# Tagged Questions

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### A Direct Sum of Members of a Certain Class of Modules

Let $S$ be a class of $R$-modules and let an $R$-module $M$ be countably generated. Suppose that, for every direct summand $K$ of $M$, each element of $K$ belongs to a direct summand of $K$ that is ...
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### Direct product of direct sum of a flat module

I have a problem concerning flat modules: Let $M$ be an $R$-module such that the direct product $M^A$ is flat for all sets $A$. I want to prove that $(M^{(B)})^A$ is also flat for any sets $A$ ...
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### Localization of a direct product

Is the localization of a direct product of two rings at a maximal (or prime) ideal identified with a localization of one of them? I would appreciate for any detailed answer.
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### An inverse limit of a certain inverse system

Let $∆$ be a directed set and $(N_i,f_{ji})_{i∈∆}$ be an inverse system of $R$-modules. Fix $α \in∆$ and consider $(M_i,g_{ji})_{i\in∆}$ as follows: $M_i=N_i$ for $i≥α$, $M_i=0$ for $i<α$, and ...
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### Direct limits commute with $\mathrm{Tor}$ functor

How one could prove that direct limits commute with the functor $\mathrm{Tor}$? Of course, I know that $\mathrm{Tor}$ with its first $0$ index is the same as tensor product which does commute ...
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### Any counterexample for inverse limit functor not to be right exact [duplicate]

We know that inverse limit is a "left" exact functor on the category of modules in the sense that whenever $r:(A_i,α_j^i )→(B_i,β_j^i )$ and $s:(B_i,β_j^i )→(C_i,γ_j^i )$ are transformations of ...
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### Injective resolution for an integral domain

How could one write an injective resolution for an arbitrary commutative integral domain $R$? Thanks in advance!
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### A statement equivalent to flatness

If $R$ is a ring with identity and $P$ is a flat right $R$-module, it is a fact that any $R$-homomorphism $f$ from a finitely presented right $R$-module $M$ to $P$ factors through a finitely generated ...
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### Examples of algebras having a module basis

I'm looking for examples of associative $R$-algebras, for which an $R$-module basis can be specified. Of course, if $K$ is a field, then any $K$-algebra admits such a basis, but this dis not what I'm ...
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### Characteristic of ring $R_1\otimes\ldots\otimes R_k$

Let $R_1,\ldots, R_k$ be unital rings and $\otimes=\otimes_\mathbb{Z}$ and $\mathrm{chr}$ the characteristic. How can one see that \mathrm{chr}(R_1\otimes\ldots\otimes ...
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### divisible modules

In surveying LMR of T.Y.Lam, I reached a paragraph stating that "when R is a domain every direct sum or direct product of divisible modules is divisible." My question is that "should R is not a ...
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If $R$ is a unital associtaitve commutative ring, then can any $R$-algebra $A$ may be filtered as $A_0:=A$ and $A_{i+1}:=(A_i,A_i)$ where $(-,-)$ is the commutator of $A$ with respect to $A$'s ...
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### Definition of a filtration on a ring, module, algebra

In most books, a graded ring/module/algebra means either a $\mathbb{N}$- or $\mathbb{Z}$-graded ring/module/algebra. But often, different gradings appear: doubly graded (spectral sequences) = ...
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### Projective objects in the category of rings

What are the projective objects in the category of rings with identity ? Remarks: The only projectives I could find so far are $\{ 0\}$ and $\mathbb{Z}$. If $R$ is projective and ...
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### tensoring a polynomial algebra: $R[x_i]/\mathfrak{a}\otimes A \cong A[x_i]/\mathfrak{a}$

Let $R$ be a commutative unital ring, $A$ an associative unital $R$-algebra, $I$ an arbitrary set, and $\mathfrak{a}$ an ideal of $R[x_i; i\!\in\!I]$. If $A$ is commutative, then there is an ...
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### Non-commutative integral extensions?

In Commutative algebra there is a notion of an integral extension: Let $P$ be a subring of $R$. Then $R$ is the integral extension of $P$ if each element of $R$ is a root of a monic polynomial with ...
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### Applications of Govorov-Lazard Theorem?

The Govorov-Lazard Theorem states that a (right) module over an unital ring is flat iff it is a direct limit of finitely generated free (right) modules. I wonder if this theorem has interesting ...
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### An $(R,S)$-bimodule is a left $R \otimes_k S^{\text{op}}$-module

Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a ...
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### Annihilator of a simple module

Let $R$ be a finitely generated commutative ring and $C$ an $R$-algebra ($C$ is not necessarily commutative). Assume that $C$ is a finitely generated $R$-module. If $S$ is a simple $C$-module, then ...