Download Math 547 Final Exam Solutions, Spring 2005 - Prof. A. Vraciu and more Exams Mathematics in PDF only on Docsity!
Math 547, Final Exam, Spring , 2005 The exam is worth 100 points. Each problem is worth 11 1/9 points.
Write your answers as legibly as you can on the blank sheets of paper provided. Use only one side of each sheet. Take enough space for each problem. Turn in your solutions in the order: problem 1, problem 2,... ; although, by using enough paper, you can do the problems in any order that suits you.
I will e-mail your grade to you as soon as I finish grading the exams.
I will post the solutions on my website later today.
- Let K ⊆ L be fields, f (x) be a polynomial in K[x] , σ ∈ AutK L , and
∈ L. Suppose that f () = 0. Prove f (σ(`)) = 0. Give all details.
Let f (x) =
∑n j=
kj xj^ , with each kj ∈ K. We have 0 = f (`). Apply the ring
homomorphism σ to both sides to get
0 = σ(0) = σ(f (`)) = σ
∑^ n
j=
kj `j
∑^ n
j=
σ(kj )(σ(`))j.
The hypothesis also tells us that σ(kj ) = kj for all j ; so
∑^ n
j=
kj (σ())j^ = f (σ()).
- Let K ⊆ L be fields, f (x) be an irreducible polynomial of K[x] , and α 1 and α 2 be elements of L with f (α 1 ) = f (α 2 ) = 0. Prove that there exists a ring isomorphism σ : K[α 1 ] → K[α 2 ] with σ(α 1 ) = α 2 and σ(k) = k for all k ∈ K. Give all details.
There is a surjective ring homomorphism φ 1 : K[x] → K[α 1 ] with φ 1 (g(x)) = g(α 1 ) for all g(x) ∈ K[x]. The kernel of φ 1 is generated by the minimal polynomial f (x) of α 1. The first isomorphism theorem ensures the existence of a ring isomorphism with φ¯ 1 (¯g) = φ 1 (g) = g(α 1 ) for all g ∈ K[x]. We repeat the above procedure to produce a ring isomorphism φ¯ 2 : K[x]/(f (x)) → K[α 2 ] , with φ¯ 2 (¯g) = g(α 2 ) for all g ∈ K[x]. It follows that φ¯ 2 ◦ φ¯− 1 1 : K[α 1 ] → K[α 2 ] is a ring isomorphism. It is clear that φ¯ 2 ◦ φ¯−^1 1 (α^1 ) = φ¯ 2 (¯x) = α 2.
- State the Fundamental Theorem of Galois Theory. Please give hypotheses and conclusions.
Let K be a field with Q ⊆ K ⊆ C , let f (x) be a polynomial in K[x] , and let F be the splitting field of f over K. Then a. | AutK F | = dimK F. b. There is a one-to-one, inclusion reversing, correspondence between the subgroups H of AutK F and the intermediate fields E with K ⊆ E ⊆ F.
The correspondence is given as follows. If H is a subgroup of AutK F , then the corresponding field is F H^ , which is defined to be
{α ∈ F | σ(α) = α, for all σ ∈ H }.
If E is a field with K ⊆ E ⊆ F , then the corresponding group is
AutE F = {σ ∈ AutK F | σ(e) = e for all e ∈ E }.
c. If F H^ is one of the fields with K ⊆ F H^ ⊆ F for some subgroup H of AutK F , then dimF H F = |H|.
- Let I be an ideal of the ring R. Prove that I is a maximal ideal of R if and only if R/I is a field.
⇒ We need show that each non-zero element of R I has a multiplicative inverse in R I. Pick a non-zero element of^
R I. This element has the form^ ¯a^ where^ a^ is an element of R which is not an element of I. We must show that the element ¯a of R I has an inverse in^
R I. Let (I, a) denote the smallest ideal of R which contains I and a. Observe that (I, a) = {m + ra | m ∈ I and r ∈ R}. The hypothesis ensures us that (I, a) = R. In other words, there exist elements m ∈ I and r ∈ R with 1 = m + ra. We conclude that ¯r is the inverse of ¯a in R I.
