This is Exercise 1.2.6(a) in Liu, Algebraic Geometry and Arithmetic Curves Let $B$ be a flat $A$-algebra. Show that for any finite family $\{I_\lambda\}_{\lambda\in \Lambda}$ of ideals of $A$, we have $\cap_{\lambda\in\Lambda}(I_\lambda B)=(\cap_{\lambda\in \Lambda}I_\lambda)B$. How can I prove that? Also does the following holds: Let $B$ be a flat $A$-algebra. Does there exist an infinite […]

Given two submodules $U,V \subseteq M$ over a (commutative) ring $R$, and a flat $R$-module $A$, I can interpret $U \otimes_R A$ and $V \otimes_R A$ as submodules of $M \otimes_R A$. Is it necessarily true that $$(U \cap V) \otimes_R A \cong (U \otimes_R A) \cap (V \otimes_R A) ?$$ I think it should […]

$M$ is a flat $R$-module if and only if $\hom_{\mathbb{Z}}(M,\mathbb{Q/Z})$ is injective. One direction is easy. Suppose $M$ is flat. We know that $$ \hom_\mathbb{Z}(-\otimes M, \mathbb{Q/Z}) \cong_{\mathbb{Z}} \hom_{\mathbb{Z}}(-,\hom_{\mathbb{Z}}(M,\mathbb{Q/Z}))$$ Since $- \otimes M$ is exact and $\mathbb{Q/Z}$ is injective, the left functor is exact which shows that the right functor is exact, i.e. $\hom_{\mathbb{Z}}(M,\mathbb{Q/Z})$ is […]

So when you work over a commutative ring, this result is quite well known. I am wondering if the same holds true for an arbitrary ring; that is, if $R$ is some (possibly noncommutative) ring, does the following implication hold: $$\textit{Flat module} \implies \textit{Torsion free}\ ?$$ In particular, I am considering a ring which has […]

I’m studying some results about flatness and faithful flatness and I’d like to keep in my mind some examples about faithfully flat modules. In general, free modules are the typical examples. Another (unusual) example of faithfully flat module is the “Zariski Covering”. (Let $R$ be a ring, $(f_1,\dots,f_n)=R$, and let $R_{f_i}$ be a localization $\forall […]

Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$N = R[x]/(tx-t).$$ I claim that $N$ is not a flat $R$- module. If we consider the exact sequence $$0 \rightarrow(t) \rightarrow R$$ such that the ideal $(t)$ is viewed as an […]

Let $R=k[x_1,\ldots,x_n]$. According to the first answer, every finitely generated flat module over an integral domain is necessarily projective. Therefore, the only hope to find a flat non-projective $R$-module $M$ is when $M$ is not finitely generated. Can anyone please suggest such modules? Actually, here there are some examples of flat non-projective modules (but not […]

I’m studying recently faithfully flat modules and I’d like to know the following: Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of nilpotent elements of $R$?

This question already has an answer here: Direct proof of non-flatness 1 answer

Let $R$ be a commutative ring and $M$ an $R$-module such that for every ideal $I \subset R$ the natural map $I \otimes_R M \rightarrow IM$ is an isomorphism. Why is $M$ flat ? This result is taken from Wikipedia.

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