20 votes
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Are there (enough) injectives in condensed abelian groups?

Indeed, there are no nonzero injective condensed abelian groups. Let $I$ be an injective condensed abelian group. We can find some surjection $$ \bigoplus_{j\in J} \mathbb Z[S_j]\to I$$ for some index ...
Peter Scholze's user avatar
6 votes
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On definitions and explicit examples of pure-injective modules

This is not pure-injectivity. The relevant concept is that of an fp-injective ("finitely presented-injective") module, otherwise known as an "absolutely pure" module. A left $R$-module $J$ is said ...
Leonid Positselski's user avatar
5 votes
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Projective (or injective) object in a subcategory

You need more assumptions for this to be true. Consider the ring $$A = \begin{bmatrix} \mathbb k & \mathbb k \\ 0 & \mathbb k \end{bmatrix},$$ where $\mathbb k$ is some field, and let $\...
Dag Oskar Madsen's user avatar
5 votes
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Nuclearity of $C(\partial_F \mathbb{G})$

If there is a bound on the size of matrix blocks in $\ell_\infty(G)$, then this algebra is nuclear by a result of S. Wassermann [1], which would imply the nuclearity of $I_G(\mathbb{C})$. Without such ...
Makoto Yamashita's user avatar
5 votes
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On some sense of representing an endofunctor of the category of modules over polynomial rings

Replacing $R$ by $R[x_1,\ldots,x_n]$ if necessary, we may assume $n = 0$. Note that preserving all finite limits is equivalent to being (additive and) left exact (see e.g. Tag 010N). Lemma. Let $R$ ...
R. van Dobben de Bruyn's user avatar
5 votes

Why does there exist a non-split sequence with the condition that $\mathrm{pd} M=\infty$?

Since $S$ has infinite projective dimension, there is some indecomposable summand $M$ of $\Omega^n(S)$ that has infinite projective dimension. For simple modules $T$ of finite injective dimension, $\...
Jeremy Rickard's user avatar
5 votes

Question on injective hulls

This is false: Let $A=K[x]/(x^3)$ and $S$ the unique simple $A$-module. The injective hull is $\pi : S \rightarrow A$ with cokernel of dimension 2 and thus not simple. Is there an assumption missing? (...
Mare's user avatar
  • 25.8k
4 votes
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injective hull and projective cover of simple modules are indecomposable

One definition of "projective cover" of $S$ is that it is a projective module $P$, together with an epimorphism $\phi\colon P\to S$ such that the kernel $K$ is a superfluous submodule of $P$,...
Alex B.'s user avatar
  • 12.8k
4 votes
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When is the pullback of an injective sheaf injective?

Let's say that $X={\rm Spec\,} k$ for a field $k$. Then it is certainly Gorenstein and $k$ is an injective sheaf on $X$. For any $S$ and $p$ as defined in the question, $p^*k\simeq \mathscr O_S$. If ...
Sándor Kovács's user avatar
4 votes
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Example of non vanishing Ext

Take $R$ a local artinian non-Gorenstein ring, $I=0$, $M$ the residue field of $R$. The vanishing of $Ext^i_R(R/I,R/I)=Ext^i_R(R,R)$ for $i>0(=:n)$ is immediate, while for the non-vanishing of $Ext^...
A.G's user avatar
  • 533
3 votes
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Question on simple modules and projective covers

Here, $P_S$ is a projective cover of a simple module $S$. This means that there is a surjection $\sigma\colon P_S\to S$ from projective $P_S$ onto $S$, which has superfluous kernel $K$. The fact that $...
Keith Kearnes's user avatar
3 votes
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Question on injective hulls

Let me explain the underlined portion of Lemma 2.2. Lemma. If $h\colon A\to B$ is a nonzero module homomorphism, then there are a simple module $S$, its injective hull $I_S$, and a map $q\colon B\to ...
Keith Kearnes's user avatar
3 votes
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Direct sum of K-injectives over a noetherian ring

No. The $K$-injective complexes form a triangulated subcategory of the homotopy category of complexes, and so if they were also closed under coproducts, the homotopy colimit of a sequence of $K$-...
Jeremy Rickard's user avatar
2 votes
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Can we extract an injective envelope from a monomorphism?

