LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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

Statement

Let $A \subseteq B$ be an integral extension of domains. Then $A$ is a field if and only if $B$ is a field.

Facts & Assumptions

Given: An integral extension of domains $A \subseteq B$ .

[L1]

In an integral ring map, every element of the target ring satisfies a monic polynomial over the source ring (Integral ring maps and integral extensions).

[L2]

An integral domain is a nonzero commutative ring with no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

[L3]

A field is a nonzero commutative ring in which every nonzero element is invertible (Field).

Proof

technique · direct

1.1L1L2L3givenalgebra

Assume $A$ is a field, and let $0 \neq b \in B$ . By [L1] there is a monic relation $b^{n} + a_{n - 1} b^{n - 1} + \dots + a_{0} = 0$ over $A$ of minimal degree. The constant term cannot vanish: if $a_{0} = 0$ , then $b (b^{n - 1} + a_{n - 1} b^{n - 2} + \dots + a_{1}) = 0$ , and [L2] with $b \neq 0$ would give a smaller monic relation, contradicting minimality. Since $a_{0} \neq 0$ and $A$ is a field, $a_{0}^{- 1} \in A$ , and rearranging gives $b^{- 1} = - a_{0}^{- 1} (b^{n - 1} + a_{n - 1} b^{n - 2} + \dots + a_{1}) \in B$ . Thus every nonzero element of $B$ is invertible.

1.2L1L3givenalgebra

Assume $B$ is a field, and let $0 \neq a \in A$ . Then $a^{- 1} \in B$ , and [L1] gives a monic equation $(a^{- 1})^{n} + c_{n - 1} (a^{- 1})^{n - 1} + \dots + c_{0} = 0$ with $c_{i} \in A$ . Multiplying by $a^{n}$ yields $1 + c_{n - 1} a + \dots + c_{0} a^{n} = 0$ , so $a (- c_{n - 1} - c_{n - 2} a - \dots - c_{0} a^{n - 1}) = 1$ . Hence $a$ is invertible in $A$ .

2.1step 1.1step 1.2∎

Step 1.1 proves that $A$ field implies $B$ field, and step 1.2 proves the converse. Therefore $A$ is a field if and only if $B$ is a field.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources