Artin M. Galois theory (2ed, London, 1944)(86s).pdf: eKnigu

Home / lib / M_Mathematics / MA_Algebra / MAa_Abstract algebra /

Artin M. Galois theory (2ed, London, 1944)(86s).pdf

Size 1.5Mb
Date Nov 3, 2002

tative in every field, and 2) a field may have only a finite number of elements. More exactly, a field is a set of elements which, under the above mentioned operation of addition, forms an additive abelian group and for which the elements, exclusive of zero, form a multiplicative group and, finally, in which the two group operations are connected by the distributive law. Furthermore, the product of o and any element is defined to be o. If multiplication in the field is commutative, then the field is called a commutative field. B . Vector Spaces. If V is an additive abelian group with elements A, B, . . . , F a field with elements a, b, . . . , and if for each a c F and A e V...

5 linearly in terms of A,, . . . , Am, i.e., A = Ca.A. for a suitable choice i=ll 1 ofa,, i = l , . . . , m , i n F . THEOREM 2. In any generating system the maximum number of i n d e p e n d e n t vectors is equal to the dimension of the vector space. L e t A,, . . . , A , , , be a generating system of a vector space V of d i m e n s i o n n. Let r be the maximum number of independent elements in t h e generating system. By a suitable reordering of the generators we may assumek,, . . . , Ar independent. By the definition of dimension it follows that r < n. For each j, A,, . . . , A , . A,+j a,A, + a,A, are dependent, and in the relation...

anlxl + an2x2 + . . . + annxn = 0 have only the trivial solution. If they have only the trivial solution, then the column vectors are independent. It follows that the original n equations in n unknowns Will have a unique solution if they have any solution, since the difference, term by term, of two distinct solutions would be a non-trivial solution of the homogeneous equations. A solution would exist since the n independent column vectors form a generating system for the n-dimensional space of column vectors. Conversely, let us suppose our equations have one and only one solution. In this case, the homogeneous equations added term by term to a solution of the original equations would yield a new solution to the original equations. Hence, the homogeneous equations have only the trivial solution. F. Qeterminants. l) The theory of determinants that we shall develop in this chapter is not needed in Galois theory. The reader may, therefore, omit this section if he
SO...

16 Next we specialize (10) in the following way: If i is a certain subscript from 1 to n-l we put A, = U, for k f i, i+ 1 Ai = Ui + Ui+r , Ai+, = 0. Then D( A,, A,, +. . , A, ) = 0 since one column is Q, Thus, D(Ai ,A;, , . . , An) = 0; but this determinant differs from that of the elements bj, only in the respect that the i+l-st row has been made equal to the i-tb. We therefore see: A determinant vanishes if two adjacent rows are equal. I&ch term in (9) is a product where precisely one factor cornes...

25
which guarantees for a given polynomial in F the existence of an extension field in which the polynomial has a root. We shall also show that, in a given field, a polynomial cari net only be factored into irreducible factors, but that this factorization is unique up to a constant factor. The uniqueness depends C. Algebraic Elements. Let F be a field and E an extension field of F. If a is an element of E we may ask whether there are polynomials with coefficients in F which have a as root. a ia çalled algebraic- with respect to F if . tkere are such polynomials. New let a be algebraic and Select among ail polynomials in F which have a as root one, f(x), of lowest degree. We may assume that the highest coefficient of f(x) La 1. We contend that this f(x) ia uniquely determined, that it ts trreducible and that each polynomial in F w#r the root o is divisible by f (x ). If, indeed, g ix ) !w a palynomial in F with g(a) = 0, we may divide on the theorem of Kronecker....

27 a new kind of multiplication of two elements g (5) and h (4) of E i denoted by g ([) x h (5). It is defined as the remainder r (6) of the g (6) h(c) un d er d ivision by f (4‘ ). We first remark of m terms gi( c), gz( t), . . . , g,( 0 is again the reg i( 5) g,( 5). . . g,( 5). This is true by of two remainders of these two is associative and ordinary product that any product...

factors of p(x) and collecting to get the original form. Since p ’ (x) cari be split in F’ , we must have F ’ = E ’ . In this case, o itself is the required extension and the theorem is proved if a11 roots of p(x) are in F. We proceed by complete induction. Let us suppose the theorem proved for a11 cases in which the number of roots of p(x) outside of F is less than n > 1, and suppose that p (x ) is a polynomial having n roots outside of F. We factor p (x ) into irreducible factors in F; p ( x ) = f,(x) fJx) . . f,(x). Not a11 of these factors cari be of degree 1, since otherwise p (x ) would split in F, contrary to assumption. Hence, we may suppose the degree of f 1( x) to be r > 1. Let f’,(x).f\(x) . . . f;(x) = p’(x) be the factorization of p’(x) into the polynomials corrrespondng to f 1( x ) , . . . , fm( x ) under O. fi (x ) is irreducible in F ’ , for a factorization of fi (x) in F ’ would induce 1) under 0-l a factorization of f,(x), which was however taken to be irreducible. By Theorem 8, the isomorphism o cari be extended to an isomorphism ol, between the fields F(a) and F ’ (a’ ). Since F C F(a), p(x) is a polynomial in F(U) and E is a splitting field for p(x) in F(a). Similarly for p ’ (x). There are now less than n roots of p (x ) outside the new ground field F (a). Hence by our inductive assumption o1 cari be extended from an isomorphism between F(a) and F ’ (a ’ ) to an isomorphism o2 between E and E ’ . Since u, is an extension of (T, and o2 an extension of o,, we conclude u2 is an extension of u and the theorem follows....

(If o(a) = 0 for one element a, then o(x) = 0 for each x t G, since o( ay) = o(a). o(y) = 0 and ay takes a11 values in G when y assumes a11 values in G). T h e characters or, 02,. . . , onare called dependent if there exist e l e m e n t s a r, a,, . . . , a , , not a11 zero in F such that a,o,(x) + a202(x) + . . . + anon = 0 for each x t G. Such a de-...

n o n - t r i v i a l dependence a ru r (x ) + . + anun (x ) = 0 which holds for e v e r y x 6 E). T h i s follows from Theorem 12, since E without the 0 is a group...

[E.g., u~(x.Y) =o(r(~.~)) = o(r(~>.~(y)> = a(~(x))~o(s(y))l.
We shall cal1 UT the product of o and r. If o is an automorphism (o(x) = y), then we shall cal1 0-l the mapping of y into x, i.e.,o-‘(y) the inverse of o. The reader may readily verify that o-r is an automorphism. The automorphism 1 (x ) = x shall be called the unit - -- automorphism. Le ~ -mma. If E is an extension field of F, the set G of automorphisms which leave F fixed is a group. The product of two automorphisms which leave F fixed clearly =x...