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: $$h_1(I) = \sup \{ n \geq 0 : \exists\, \mathfrak p_0 \subsetneq \cdots \subsetneq \mathfrak p_n \subset I, \mathfrak p_i \text{ prime} \}$$
I think that $h_1(I) = \sup_{\mathfrak p \subset I} \mathrm{ht}(\mathfrak p) =: h_2(I)$, am I right?
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!
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.