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annihilated
aˈnnihilated, ppl. a. [f. prec. + -ed.] Reduced to nothing, utterly destroyed.1769 Burke Pres. St. Nat. Wks. II. 82 The credit of France was low; but it was not annihilated. 1843 Mill Log. ii. v. §6 Imagining a portion of matter annihilated.
Oxford English Dictionary
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Heroes of Annihilated Empires
Heroes of Annihilated Empires is a real-time strategy role-playing video game developed by GSC Game World and released in October 2006. Overview
Heroes of Annihilated Empires is the first installment of a planned trilogy.
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Why are ASX 200 bank shares like CBA being annihilated today?
Mar 10, 2023Here's a summary of what's happening with ASX 200 bank shares today: The ANZ Group Holdings Ltd ( ASX: ANZ) share price is down 3%. The Bank of Queensland Ltd ( ASX: BOQ) share price has dropped 2 ...
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what is the meaning of "torsion group annihilated by integer"? Provide examples for: (1) a group that is neither torsion-free nor torsion (2) a torsion group which is not annihilated by any integer (3) a torsion-fr...
(1) is good: the group $\mathbb{Z}_p\times\mathbb{Z}$ is neither torsionfree because $(1,0)\ne(0,0)$ is annihilated by $p$, while $(0,1)$ has not finite
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Submodule of an injective $R$-module annihilated by an ideal $I$ of $R$ is an injective $R/I$-module. I want to solve this, but I have no idea: > Let $Q$ be an injective left $R$-module and $I\subset R$ is an ideal. ...
More detail:
Let $S$ denote the submodule of $Q$ annihilated by $I$. We'll try to use Baer's Criterion.
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Is an element in a module that is annihilated by all the scalars necessarily zero? Let $R$ be a _non-unital_ ring and $M$ a $R$-module. If $x \in M$ has the property that $rx=0$ for all $r \in R$, does it follow that ...
$\newcommand{\Z}{\mathbb{Z}}$Consider the Prüfer group $M = Z(2^{\infty})$. It is an abelian group, thus a $\Z$-module. It is also a module over $R = 2 \Z$. If $x \ne 0$ is the unique involution of $M$, then $R x = 0$.
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Example of a (non-simple) right $R$-module that is not annihilated by the Jacobson radical $J(R)$. According to Wikipedia, > the Jacobson radical of a ring $R$ is the ideal consisting of those elements in $R$ that an...
Any ring $R$ with nonzero Jacobson radical (e.g., $R=\mathbb{Z}/4\mathbb{Z}$) and the regular right module $R_R$.
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Natural $(R/I)$-module structure for an $R$-module $M$ annihilated by $I$ Suppose that $I$ is a two-sided ideal in the ring $R$, and that $M$ is a module over the quotient ring $R/I$. Why can we naturally regard $M$ a...
The obvious module action is $rm:=(r+ I)m$. Conversely, given an ideal $I$ and an $R$ module $M$ we would like to use $(r+ I)m:= rm$, but this is _not well-defined_ unless $I$ is contained in the annihilator of $M$. Explicitly, given $r+I=r'+I$ and $m=m'$ it needs to be established that $(r+I)m=(r'+...
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Can a vector separating for both $\mathcal{R}$ and $\mathcal{R}'$ be annihilated by successively applying projections from each? If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$, and $v \in H$ is ...
Answering the original version of the question. Let $H=\mathbb C^2$, and $$ \mathcal R=\left\\{\begin{bmatrix} a&0\\\0&b\end{bmatrix}:\ a,b\in\mathbb C\right\\}\subset M_2(\mathbb C),\ \ \ \ \ \ v=\begin{bmatrix} 1\\\1\end{bmatrix}. $$ Then $v$ is cyclic and separating, and $\mathcal R'=\mathcal R$....
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Is group cohomology killed by exponent of group? Let $G$ be a finite group, the exponent $e(G)$ is defined to be the lcm of order of elements in $G$. Let $M$ be a $G$ module, we know by restriction corestriction that ...
Yep. For every finite group, there's $M$ such that $H^2(G, M) = \Bbb Z/|G|$. For every group (not necessary finite) augmentation ideal $I$ can be covered by free module of rank equal to rank of group: suppose $G$ generated by $g_i$, then map goes like $$\Bbb Z[G]^{\mathrm{rk} G} \to I, x_i \mapsto (...
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Viewing $m/m^2$ as a vector space I am reading Atiyah MacDonald`s Commutative Algebra textbook. In Page 91, next paragraph, I do not understand part of the following: > If $A$ is a local ring, $m$ its maximal ideal,...
$\mathfrak{m}/\mathfrak{m}^2$ is annihilated by $\mathfrak{m}$ because if $x,y\in \mathfrak{m}$ then $xy\in \mathfrak{m}^2$, hence $x\cdot(y+\mathfrak{
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an equivalence of functors involving tensor and hom Let $\phi:(R,m,k) \rightarrow (S,n,l)$ be a morphism of local Noetherian rings. Let $M$ be a finite $R$-module and $N$ a finite $S$-module that is flat over $R$. >...
Ok, i figured it out: > Theorem 1 (Rotman, Introduction to Homological Algebra, Thm 4.86): Let $L$ be flat $R$-module, $E$ finitely-presented $R$-module and $M$ any $R$-module. Then $L \otimes_R Hom_R(E,M) \cong Hom_R(E,M \otimes L)$.
> > Theorem 2 (Rotman, Introduction to Homological Algebra, Thm 4...
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Two armies in the field Assume two armies of unequal size are fighting, where each soldier on both sides is of equal ability. Assume that each soldier is capable of taking out an enemy soldier each [unit of time], an...
You appear to be describing a situation modeled by Lanchaster's linear law, where the result is the larger force has the difference in numbers surviving. With firearms, this corresponds to the results of unaimed fire. You should also be aware of Lanchester's square law, which applies when there is a...
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Module of dimension zero over a local ring. > Let $(A,m)$ be a Noetherian local ring and let $M$ be an $A$-module of dimension $0$ ($\dim M=\dim A/\operatorname{Ann}M$). Then every element of $M$ is annihilated by a p...
Since $\dim M=0$ we have $\dim A/\operatorname{Ann}M=0$, so $A/\operatorname{Ann}M$ is a local artinian ring and you are done.
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A prime is minimal among primes containing an ideal Let $I$ be an ideal in a noetherian ring $R$, and $P$ prime containing $I$. I must prove that if in the localization $R_P$, $R_P/I_P$ is annihilated by a power of $P...
**Proposition:** Let $(R, m)$ be a local ring. If $m$ is nilpotent, i.e. $m^n = 0$ for some $n$, then $m$ is the only prime ideal of $R$. Proof: Let $p$ be any prime ideal of $R$, then $m^n = 0 \subseteq p \implies m \subseteq p$, so $m = p$ (since $m$ is maximal). Now use the correspondence between...
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