Skip to main content
2 of 2
grammar; blackboard bold -> bold; romanized symbols
David Handelman
  • 4.7k
  • 2
  • 23
  • 35

Proof of Krull's intersection theorem with Taylor expansion

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?

Seewoo Lee
  • 2.2k
  • 12
  • 16