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Injective integral extensions preserve Krull dimension

Statement

Assume the Axiom of Choice.

Let $A \subseteq B$ be an injective integral extension of nonzero commutative rings. Then $\dim A = \dim B$ .

Facts & Assumptions

Given: An injective integral extension of nonzero commutative rings $A \subseteq B$ .

[L1]

The Krull dimension of a nonzero commutative ring is the supremum of the lengths of its strict chains of prime ideals (Krull dimension of a nonzero ring).

[L2]

Assuming the Axiom of Choice, every prime of $A$ has a prime above it in $B$ under an integral extension (Lying over for integral ring maps).

[L3]

Assuming the Axiom of Choice, every finite prime chain in $A$ lifts to one in $B$ once its first prime has been chosen (Integral extensions lift finite prime chains from the base).

[L4]

Every strict prime chain in $B$ contracts to a strict chain in $A$ under an integral map (Strict prime chains contract strictly under integral extensions).

Proof

technique · direct

1.1L1given

Both dimensions in the statement are defined by [L1] because $A$ and $B$ are nonzero.

1.2L1L2L3givenalgebra

Let $p_{0} ⊊ \dots ⊊ p_{n}$ be any strict prime chain in $A$ . By [L2], choose a prime $q_{0}$ of $B$ over $p_{0}$ . Then [L3] gives primes $q_{0} \subseteq \dots \subseteq q_{n}$ of $B$ over the whole chain. These inclusions are strict, because $q_{i} = q_{i + 1}$ would force their contractions $p_{i}$ and $p_{i + 1}$ to agree. Hence $B$ has a strict prime chain of length $n$ , so $\dim B \geq \dim A$ .

1.3L1L4given

Conversely, every strict prime chain in $B$ contracts to a strict prime chain of the same length in $A$ by [L4]. Therefore $\dim A \geq \dim B$ .

2.1step 1.2step 1.3∎

The two inequalities of steps 1.2 and 1.3 imply $\dim A = \dim B$ .

Depends on

Used by

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Exercises (14.4)-(14.6) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 7.7 (standard reference, not scraped)