1. a) Let ω = e 2πi/p with p an odd prime. Use that disc(ω p ) = ( 1) p 1

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1 Number Theory Mat 6617 Homework Due October 15, 018 To get full credit solve of the following 7 problems (you are welcome to attempt them all) The answers may be submitted in English or French 1 a) Let ω = e πi/p with p an odd prime Use that disc(ω p ) = ( 1) p 1 p p to show that Q(ω p ) contains p if p 1 mod and p if p 1 mod b) Show that the 8 th cyclotomic field contains,, and i c) Show that every quadratic field is contained in a cyclotomic field Solution: a) We know that the discriminant is equal to disc = ±p p Now the construction for the discriminant implies that det{σ i (ωp)} j = ±p p Taking the square root we get p p 3 ±p is an element in Q(ωp ), and thus (since p p 3 is rational, as p is odd and greater or equal than 3), ±p is an element in Q(ω p ) and must generate the quadratic subfield For the sign, notice that we get + if p 1 mod and if p 1 mod b) First ω 8 = i Q(ω 8 ) Also ω 8 = 1+i implies ω 8 + ω 1 8 = Q(ω 8 ) and then clearly = i Q(ω 8 ) Note: The Galois group is isomorphic to (Z/8Z) = Z/Z Z/Z and therefore there are 3 subextensions of degree, so those are all the three quadratic subextensions c) Let Q( m) be the quadratic extesion to consider Write m as a product of primes Without loss of generality we can assume that m is square free (otherwise, write m = s n with n square free and use that Q( m) = Q( n)) Thus m = ±p 1 p k From points (a), (b), we have that ±p Q(ω 8, ω p ) We have that if r s, then Q(ω r ) Q(ω s ) since ω r = ωs s/r From this and the previous observation, m Q(ω8, ω p1,, ω pk ) Q(ω 8m ) (the factor 8 is added to guarantee inclusion in the case of m even and or m 1 mod ) The cyclotomic field is Q(ω 8m ) Let L = Q( m, n) with m, n square-free integers, different from 1, relatively prime a) Let m n 1 mod Show that any α O L is of the form a + b m + c n + d mn a b c d mod b) Find an integral basis for the case m 1 mod, n, 3 mod

2 Solution: By using the same trick as in problem 5 from homework 1, one can see that n Q( m) if n m are integers This automatically implies that Q( m, n) has degree over Q since x n, x m are irreducible by Eisenstein criterion (one could add more details here) We will use that Z[ m], disc = m, m, 3 mod O Q( m) = [ ] Z 1+ m, disc = m, m 1 mod In both items m and n or m and n are relatively prime and that implies that (disc Q( m), disc Q( n) ) = 1 as long as n m and at least one of them is congruent to 1 modulo Then we can apply the theorem for integral basis of extensions that in this case implies that O Q( m) O Q( n) = O Q( m, n) a) By the previous observation, in this case the integral basis is given by {1, 1+ m, 1+ n, (1+ m)(1+ n) } From here, any integral is of the form x + y 1 + m + z 1 + n + w (1 + m)(1 + n) = (x + y + z + w) + (w + y) m + (w + z) n + w mn Clearly, all the coefficients are the same modulo Notice that the converse is false If we take a = 0, and b = c = d =, then, we can not find x, y, z, w satisfying the equation b) By the previous observation, the integral basis is given by {1, 1+ m, n, 1+ m n} 3 Let f(x) = x 3 + ax + b, with a, b Z, a 0, and assume that f is irreducible in Q Let α be a root of f a) Show that f (α) = aα+3b α b) Show that aα+3b is a root of ( x 3b a 3b) c) Show that disc(α) = (a 3 + 7b ) ) 3+a ( x 3b ) a +b Use this to find NormQ(α)/Q (aα+ Solution: a) Notice that α 0, since f(x) is irreducible Also, f (x) = 3x + a Thus f (α) = 3α + a Since α 3 + aα + b = 0, we have α = a b α Thus f (α) = 3 ( a b α) + a = aα+3b α Page