⇐ Suppose J is an ideal of R with I ( J. Let j ∈ J with j ∈/ I. The hypothesis that R/I is a field ensures that there exists an element r of R with rj − 1 ∈ I. It follows that 1 is equal to rj plus an element of I. Thus, 1 ∈ J and J is equal to all of R.
- Prove that Q[x] is a Principal Ideal Domain.
Let I be a non-zero ideal in Q[x]. Let f be a non-zero polynomial in I of least degree. We show that I = (f ). It is clear that (f ) ⊆ I. We show that I ⊆ (f ). Let g be an arbitrary element of I. Divide f into g and get g = hf + r for polynomials h and r of Q[x] where either r is the zero polynomial or r has degree less than the degree of f. We see that r = g − hf ∈ I. We chose f to be a non-zero polynomial in I of least degree. It follows that r is the zero polynomial and g ∈ (f ).
- Let I be an ideal in a Principal Ideal Domain R. Prove that the following statements are equivalent. (That is, if one of the statements is true, then they all are true. If one of the statements is false, then they all are false.) (a) There is an irreducible element r of R with I = (r). (b) The ideal I is a non-zero prime ideal. (c) The ideal I is a maximal ideal.
(a) =⇒ (c) Suppose J is an ideal of R with I ( J. The ring R is a Principal ideal domain, so J = (j) for some element j of R. The fact that f ∈ I ⊂ J = (j) tells us that f = jr for some r in R. The fact that I 6 = J tells us that r is not
We see that
H = {(1), (1, 2 , 3 , 4), (1, 3)(2, 4), (1, 4 , 3 , 2), (1, 3), (1, 2)(3, 4), (2, 4), (1, 4)(2, 3)},
(1, 2)H = {(1, 2), (2, 3 , 4), (1, 3 , 2 , 4), (1, 3 , 2 , 4), (1, 4 , 3), (1, 3 , 2),(3, 4), (1, 2 , 4),
and
(1, 4)H = {(1, 4), (1, 2 , 3), (1, 3 , 4 , 2), (2, 4 , 3), (1, 3 , 4), (1, 2 , 4 , 3), (1, 4 , 2), (2, 3)}
It is clear that the orbit of H is {H}. We see that (1), (1, 3)(2, 4), (1, 2)(3, 4) , and (1, 4)(2, 3) all carry (1, 2)H to (1, 2)H , and (1, 2 , 3 , 4), (1, 4 , 3 , 2), (1, 2 , 3) , and (2, 4) all carry (1, 2)H to (1, 4)H. We conclude that the orbit of (1, 2)H is {(1, 2)H, (1, 4)H}. The normalizer of H in S 4 is the union of all of the cosets which have orbits consisting of only one element. So, the normalizer of H in S 4 is simply H. (Of course, you know that NS 4 (H) = H without doing any work. Indeed, NS 4 (H) is a subgroup of S 4 , with H a normal subgroup of NS 4 (H). The only subgroups of S 4 which contain H are S 4 and H. It is easy to see that H is not a normal subgroup of S 4 ; so we see that NS 4 (H) = H .)
- Let K be the splitting field of x^17 − 1 over Q. We know that K = Q[ω] ,
for ω = e 217 πi
. We also know that AutQ K is the cyclic group of order 16 which is generated by the automorphism σ where σ(ω) = ω^3. You do not have to re-prove any of the above facts. However, I do want complete details for the following things: Find a subgroup H of AutQ K with 8 elements. Find the field KH^. (“Find” means tell me generators.)
The element σ^2 of AutQ K has order 8 ; so H is generated by σ^2. We see that σ^2 carries ω 7 → ω^9 7 → ω^13 7 → ω^15 7 → ω^16 7 → ω^8 7 → ω^4 7 → ω^2 7 → ω.
It follows that if
u = ω + ω^9 + ω^13 + ω^15 + ω^16 + ω^8 + ω^4 + ω^2 ,
then σ^2 (u) = u. Thus, u ∈ KH^. On the other hand, σ moves u , so u /∈ Q. The fact that dimQ KH^ = 2 ensures that there do not exist any field properly between Q and KH^ ; and therefore, KH^ = Q[u].
