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quot
▪ I. quot, pa. pple. dial. (kwɒt) Also 7 quotted, 8 quott. [f. quot quat v.1 1 b.] Sated, cloyed.1674–91 Ray S. & E.C. Words, Quotted, cloyed, glutted. Suss. c 1741 E. Carter Let. in Mem. (1808) I. 27, I believe I am grown quott of assemblies, &c. 1887 Kentish Gloss., Quot, cloyed, glutted.▪ II. † q... Oxford English Dictionary
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Quot scheme
scheme called the quot scheme associated to a Hilbert polynomial . This shows represents the quot functor Projective space As a special case, we can construct the project space as the quot schemefor a sheaf on an wikipedia.org
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Quotient field of a domain Let $A$ be a commutative domain and $K=Quot(A)$, its field of fractions (quotient field). Prove that $K$ is a f.g. $A$-module if and only if $A=K$.
Let $\\{\frac{1}{a_1},\ldots,\frac{1}{a_n}\\}$ generate $K$ as an $A$ module, with each $a_j\in A$, and let $x$ be an element of $K$. Then there exist $b_1,\ldots,b_n$ in $A$ such that $\frac{x}{a_1a_2\cdots a_n}=\sum\limits_{j=1}^n\frac{b_j}{a_j}$. Multiplying both sides of the equation by $a_1a_2\...
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$\mathbb{Q}[t]$ is integrally closed in $Quot(\mathbb{Q}[t])$ I'm having trouble trying to show that $\mathbb{Q}[t]$ is integrally closed in $Quot(\mathbb{Q}[t])$. Where $Quot(\mathbb{Q}[t])$ is the field of fractions...
A unique factorization domain is always integrally closed in its field of fractions. The proof is exactly the same as for $\mathbb Z$.
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Degree of the field extension $K(x)\hookrightarrow\operatorname{Quot}\left(K[x,y]/(f)\right) $ > Let $K$ be a field, and $f\in K[x,y]$ an irreducible degree $d$ polynomial. I want to prove that the field extension $$ ...
I've proved in this thread that the field of fractions of $KX,Y$ is $K(X)[Y]/(f)$, so $[K(X)[Y]/(f):K(X)]=\deg_Yf$.
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Extensions between integral domains give extensions of fields of the same degree. Assume that $S \subset R$ is a ring extension where, both $S$, $R$ are integral domains. Furthermore, assume that $R$ is a free $S$-mod...
_Hint:_ Show that $\text{Quot}(R) \xleftarrow{\cong} R\otimes_S \text{Quot}(S)$; in words, it suffices to have all elements of $S\setminus\\{0\\}$ invertible
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Rings with same quotient field Let $R$ be an integral domain and $0 \neq I$ an ideal of $R$. Denote by $\phi: R \rightarrow R/I$ the canonical homomorphism. Let $S$ be a subring of $R/I$ such that $R/I$ is integral ov...
Now we have $r = \frac{rf}{f} \in Quot(T)$, which shows $R \subset Quot(T)$, which is literally all we need.
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Quot digiti praefixae sunt in principio mobilis ... - 陈沩亮博客
May 4, 2022Cum vocationem internationalem ad Taiwan peregre faciendam facit, debes Taiwan internationalem praepositionem 886 dial.Quot digiti praefixa sunt in principio mobilis numeri telephonici in Taiwan?
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比"过山风"毒性强100倍，30秒便可致死的动物，号称"海洋杀手"_水母
澳洲方水母也叫箱形水母，是一种淡蓝色的透明水母。 也有人称它为"海胡蜂" ，主要原因是它外形微圆，像一只方形的针，因此而得名。 澳洲方水母，有4个明显的侧面。 据澳大利亚海洋科学研究所人员表示， 方水母大约有15条触须，每条触须上都布满了释放毒液的刺细胞 。 谁都想不到，这个看起来轻灵飘逸的美丽的水中仙子竟然是世界十大毒王之首， 拥有致命的毒液，一旦被蜇到，三十秒内就会死亡。 --生活区域 方水母一般都生活在热带海域，多在澳大利亚海湾浅水带 ，在风和日丽的时候可能会游向海滨浅滩处。 要是有人碰到它身上的刺细胞，几十秒就会死亡，抢救都来不及。 在澳大利亚沿海地区 。 25年来，因为方水母中毒而死亡的就有多达60人，相比较之下，死于鲨鱼腹中的人只有16人。
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Is every field the field of fractions for some integral domain? Given an integral domain $R$, one can construct its field of fractions (or quotients) $\operatorname{Quot}(R)$ which is of course a field. Does every fie...
Yes. $F$. (You can't hope to do better than this in general; consider the finite fields.) Here's a cute example, though. It turns out that $\mathbb{C}$ is isomorphic to $\overline{ \mathbb{C}(t) }$. From this it follows that $\mathbb{C}$ is isomorphic to the fraction field of the integral closure of...
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70+ "王集" profiles | LinkedIn
View the profiles of professionals named "王集" on LinkedIn. There are 70+ professionals named "王集", who use LinkedIn to exchange information, ideas, and opportunities.
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灵魂筹码 - 荣氏彩云之端套装 Soul at Stake - "In the Cloud" Rose's ...
灵魂筹码 - 荣氏彩云之端套装 Soul at Stake - "In the Cloud" Rose's Outfitの開発者からのアップデート、イベントやニュースを表示します。 ログイン ストア
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八角街（八角街是"帕廓街"的音误）_百度百科
Barkhor Street 原 名 帕廓街 地理位置 古城 拉萨 的中心 目录 1 简介 街道外况 得名 传说 2 历史背景 大昭寺 转经道 街内古迹 3 周围古迹 旧木鹿寺 新木鹿寺 吐巴 4 西藏手工艺品集散地 商品集散地 名店 旅游购物
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给观赏鱼治病的黄粉,真能"包治百病"吗?...
2022年4月14日 黄粉,一度被水族商家神话为"包治百病"的“万能”鱼药!你别说,黄粉对于很多观赏鱼疾病的预防和治疗确实是有特效的,这也是黄粉得到很多鱼友认可的原因之一。 黄粉是什么? 黄粉属于复...搜狐网 news.sohu.com
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On Transcendence Degrees Let $A$ and $B$ be integral domains with $A\subset B$. The transcendence degree of $B$ over $A$ is defined as the transcendence degree of $Quot(B)$ over $Quot(A)$. Denote it by $trdeg(B/A)$. L...
Therefore every element of $\operatorname{Quot}(B)$ is algebraic over $A[S]$, hence over $\operatorname{Quot}(A[S]) = \operatorname{Quot}(A)(S)$. Thus $S$ is a maximal algebraically independent set in $\operatorname{Quot}(B)$ over $\operatorname{Quot}(A)$.
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