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An Artinian integral domain is a field

Statement

Let $D$ be a commutative Artinian integral domain. Then $D$ is a field.

Facts & Assumptions

Given: A commutative Artinian integral domain $D$ and a nonzero element $a \in D$ .

Proof

technique · direct

1.1

By The ideal generated by a subset and principal ideals, the principal ideals $(a) \supseteq (a^{2}) \supseteq (a^{3}) \supseteq \dots$ form a descending chain of ideals. Since $D$ is Artinian, this chain stabilizes, so $(a^{n}) = (a^{n + 1})$ for some integer $n \geq 1$ .

givenalgebra

2.1

Because $a^{n} \in (a^{n + 1})$ , there is $b \in D$ with $a^{n} = a^{n + 1} b$ , hence $a^{n} (1 - a b) = 0$ . As $D$ is an integral domain, every power of the nonzero element $a$ is nonzero, so $a^{n} \neq 0$ and therefore $1 - a b = 0$ . Thus $a b = 1$ , and the chosen nonzero element $a$ is a unit.

step 1.1givenalgebra

3.1

Every nonzero element of $D$ is a unit, so Field makes $D$ a field.

step 2.1∎

Depends on

Used by

Every prime ideal of an Artinian ring is maximal Theorem

Dependency tree · two levels

11 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 16.1 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Section 19 (standard reference, not scraped)