Inequality of Height of Proper Ideal

Theorem
Let $A$ be a commutative ring with unity.

Let $I$ be a proper ideal in $A$.

Then:
 * $\map {\operatorname {dim_{Krull} } } {A / I} + \map {\mathrm {ht} } I \le \map {\operatorname {dim_{Krull} } } A$

where:
 * $A/I$ is the quotient ring of $A$ by $I$
 * $\operatorname {dim_{Krull} }$ denotes the Krull dimension
 * $\map {\mathrm {ht} } I$ is the height of $I$

Proof
Let:
 * $n := \map {\operatorname {dim_{Krull} } } {A / I}$

Then there are $\mathfrak q_0, \ldots, \mathfrak q_n \in \Spec {A / I}$ such that:
 * $\mathfrak q_0 \subsetneqq \cdots \subsetneqq \mathfrak q_n$

Let $\pi : A \to A / I$ be the quotient epimorphism.

Let:
 * $\tilde {\mathfrak q_i} := \pi^{-1} \sqbrk {\mathfrak q_i}$

for $i=0, \ldots, n$.

Then:
 * $\tilde {\mathfrak q_0}, \ldots, \tilde {\mathfrak q_n} \in \Spec A$

and:
 * $(1):\quad I \subseteqq \tilde {\mathfrak q_0} \subsetneqq \cdots \subsetneqq \tilde {\mathfrak q_n}$

On the other hand, let:
 * $m := \map {\mathrm {ht} } {\tilde {\mathfrak q_0} }$

Then there exist $\mathfrak p_0, \ldots, \mathfrak p_m \in \Spec A$ such that:
 * $(2):\quad \mathfrak p_0 \subsetneqq \cdots \subsetneqq \mathfrak p_m = \tilde {\mathfrak q_0}$

Hence: