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If $A$ is a $\mathbb{Q}$-algebra, then there is a group $G$ such that $A$ is a quotient of $\mathbb{Q}[G]$ if and only if $A$ is generated by units. For the "if" direction, take $G$ to be the group of …

No, not always. In Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID …

There's a nice, short proof in Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, …

If $\mathcal{T}_{1}=\langle M_{1}\rangle_{d_{1}+1}$ and $\mathcal{T}_{2}=\langle M_{2}\rangle_{d_{2}+1}$, then $\mathcal{T}_{1}\ast\mathcal{T}_{2}\subseteq\langle M_{1}\oplus M_{2}\rangle_{d_{1}+d_{2} …

The answer is "no" unless $A=k$. Let $a\in A\setminus k$, and let $l_2$ and $l_1$ be the first and second rows of the commutative diagram $$\require{AMScd} \begin{CD} 0@>>>A@>\begin{pmatrix}1\\0\end{p …

Every short exact sequence $0\to X\stackrel{\alpha}{\to}Y\stackrel{\beta}{\to}Z\to0$ is the quotient of a split short exact sequence by a split short exact subsequence. This answer is copied from my a …

No. Let $k$ be a field, and let $A$ be the algebra of upper triangular $2\times 2$ matrices over $k$, with augmentation map $\pmatrix{a&b\\0&c}\mapsto a$. $A$ and $A^e$ have finite global dimension, s …

Let $G$ be infinite cyclic, generated by $x$. Let $M_\bullet$ be a free resolution of the $\mathbb{Z}[G]$-module $U=\mathbb{Z}/3\mathbb{Z}$ with $x$ acting by multiplication by $-1$. For example, tak …

Without extra finiteness assumptions, this is not true in general. Even for $A=\mathbb{Z}$, there are infinitely generated $A$-modules $M$ that are locally free (in the sense that $M_\mathfrak{p}$ is …

Here's a counterexample for unbounded derived categories (this doesn't answer the revised question with the "$t$-left exact" condition). Suppose $\mathcal{A}$ is a Grothendieck category with enough p …

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 …

If I understand correctly what Question 1 is asking, then there are easy counterexamples even using commutative fields. Let $K=\mathbb{R}$ and $L=\mathbb{C}$. Then $\begin{pmatrix}1&i\\1&i\end{pmatri …

This isn't really an answer to the question, but an example to show how badly things can go wrong. Let $M$ be any indecomposable object of $D^b(X)$. The functor $\operatorname{Hom}(M,-)$, from $D^b(X …

Yes. The perfect objects in $\mathbf{D}(R)$ are the objects isomorphic to bounded complexes of finitely generated projective modules, and $\mathbb{k}$ is isomorphic in $\mathbf{D}(R)$ to its minimal p …

Even for $4$-dimensional algebras with identity it's not true. For $a\in F$ let $B(a)=F\langle x,y|x^2=y^2=0,xy=ayx\rangle$. Then $B(a)\not\cong B(b)$ unless $a=b$ or $a=b^{-1}$. This is quite easy t …