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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. Let $R$ be the ring of eventually constant sequences of integers, and let $e_i\in R$ be the sequence whose $i^{th}$ term is 1, with all other terms zero. Then if $I$ is the annihilator of an eleme …

Let $k$ be a field, let $R$ be the path algebra over $k$ of the quiver $$1\xrightarrow{\gamma}2\xrightarrow{\delta}3$$ modulo the relation $\gamma\delta=0$, and let $e=e_2$ be the idempotent associate …

If $\Lambda$ is a ring, then the isomorphism classes of finitely generated projective $\Lambda$-modules form a commutative monoid $(A,+)$, with $[P]+[Q]=[P\oplus Q]$. This monoid contains a distinguis …

$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need …

In this answer, I was assuming the naive definition of a simple module $M$ for a nonunital ring $R$ as one with no proper nonzero submodules. It seems to be common to also insist that $RM\neq0$, in wh …

No, even if $S$ is commutative. There may be easier counterexamples, but ... There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …

There's a four-dimensional counterexample over any field. $A$ has basis $\{e,a,b,c\}$, with all products of basis elements zero except for $$e^2=e,\quad ab=c,\quad ea=a,\quad ec=c,\quad be=b,\quad ce= …

I'm no expert, but I think it follows from Theorem 11 and the rest of the discussion in Section 6 of Bergman, George M.; Lewin, Jacques, The semigroup of ideals of a fir is (usually) free, J. Lond. Ma …

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book Prest, M …

I haven't thought about base changes, but the original problem for $\alpha=1$ (so $\sigma$ is the identity map and $R=\mathbb{Z}_p[[t]][F]$ is just a polynomial ring over $\mathbb{Z}_p[[t]]$, and clas …

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\ …

There’s a natural injective module homomorphism $$R\to\bigoplus_iR/I_i$$ that takes $x$ into the semisimple submodule $\bigoplus_i\text{soc}(R/I_i)$, so the right ideal generated by $x$ is semisimple …

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, $\ …

The answer to (2) is "no" even for hereditary rings. For example, if $S=T$ is the algebra of upper triangular $2\times 2$ matrices (or, more generally, pretty much any finite dimensional hereditary al …