prime ideal $P \subseteq R$ with $d \in P^{(2)}$ but $d \not\in P^2$.

(1) The proof of Th. 1.2 in the linked pdf uses in an essential way following property of $P^{(t)}$:

If $xr\in P^{(t)}$ and $x \not\in P$ then $r \in P^{(t)}$.

In general, $P^t$ doesn't have this property. The following example is taken from Atiyah-MacDonald: Let $R=k[X,Y,Z]/(XY-Z^2)$ and denote by $x,x,z$ the images of $X,Z,Z$ in $R$. Then $P=(x,z)$ is prime, $xy=z^2\in P^2, y \not\in P$, but $x\not\in P^2$.

BTW: I don't think it's a dumb question. Scrutinizing proofs from literature is quite helpful to get a thorough understanding of the stuff!