Let $R$ a commutative ring and $M$ a $R-$module. Show that the following statement are equivalent: 1) $M=0$, 2) $M_{\mathfrak p}=0$ for all $\mathfrak p\in Spec(R)$, 3) $M_{\mathfrak m}=0$ for all $\mathfrak m\in Specm(R)$. Proof : $\bullet1)\Rightarrow 2)$: Let $M=0$. Let $\frac{a}{b}\in M_{\mathfrak p}$ (i.e. $b\notin \mathfrak p$). Then, $$\frac{a}{b}=\frac{a}{1}\frac{1}{b}=\frac{0}{1}$$ since $\frac{a}{1}=\frac{0}{1}$ and $\frac{0}{1}\frac{1}{b}=0$. $\bullet […]

Given a ring $R$, I want to show that the localization of $R$ at the prime ideal $P$ of $R$ (denoted as $R_P$) is isomorphic to the set of prime ideals of $R$ contained in $P$. That is: $$ \text{Spectrum}(R_P)\cong \{I\subseteq P \mid \text{$I$ is an ideal of $R$}\} $$ From the statment, I can […]

Let $R$ be a ring, $S\subset R$ a multiplicative subset, and let $M$ be a Noetherian $S^{-1}R$-module, then does there exist some Noetherian $R$-module such that $S^{-1}N\cong M$? What about if we only consider localizations of the form $R_f$? What if we also require $R$ to be Noetherian? If we let $N={_R M}$ then clearly […]

Suppose $R$ is a ring and $M,N$ are $R$-modules and $f:M\to N$ is an $R$-linear map. If $\frak{p}$ is a prime ideal of $R$, then we have a $R$-linear map $f_{\frak{p}}:M_{\frak{p}}\to N_{\frak{p}}$ (where $M_{\frak{p}}:=M/\mathfrak{p}M$ and $N_{\frak{p}}:=N/\mathfrak{p}N$) defined by $f_{\frak{p}}([m])=[f(m)]$. I’m trying to prove that $(\ker (f))_{\frak{p}}=\ker (f_{\frak{p}})$. It is obvious to check inclusion $\subset$, but […]

Let $R$ be a UFD and $D \subseteq R$ multiplicative set. What are the units in $D^{-1}R$? I assume the answer should be $D^{-1}R^{\times}$, but I get stuck: If $a/b$ is a unit, then there exists $c/d$ so that $$\frac{a}{b} \cdot \frac{c}{d} = \frac{1}{1} \Longleftrightarrow ac = bd,$$ but I don’t see what this tells […]

I’m a little bit stuck/confused on this abstract algebra problem: Let $R$ be a nonzero commutative ring, and let $T$ be a nonempty subset of $R$ closed under multiplication and containing neither $0$ or zero divisors. Starting with $R \times T$, we can show that $R$ can be enlarged to a partial ring of quotients, […]

Let $Q(R)$ denote the quotient field (or field of fractions) of an integral domain, $R$. If $R$ and $S$ are integral domains such that $Q(R)\cong Q(S)$, does this imply that $R\cong S$? I am not sure about this. I was thinking the answer is yes, but I couldn’t prove it. Perhaps there exist a counterexample?

Let $A$ be a ring, $S\subset A$ a multiplicative set, and $I\subset A$ an ideal not intersecting $S$. For any ideal $J$, let $r(J)$ denote the radical of $J$. Is $S^{-1}r(I) = r(S^{-1}I)$? Certainly $S^{-1}r(I)$ is generated by elements of the form $\frac{x}{s}$, where $x^n\in I$. This implies that $\left(\frac{x}{s}\right)^n = \frac{x^n}{s^n}\in S^{-1}I$, so $\frac{x}{s}\in […]

I want to characterize the total classical quotient rings of the (commutative) rings $R=\mathbb Z_4$, $R=\mathbb Z_8$ or any $R=\mathbb Z_{2^n}$. In fact, if we get $S$ to be the regular elements of $R$, by the classical quotient ring we mean $RS^{-1}$ (which is the ring of fractions of $R$). Thanks for any good answer!

I’m working on the following problem with no luck. Let $A$ be a domain. If $0 \neq f \in A$, then prove that $$A_{f} = \bigcap_{P \in\operatorname{Spec}(A), f \not\in P} { A_{P}}$$ and conclude that $A_{f}$ depends only on the open subset of $\operatorname{Spec}(A)$ that contains $f$. I’m thinking about the fact that the primes […]

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