Download Math 594, HW7

Math 594, HW7- Solutions
Gilad Pagi, [email protected]
March 18, 2015
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a). Since the set of constructible numbers, denoted C, is a subset of R, containing 1 and
0, suffices to show that C is closed under addition, multiplication, and inverse, all of
which were proved in class. We shall give a proof for the inverse of a > 0, given that a is
constructible.
I. Given length a on the line AB, |AB| = a.
II. Construct the point C s.t. |BC| = 1.
III. Construct a perpendicular line to AB at point B.
IV. Construct the point D s.t. |BD| = 1.
V. Construct a similar triangle to ABD on with the edge BC correspond to AB.
VI. Get |CE| = 1/a.
D
•
•
1
•
A
a
B
•
1
•
E
C
b). We shall construct the coordinate system as follows: Given the (initial) points, denoted
as A and B, construct the line AB and a perpendicular line to AB through A. Let the
A point denote the origin, and B be (1, 0), so the corner starting at A, going through B,
is the positive x-axis. Any length a on AB, starting from A, will be treated as (a, 0) or
(−a, 0) depending if the corresponding end point ( (C) in the figure) is on the positive
x-axis or not. The positive y-axis is set according the the ”right hand rule”. Given
constructible length p0 , p1 can position the p0 on the positive x-axis, and p1 on the positive
y-axis s.t. the length from A Correspond. By constructing two perpendicular lines to
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the axes we get the point P = (p0 , p1 ) of constructible coordinates. Conversely, given
P = (p0 , p1 ), we construct the perpendicular line to the axes through P and obtain the
coordinates. This proves the statement.
y
p1
•
•
•
•
(A) (B)
•
(P )
•
(C) p0
x
c). Given A = (ca , da ), B = (cb , db ) the line equation is (cb −ca )(y−da ) = (db −da )(x−ca ) , and
the circle equation is (WLOG - the center is at A) (x−ca )2 +(y−da )2 = (cb −ca )2 +(db −da )2
d). Intersection (x0 , y0 ) of 0 = ax + by + c and 0 = dx + ey + f , if exists, is the solution of a
linear system, which is in the field k since it consist of addition, subtraction, multiplication
and division of the coefficients. For a line in the form y = ax + b, and a circle (WLOG centered at the origin) x2 + y 2 = r2 we get x20 + (ax0 + b)2 = r2 . So the solution for x0 is
either in k or in a quadratic extension of k (See HW6, Q1). Being on the same k-line, x0
and y0 are in the same extension. For a vertical line x = a, the statement is obvious.
e). For two circles: (x−a)2 +(y −b)2 = r2 and (x−c)2 +(y −d)2 = s2 , notice that subtracting
one from the other eliminates the term x2 , y 2 so this problem is reduced to the previous
one. Thus, the solution, if exists, is in k or a quadratic extension of k.
f). From previous parts, and since any additional constructible point is either an intersection
of 2 lines, 2 circles or a line and a circle in k, the statement is true - any additional
constructible point lies in either k or a quadratic extension of k.
g). Suppose that θ is constructed by a finite process of constructing P0 = 0, P1 = 1, P2 , P3 , ..., Pt =
θ. From previous parts, starting from P0 = 0 and P1 = 1 in Q, P2 is either in Q or in
quadratic extension of Q, denoted k2 . P3 is either in k2 or a quadratic extension of k2 ,
denoted k3 . There for, Pt = θ is in kt ⊃ kt−1 ⊃ ... ⊃ k2 ⊃ Q, each extension is of degree
2
2 or 1. Thus [kt : Q] = 2m for some m. Specifically, Q(θ) ⊃ Q is finite.
√
h). Q(π) ⊃ Q is infinite and therefor π is not constructible. Likewise, nor is π. The side
length of any constructed square is constructible. But if we could construct √
a square
whose area is the same a circle with unit radius, then its side length would be π.
i). Q(21/3 ) ⊃ Q is an extension of degree 3 6= 2m , so 21/3 is not constructible. Any attempt to
construct a cube with volume 2 will result an edge of length 21/3 which is not constructible.
j). The matrix for rotating though θ is (notice the used abbreviation)
cos θ − sin θ
c −s
A=
=
sin θ cos θ
s c
Rotation through 3θ is the composition of this linear transformation with itself three
times; so its matrix is
2
cos 3θ − sin 3θ
c −s
c − s2 −2cs
c(c2 − s2 ) − 2cs2 ∗
3
=
=A =
sin 3θ cos 3θ
s c
2cs
c 2 − s2
∗
∗
So cos(3θ) = c3 − 3cs2 = c3 − 3c(1 − c2 ) = 4c3 − 3c as desired.
k). Denote cos(π/9) as c. Since constructing an angle and constructing the cosine of the
angle is equivalent, we shall prove that c is not constructible. Notice that from the
previous part, c satisfies cos(π/3) = 4c3 − 3c i.e. 0.5 = 4c3 − 3c i.e. c is a root of the
polynomial p(x) = 8x3 − 6c − 1 ∈ Q[x]. Want to show that this is irreducible over Q,
then the extension is of degree 3 and thus c is not constructible. By Gauss’s Lemma,
p(x) is irreducible over Q if and only if it is irreducible over Z. But, if it factors over
Z, then it has a degree one factor, it would have a root which would necessarily be one
of the following {± 11 , ± 12 , ± 14 , ± 18 }, none of which is a root (by inspection). Therefor
p(x) = 8x3 − 6c − 1 ∈ Q[x] is irreducible.
