Search Results

Let $\mathcal{A}$ be the two-dimensional Lie algebra over a field $k$ with basis $\{x,y\}$ and relation $[x,y] = y$. Let $\mathcal{I}$ be the ideal of $\mathcal{A}$ spanned by $y$. Let $D' : \mathcal{ …

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 …

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 …

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 …

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 …

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 $H^i( X, \lim\limits_\longleftarrow \mathcal{F}_k )$ …

It's true if $A$ is Noetherian. For any $A$-algebra $C$ and any ideal $J$ in $A$, note that $C/JC$ is isomorphic to the tensor product algebra $C \otimes_A A/J$. Now for any $n \geq 0$, $A/I^n$ is a …

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 $A = k[x_1,x_2, \ldots]$ 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 …

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 …

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 …

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 …

I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left idea …

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 $R$, …