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Choose a bijection $\alpha:\mathbb{N}\to\mathbb{Q}$, and for each $x\in\mathbb{R}$ let $$a(x)_i=\begin{cases}0&\mbox{ if $\alpha(i)<x$}\\1&\mbox{ if $\alpha(i)\geq x$}\end{cases}.$$ Then the set of s …

Any space with at least one positive vector $w$ is spanned by positive vectors: if $v$ is any vector then $v-\lambda w$ is positive for sufficiently large $\lambda$, and so $v=(v-\lambda w)+\lambda w$ …

If $A$ is finitely generated, as in your first edit, then this is much more elementary than the result of Dixmier that Faisal mentioned. A simple $A$-module would have the structure of a field extensi …

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 …

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{ …

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 …

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 …

The answer of user39385 to the other question generalizes. The minimal left ideals of $M_n(R)$ are in bijection with the minimal submodules of $R^n$. Let $K$ be the quotient field of $R$, and let $d$ …

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 …

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 …

This is probably an absurdly over-complicated answer, but ... Let $$J(X)=\left\{\sum_{x\in X}a_xx\in GX: \sum_{x\in X}a_x=0\right\}.$$ I claim that $J$ is not of the form $H\circ G\circ F$. Suppos …

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 …

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 …

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 …

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 …