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This answer to a related question gives a way of constructing counterexamples. For a similar but more concrete example, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a $3$-dimensional alge …

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 $\alpha$ …

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{ …

Example 5.10 of Towers, Matthew, Endomorphism algebras of transitive permutation modules for $p$-groups., Arch. Math. 92, No. 3, 215-227 (2009) (whose author you might know) gives a positive answer to …

Countable torsion abelian groups are better behaved than uncountable ones. For example, Kaplansky’s “test problems” If $G$ and $H$ are isomorphic to direct summands of each other, is $G\cong H$? If …

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 …

Without the axiom of choice, it is possible that there is a vector space $U\neq 0$ over a field $k$ with no nonzero linear functionals. Let $V$ be the direct sum of countably many copies of $U$, and $ …

As YCor notes in comments, (2) is a special case of (1), so I'll only address (1). Suppose $\Lambda^k\varphi:\Lambda^kV\to\Lambda^kW$ is not injective, and let $x\neq0$ be in the kernel. Then $x$ ca …

I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimension …

No, it’s not consistent. Let $V=k^{(\omega)}$ be the vector space of finite sequences of elements of $k$. Then $V^*$ can be identified with the vector space $k^\omega$ of all sequences, and elements …

Even for a $2$-dimensional vector space (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings. Bo …

I'm really just expanding on other people's comments, so I've made this answer community wiki. If $R$ is a subring of a finite dimensional algebra over a field $K$ ($d$-dimensional, say), then $R$ em …

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 question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?. For any field $k$, the field $k(x)$ of rational functions in one variable has an explic …

This is a very incomplete answer, but maybe others can fill in the gaps (and I'll try to). [Edit: I've not been able to make this idea work, although the ideas may lead somewhere, so I'll leave this …