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Here's a more concrete version of the Ohdarkdevil's argument. Since every module is free, every short exact sequence splits. Suppose R is not a division ring, so it has some nontrivial left ideal I. …

Jose has already given the answer, but here's a quick proof of it. First, note that any diagonal matrix D is normal, since its adjoint is also diagonal and diagonal matrices commute. Now suppose you …

As Angelo points out, this statement is trivial if by "rows" you mean rows with respect to a fixed basis. A generalization would be to allow the "rows" to come from any basis. Up to some duality, th …

First, some easy observations: $T$ must be injective since for any $A$, there is some $B$ such that $B$ and $A+B$ have different determinants (easy exercise). By multiplying $T$ by $T(1)^{-1}$, it ma …

As explained in the comments, if $k$ is a field, then $k$-algebra homomorphisms $k^X\to k$ are in bijection with $|k|^+$-complete ultrafilters on $X$ (that is, ultrafilters closed under $|k|$-fold int …

As a generalization of David Speyer's argument, here is a proof that if $k(X)$ always has a basis, then the axiom of choice for finite sets of bounded cardinality holds. In fact, to get the axiom of …

I don't have anything to say about specific examples, but here are some general remarks. A way to construct the abelianization of any compact group is to consider its image under the product of all i …

For infinite-dimensional vector spaces this not possible. According to this paper of Brenner and Ringel, if $A$ is any principal ideal domain that is not a field or a complete valuation ring, then th …