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When $A=\mathbb Z$ the condition is equivalent to $\mathrm{Ext}^1_{\mathbb Z}(A,\mathbb Z)=0$ and the problem as to whether this implies that $A$ is free is the Whitehead problem and was shown by She …

I think all non-archimedean locally compact fields are homeomorphic: Their rings of integers are compact, metric and totally disconnected and hence are all homeomorphic (to the Cantor set). The same i …

I think this works. Suppose we have a ring $R$ and an $R$-module $M$ all of whose localisations are projective and consider $S=S^\ast_RM$, the symmetric algebra on $M$. Then $R \rightarrow S$ is forma …

This is an attempt to complete Tyler's argument. We first note that $KO^0(S^5)=\mathbb Z$ (note this true for all spheres of dimension $\equiv 5,6,7 \bmod 8$). This means that every topological vector …

A common generalisation covering the two examples of (1) is an Artinian self-injective (local) algebra $A$. Then the product is injective and any injective is a sum of indecomposable injectives and th …

I think that $\mathcal F$ is indeed locally free of rank $n$: Pick a point $x\in X$. It will be enough to show that there is a neighbourhood of $x$ on which $\mathcal F$ is free of rank $n$. Now, the …

For the first question you already have had an answer in Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective? if $\mathrm{Ext}^1_{\mathbb Z}(P,M)=0$, then it …

If you accept the fact that a $2$-dimensional (local) ring has global dimension $2$, the following is a (somewhat) alternative proof. Choose a free f.g. presentation $F_1 \to F_0 \to M^\ast \to 0$ and …

I do not know the answer to your first question. As for the next two the answer is positive; one need only slightly modify standard proofs for the usual polynomial ring: If $R$ is a graded ring then …

The point is that you have the more general formula $f(g(t)+\epsilon h(t)) = f(g(t))+f'(g(t))h(t)\epsilon$. From that the chain rule follows: $$(f\circ g)(t+\epsilon) = f(g(t+\epsilon)) = f(g(t)+g'(t) …

First when it comes to comparison with the simplicial complex it should be realised that the Stanley-Reisner ring corresponds to the cone over the complex. There is a non-homogeneous version of it whe …

$\newcommand{\C}{\mathbb C} $I think this is OK. The first step is the inclusion of $\C[X,Y]$ into its fraction field which is $\C(X,Y)$. For each irreducible polynomial $f$ (normalised so that the to …

The answer to your first question is a resounding no. An example (among many) is given by $X=\mathrm{Spec} k$, $R=k[x]/(x^2)$ and $M=k$ considered as an $R$-module through the $k$-algebra homomorphism …

Any $R$-homomorphism (in fact any $\mathcal O_S$-homomorphism) $M \to M^{**}$ extends to a morphism $M^{**}\to M^{**}$ (as $M$ is locally free in codimension $1$ and $M^{**}$ is the maximal extension …

Depending on what you mean by cover your statement isn't true in the DVR case. To make it true in that case you can throw in the condition that the cover be normal and it may also be a good idea to as …