√
Note that, for example, we can trisect π (cos(π/3) = 0.5) and π/2 (cos(π/6) = 3/2).
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a). Suppose we have some polynomials g1 , ..., gt ∈ R s.t.
g1 f1 (xf1 ) + ... + gt ft (xft ) = 1
in R. Let L be the algebraic extension of F obtained by adjoining roots α1 , . . . , αt of
the polynomials f1 (x), . . . , ft (x), say, a splitting field for the product f1 f2 . . . ft . There
is an obvious ring homomorphism from R to a free L-algebra S given by sending
xf1 7→ α1 , ..., xft 7→ αt , and the remaining xf to indeterminates. This map applied to the
relation above gives 0 = 1 in S, a contradiction.
b). First note (using Zorn’s Lemma) that the proper ideal I is contained in a maximal ideal
M. The maps F ,→ R → R/M = K1 produce a field extension F ,→ K1 of F . Now
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every (monic, non constant) polynomial f (y) ∈ F [y] can be considered a polynomial in
K1 [y]. But the image of xf in R/M, call it αf , is a root of f (y) in K1 , since f (xf ) ∈ M
means f (αf ) = 0 in R/M.
c). Continue the previous procedure, creating a Kn+1 , an extension of Kn (and of F as well)
in which for every polynomial f (y) ∈ Kn [y] there’s at least one root in Kn+1 [y]. Denote
K = ∪∞
1 Kn . Every polynomial f [y] ∈ K[y] has finitely many coefficient so exist N , large
enough, s.t. f (y) ∈ KN [y], therefore f has a root in KN +1 ⊂ K. Thus, K is algebraically
closed.
d). In the last problem set, we proved that addition/multiplication/division of algebraic
numbers is algebraic (we used that when we reasoned why the ”algebraic-ness” of an
extension is determined by its generators). So, the set of all algebraic numbers over F
forms a field that contains F . Therefore, the set of all algebraic numbers over F in the
above K form a field, denoted as F . We claim that this is an algebraic closure of F : for
every f (y) ∈ F [y] there is a root a in K, but such root is algebraic so it must be in F .
Divide (in F [x]) f by (x − a) to get g(x). Since K is algebraically closed, g has a root b
in K. So F (b)/F is algebraic, and F /F is algebraic, so F (b)/F is algebraic(HW6Q4a)
⇒ b ∈ F . Now repeat. Since f is of finite degree the procedure ends and we show that
all the roots of f are in F .
Note here that we proved that for any F ⊂ K ⊂ F , and g(x) ∈ K[x], then g(x) splits
completely in F .
e). Suppose that F has two algebraic closures F , Fe. Observe that each contains (an isomorphic
copy of) the splitting field of every polynomial in F [x]. Indeed, both F , Fe can be viewed
as the union (or directed limit) of all splitting fields they contain. In class, we proved
the uniqueness of a spitting field over F . As a result, for any splitting field K1 over F
f1 in Fe. From the same result, the splitting field
contained in F , exists an isomorphic copyK
f2 in Fe which extends K
f1 . These
of K2 of a polynomial g ∈ K1 [x] is isomorphic to some K
isomorphisms induce an isomorphism of the direct limits, and hence an F -isomorphism
F → Fe. The image of F in Fe is F -isomorphic to F , hence is an algebraic closure of F .
But if the inclusion is proper, this would contradict the claim that Fe is an algebraic
closure of F , since it shouldn’t contain a proper subfield in which all polynomials have a
root.
If you prefer not to use the direct limit, you can also argue using Zorn’s lemma: Define
a collection A of all subfields of F which have an isomorphic (over F ) copy in Fe. Note
that A contains all the splitting fields. Ordinary inclusion of sets is a partial order on A.
Given a chain K1 ⊂ K2 ⊂ ... in A, observe that the chain has an upper bound K = ∪∞
1 Ki
e
which also has an isomorphic copy in K, namely the union of a corresponding chain of
e i ’s in Fe. By Zorn’s lemma, the set A has a maximal element, call it K. So K is the
K
largest field between F and F which is F -isomorphic to some subfield of Fe. We will be
done if we can show that K = F . Suppose a ∈ F but a ∈
/ K. But a is an element of the
splitting field of its mimimial polynomial, which is a subfield in A. Hence a ∈ K, and
the proof is complete.
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a). The complex numbers ζnk = (e2πi/n )k where k = 0, 1, . . . , n − 1, are n distinct numbers
(situated at equally spaced intervals around the unit circle in C), all of which satisfy the
polynomial xn − 1. This means that xn − 1 factors as (x − ζn0 )(x − ζn1 ) · ... · (x − ζnn−1 ),
so that Q(e2πi/n ) is the unique splitting of xn − 1 contained in C. The set of roots is
µn = {ζn0 , ζn1 , ..., ζnn−1 }, which is cyclic of order n, generated (for example) by ζn .
b). Clear from the identification ζn = e2πi/n .
c). Since xn − 1 = (x − ζn0 )(x − ζn1 ) · ... · (x − ζnn−1 ), the point is to show that µn is the disjoint
union of all the primitive roots in µd s.t. d | n. But ζnk is a primitive generator for µd if
and only if its order (in C, hence in µn ) is d. Now, the number of members of order d in
n
µn is φ(d) (known results coming from the lemma: xa ∈< x >∼
.
= Cn has an order gcd(a,n)
P
The partition above give us the following partition for n: n = d|d φ(d).
d).
x8 − 1 = Φ1 · Φ2 · Φ4 · Φ8 =
(x − 1) · (x + 1) · (x − i)(x + i) · (x − eiπ/4 )(x − e3iπ/4 )(x − e5iπ/4 )(x − e7iπ/4 )
where
Φ1 = (x − 1), Φ2 = (x + 1), Φ4 = x2 + 1,
Φ8 = x4 + 1.
e). For n = 1, 2 see above.
Assume that the statement is true for 1, 2, ..., n − 1. By previous
Q
n
part x − 1 = Φn d|n,d<n Φd . By the induction assumption, xn − 1 = Φn · g(x) where both
xn − 1, g(x) ∈ Z[x], which imply that Φn ∈ Q[x]. Gauss’s Lemma not implies that any
factorization in Q[x] induces a factorization in Z[x] simply by multiplying by constants,
however all the polynomials are monic. Therefore Φn ∈ Z[x].
f). f (x) irreducible ⇐⇒ f (x + 1) is irreducible is clear, since h(x)g(x) = f (x) ⇐⇒
p
p −1
h(x + 1)g(x + 1) = f (x + 1). Now, Φp (x) = xx−1
, so Φp (x + 1) = (x+1)x −1 . Using
i−1
Pp
p
the binomial formula we get the following polynomial:
. We get a monic
i=1 pi x
polynomial where all the rest of the coefficient are a multiple of p, but the constant
coefficient is not a multiple of p2 . By Eisenstein, the polynomial is irreducible.
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NOTE: we use the notation from the previous question.
√
√ 1
√
0 p
p
p
The roots of the polynomial
(in
C)
are:
aζ
,
aζ
,
...,
aζpp−1 . Denote
K as the splitting
p
p
√
√
p
p
field. Clearly,
K ⊂ Q( a, ζp ). For the opposite including observe ( aζp )p = aζp ⇒ ζp ∈
√
p
K ⇒ a ∈ K. Without loss of generality, we may
assume that a is an integer, since if
√
√
√
√
p
p √
p √
p−1
p−1
p−1
p
√
p
p
(bc
)
b c
b
a = b/c, then p a = √
= b cc
=
where c, b ∈ Z. This means that
pc =
c
c
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the splitting field of xp − a is the same as the splitting field of xp − d where d = bcp−1 . Also d
is not a p-th power since b/c is not. Note that if d has a prime factor q which divides d exactly
once, then the irreducibility of xp − d follows from Eisenstein and Gauss’s lemma. If every
prime factor of d appears with multiplicity two or more, we can not directly apply Eisenstein’s
criterion; we need to replace the polynomial xp − d by another polynomial xp − d0 which has
the same splitting field but in which some prime factor of d0 appears with multiplicity exactly
one. Since d is not a p-th power, there is a prime factor q which appears in d with multiplicity
k, where (k, p) = 1. Write Say d = aq k . Since k is a generator for the cyclic group Z/pZ,
there is an m such p
that km = 1pmodulo p. Note that dm = am q 1+pr = (am q)(q r )p for some
r. Observe that Q( p (d)) = Q( p (d0 )) where d0 = (am q). So we can work with the splitting
field of xp − d0 instead.
√ 0
√ 0
In order to find the degree [Q( p d , ζp ) : Q], notice that [Q( p d ) : Q] = p as simple extension
with minimal irreducible polynomial xp − d0 and [Q(ζp ) : Q] = p − 1 as a simple extension
with minimal irreducible polynomial Φp . Thus, these intermediate fields have degrees p and
√ 0
p − 1 relatively over Q, showing that p and p − 1 both divide [Q( p d , ζp ) : Q]. Since p and
√ 0
p − 1 are relatively prime, we see that p(p − 1) divides [Q( p d , ζp ) : Q]. On the other hand,
√ 0
√ 0
clearly p(p − 1) is also an upper bound for [Q( p d , ζp ) : Q], since adjoining p d to Q(ζp ) is a
simple extension of degree at most p, since its minimal polynomial must divide xp − d0 .
For QR problems, see the Math department website, where answers are posted under Past
Exams.
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