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Math 114 - Spring 1994 - Bergman - Final

Math 114 - Spring 1994 - Bergman - Final - SUN George M...

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Unformatted text preview: 11/11/2001 SUN George M. Bergman 5 Evans Hall MOFFITT LIBRARY Spring 1994, Math 114 Final Exam 15:52 FAX 6434330 20 May, 1994 3 hours, between 4 and 8 PM 1. (40 points) Mark statements T (true) or F (false). Each correct answer will count 1 point, each incorrect answer +1 pomt, each unanswered item 0 points. Zn is an integral domain if and only if it is a field. Every polynomial over a field K has a root in K. Every polynomial of prime degree is irreducible. All simple transcendental extensions of a given field are isomorphic. All simple algebraic extensions of a given field are isomorphic. If [((oc) '=' K06) as extension-fields of K, then a and ,6 have the same minimal polynomial. Given a line’zsegment of length 1, one can construct by ruler and compass a line-segment of length 3 . If L is a field, the only L-automorphism of L is the identity. If S g R2 is the circle of radius 1 centered at the origin, then for every point (a, the S, the field Q(or, [3) is algebraic over Q, of degree a power of 2. The Galois group F(C:R) is abelian. Every separable field extension is normal. If L is generated over K by elements separable over K, then every element of L is separable over K. If K and L are fields of the same characteristic, then there exists a homomorphism K —-> L. If K and L are fields and there exists a homomorphism K—) L, then K and L have the same characteristic. Distinct automorphisms of a field K are linearly independent over K. The extension 0(21/2):Q is normal. The extension (2(21/3 ):Q is normal. The extension (10” 4):Q is normal. as" In the next six parts, let K Q L g E, with E finite over K. “m If L:K and E:L are both normal, then EzK is normal. If E: K is normal, then EzL is normal. If EzK is normal, then LzK is normal. If L:K and E21. are both separable, then E: K is separable. If E:K is separable. then EzL is separable. lf EzK is separable, then LzK is separable. Every group of order 96 has a subgroup of order 8. If G is a group, p is a prime, and H1, H2 are p—subgroups of G, then the subgroup of G generated by H1 and H2 is alsoap—subgroup. If G is agroup and N1, N2 are normal groups of G, then the subgroup of G generated by N1 and N2 is also normal. Every simple solvable group is cyclic. Every finite nontrivial p-group has nontrivial center. Every reducible quintic polynomial over a field of characteristic 0 can be solved by radicals. 001 11/11/2001 SUN 15:52 FAX 6434330 MOFFITT LIBRARY 002 _ If K Q L is a finite extension, and tr.deg.(L:K) = 0, then L is algebraic over K. _ Every finitely generated field extension is algebraic. _ There exists a field with 99 elements. The regular 32-gon is constructible with ruler and compass. The regular 33-gon is constructible with ruler and compass. The regular 34-gon is constructible with ruler and compass. If the discriminant of a cubic polynomial fe K [t] is a square in K, then f is reducible. Every algebraic extension of the field R of real numbers is normal. For every prime p there exists an ordered field of characteristic p. Every field isomorphic to the field C of complex numbers is algebraically closed. 2. (45 points) Definitions and examples. In this question, when you give an example you do not have to prove that your example has the properties asked for. (a) (10 points) Define what is meant by a solvable group, and give examples of a solvable and a nonsolvable group. (b) (15 points) Define what is meant by the transcendence degree of a finitely generated extension L:K, and for every positive integer n, give an example of a finitely generated extension of transcendence degree n. (In giving the definition, you may assume the concept of algebraically independent family of elements already to have been defined, but not the concept of transcendence basis.) (c) (10 points) Define what is meant by an algebraically closed field, and give an example of an algebraically closed field, and an example of a field that is not algebraically closed. (d) (10 points) Define what it means for a group G of permutations of a set X to be transitive. For some set X, give both a transitive group of permutations of X and a nontransitive group of permutations of X. 3. (50 points) Suppose L: K is a finite separable normal extension, whose Galois group is cyclic, with cyclic generator1 r of order n; thus the norm map N: L —> K is given by N(a) = ac(a)...rn_ (a). (a) (5 points) Show that if a = b/r(b) for some bEL — {0}, then N (a) = 1. (b) (25 points) Prove the converse statement: if ae L satisfies N (a) = 1, then there exists be L — {0} such that a = b/r(b) (Hilbert‘s Theorem 90). (If you remember the proof in the book and want to give that, fine. In case you don’t. I will remind you of the key idea of the version of that proof I showed in class: consider the properties of the K—linear map 9: L —) L given by 9(1)) 2 0107).) (c) (10 points) Describe briefly the role of I-Iilbert‘s Theorem 90 in the proof that a polynomial over a field of characteristic 0 is solvable in radicals if and only if it has solvable Galois group. (Fig, is it used in proving the “if” or the “only if " direction? To what case of the desired result does one apply the Theorem?) ((1) (10 points) Assuming the general hypotheses on LzK given at the beginning of this question, suppose that char K :e 2, that L contains a root of t - 2, which we will denote 42, and that 10.12) = -J2. Show that if r has order 2. there does not exist a nonzero element be L such that r(b) = (1 +42)b, but that if T has order 4, there does exist such an element. 4. (15 points) We have seen that if K is a field of prime characteristic p, then the map a -—> up is an endomorphism of K (meaning a homomorphism of K into itself; in this case called the “Frobenius endomorphism”). In this problem we shall see that “such things only happen in prime characteristic”. (3.) (7 points) Show that for any field K, if fEK[t] and the map a n—af(a) gives an endomorphism of K, then either f : t, or the fixed field of this endomorphism is finite. (You may take for granted that the fixed set of an endomorphism of a field is a subfield.) (b) (8 points) Deduce that in the above situation, if f :t t, then K has prime characteristic. ...
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