Inequality of Height of Proper Ideal

Theorem

Let $A$ be a commutative ring with unity.

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

Then:

$\map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } I \le \map {\operatorname {dim_{Krull} } } A$

where:

$A/I$ is the quotient ring of $A$ by $I$

$\operatorname {dim_{Krull} }$ denotes the Krull dimension

$\map {\operatorname {ht} } I$ 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:

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

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On the other hand, let:

$m := \map {\operatorname {ht} } {\tilde {\mathfrak q_0} }$

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:

\(\ds \map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } I\)

\(\le\)

\(\ds \map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } {\tilde {\mathfrak q_0} }\)

as $I \subseteqq \tilde {\mathfrak q_0}$

\(\ds \)

\(=\)

\(\ds n + m\)

\(\ds \)

\(\le\)

\(\ds \map {\operatorname {dim_{Krull} } } A\)

by $(1)$ and $(2)$

$\blacksquare$

Sources

• 1980: Hideyuki Matsumura: Commutative Algebra $12:$ Dimension