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Math 256 (11) - Midterm Test 1
Spring Quarter 2016
Friday April 22, 2016 - 08:30 am to 09:20 am
Instructions:
Prob.
Points Score
possible
1
22
2
14
3
14
TOTAL
50
• Read each problem carefully.
• Write legibly.
• Show all your work on these sheets. Feel free to
use the opposite side.
• This exam has 4 pages, and 3 problems. Please
make sure that all pages are included.
• You may not use books, notes, calculators, etc.
Cite theorems from class or from the texts as appropriate.
• Proofs should be presented clearly (in the style
used in lectures) and explained using complete
English sentences.
Good luck!
Math 256 (11) - Midterm Test 1
Spring Quarter 2016
Page 2 of 4
Question 1. (Total of 22 points)
a) (4 points) Let E : F and α ∈ E be algebraic over F . State the definitions of the
minimal/irreducible polynomial irr(α, F ) and the degree deg(α, F ) of α over F .
√
b) (5 points) Find the minimal polynomials of α := 1 + 3 2 over Q and over R (here
√
3
2 denotes the real cube root of two).
c) (4 points) State the definition of a finite extension, the degree of a finite extension
and the tower law.
√
√
d) (5 points) Determine whether 5 ∈ Q(1 + 3 2).
√
√
√
√
e) (4 points) Find [Q( 5, 1 + 3 2) : Q] and a basis for Q( 5, 1 + 3 2) over Q.
Math 256 (11) - Midterm Test 1
Spring Quarter 2016
Page 3 of 4
Question 2. (Total of 14 points)
a) (6 points) Suppose E : F is a finite extension. Show that E : F is an algebraic
extension and, furthermore, every β ∈ E satisfies deg(β, F ) ≤ [E : F ].
b) (8 points) Let E : F and α ∈ E be algebraic over F and suppose deg(α, F ) is odd.
Show that F (α2 ) = F (α).
Math 256 (11) - Midterm Test 1
Spring Quarter 2016
Page 4 of 4
Question 3. (Total of 14 points)
To construct a regular heptagon by straightedge and compass requires the construction
of the angle cos(2π/7).
a) (4 points) Let n ∈ N be odd, n ≥ 3, and ω := cos 2π/n + i sin 2π/n. Show that
n−1
X
ω i = 0 and deduce that
i=0
(n−1)/2
X
i=0
(n−1)/2
i
ω +
X
ω̄ i = 0
i=1
where ω̄ = ω −1 denotes the complex conjugate.
b) (3 points) Using part a), show that 2 cos(2π/7) is a root of f (x) = x3 +x2 −2x−1.
c) (7 points) Conclude that it is impossible to construct a regular heptagon using a
straightedge and compass.