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I asked this question last year in MSE, but I didn't get an answer.

I took a commutative algebra course last semester (using Kaplansky's book), and I learned about Krull's intersection theorem. In the course, we proved it without using the Artin-Rees Lemma. I have heard that the standard proof uses the lemma.

Are there any other simple or intuitive proofs for the theorem? I have heard that there is an intuitive reason for coordinate rings in this post, that the only function which vanishes to arbitrarily high order at a point is the zero function. In case of polynomials, this is easy to accept if we consider the polynomial as a Taylor series expansion (with finitely many terms). Using this, I tried to prove it for a local ring, as follows.

Let $R$ be a Noetherian ring. Suppose there exists $D:R\to R$ satisfying $D(ab)=(Da)b+a(Db)$ and $D(a+b)=Da+Db$, that is, it is a derivation on $R$. Define $\mathrm {Poly}(R):=\{r\in R\,| \,D^{k}r=0$ for some $k>0\}$, which means that the set of elements in $R$ on which $D$ behaves as a polynomial.

We check that $\mathrm {Poly}(R)$ is a subring of $R$, and I wonder whether $R=\mathrm {Poly}(R)$. Then I want to approximate elements in $R$ via their Taylor expansions, but it was not easy to formalize this. I think that the evaluation map corresponds to quotients by maximal ideal; and then it would be possible to find a function $r^{*}:k\to k$ where $k=R/\mathfrak{m}$ is a field and $r^{*}$ is induced by an element $r\in R$.

All these things are possible in the case of polynomial ring, but it is very hard to do it for general Noetherian rings (or Noetherian local rings). I also noticed that there aren't any nontrivial derivations on the ring $\mathbf{Z}$. Is there a proof that uses these ideas or versions of them?

David Handelman
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Seewoo Lee
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  • This is not an answer to your question, but you may want to know, just in case that you don't already do, that there is a short, neat, slick, and amazingly beautiful proof of Krull's intersection theorem by Hervé Perdry on the AMM: An Elementary Proof of Krull's Intersection Theorem, Amer. Math. Monthly 111 (Apr., 2004), No. 4, 356-357. – Salvo Tringali Apr 25 '17 at 09:09
  • Your definition of $\mathrm{Poly}(R)$ depends on $D$ and for different $D$ you may get different subrings. So, even for polynomial rings, you will have to carefully choose your $D$ (Think of Euler derivation). What that would mean for an arbitrary Noetherian ring is anybody's guess. – Mohan Apr 25 '17 at 12:38
  • To answer your question about whether $\text{Poly}(R) = R$: consider the ring $\mathbb{C}[x, e^x]$ with the derivation $\partial_x$. – user44191 Apr 25 '17 at 23:32
  • Concerning the OP's statement, "I wonder whether $R = {\rm Poly}(R)$", the answer, of course, is no: Consider the trivial derivation on $R$ mapping everything to $0_R$. – Salvo Tringali Apr 26 '17 at 08:42
  • @SalvoTringali I'll check it later since I don't have any authority to see the paper now. Also, I want to find nontrivial derivation which gives $R=\mathrm{Poly}(R)$. – Seewoo Lee Apr 26 '17 at 09:11
  • @Mohan Yes I agree, and It will be hard to find a canonical way to construct nontrivial derivation for a given (Noetherian, and even local) ring. – Seewoo Lee Apr 26 '17 at 09:16
  • @See-WooLee I think you got an email. ;-) As for the rest, why don't specify in the OP that $D$ is a non-trivial derivation? Also, your comment sounds like your question is different than what is asked in the OP right now: Do you wonder whether there exists at least one derivation on $R$ such that ${\rm Poly}(R) = R$? Or whether ${\rm Poly}(R) = R$ for every non-trivial derivation $D$ on $R$? – Salvo Tringali Apr 26 '17 at 09:21
  • Whatever the case, let $p$ be a prime number, $R$ the ring of polynomials in one variable $x$ with coefficients in $\mathbb F_p$ (the field with $p$ elements), and $D$ any derivation on $R$. Then $R$ is Noetherian (you ask for Noetherian rings), and $x^{p-1} \notin {\rm Poly}D(R)$: If $x^{p-1}=D(f)$ for some $f \in R$ and we write $f(x)=\sum{i=0}^na_ix^i$, then $n \ge p$, and we must have $ia_i = 0$ for every $i \in [![1,n]!] \setminus {p}$ and $pa_p = 1$, which is impossible. – Salvo Tringali Apr 26 '17 at 11:22

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