18 Find \u03b1 E that is not in F Now \u03b1 is algebraic over F and must be of degree 2

# 18 find α e that is not in f now α is algebraic

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18. Find α E that is not in F . Now α is algebraic over F , and must be of degree 2 because [ E : F ] = 2 and [ F ( α ) : F ] = deg( α, F ). Thus irr( α, F ) = x 2 + bx + c for some b, c F . Because α E , this polynomial factors in E [ x ] into a product ( x - α )( x - β ), so the other root β of irr( α, F ) lies in E also. Thus E is the splitting field of irr( α, F ). 19. Let E be a splitting field over F . Let α be in E but not in F . By Corollary 50.6, the polynomial irr( α, F ) splits in E since it has a zero α in E . Thus E contains all conjugates of α over F . Conversely, suppose that E contains all conjugates of α E over F , where F E F . Because an automorphism σ of F leaving F fixed carries every element of F into one of its conjugates over F , we see that σ ( α ) E . Thus σ induces a one-to-one map of E into E . Because the same is true of σ - 1 , we see that σ maps E onto E , and thus induces an automorphism of E leaving F fixed. Theorem 50.3 shows that under these conditions, E is a splitting field of F . 20. Because Q ( 3 2 ) lies in R and the other two conjugates of 3 2 do not lie in R , we see that no map of 3 2 into any conjugate other than 3 2 itself can give rise to an automorphism of Q ( 3 2 ); the other two maps give rise to isomorphisms of Q ( 3 2 ) onto a subfield of Q . Because any automorphism of Q ( 3 2 ) must leave the prime field Q fixed, we see that the identity is the only automorphism of Q ( 3 2 ). [For an alternate argument, see Exercise 39 of Section 48.] 51. Separable Extensions 167 21. The conjugates of 3 2 over Q ( i 3 ) are 3 2 , 3 2 - 1 + i 3 2 , and 3 2 - 1 - i 3 2 . Maps of 3 2 into each of them give rise to the only three automorphisms in G ( Q ( 3 2 , i 3 ) / Q ( i 3 )). Let σ be the automorphism such that σ ( 3 2 ) = 3 2 - 1+ i 3 2 . Then σ must be a generator of this group of order 3, because σ is not the identity map, and every group of order 3 is cyclic. Thus the automorphism group is isomorphic to Z 3 . 22. a. Each automorphism of E leaving F fixed is a one-to-one map that carries each zero of f ( x ) into one of its conjugates, which must be a zero of an irreducible factor of f ( x ) and hence is also a zero of f ( x ). Thus each automorphism gives rise to a one-to-one map of the set of zeros of f ( x ) onto itself, that is, it acts as a permutation on the zeros of f ( x ). b. Because E is the splitting field of f ( x ) over F , we know that E = F ( α 1 , α 2 , · · · , α n ) where α 1 , α 2 , · · · , α n are the zeros of f ( x ). As Exercise 33 of Section 48 shows, an automorphism σ of E leaving F fixed is completely determined by the values σ ( α 1 ) , σ ( α 2 ) , · · · , σ ( α n ) that is, by the permutation of the zeros of f ( x ) given by σ . c. We associate with each σ G ( E/F ) its permutation of the zeros of f ( x ) in E . Part( b ) shows that different elements of G ( E/F ) produce different permutations of the zeros of f ( x ). Because multiplication στ in G ( E/F ) is function composition and because multiplication of the permutations of zeros is again composition of these same functions, with domain restricted to the zeros of f ( x ), we  #### You've reached the end of your free preview.

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• Algebra, φ, Identity element, Inverse element
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