Yes: given a monomorphism $f\colon X\to I$ with $I$ injective, as in the question, you can find a decomposition $I=I_0\oplus I_1$ such that $f=\begin{pmatrix}f_0&0\end{pmatrix}$ and $f_0\colon X\...
Matthew Pressland's user avatar
2 votes

injective hulls in mixed characteristic

I believe that the following works in reasonable generality, at least if $R$ is regular, although I am not sure of the precise minimal hypotheses. Let $n$ be the Krull dimension of $R$. The top ...
Neil Strickland's user avatar
2 votes
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Injective Change of Rings

The underlying idea of the below argument comes from the technique of spectral sequence presented in Chapter five of Weibel's book. Let $A$ be an injective $R/xR$-module. Let $M$ be an arbitrary $R$-...
G.-S. Zhou's user avatar
2 votes
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Looking for example of quotient of group algebra by ideal of group ring which fails to be injective

So long as $G$ is nontrivial, the augmentation ideal of $\mathbb{Z}[G]$ still works. If $I$ is any submodule of $\mathbb{Z}[G]$ then there is a short exact sequence of $\mathbb{Z}[G]$-modules $$0\to\...
Jeremy Rickard's user avatar
2 votes
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A generating set for injective envelope

$R/\mathfrak m$ is simple, so $\langle x_i \rangle \cap R/\mathfrak{m}$ is either $0$, or $R/\mathfrak{m}$. But $R/\mathfrak{m}$ is an essential submodule, so the former cannot happen, assuming that $...
Pavel Čoupek's user avatar
2 votes

Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?

This is a partial answer that solves the commutative analogue of the original question. Consider $\ell_\infty$-morphisms $$ \rho:c_0\to\mathcal{B}(\ell_\infty, c_0):x\mapsto(a\mapsto x\cdot a), $$ $$ \...
Norbert's user avatar
  • 1,687
2 votes
Accepted

Is $Hom_R(S_X^{-1}R, E)$ the minimal injective cogenerator of $S_X^{-1}R$?

I give a counter-example. Let $R$ be a semi-local domain with two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$. So the minimal injective generator module is $E = E(R/\mathfrak{m}) \oplus E(R/\...
Pham Hung Quy's user avatar
2 votes
Accepted

When is every injective module $\Sigma$-injective?

A result of Faith and Walker (page 205 in C. Faith, E. A. Walker, Direct-sum representations of injective modules, J. Algebra 5 (1967), 203-221) answers your question: If each injective left $R$-...
Fred Rohrer's user avatar
  • 6,660
2 votes

injective modules and divisible modules

Let $A$ be a domain. Then an $A$-torsion-free $A$-module is injective if and only if it is divisible. This is well-known. As mentioned in one of the comments, an arbitrary divisible $A$-module is ...
Jesse Elliott's user avatar
2 votes
Accepted

For a pure-injective module $M$ does the property "$\operatorname{Hom}(-,M)$ is surjective" commute with certain limits?

$\DeclareMathOperator\Hom{Hom}$The answer seems to be positive. Let $\psi_i: X_i \rightarrow Y_i$. Consider the directed system of exact sequences $$ 0 \longrightarrow K_i \longrightarrow X_i\otimes- \...
kevkev1695's user avatar
  • 1,023
2 votes

Pontrjagin dual of modules

I am not sure either that the question is appropriate for MO, but this is too long for a comment. (1): Note that this is just a question about $\mathbb{Z}$-modules. You have to prove that given $m\neq ...
abx's user avatar
  • 37.1k
2 votes
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On the finiteness of an Auslander-Reiten component

Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$. In a locally finite quiver, given any $d\...
Matthew Pressland's user avatar
1 vote
Accepted

When do faithfully semiinjective complexes exist?

Turns out it isn't that hard. We can do better and always find faithfully injective modules (I'm certain this is written down somewhere but I'm a "noob" and (i) didn't already know it and (...
FShrike's user avatar
  • 569
1 vote

infinite left degrees

The fact you asking for is proved in Lemma 1.2 (and stated explicitly in the corollary to this lemma) in Liu, S. (1992). Degrees of Irreducible Maps and the Shapes of Auslander-Reiten Quivers. Journal ...
Ivan Yudin's user avatar
1 vote

About composition factors

Let $A$ be an Artin algebra, $S$ a simple $A$-module, and $M$ a finitely generated $A$-module. Then $\operatorname{dim}\operatorname{Hom}(P_s,M)$ and $\operatorname{dim}\operatorname{Hom}_A(M,I_s)$ ...
Matthew Pressland's user avatar

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