Number Fields Flashcards
Define an alg integer. Show TFAE
(i) alpha an alg integer
(ii) m_alpha / Q is an integer poly
(iii) Z[alpha] is finitely generated as a Z-module.
Show that sums, products and inverses are still algebraic.
An alg integer is a root of some monic integer polynomial.
(i)==>(ii). Let m_alpha the monic, min poly of alpha / Q and f a monic integer poly with alpha a root. Thus m_alpha divides f in Q[X], so m_alpha * h = f for some monic h. Clear the denominators of m_alpha and h. It is clear that m_alpha, h are integer.
(i) ==>(iii). 1, alpha, …, alph^{n-1} generate Z[alpha] = {f(alpha)}.
(iii) ==>(i) If g_1,…, g_r in Z[T] generate Z[alpha], then if k=max deg g_i, Z[alpha] is generated by 1, alpha, …, alpha^k. Thus alpha^{k+1} is some Z-lin combo of the previous powers, so done.
If alpha, beta algebraic,
Z[alpha,beta] Z-generated by alpha^ibeta^j, 0 <= i < deg_alpha, 0 <= j < deg_beta. So it is finitely generated, hence so are the Z-submodules (as Z a pid). Z[alphabeta], Z[alpha+beta], Z[alpha^{-1}] are submodules.
State and prove a theorem about integers in quadratic extensions of Q.
Let d in Z{0,1} square free, K=Q(sqrt(d)).
(i) if d!=1 mod 4, O_K=Z[sqrt(d)]
(ii) if d=1 mod 4, O_K=Z[(1+sqrt(d))/2].
Proof:
For v!=0, the min poly of u+vsqrt(d) is x^2-2u*x+(u^2-dv^2)=0. So it is an alg integer iff 2u in Z and u^2-dv^2 integral.
Case 1: u in Z. So dv^2 in Z. But d square free. So v in Z.
Case 2: u = (2a+1)/2, a in Z. Then iff (4a^2+ra+1)-4dv^2 in 4Z. iff 4dv^2-1 in 4Z. So v=k/2, k in Z and dk^2 is 1 mod 4, and k odd, d=1 mod 4. So the integers are
d!=1 mod 4: {u+vsqrt(d): u,v integral} = Z[sqrt(d)].
d=1 mod 4, {u+vsqrt(d): u,v in Z/2, u-v in Z} = Z[(1+sqrt(d))/2}
Define the complex embeddings of K into C. Show that there are n of them if [K:Q]=n. Call them sigma_1, …, sigma_n. Show that if alpha in K of degree d/Q, then the roots of the min poly of alpha are precisely sigma_i(alpha), each root occuring n/d times in the list.
Define the norm N_{K/Q}(alpha) and trace Tr_{K/Q}(alpha). Define and give two formulas for the char poly of alpha. Give a formula the norm and trace of alpha. Show that they are integers. Show that norm=+/-1 iff alpha is a unit.
A complex embedding of K in C is just a Q-homo K–>C. By the prim element theorem, K=Q(gamma) where gamma is deg n. So the Q-homos are precisely the assignment of gamma to the n (distinct) roots of its min poly / Q. If Q <= Q(alpha) <= K, we see that every C-embedding of Q(alpha) extends to precisely [n/d] C-embeddings of K.
Define u_alpha : K–>K the linear map which is multiplication by alpha. Then N_{K/Q}(alpha) is the det of this map and trace is the trace.
Note that the min poly of alpha kills u_alpha. If K=Q(alpha), we see that m_alpha must be the char poly of u_alpha by degrees. In general, K is just n/d copies of Q(alpha) as a Q(alpha) vector space. So by considering the form of u_alpha in a Q-basis of Q(alpha)^{n/d}, we see that the char poly is simply m_alpha^(n/d) which equals the product of (X-sigma_i(alpha)). Then the norm is just the product of sigma_i(alpha) and the trace the sum.
Define disc(alpha1,…,alpha_n) for an extension [K:Q] of degree n. Why is this natural?
(i) Give another formula for disc
(ii) Show that alpha_1,…,alpha_n a Q-basis of K ifff disc(alpha_i)!=0.
(iii) If beta_i=A_ji alpha_j, what is Disc(beta_i)?
