20
votes
Accepted
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 ...
6
votes
Accepted
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 ...
5
votes
Accepted
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 $\...
5
votes
Accepted
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 ...
5
votes
Accepted
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 $R$ ...
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, $\...
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? (...
4
votes
Accepted
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 $P$,...
4
votes
Accepted
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 ...
4
votes
Accepted
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^...
3
votes
Accepted
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 $...
3
votes
Accepted
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 ...
3
votes
Accepted
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 $K$-...
2
votes
Accepted
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\...
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 ...
2
votes
Accepted
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 $R$-...
2
votes
Accepted
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
$$0\to\...
2
votes
Accepted
A generating set for injective envelope
$R/\mathfrak m$ is simple, so $\langle x_i \rangle \cap R/\mathfrak{m}$ is either $0$, or $R/\mathfrak{m}$. But $R/\mathfrak{m}$ is an essential submodule, so the former cannot happen, assuming that $...
2
votes
Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
This is a partial answer that solves the commutative analogue of the original question. Consider $\ell_\infty$-morphisms
$$
\rho:c_0\to\mathcal{B}(\ell_\infty, c_0):x\mapsto(a\mapsto x\cdot a),
$$
$$
\...
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/\...
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 $R$-...
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 ...
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- \...
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 ...
2
votes
Accepted
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\...
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 (...
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 ...
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 $\operatorname{dim}\operatorname{Hom}_A(M,I_s)$ ...
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