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?

  • $\begingroup$ Not that i care too much about internet points, but there is a reason why my question has been downvoted? It's in some way against the rules? $\endgroup$ – karmalu Aug 2 '15 at 12:33

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, $P^2\ne P^{(2)}$.

  • $\begingroup$ The symbolic powers are $P$-primary. But is this their only property that is used in the proof concerned by the question ? Explicitly: Suppose $Q$ is $P$-primary and we have $xy \in Q, x \not\in P$. Does it follow $y \in Q$ ? To my understanding we could only say that a power of $y$ lies in $Q$. But this wouldn't suffice in the proof in question. $\endgroup$ – tj_ Aug 1 '15 at 19:46
  • $\begingroup$ The point is: I don't think your answer actually answers the question because it is missing an important property of $P^{(t)}$ that is used in the proof. $\endgroup$ – tj_ Aug 1 '15 at 20:03
  • $\begingroup$ "Suppose Q is P-primary and we have xy∈Q, x∉P. Does it follow y∈Q ?" Yes. (Read carefully the definition of primary ideals.) $\endgroup$ – user26857 Aug 1 '15 at 20:06
  • $\begingroup$ @tj_ The point is: you don't know the definition of primary ideals. (Helpful: math.stackexchange.com/questions/264879/…) $\endgroup$ – user26857 Aug 1 '15 at 20:07
  • $\begingroup$ @ user26857: If one gives an answer and someone says, I don't understand this part, one should be able to give an exact proof of this part. This is how mathematics works. $\endgroup$ – tj_ Aug 1 '15 at 20:12

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 ?


(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!

  • $\begingroup$ Thanks. For some reason i thought that $P^n$ was P-primary, but both counter-example are interesting on their own $\endgroup$ – karmalu Aug 1 '15 at 23:01
  • $\begingroup$ @karmalu As far as I can see this answer doesn't use the word "primary". $\endgroup$ – user26857 Aug 2 '15 at 6:06
  • $\begingroup$ Well, the property highlighted in blue derives from $P^{(n)}$ being P-primary. So if $P^n$ was P-primary you could have used them instead with no problem. $\endgroup$ – karmalu Aug 2 '15 at 8:01

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