(iv) What is the discriminant of an additive subgroup of K (i.e., a Z-submodule of the Q-module K)? Why is it well-defined.
(v) If H’ <= H two subgroups of K of rank n, what is the index of H’ in H in terms of the discriminants?
(vi) Define an integral basis and prove that it exists.
We define delta(alpha1,…,alphan) = det (Tr_{K/Q}(alpha_ialpha_j) : 1<=i,j <=n)). This is just the discriminant of the natural bilinear form on K, Tr(alphabeta).
(i) Note that Tr_{K/Q}(alpha_i *alpha_j)=sum sigma_k(alpha_i) sigma_k(alpha_j)= (D^tD)_ij, where D_ij=sigma_i(alpha_j). Then disc(alpha_i)=det(D^tD)=det(D)^2.
(ii) Suppose that sum b_i alpha_i=0, for b_i in Q. So b_i sigma_j(alpha_i)=0. This is a linear relation on the columns of D. Thus delta is singular, so disc=0.
Now suppose alpha_i a Q-basis. We note that (alpha,beta) |—> Tr(alphabeta) is a non-degenerate bilinear form on K/Q (non-degenerate because we can just pick alpha=1/beta). The matrix of this bilinear form in our basis is simply Tr(alpha_ialpha_j), which must then by non-singular.
(iii) Its just scaled up by det(A)^2, which is immediate from (i).
(iv) If alpha_i and beta_i generate the same Z-module, they are related linearly by a Z-invertible matrix, which then must have det +/- 1. So they have the same disc by (iii), and so we define the discriminant of an additive subgroup to be the discriminant of any Z-generating set.
(v) We’ve seen that if H’=H and alpha_1,…,alpha_n basis of H, beta_1,…, beta_n basis of H’, we can write beta_i=A_ji alpha_j and then [H:H’]=|detA|. So the [H:H’]^2=disc(H’)/disc(H).
(vi) An integral basis for K is a simultaneous Q-basis of K and Z-basis of O_K. Given a Q-basis omega_n of K, we can multiply omega_n by a least positive integer so that they are then in O_K, then form the subgroup of K generated by these omega_n, say H. Consider out of all subgroups generated as such, one with minimal disc(H). Then if alpha is in O_K, we consider H’=H+Zalpha. Then note H’ is f.g. and since H is rank n, so is H’. Thus by an easy lemma which says all f.g. rank r subgroups of a Q-vector space are generated (freely) by r Q-lin. indep elements, we see that H’ can be obtained in our process (to prove this, just use the structure theorem). Thus H=H’ by minimality of disc.
When can we immediately conclude some linearly indep alg integers are an integral basis from the disc? What happens if this nice condition doesn’t hold? Show that the condition is not necessary.
The condition is that disc(omega_1,…,omega_n) is squarefree. Then since (O_K:H)^2=disc(omega_i)/d_K, we get that O_K=H.
Show that nonzero ideals are f.g over Z and have rank n. (btw, define rank as well).
Define the norm of an ideal, and relate it to the discriminant of the ideal and d_K. What is the norm of a principal ideal?
Show that every ideal contains its norm, and there are only finitely many ideals of a given norm.
ideals I are sub-Z-modules of O_K, which is f.g. as a Z-module, hence I is f.g. If omega_1,…,omega_n an integral basis and alpha nonzero in I, alpha*omega_i is lin indep. Hence I has rank n (the dim span of I).
The norm of an ideal is the index in O_K. Since ideals are a special type of additive subgroup of K, we get that N(I)^2=(O_K:I)^2=disc(I)/d_K.
If I=(alpha) is principal, alphaomega_i is a Z-basis of I. So disc(I)=disc(alphaomega_i)=det(sigma_i(alphaomega_j)) ^2= det(sigma_i(alpha) sigma_i(omega_j))^2=det(sigma_k(alpha)delta_ki)^2det(sigma_i(omega_j))^2=N_{K/Q}^2(alpha)d_K. So N(I)=|N_{K/Q}(alpha)|.
Consider the additive group O_K/I. Then it has size N(I). Hence by Lagrange, N(I)*x=0 for all x in O_K/I. In particular, take x=1 to get N(I) in I.
