Let $I$ be an ideal of a commutative ring $A$. Usually the height is defined as $$\operatorname{ht}(I) = \min_{\mathfrak p \supseteq I} \operatorname{ht}(\mathfrak p), $$ where $\mathfrak p$ runs over the prime ideals of $A$.

But we could define also: $$h_1(I) = \sup \{ n \geq 0 : \exists\, \mathfrak p_0 \subsetneq \cdots \subsetneq \mathfrak p_n \subset I, \mathfrak p_i \text{ prime} \}$$

  1. I think that $h_1(I) = \sup_{\mathfrak p \subset I} \mathrm{ht}(\mathfrak p) =: h_2(I)$, am I right?

  2. Is there an example where $\operatorname{ht}(I) \neq h_1(I)$ ? I don't really see why we made this definition of $\operatorname{ht}(I)$. Even geometrically, I don't really have an example where $\operatorname{codim}(Y)$ is not the same as the analogue of $h_1(I)$ for $Y=V(I)$.

Thank you!

  • $\begingroup$ Yes, you are right that $h_1(I)=h_2(I)$. I have provided your requested counter-example and the geometric justification for the standard definition of height of an ideal in my answer below. $\endgroup$ – Georges Elencwajg Mar 13 '18 at 19:57
  • $\begingroup$ Dear @GeorgesElencwajg, I thank you very much for your comment, as well as your pleasant answer. $\endgroup$ – Alphonse Mar 13 '18 at 20:23
  • $\begingroup$ You are welcome, Alphonse, and yours is a very sensible and natural question, which should be addressed in introductory books. $\endgroup$ – Georges Elencwajg Mar 13 '18 at 21:48

In the polynomial ring $A=k[X]$ over the field $k$ consider the ideal $I=\langle X^2\rangle$.
The only prime ideal of $A$ containing $I$ is $\mathfrak p=\langle X\rangle$, so that $ht(I)= ht(\mathfrak p)=1$.
This is pleasant geometrically since $V(I)\subset \operatorname {Spec }k[X]=\mathbb A^1_k$ is the double point at the origin, whose codimension can reasonably only be thought of as $1$.
However your suggested height would be $h_1(I)=0$ since the only prime $\mathfrak p\subset I$ is $\mathfrak p=\langle0\rangle$ .
This would lead to the attribution of codimension $0$ to our double point $V(I)\subset \mathbb A^1_k$, a not so felicitous choice.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.