Math 202 - Assignment 6
نویسنده
چکیده
Proof. The discriminant of x + 1 is D = 256 = 2. We have x + 1 ≡ (x + 1) (mod 2). Let p be an odd prime (so p D), and suppose the irreducible factors of x + 1 have degrees n1, n2, . . . , nk. By Corollary 41, the Galois group of x + 1 contains an element with cycle structure (n1, n2, . . . , nk). Since the Galois group of x +1 over Q is the Klein 4-group, in which every element has order dividing 2, it follows that each ni = 1 or 2. This gives the possibilities (1, 1, 1, 1), (1, 1, 2), (2, 2). However, D is a square and so the Galois group in contained in A4; in particular it contains no transpositions, so (1, 1, 2) is ruled out. This leaves the possibilities (1, 1, 1, 1), and (2, 2), which correspond to the factorization into 4 linear factors or 2 quadratic factors, respectively.
منابع مشابه
Semidefinite Programming
3 Why Use SDP? 5 3.1 Tractable Relaxations of Max-Cut . . . . . . . . . . . . . . . . . . . . . . . . 5 3.1.1 Simple Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.1.2 Trust Region Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.1.3 Box Constraint Relaxation . . . . . . . . . . . . . . . . . . . . . . . . 6 3.1.4 Eigenvalue Bound . . . . . . . . . . . . ...
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