The ideals of O_K containing (M) are precisely the ideals of O_K/(M). So any ideal of norm M contains (M), and so there are finitely many of them.
Let P a nonzero prime ideal in O_K. Recall in particular this means that P is not O_K. Show that
(i) P is maximal,
(ii) P intersects Z is pZ some prime p
(iii) N(P)=p^f for some f with 1 <= f <= n=[K:Q]
(i) O_K/P is a finite integral domain, hence a field. Hence P is maximal.
(ii) P intersect Z is an ideal in Z so is of the form aZ. But clearly a cannot be composite. So a=1 or a is prime. But if a=1, the norm of P is 1, contradiction. Thus a=p prime.
(iii) pO_K subset of P proper subset of O_K. Since O_K iso to Z^n as a Z-module, (O_K:pO_K)=p^n. So (O_K:P) | p^n and is not 1.
The HARD technical Lemmas which are probably not very important to remember.
(i) If I a nonzero ideal in O_K, alpha in K with alphaI contained in I, then alpha in O_K
(ii) If I a nonzero ideal, then there exist nonzero prime ideals P_1,…,P_r such that I contains P_1 * … * P_r
(iii) If P prime, P contains P_1P_2…P_r, then P=P_i for some i.
(iv) If I a nonzero proper ideal of O_K, and alpha in I, then there exist beta in O_K(alpha) such that beta*I is contained in (alpha).
(v) (Key Step) If I a nonzero ideal, then there exists nonzero ideal J such that IJ is principal.
(i) For every integer k>=0, alpha^k*I in I. Then Z[alpha] * beta subset of I, for any fixed beta in I. So Z[alpha] subset of beta^{-1} * I, which is f.g. as a Z-module. So Z[alpha] is finitely generated. So alpha is in O_K.
State and assume the Key Step. Show that if I, I’, J are nonzero ideals,
(i) IJ=I’J ==> I=I’
(ii) I contains J iff there exists H, IH=J
(iii) There exist unique up to reordering distinct nonzero prime ideals P_1…P_r and integers a_1…a_r>=1 s.t. I=P_1^a_1…P_r^a_r
Key Step: If I a nonzero ideal, there exists nonzero J, IJ principal.
(i) Let K nonzero, JK = (alpha). So IJK=I(alpha)=I’(alpha), alpha nonzero. So alpha^{-1}(alpha)I=I=I’.
(ii) If IH=J, then I contains J.
Let I contain J. Let I*I’=(alpha). So I(alpha^{-1}I’J)=J, where alpha^{-1}I’J is an ideal in O_K.
(iii) Existence by induction on N(I). If I=O_K it’s trivial by letting r=0.
If I!=O_K, choose P containing I, P prime. By (ii), PJ=I, some J. So J contains I. If I=J, then I=PI, so by cancellation O_K=P, contradiction. Hence we are done by induction.
For uniqueness, need one of the technical lemmas. Say I=P1…Pr=Q1…Qs. Then r=s and P_i=Q_i (after reordering).
We have P_1 contains Q_1…Q_s so P_1=Q1 wlog, then use cancellation.
Define the ideal classes. Show that it is an abelian group.
Show that O_K a PID iff CL(K) is trivial iff O_K is a UFD.
I,J are equiv if there exist alpha,beta nonzero in O_K, alphaI=betaJ.
Multiplication is well-defined. Indeed, [I][J]=[IJ]. If I,I’, J,J’ are equiv, then
alphaI=alpha’I’, betaJ=beta’J’, so alphabetaIJ=alpha’beta’I’J’.
The identity is [O_K]. I is principal iff I=(alpha) ==> I equiv to O_K. Conversely, if I equiv to O_K, alpha*I=(beta). So alpha divides beta in O_K, and we have I=(beta/alpha).
Inverses exist in the class group since if I an ideal, there exists J such that IJ is principal.
O_K a PiD iff CL(K) is trivial is immediate. O_K a PID ==> O_K a UFD. But there are UFD’s that aren’t PID’s (e.g., Z[X]). If O_K a UFD, then let P a nonzero prime ideal. Pick alpha nonzero in P and factor it into irreds alpha=pi_1…pi_m. P is prime so pi_j in P some j. Then P contains (pi_j). Hence P=(pi_j).
