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If $R$ is a field of characteristic $0$, then the differentials of a transcendence basis of $R((x))$ over $R$ constitute a basis of $\Omega^1_{R((x))/R}$ over $R((x))$. In this sense this module is …

As you already mentioned, it is enough to show that every local artinian ring $A$, which is of finite type over $Z$, is finite. Let $m$ be the maximal ideal of $A$. By a standard filtration argument, …

Edit: Well I realize that the prime ideals of $Ae$ should be of the form $\mathfrak{p}e$ for some prime ideal $\mathfrak{p} \in$ Spec $A$. But how much do we know about the topology of Spec $Ae$? …

For a locally noetherian topological space $X$, the connected components of $X$ are open (EGA I, 6.1.9), in particular $X$ is a disjoint union of connected spaces (namely its connected components). Ev …

Such an isomorphism is worthless if it is written down with a choice of a basis, because then naturality is unclear (which is, of course, very important if you need this isomorphism not just as an iso …

A counterexample for the first question is any DVR $R$. Clearly, $R$ is not Jacobson. But if $\pi$ is the uniformizer, then $Q(R) = R[\frac{1}{\pi}]$ is a finitely generated $R$-algebra and a field, h …

There are various forms of the Nakayama lemma. Here is a rather general one; note that it does not involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2. …

More generally, we can ask: if $M$ is some $R$-module which is not the union of a chain of proper submodules, is $M$ finitely generated? This is in fact true. This was already asked at SE/246182, and …

Take any compact totally disconnected Hausdorff space $X$ (for example the Cantor set, or the one-point compactification of $\mathbb{N}$). Then $\mathcal{C}(X,\mathbb{F}_2)$ is a ring whose spectrum i …

Here's a proof by Karl Dahlke: Math Reference: A Free Submodule Embeds The result can be generalized to infinite ranks.

Perhaps the "Rabinowitz trick" is more clear if one writes down the proof backwards in the following way: Let $I \subseteq k[x_1,\dotsc,x_n]$ be an ideal and $f \in I(V(I))$, we want to prove $f \in …

A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. …

I doubt that there is a nice description which will satisfy you. As a $\mathbb{C}$-algebra, $R_{(x)}$ is not finitely generated. Anyway every localization of a factorial domain at a principle prime id …

When $R$ is noetherian, yes: By Baer's criterion it suffices to prove that the map $\hom_{S^{-1} R}(S^{-1} R,S^{-1} M) \to \hom_{S^{-1} R}(J,S^{-1} M)$ is surjective for every ideal $J \subseteq S^{ …

Generalizing the answer by wccanard: Let $R$ be a commutative ring. Then the kernel of $R[x] \twoheadrightarrow R_{red}[x]$ consists of nilpotent elements, hence this map induces a homeomorphism $\mat …