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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 …

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 …

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 …

Here's an elementary proof that $2n-2$ is a lower bound. Suppose that $$V_1\xrightarrow{\alpha_1}V_2\xrightarrow{\alpha_2}\dots\xrightarrow{\alpha_{n-2}}V_{n-1}\xrightarrow{\alpha_{n-1}}V_n$$ is a rep …

Interpreting the various matrices as maps between free modules in the usual way, the question becomes: If $M$ is a submodule of $R^m$, and $\alpha,\beta: R^k\to M$ are epimorphisms, then must $\alpha$ …

Yes. Let $M$ be any $A$-module. Then its socle is a direct sum of simple modules: $\operatorname{soc}A=\bigoplus_iS_i$. $A$ is a finite dimensional algebra, so the dual $\mathrm{Hom}_k(A,k)$ of $A$ is …

Not in general, no. Let $\mathbb{F}$ be the algebra of upper triangular $2\times 2$ matrices, let $n=2$, and let $$p_1=(x_1,y_1)=\left(\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{ …

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 …

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 …

Let $A=\mathbb{C}[x,y]$, $I=(x,y)$, and $M$ the $3$-dimensional $A$-module $(x,y)/(x^2,y^2)$ (with basis $\{x,y,xy\}$). Then $M/IM$ is isomorphic to a direct sum of two copies of $\mathbb{C}=A/I$, so …

I don't see how commutativity matters. Suppose $A$ is the filtered colimit of algebras $A_i$ and $x,y\in A$ with $xy=0$. Then $x$ is represented by $x_j\in A_j$ and $y$ by $y_k\in A_k$ for some $j$ a …

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 an area that I'm expert on, and it's quite possible there's a much more elementary and/or more general answer. But if the Bass Conjecture on Hattori-Stallings ranks for group rings is true …