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Not necessarily. See Examples of common false beliefs in mathematics and the answer by JSE and the reply to his example by Simon Wadsley.

Let $R = \mathbb{Z}+X\mathbb{Q}[X]$ and $I = X\mathbb{Q}[X] = (X, X/2, X/3, \ldots)$, and let $S = \{1,2,3,\ldots\}$. Then $K = \mathbb{Q}(X)$, $R_S = \mathbb{Q}[X]$, and $I_S = X \mathbb{Q}[X]$, whe …

I have a paper giving an irreducibility test (and factoring "algorithm", in some sense of the word) for formal power series over a PID, which should eventually appear in Trans. of the AMS. In particu …

Although Pete Clark's answer is great, I thought I'd post a partial answer that addresses the UFD question in a different direction. My favorite ideal-theoretic characterization of UFDs is that a doma …

If $R$ is a UFD with field of quotients $K$, then the group of units $K^*$ is the direct product of $R^*$ with the free abelian group generated by the irreducible/prime elements of $R$ (one for each c …

Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can there …

This property has been studied before: see http://www.kkms.org/kkms/vol08_1/08110.pdf. In particular, the author G. W. Chang says that an integral domain $R$ satisfies APIT (i.e., the associated prime …

For any field $k$, the ring $A = k(U)[[X,Y,Z]]/(X^2+X^3+UX^6)$ is a UFD such that $A[[T]]$ is not a UFD. For a proof of this, see Salmon, Su un problema posto da P. Samuel. For further developments, …

If you remove "perhaps at the cost of extending the initial [finite] set to a larger [finite] set of prime ideals" from your question, then the answer is yes for a fixed Krull domain $R$ if and only i …

I believe I have answered my question in the positive. Let $R = \mathbb{Z}+\varepsilon\mathbb{Q}[\varepsilon]$, where $\mathbb{Q}[\varepsilon]$ denotes the ring of dual numbers over $\mathbb{Q}$. Th …

An overring of an integral domain is a domain lying between it and its quotient field. Is it possible to have an overring $S$ of an integral domain $R$ such that $S$ has two maximal ideals $\mathfrak …

Let $R$ be a commutative ring with identity. If $P$ is a prime ideal of $R$ that is minimal over some zerodivisor of $R$, then must $P$ consist only of zerodivisors? I suspect not but I can't figure …

A domain $D$ satisfies your condition if and only if it is an LPI domain with trivial Picard group (which can be realized as the group of invertible fractional ideals mod nonzero principal fractional …

A commutative ring is said to be r-Noetherian if every regular ideal is finitely generated, where an ideal is said to be regular if it contains a non-zerodivisor. Does there exist a non-Noetherian r- …

This is a partial answer of a more general problem. If $A$ is a commutative ring with few zerodivisors (which holds iff $\text{Quot}(A)$ is semilocal), then the answer is yes: in that case, an invert …