Pub Date : 2022-01-05DOI: 10.1201/9780203719947-15
R. Almond
In all our examples the ground field shall be Q and the extension field will be a subfield of the complex numbers C. We take as basic that the only nonzero rational numbers c for which √ c ∈ Q are those positive c for which each prime factor p appears an even number of times. In particular, √ 2, √ 3, 3/2 are all irrational. Example I: K = Q(√ 2, √ 3). As √ 2 has minimal polynomial x 2 − 2, Q(√ 2) has basis 1, √ 2 over Q. Now we need a simple result: Theorem 0.1 √ 3 ∈ Q(√ 2) Proof: If it were we would have √ 3 = a + b √ 2 with a, b ∈ Q. Squaring both sides 3 = a 2 + 2b 2 + 2ab √ 2 As 1, √ 2 is a basis the coefficient of √ 2 would need be zero. That is, 2ab = 0. So either a = 0 or b = 0. 1. b = 0: Then √ 3 = a ∈ Q, contradiction. 2. a = 0. Then √ 3 = b √ 2 so 3/2 = b ∈ Q, contradiction. From Theorem 0.1 and that √ 3 satisfies a quadratic (namely, x 2 − 3) over Q(√ 2), 1, √ 3 is a basis for Q(√ 2, √ 3) over Q(√ 2) and so 1, √ 2, √ 3, √ 6 is a basis for K over Q. That is, we may write K = Q(√ 2, √ 3) = {a + b √ 2 + c √ 3 + d √ 6 : a, b, c, d ∈ Q} (1) and every α ∈ K has a unique expression in this form. Now we turn to the Galois Group Γ(K : Q). Any σ ∈ K has σ(√ 2) = ± √ 2 and σ(√ 3) = ± √ 3 which gives four (Caution: this is 2 times 2) possibilities. The value of σ on √ 2, √ 3 determines the value on all of K. The four elements of the Galois Group are e, σ 1 , σ 2 , σ 3 where e(a + b √ 2 + c √ 3 + d √ 6) = a + b √ 2 + c √ 3 + d √ 6 σ 1 (a + b √ 2 + …
在我们所有的例子中，基域都是Q，扩展域是复数c的子域。我们基本认为，只有√c∈Q的非零有理数c是那些正的c，其中每个素数因子p出现偶数次。特别是√2√3 3/2都是无理数。例子:K = Q(√2,√3)。随着√2 x 2−2最小多项式,问(√2)基础1,√2 /问:现在我们需要一个简单的结果:定理0.1√3 Q∈(√2)证明:如果我们有√3 = a + b√2 a, b∈两边平方问:3 = 2 + 2 b 2 + 2 ab√2是1,√2是一个基础√2需要的系数为零。也就是2ab = 0。所以a = 0或者b = 0。1. b = 0:则√3 = a∈Q，矛盾。2. A = 0。那么√3 = b√2所以3/2 = b∈Q，矛盾。从定理0.1和√3满足二次(即x 2−3)/ Q(√2),1,√3是一个依据Q(√2√3)/ Q(√2)1,√2√3√6是一个依据K /问:也就是说,我们可以写K = Q(√2√3)= {a + b√2 + c d√√3 + 6:a, b, c, d∈Q}(1)和每一个α∈K有独特的表达形式。现在我们转到伽罗瓦群Γ(K: Q)。任何σ∈K都有σ(√2)=±√2和σ(√3)=±√3，这有四种可能性(注意:这是2乘以2)。伽罗瓦群的四个元素分别是e， σ 1， σ 2， σ 3，其中e(a + b√2 + c√3 + d√6)= a + b√2 + c√3 + d√6 σ 1 (a + b√2 +…
{"title":"Three Examples","authors":"R. Almond","doi":"10.1201/9780203719947-15","DOIUrl":"https://doi.org/10.1201/9780203719947-15","url":null,"abstract":"In all our examples the ground field shall be Q and the extension field will be a subfield of the complex numbers C. We take as basic that the only nonzero rational numbers c for which √ c ∈ Q are those positive c for which each prime factor p appears an even number of times. In particular, √ 2, √ 3, 3/2 are all irrational. Example I: K = Q(√ 2, √ 3). As √ 2 has minimal polynomial x 2 − 2, Q(√ 2) has basis 1, √ 2 over Q. Now we need a simple result: Theorem 0.1 √ 3 ∈ Q(√ 2) Proof: If it were we would have √ 3 = a + b √ 2 with a, b ∈ Q. Squaring both sides 3 = a 2 + 2b 2 + 2ab √ 2 As 1, √ 2 is a basis the coefficient of √ 2 would need be zero. That is, 2ab = 0. So either a = 0 or b = 0. 1. b = 0: Then √ 3 = a ∈ Q, contradiction. 2. a = 0. Then √ 3 = b √ 2 so 3/2 = b ∈ Q, contradiction. From Theorem 0.1 and that √ 3 satisfies a quadratic (namely, x 2 − 3) over Q(√ 2), 1, √ 3 is a basis for Q(√ 2, √ 3) over Q(√ 2) and so 1, √ 2, √ 3, √ 6 is a basis for K over Q. That is, we may write K = Q(√ 2, √ 3) = {a + b √ 2 + c √ 3 + d √ 6 : a, b, c, d ∈ Q} (1) and every α ∈ K has a unique expression in this form. Now we turn to the Galois Group Γ(K : Q). Any σ ∈ K has σ(√ 2) = ± √ 2 and σ(√ 3) = ± √ 3 which gives four (Caution: this is 2 times 2) possibilities. The value of σ on √ 2, √ 3 determines the value on all of K. The four elements of the Galois Group are e, σ 1 , σ 2 , σ 3 where e(a + b √ 2 + c √ 3 + d √ 6) = a + b √ 2 + c √ 3 + d √ 6 σ 1 (a + b √ 2 + …","PeriodicalId":222141,"journal":{"name":"Graphical Belief Modeling","volume":"87 1","pages":"0"},"PeriodicalIF":0.0,"publicationDate":"2022-01-05","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":null,"resultStr":null,"platform":"Semanticscholar","paperid":"126515401","PeriodicalName":null,"FirstCategoryId":null,"ListUrlMain":null,"RegionNum":0,"RegionCategory":"","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":"","EPubDate":null,"PubModel":null,"JCR":null,"JCRName":null,"Score":null,"Total":0}
京公网安备 11010802042870号
京ICP备2023020795号-1
扫码关注我们
求助内容：
应助结果提醒方式：