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Here is a proof that $$\ker \lambda = \{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \quad (LC_3)$$ holds true assuming that the following definition is in use: $$A[S^{-1}] = A[T_s \vert s …

Let $A$ be an integral domain and let $N \subseteq M$ be two modules over $A$. The condition $S^{-1}N \cap M = N$ for $S = A \setminus \{0\}$ is equivalent to $N \cap aM = aN$ for every $a \in A$, pro …

The answer is no in general. Take $R = \prod_{n \in \mathbb{N}_{> 0}} \mathbb{Z}/2^n\mathbb{Z}$. Then the Jacobson radical of $R$ is $\prod_{n \in \mathbb{N}_{> 0}} 2\mathbb{Z}/2^n\mathbb{Z}$, and it …

A structure theorem and a sketch proof are presented in Exercise 3.7 of The K-book: An introduction to algebraic K-theory (Graduate Studies in Mathematics 145) by Charles A. Weibel. I found this re …

As a complement to the minimal example of Matthé van der Lee, that is, an example for which $$1 = \operatorname{height}(\mathfrak{p}) > \operatorname{height}(P) = 0,$$ it is easy to check that Nagata' …

The "local Artinian" assumption in OP's question can be removed as the following more general result holds (rings are supposed unital and commutative): If $M$ is a finitely generated module over a …

I'll assume that elementary row and column transformations have determinant one. Otherwise the question is trivial, as darij grinberg pointed out in his comment above. My answer expands on Mohan's com …

The answer is yes if $R$ is any atomic domain, e.g., $R$ is a Noetherian domain. Claim 1. Let $R$ be any integral domain. The set $S = S_R$ is saturated in the sense that if $ab \in S_R$ for …

This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated. Let us denote by $D^{\times}$ the unit group of $D$. We have $i(D) = D^{\times}$ for $D = k[x, y, z]$. In …

This answer addresses the trivial part of the question, that is multiplicative stability, while providing non-trivial examples of saturated sets $k(R)$ under two rather restrictive conditions: $R$ is …

Let us first observe that an ideal $I$ of a commutative ring $R$ with identity is minimal in OP's sense if and only if $I$ is a non-zero simple module over $R$. From now on, we will favour this termin …

The following claim characterizes circumstances under which $\ker(\text{eval}_q)$ for $q \in K = \text{Frac}(R)$ is a principal ideal of $R[X]$. In particular, this answers question $2$ in the positiv …

For the first question, consider $R = \mathbb{Z}[X]/(X^3 - 1)$, $P = (x+ 1)$ and $a = (x + 1)(1 + x + x^2)$ where $x$ denotes the image of $X$ in $R$. In order to see that it provides us with a redu …

Let us first observe that $M \Doteq \text{coker}(\phi_n)$ is naturally an $R/\text{rad}(R)$-module with $$\text{rad}(R) \Doteq P_1 \cap \cdots \cap P_n.$$ So, there is no actual loss in generality if …

Indeed, it seems that the situation gets nicer, but certainly not as nice as what I depicted in my first and very flawed answer. (See the remains below and the enlightening counter-example of Wilberd …