By Emil Artin, Arthur N. Milgram, Arthur N. Milgram
Truly provided components of 1 of the main penetrating ideas in smooth arithmetic comprise discussions of fields, vector spaces,homogeneous linear equations, extension fields, polynomials,algebraic parts, in addition to sections on solvable teams, permutation teams, answer of equations by way of radicals, and different suggestions. 1966 ed.
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The illustration idea of finite teams has noticeable speedy progress in recent times with the advance of effective algorithms and computing device algebra structures. this can be the 1st ebook to supply an creation to the standard and modular illustration concept of finite teams with specific emphasis at the computational points of the topic.
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Proof: If a does not divide c, the greatest common multiple of a and c would be larger than c and we could find an element of that order, thus contradicting the choice of c. We now prove Theorem 17. Let n be the order of S and r the largest order occuring in S. Then x’ - 1 = 0 is satisfied for a11 ele- 51 ments of S. Since this polynomial of degree r in the field cannot have more than r roots, it follows that r -> n. On the other hand r -< n because the order of each element divides n. S is therefore a cyclic group consisting of l,~, ~2,.
The characteristic of a field is always a prime number. a). a = b # Oif a f 0 and r b + 0 since both r and s are less than p, SO that pa f 0 contrary to the definition of the characteristic. If na = 0 for a f 0, then p divides n, for n = qp + r where 0 -< r < p and na = (qp + r)a = qpa + ra. Hence na = 0 implies ra = 0 and from the definition of the characteristic since r < p, we must have r = 0. If F is a finite field having q elements and E an extension of F such that (E/F) = n, then E has q” elements.
N. Suppose that (E/F ) > n. Then there exist n + 1 elements al, a*, . . , a,,, of E which are linearly independent with respect to F. By Theorem 1, there exists a non-trivial solution in E to the system of equations x1 q (a1 > + x2 q (a, > + . . + xn+, q (a,,, > = 0 (’ > x1 u2 (a1 > + x2 o2 (a2 > + . . + xn+l a2 ( a,+l > = 0 . . . . . . . . . . . . . . . . . xlun(al) + x2un(a2) + . . e that the solution cannot lie in F, otherwise, since u1 is the identity, the first equation would be a dependence between a,, .