This preview shows page 1. Sign up to view the full content.
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 extensionﬁelds 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 linesegment of
length 3 . If L is a ﬁeld, the only Lautomorphism 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 pgroup 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 32gon is constructible with ruler and compass. The regular 33gon is constructible with ruler and compass. The regular 34gon 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 IIilbert‘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. ...
View
Full
Document
This note was uploaded on 04/05/2008 for the course MATH 114 taught by Professor Borcherds during the Spring '03 term at University of California, Berkeley.
 Spring '03
 Borcherds
 Math

Click to edit the document details