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The answer to your first question is "yes". You can find a calculation of the full ring of differential operators on a suitably nice scheme here : Theoreme 16.11.2 on page 54 of EGA 4 IV, PIHES 32 (19 …

Take $R = \mathbb{Z}_p \times \mathbb{F}_q$ for distinct primes $p,q$ and $\mathfrak{m} = qR$. Then $\hat{R} \cong 0 \times \mathbb{F}_q$ is isomorphic to an ideal in $R$, so $Hom_R(\hat{R}, R) \neq 0 …

Let $A \to B$ be a map of rings such that $B$ is a finitely generated $A$-module. Suppose $A$ is semi-local. Then $B$ is also semi-local. To see this, note that every simple $B$-module is finitely gen …

Let $A = k[x_1,\ldots,x_r]$, let $\mathfrak{m}$ be the ideal $(x_1,\ldots,x_r)$ in $A$ and let $I$ be the ideal $(p_1,\ldots,p_t)$ in $R$. The assumption $\sqrt{AI} = \mathfrak{m}$ implies that $\math …

To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F \to \m …

No. If $A = k[x,y]$ is the polynomial ring in two variables, $\mathfrak{p}$ is the zero ideal, $N = A_{\mathfrak{p}} = k(x,y)$ is the field of fractions of $A$, $M := (x,y) \subsetneq M' := A \subsete …

Let $A = k[x_1,x_2, \ldots]$ be the polynomial ring over a field $k$ in infinitely many variables and let $\mathfrak{m} = \langle x_1,x_2,\ldots \rangle$. Then $(A / \mathfrak{m}^2)^2$ satisfies all o …

The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma". In the case you are interested in, the "short Schanuel's Lemma", as explained here http://en.wikipedia.org/wiki/ …

Instead of updating my previous answer, I've decided to add a new answer in order to keep it short(ish). In the comments following his original question, TonyS added the extra assumption that $R$ is …

Since $A$ is regular local commutative ring, it's an integral domain. Let $S = A - 0$ so that the localisation $A_S$ is the field of fractions $F$ of $A$. Assuming the algebra $A$ is central in $R$, …

I would guess not. Here is some evidence towards this guess. Let $S$ be a $p$-adically complete ring of characteristic $0$ and let $M$ be $p$-adically complete $S$-module which is free over $S$. Then …

Suppose that $A$ is a not necessarily commutative ring, which is both left and right Noetherian. Given a left $A$-module $N$, and a right $A$-module $M$, there is a general (fourth quadrant, cohomolog …

Yes --- this follows from EGA3 I, Chapter 0, Proposition 13.3.1 . This general result gives conditions under which it is possible to conclude that $H^i( X, \lim\limits_\longleftarrow \mathcal{F}_k )$ …

No. Let $R = \mathbb{Z}_p[C_p]$ and consider the $R$-modules $M = R$ and $N = \mathfrak{m}$, the maximal ideal of the local ring $R$. Let $A,B$ the matrices in $GL_p(\mathbb{Z}_p)$ giving the action o …

Over a general ring (not necessarily commutative), every finitely presented flat module is projective. So, one simple condition on the ring that gives you what you want is that every finitely generat …