# Definition of height of an ideal

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!

• 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. – Georges Elencwajg Mar 13 '18 at 19:57
• Dear @GeorgesElencwajg, I thank you very much for your comment, as well as your pleasant answer. – Alphonse Mar 13 '18 at 20:23
• You are welcome, Alphonse, and yours is a very sensible and natural question, which should be addressed in introductory books. – 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.