数魂 数魂
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【Diary】Tuesday,April 19,2011 Today I got up (very)^n[denotes to very very....very and the total number of "very" is n which is a natural number that is not zero]....late,because I slept (very)^n   late last night.A SB[abbreviation for Silly Boy,oh I just remember that B is also pronounced "boy",above all,these two words accurately present that guy.]introduced me a short animation series(just 12 episodes) named Ayakashi(made in Japan) so that I abondoned my sleep to watch it and of course didn't abide by the shedule in which I required myself to go to bed early.Apparently this production is not able to be called classic yet there are some questions in it about Humanity deserves to be rethought. This morning,my dream came to an abrupt end when sb knocked on the door.It was Dongdong whom reminded me that familiar yet lack-of-beating sarcasm: so why?You were absent again this morning.....But we hadn't any class and I needn't apply him for the message that the teacher didn't call our names. All this afternoon I was absorbed in the vocabulary learning with the newly created method and also I found that these days "energetic" was more than an abstrack definition to me because this is the actual statement when I was doing my assignment. Though explicit reasons for this has not yet been found,sounds a little absurd?I believe I'll find some tips of it in the short future. This evening it was my turn to help tutor's undergraduate class with their Advanced Algebra.It is a positive and happy class with its youthful and alive class mumbers.Also I joined their QQ group and made the preparation for providing abundant files related to acaedemic and study.
【问题整理5】TEX 【1翻译有源码文件的步骤:1、看pdf原文,建立中英单词对照表——2、“全字匹配”“全部替换”(一定要全字匹配啊啊啊!)——3、在tex中翻译。在进行新的文章时,可以是2——5、编译——1,再按照上面的步骤进行】 附:昨天建立的词表 so we have 于是我们有 such that 使得 exchange 交换 progenerator 有限投射生成元 Hence 因此 As 由于 since 因为 So 所以 Likewise 同样地 direct summand 直和项 Assume 假定 Thus 因此 we have 我们有 deduce 得出 Therefore 因此 refinement 加细 matrix 矩阵 If 如果 regular 正则的 We say that 我们称 Let 令 there exists 存在 Lemma 引理 equivalent 等价的 ring 环 modules module 模 then 则 orthogonal 正交的 ideals ideal理想 Suppose that 假定 decomposition 分解 obtain得到 projection 投影映射 injection 单射 given by 规定 for any 对任意的 Clearly 显然 homomorphism 同态 Corollary 系 verify 证明 quasi-inverse 准可逆的 for some 对一些 we see that 我们得到 This means that 这意味着 we get 我们得到 as desired 即为所求 By hypothesis 由假设 right 右 left 左 a pair of 一对     In view of 根据 idempotents 幂等元 we can find 可以找到 regular 正则的 Furthermore 此外 it follows from 通过 Obviously 显然 we deduce that 我们得到 Theorem 定理 we prove that 我们来证明 it is easy to verify that 很容易证明 there exist 存在 proof 证明 The following result 下面的结果 shows that 表明 characterized 刻画 substitution 代入 properties 性质 progenerstors 有限投射生成元 finitely 有限 generated 生成 projective 投射 In addition 另外 One easily checks that 容易证明 full 满的 It is easy to verify that 容易证明 By the consideration above 综上所述 By virtue of 通过 Using 利用 Consequently 因此 one checks that 我们有 equivalent 等价的 Given any 对于任意的 Analogously 类似地 we proved that 我们证明了 we know that 我们有 we can find 可以找到 as required 即为所求 Inner Substitution 内代入 By virtue of 通过 According to 根据 Construct a map 建立映射 map 映射 epimorphic 满的 if and only 当且仅当 isomorphic同构 satisfy 满足 satisfying 满足 in case 如果 for any 对任意的 central 中心 As well known 我们知道 provided that 如果 As a result 从而 unique 唯一的 diagram commutes 图表可交换 morphism 态射 denote 规定 split epimorphism 裂满同态 split monomorphism 裂单同态 right invertible 右可逆 left invertible 左可逆 This implies that 这意味着 we conclude that 我们得到 Cancellation 消去律 projective modules 投射模 positive integer 正整数 These infer that 这意味着 Applying 利用 This means that 这意味着 element 元素 nonzero 非零 algebras 代数 Proposition 命题 REFERENCES 参考文献
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