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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)

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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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