Regular local ring implies integral domain
This article gives the statement and possibly, proof, of an implication relation between two commutative unital ring properties. That is, it states that every commutative unital ring satisfying the first commutative unital ring property must also satisfy the second commutative unital ring property
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Any regular local ring is an integral domain. In other words, if a ring has a unique maximal ideal and that ideal is generated by a set whose size is the Krull dimension of the ring, then the ring is an integral domain.
Base case for induction
Suppose the result is true for dimensions up to . We need to prove that the result is true for of Krull dimension .
We know the following:
- By Nakayama's lemma,
- The set of minimal prime ideals of is finite (This needs to be clarified/explained properly)
Now, suppose were contained in the union of and the minimal prime ideals. Then, by the prime avoidance lemma, must be contained either in or in one of the minimal prime ideals. thus forces to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.
Thus, there exists an element in which is outside the union of and all the minimal prime ideals. Let and . Clearly is the unique maximal ideal in . By the choice of , , and in fact we can conclude that This needs to be clarified/explained properly.
Now is a proper homomorphic image of so it can be generated by elements. By Nakayama's lemma, can also be generated by elements.
Thus is a regular local ring, and hence, by the induction step assumption, is an integral domain. Hence is a prime ideal of . But since lies outside every minimal prime ideal, there is a minimal prime ideal properly contained inside . Call this minimal prime ideal .
If is any element, then we may write for some . But then since , . Thus, . Nakayama's lemma now yields , and hence is an integral domain.