crimboi[19] 吃屎绕大弯

题目背景

为证一个非零吃大师

题面

Assuming AC, prove that for a integral domain R, if any nonzero prime ideal P contains a nonzero prime (p), R is UFD.

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提示

Let S be the set of elements that are prime factorizable (note that it's not conventionally factorizing into irreducibles!!) except for 0. Clearly S is multiplicative. If the conclusion does not hold, take nonzero x\in R-S. Firstly it's easy to demonstrate (x)\subset R-S, in the same way of proving irreducible elements are prime in UFD. Now we need to prove the following lemma: if R is commutative with 0\notin S multiplicative, then for any ideal I\subset R-S we can find a prime I\subset P\subset R-S using AC. Therefore take (x)\subset P\subset R-S. We finally demonstrated P is fucking nonzero. 😮

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提示给的很足热