CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

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Under an integral extension, a prime is maximal if and only if its contraction is maximal

Statement

Let $A \subseteq B$ be an integral extension, let $q$ be a prime ideal of $B$ , and let $p : = q \cap A$ . Then $q$ is maximal if and only if $p$ is maximal.

Facts & Assumptions

Given: An integral extension $A \subseteq B$ , a prime ideal $q \subseteq B$ , and its contraction $p : = q \cap A$ .

[L1]

In an integral extension of domains, the upper ring is a field if and only if the lower ring is a field (For an integral extension of domains, the upper ring is a field if and only if the lower ring is).

[L2]

A quotient by a prime ideal is a domain ( $R / P$ is an integral domain if and only if $P$ is a prime ideal).

[L3]

A quotient by a maximal ideal is a field ( $R / M$ is a field if and only if $M$ is a maximal ideal).

[A1]

The induced map $A / p \to B / q$ is injective and integral.

Proof

technique · direct

1.1L2A1given

Because $q$ is prime, [L2] makes $B / q$ a domain. The map $A / p \to B / q$ is injective by definition of $p$ , so $A / p$ is a subring of a domain and is therefore a domain. Thus [L2] also shows that $p$ is prime.

2.1L1L3step 1.1∎

By [A1], $A / p \subseteq B / q$ is an integral extension of domains. Therefore [L1] says that $A / p$ is a field if and only if $B / q$ is a field. Using [L3] on both quotients, this is exactly the statement that $p$ is maximal if and only if $q$ is maximal.

Under an integral extension, a prime is maximal if and only if its contraction is maximal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources