Math202hw6asols
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Transcript of Math202hw6asols
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Math 202 Homework #6,
14.7 Solutions
May 19, 2015
Sec 14.7, #3. If is a square in F, then F() = F() = F if and only if is a square in F. So assume is not a square in F, in particular = 0. Claim:F(
) = F(
) if and only if / (F)2. One direction is clear: if = x2with x F, then F() = F(x) = F(). For the converse, suppose thatF(
) =F(
). Then
= x+y
for some x, y
F, so = x2 + 2xy
+y2.
Therefore 2xy= 0. Since char(F)= 2, this implies x= 0 or y = 0. But y = 0 implies = xF, hence is a square in F; as we saw in the beginning of the proof, if
is a square in F then F(
) =F(
) =F if and only if is also a square in F (inwhich case / (F)2). Hence were left with the case x = 0, which implies that
= y
, so/= y2 (F)2 as claimed.Applying this result to F =Q(
2), we see that Q(
1 2) = Q(i,2) if and only
if (1 2)/(1) =1 + 2 is a square in Q(2). Set (x+ y2)2 =1 + 2with x, y Q. We obtainx2 + 2y2 =1 and 2xy = 1. The first equality is clearlyimpossible to satisfy, since x2 + 2y2 > 0 for any real numbers x and y. Therefore
Q(1
2)=Q(i,
2).
#7. (a) We have ( na)n = ( nan) = (a) = a, hence ( na) = na for some nthroot of unity . Since F, iterating this equality m times for any integer m yieldsm( n
a) =m n
a. Therefore the order of is exactly equal to the multiplicative order
of. Since has order d, we see that must be a primitive dth root of unity.
(b) By parta, both( n
a)/ n
aand ( n
b)/ n
bare primitivedth roots of unity. Hencethe first of these can be written as the second to the power i for some integer i rela-
tively prime to d. The equation ( n
a)/ n
a = (( n
b)/ n
b)i yields ( n
a)/( n
bi) =
n
a/ n
bi. Therefore n
a/
n
bi
is fixed by Gal(K/F) = and hence lies in F.
(c) Again one direction is easy: Ifa= bi
cn
1 and b= aj
cn
2 then n
a= c1n
bi
F( n
b)and nb= c2 naj F( na), soF( na) =F( nb). For the converse, ifF( na) =F( nb),then part (b) implies that a = bicn1 for some c1 F and by symmetry b = ajcn2 forsome j and some c2Fas well. This completes the proof.
#9. Since this problem relies on previous exercises (namely, #21 and #26 of Section 2),let us write out what we need from those exercises along the way. Write Gal(K/F) =
1
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. The linear independence of characters implies that
TrK/F = 1 ++2 + +p1 = 0
as a functionK K. Therefore, there exists an elementK such that TrK/F= 0Let
= 1
TrK/F(() + 22() + (p 1)p1()).
Then () is equal to1
TrK/F
(2() + 23() + (p 1)p()) (() + 22() + (p 1)p1())
= 1
TrK/F( () 2() 3() p1()) =1,
so () =
1. In particular, since ()
= , we see that
F and hence
F() = K (since [K : F] = p is prime, the only subfields ofK are F and K itself).Now let a = p . Note that
(a) =()p () = ( 1)p ( 1) =p 1 ( 1) =p = a.
(Here we used the freshmans binomial theorem, (1)p =p1 in characteristicp.)Since generates Gal(K/F), we find thataF. Therefore satisfies the polynomialxpxaF[x], and since this polynomial has degree p, it is the minimal polynomialof.
Note that the roots of this polynomial are + k for k Fp, and the element i Gal(K/F) acts on the roots by adding i.
#11. (a) SinceH(Z/pZ) is the subgroup of squares, it follows that0 =
a squareap
and1=
b non-squarebp.A generator of (Z/pZ)
is a non-square. SoGal(Q(p)/Q)satisfies (p) =
gp for a non-square g (Z/pZ). Now the product of a square and
non-square is a non-square, and the product of two non-squares is a square (this isjust the group law in the group (Z/pZ)/H of order 2. We therefore have (0) =
a squareagp =
b non-square
bp =1 and similarly (1) =0.
(b) Ifeis a generator of (Z/pZ), then the squares are precisely the powerse0, e2, . . . , ep3
and the non-squares are precisely the powers e1, e3, . . . , . . . ep2. Then
0+1=
p2k=0
ek
p =
p2k=0
k(p) = (p, 1)
and
0 1=p2k=0
(1)kekp =p2k=0
(1)kk(p) = (p,1)
2
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by definition. Finally, it is clear that (p, 1) =p1
j=1jp =1 since p satisfies the
polynomial p(x) =xp1 +xp2 + + 1.
(c) As i ranges from 0 to p 1, i2 modulo p covers the value 0 once and each of thesquares in (Z/pZ) twice. Therefore
g= 1 + 20= 1 +0+0 = 1 +0+ (1 1) =0 1= (p,1).
(d) Sinceg = 0 1, andHfixes0 and 1 whereasGH swaps0 and1, it is clearthat (g) =g if H and (g) =g ifH. In particular, g lies in the fixed fieldofH, which has degree 2 overQ sinceHis an index 2 subgroup ofG, andgQ sinceg is not fixed by all ofG. Therefore [Q(g) : Q] = 2. Also, since complex conjugationis equal to 1, our description above shows that g =g with the sign depending onwhether1H, i.e. whether or not1 is a square modp.(e) The problem is mostly solved in the parenthetical comment, which we dont copyover. Fork = 0 top
2, considerk(p)/p. If = eas above, then
k(p)/p = ek1p .
Ifk = 0, then clearly this is just 1. In this case we obtain
p2j=0
j(1) =p 1.But if 0< k < p1, thenek 1 (mod p) sincee is a generator of (Z/pZ), soek1p isa primitive pth root of unity . In this case we obtain
p2j=0
j() = TrQ(p)/Q =1.An explanation for this last equality is that TrQ(p)/Q is the negative of the coefficientof the second highest exponent term in the minimal polynomial for , which is p(x) =xp1 + xp2 + . . . + 1. Putting this together with the computation in the hint we obtain
gg = (p 1) +p2k=1
(1)k = (p 1) + 1 = p.
(f) Ifp1 (mod 4), then1 is a square mod p, so g =g and hence g2 =gg =p. Ifp3 (mod 4), then1 is not a square modp, sog=g and hence g2 =gg =p.These can be combined by writingg2 = (1)(p1)/2psince (1)(p1)/2 =1 dependingon p mod 4. Thereforeg =
(1)(p1)/2p and it follows that Q(
(1)(p1)/2p) is
the unique quadratic subfield ofQ(p).
Final note: Figuring out exactly what sign occurs here in the equation
g=
(1)(p1)/2p
is a problem that Gauss spend a lot of effort thinking about.
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