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Let $k$ be a field and $V$ a $k$-vector space, and let $R=k\oplus V$, where $V$ is a square zero ideal. Then the $k$-linear dual $R^\ast=k^\ast\oplus V^\ast$ is an injective $R$-module with unique mi …

If $F$ is a free direct summand of $M$ of maximal rank, then $M=F\oplus N$, where $N$ has no free direct summand. So $\mathrm{coker} \;\phi_M\cong\mathrm{coker} \;\phi_F\oplus\mathrm{coker} \;\phi_N$, …

I think your question is equivalent to asking whether there is an extension $0\to\mathbb{Z}\to B\to C\to 0$ of abelian groups where $C$ is torsionless but $B$ isn't. Given such an extension, take $A …

Let $F$ be a field, and $k=F[x,y]/(x^2,xy,y^2)$. Since $k$ is a finite-dimensional $F$-algebra, flat=projective, and for $k$-modules $M,N$, there is a natural isomorphism $\operatorname{Hom}_k(M,N^\as …

No. For example, let $R=\mathbb{Z}$, $M=\mathbb{Z}^2$, $N$ the subgroup generated by $(4,0)$ and $(2,1)$, and $L$ the subgroup generated by $(8,0)$ and $(0,2)$. Then $M/N$ and $N/L$ are both isomorph …

I assume from the tag that you intend $R$ to be commutative? In which case this is proved in Corollary 5.9 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra …

There's a clue to a counterexample at the end of Section 5 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra 5, 203-221 (1967). There are rings $R$ that have …

In "Rings of continuous functions in which every finitely generated ideal is principal" by L. Gillman and M. Henriksen (Trans. Amer. Math. Soc. 82 (1956), 366-391 link), Example 3.4 is of a topologica …

Let $A=k[x]$ and $B=k[x]/(x^2)$, let $X$ be the complex $\hskip{.1in}\dots\to 0 \to B\stackrel{x}{\to} B\to 0\to \dots$, and let $Y$ be $\hskip{.1in}\dots\to 0\to k\stackrel{0}{\to}k\to 0\to\dots$. Th …

Take the direct sum of the short exact sequences $$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ for $0\leq i\leq n$. This has the form $$0\to \bigoplus_{i=1}^n F_i\to \bigoplus_{i=0}^n F_i\to F_0\to 0$$ …

Let $R$ be the four-dimensional algebra $k[x,y]/(x^2,y^2)$, where $k$ is an infinite field. For each $\lambda\in k$, $M_\lambda=R/(x-\lambda y)R$ is a two-dimensional module with one-dimensional radi …

Let $k$ be a field, $I$ and $J$ infinite sets, and $A$ the $k$-subalgebra of $$k(t)[x_i,y_j: i\in I,j\in J]$$ generated by $$\{x_i,y_j,tx_i,t^{-1}y_j: i\in I, j\in J\}.$$ Then $$(tx_i)_{i\in I}\ot …

My previous attempt was completely wrong, as Jason Starr politely pointed out. But I think the idea I was grasping for does work, in this example: Let $R=k[x,y]$ for a field $k$, and let $$M=\frac{ …

There is at least one proof of Specker's theorem that can be adapted in an obvious way. I believe that the first half of this proof is due to Sąsiada, and the second half to Łoś. Let $A$ be a free $\m …

Let $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x^2,x^3]$, and $M=\mathbb{C}[x]$. Then $B$ and $M$ are free as $A$-modules, but $M\otimes_BM$ (and also $\wedge^2M$) has a one-dimensional $A$-submodule spanned …