Show that the ideal norm is multiplicative.
Enough to show N(IP)=N(I)N(P) for prime P. Well N(IP)=[O_K:IP]=[O_K: I][I:IP]=N(I)[I:IP]. So suffices to show N(P)=[I:IP].
Note first that there are no ideals strictly between IP and I. Indeed if J were, I contains J so II’=J. Then J contains IP so I’ contains P (by cancellation). Since P is maximal, I’=P or O_K. So J=I or IP.
Then pick some alpha in I but not IP. Consider the map (regarded as a Z-module homo) O_K/P --> I/IP x+P |---> alpha*x+IP, noting it is well-defined as alpha*P contained in IP.
Surjective: the image contains alpha+IP, and hence (alpha)+IP=P.
Injective: if alpha*x in IP, then (alpha)(x) in IP. Note that I contains (alpha) so (alpha)=IK. If P divides K, then IP divides (alpha) so alpha is in IP, contradiction. Hence P divides (x). Hence x is in P.
Alternative: Consider O_K/P and I/IP as O_K/P vector spaces, and note that the map is linear.
We have that the map is surjective by above. Note that the LHS is 1-dimensional and the RHS is non-trivial. So the RHS must have dimension 1. In other words, the map is an isomorphism.
State and prove Dedekind’s Criterion
Let K=Q(theta) where theta in O_K has min poly f. Assume (O_K: Z[theta]) is not divisible by a prime p. Let f factor into irred g_i^e_i over F_p. Then (p)=product P_i^e_i, where P_i=(p,g_i(theta)), N(P_i)=p^(deg(g_i))=p^(f_i).
Proof:
Step 1: Assume O_K=Z[theta]. Have
O_K/P_i=Z[theta]/(p,g_i(theta))=Z[T]/(f,p,g_i)=F_p[T]/(f^bar,g_i^bar)=F_p[T]/(g_i^bar) a finite field with p^(f_i) elements. So P_i is prime, N(P_i)=p^(f_i).
Step 2: g=
Define what it means for a prime p to (i) ramify (ii) totally ramify (iii) remain inert (iv) split in a number field K.
State and prove a lemma about the trace of prime powers.
State a theorem about when p ramifies and prove the necessary direction.
If alpha in O_K, then Tr(alpha^p)=Tr(alpha) mod p.
Proof: We know Tr(alpha)=Tr(alpha)^p mod p.
Tr(alpha)^p-Tr(alpha^p)= (sum alpha_i)^p-sum(alpha_i^p)
= sum_{0 <= i_1,…,i_p < p with i_1+…+i_p=p} alpha_j^(i_j)
p ramifies iff p|d_K. We prove that if p ramifies, then p|d_K.
- d_K=det(Tr(omega_iomega_j)). It suffices to find a linear relation sum a_i Tr(omega_iomega_j)=0 mod p, where not all a_i are 0 mod p. It suffices to find alpha not in (p) so that Tr(alphaomega_j)=Tr(alpha^pomega_j^p) = 0 mod p.
- Let (p)=P_1^e_1…P_r^e_r, e_1>=2. Pick alpha in P_1^(e_1-1)P_2^e_2…P_r^e_r \ (p). Then note that alpha^p in (p), as p*(e_1-1)>=e_1.
Explain how knowing there is an element x in I of norm N(x) <= N(I)*c_K implies anything about the norm of an ideal related to I.
State Minkowski’s Bound for imaginary quadratic extensions. Deduce that the class group is finite for these extensions.
State Minkowski’s Bound for general extensions.
We have that (x)=I*J some ideal J. Thus N(J) <= c_K. So [J]=[I^{-1}], and J has a rep of norm <= c_K.
If we knew that every ideal had such an element, we could apply this logic to an ideal representing [I^{-1}] to get that every ideal is rep’d by something of norm <= c_K.
Minkowski: For every nonzero ideal I, there exists x in I of norm at most N(x) <= N(I)c_K, c_K=2sqrt(d_K)/pi
The same, but now c_L = (4/pi)^s n!/n^n sqrt(d_K)
State Dirichlet’s unit theorem.
O_K^x iso to mu \times Z^{r+s-1}.