3 b) When we replace x = aα + 3b in x 3b we get α Thus, the equation is satisfied a since α is a root of f This new polynomial is irreducible since it is obtained after a linear change of variables from f which is assumed to be irreducible Notice that Q(α) = Q(aα + 3b) Then Norm Q(α)/Q (aα+3b) is equal to MINUS ( the constant coefficient of the MONIC ( version of this polynomial, which is (a) 3 3b ) ) 3 a 3ab + b = 7b 3 + ba 3 a c) We know that disc(α) = Norm Q(α)/Q (f (α)) = ( 1)3 Norm Q(α)/Q (aα + 3b) Norm Q(α)/Q (α) = 7b3 + ba 3 b = 7b a 3 (One has to be careful with the signs here) Prove that θ = 0 { 1, θ, θ+θ } is an integral basis for the ring of integers in Q(θ) where θ 3 Solution: The polynomial x 3 x is irreducible because it has no roots modulo 3 Therefore θ Q and {1, θ, θ } is a basis for Q(θ) as a Q-vector space Since θ 3 θ = 0, then θ is integral Let α = θ +θ Then (α) = θ + θ 3 + θ = θ(θ + ) + (θ + ) + θ = θ + 6θ + 8 α = θ + 3θ + = α + θ + α α θ = 0 Then α is integral over Z[θ] O L and therefore it is integral One can also see that α is integral by checking that it is a root of x 3 x 3x To find this polynomial, work with the previous equation to obtain This yields α = θ + θ = (α α ) + α α (α 1)(α 3 α 3α ) = 0 Now we compute the discriminant for B = {1, θ, α} and use problem 3: disc(b) = disc(1, θ, θ ) = 1 ( 7( ) ( 1) 3 ) = since 107 is square-free, B = {1, θ, α} is a basis 1 Page 3

4 5 Prove that Z[ω p + ωp 1 ] is the ring of integers of Q(ω p + ωp 1 ) for p odd prime Solution: We know that O Q(ωp) = Z[ω p ] Let α O Q(ωp+ω p 1 ) Then α = a 0 + a 1 (ω p + ω 1 p ) + + a N (ω p + ω 1 with N < p 1 and the a i Q (The reason for this is that [Q(ω p + ωp 1 ) : Q] = p 1 To see this, we use that [Q(ω p ) : Q] = p 1, the fact that Q(ω p + ωp 1 ) Q(ω p ), and that this subextension has degree, because Q(ω p + ωp 1 ) Q(ω p ) since the first field is contained in R and the second is not However, the degree can not be larger than, since ω p satifies the polynomial x (ω p + ωp 1 )x + 1 over Q(ω p + ωp 1 )) Without loss of generality, we can assume that a N Z (Otherwise we can always substract a N (ω p + ωp 1 ) N and still have an integral element) Multiply by ωp N and expand, we get αωp N = a N + + a N ωp N which is an algebraic integer in Q(ω p ), and therefore, an element of Z[ω p ] Thus a N Z (for this it is important that N < p 1, so that the expression is a combination of elements in the integral basis of Z[ω p ] {1, ω p,, ωp p }) Thus we get a contradiction, and the a i must all belong to Z p ) N 6 Let K be a field of the form F q (t)[x]/(g) with F q a finite field and g 0, so K is obtained from F q (t) by adjoining one algebraic element Let O K be the ring of elements of K that satisfy a nonzero monic polynomial with coefficients in F q [t] Make the further assumption that O K is finitely generated as an F q [t] module Prove that every prime ideal of O K is maximal Solution: Let p be a nonzero prime ideal in O K Let α p be a nonzero element and let m(x) F q [t][x] be its minimal polynomial Let m(α) = α n + a n 1 [t]α n a 0 [t] = 0 Then a 0 [t] = (α n + a n 1 [t]α n a 1 [t]α) p Thus, there is a surjective ring homomorphism O K /a 0 [t]o K O K /p Now O K /a 0 [t]o K = (Fq [t]/a 0 [t]f q [t]) r = (Fq d) r, where r is the (finite) rank of O K as F q [t] module Thus, O K /p is a finite ring, and it is also an integral domain (since p is prime) Therefore, it is a field, and p is maximal Page

5 7 Let a = (, 1 + 3) in Z[ 3] Show that a () but a = ()a Conclude that ideals do not factor uniquely into prime ideals in this ring Solution: To see that a (), we need to see that () Now, every element in () is given by (a + b 3) with a, b Z and, since {1, 3} is lineraly independent over Q, this representation is unique, then clearly () On the other hand, a = (, + 3, + 3) = (, + 3) = ()(, 1+ 3) If there were unique factorization, we would have that ab = ac would imply b = c But we have just found a counterexample to this Note: Problem 6 is taken from Stein s course Page 5

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