POLYNOMIALS WITH ROOTS IN Qp FOR ALL p
POLYNOMIALS WITH ROOTS IN Qp FOR ALL p
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所有 p 的根为 Qp 的多项式
DOI：
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发表时间：
2008
期刊：
影响因子：
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通讯作者：
J. Sonn
中科院分区：
文献类型：
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作者：
J. Sonn
设f(X)是Z[x]中的一元多项式，没有有理根，但对所有p都有根，或者等价地，对所有n都有根mod n.众所周知，f(X)不能是不可约的，但可以是两个或更多个不可约多项式的乘积，如果f(X)是m>；1个不可约多项式的乘积，则它的Galois群一定是m个真子群的共轭的并.我们证明了对于任意m>；1，每个有限可解群是m个真子群(其中所有这些共轭子群都有平凡交)的并的有限可解群都是这样一个多项式的Galois群，并且同样的结果(当m=2)也适用于所有Frobenius群。我们还观察到，每个不可解的Frobenius群都可以实现为Q(T)的一个几何扩张即正则扩张的Galois群。
Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Qp for all p, or equivalently, with roots mod n for all n. It is known that f(x) cannot be irreducible but can be a product of two or more irreducible polynomials, and that if f(x) is a product of m > 1 irreducible polynomials, then its Galois group must be a union of conjugates of m proper subgroups. We prove that for any m > 1, every finite solvable group that is a union of conjugates of m proper subgroups (where all these conjugates have trivial intersection) occurs as the Galois group of such a polynomial, and that the same result (with m = 2) holds for all Frobenius groups. It is also observed that every nonsolvable Frobenius group is realizable as the Galois group of a geometric, i.e. regular, extension of Q(t).