Week 3 the Ring of Integers

Home , Linear relation, Ring of integers

Week 3

We continue our study of the ring of integers of a number field. Note that if K/Q is a number field, then by definition we have a Q-basis, that is, elements x1,...,xn such that

K = a1x1 + + anxn : ai Q { ··· 2 } (and each such representation is unique). It would certainly be useful if we could choose the basis in such a way that

K = a1x1 + + anxn : ai Z . O { ··· 2 } For quadratic extensions we have seen that we can, in fact a basis of the form 1,↵ ,somaybethispropertyholdsingeneral. { } 3.1 Modules

The study of extensions of fields is aided a great deal by considering a field as a vector space over a subfield. To study extensions of rings, we need a structure over a ring which is similar to a vector space. Definition 3.1. Let R be a ring. An R-module M is an abelian group with aruleforscalarmultiplicationbyelementsinR,satisfying(foralla, b R and x, y M) 2 2 1. (a + b)x = ax + bx

2. a(x + y)=ax + ay

40 3. a(bx)=(ab)x

4. 0Rx =0M

5. 1Rx = x. A submodule of M is a subset N M containing 0 which is itself an R- ✓ M module. An R-module homomorphism is a map of R-modules ' : M1 M2 such that '(x + y)='(x)+'(y), and '(ax)=a'(x), for all a R!and x, y M (i.e., a linear transformation of modules). Two R-modules2 are isomorphic2 if there is a bijective R-module homomorphism taking one to the other. Finally, given and R-module M,andelementsx1,...,xn,thesub- module of M generated by x1,...,xn is simply

x ,...,x = a x + a x + + a x : a R 0 . h 1 ni { 1 1 2 2 ··· n n i 2 }[{ M }

We say that x1,...,xn are a generating set for M over R if the submodule they generate is M itself, and if such a set exists (we’re requiring it to be ﬁnite) we say that M is of ﬁnite type as an R-module.

Note that if F is a field, then an F -module is just a vector space over F ,andanF -module of finite type is just a finite-dimensional vector space. But modules are much more general. For example, any abelian group can be given the structure of a Z-module, by letting n Z+ act by 2 nx = x + x + + x, ··· where the sum on the right has n terms, and letting ( n)x be the inverse of nx. And this will be a Z-module of finite type if and only if the group is finitely generated. Note that this makes for some crucial di↵erences between modules in general, and vector spaces in particular. In a vector space V over afieldF ,weknowthatcx =0V implies either c =0F (the zero scalar) or x =0V (the zero vector). This is not true in modules. For example, if we let M = Z/7Z,asaZ-module, then 7 2=0M ,eventhough2=0M ,and · 6 7 =0Z. We have to be careful, then, in using our intuition about vector spaces6 in general modules. For our purposes, it is enough to think about Z-modules, which are simply abelian groups. It is useful to phrase things in terms of modules, however, for the purposes of later generalizations.

41 Another natural construction of a module over a ring R is to take the cartesian product Rn = R R R, ⇥ ⇥··· and to make this an R-module by setting

r(a1,a2,...,an)=(ra1,ra2,...,ran)

(the addition in Rn is pointwise).

Definition 3.2. An R-module M is said to be a free R-module of rank n if n and only if it is isomorphic to R .Asetx1,...,xn M is a basis for M over 2 n R if and only if there is an isomorphism ' : M R such that '(xi)isthe element (a ,...,a ) Rn such that a =1,anda!=0forj = i. 1 n 2 i j 6 Exercise 3.3. Let A be a ring, and view A as an A-module in the obvious way. Show that the submodules of A are exactly the ideals of A,andthat the submodules of finite type are exactly the finitely generated ideals.

Note that modules of ﬁnite type over Z are jsut ﬁnitely generated abelian groups. From group theory, we know that every such group is of the form

G Zr, ⇥ for G afinitegroupandr some non-negative integer. This Z-module of finite type is a free module of rank r when G is the trivial group. The following lemma is useful, and is equivalent to the statement that subgroups of finitely generated groups are always finitely generated. Note that it is easy to replace Z with any PID A in the proof, and get the analogous result for A-modules (only when A is a PID).

Lemma 3.4. Let M be a free Z-module of rank n, and let N M be a ✓ submodule. Then N is a free Z-module of rank at most n.

Proof. The main thing is to show that N is ﬁnitely generated. Let x1,...,xn be a Z-basis for M,andletN0 = 0 be the trivial Z-module. We will set { }

Nj = N (x1Z + + xjZ), \ ··· and show that Nj is a free module of rank at most j for 1 j n,by induction. The base case (j =0)istrivial.  

42 Suppose that Nj is a free module of rank at most j,andlet

a = a Z : y + axj+1 Nj+1 for some y Nj . { 2 2 2 } Then it is easy to check that a is an ideal of Z,andsomustbeaprincipal ideal. If a =(0),thenNj+1 = Nj which we have assumed free of rank at most j

Now to show that K is a free Z-module, it is enough to show that it is O contained in a free Z-module, which might be easier. In order to do that, we need a little more linear algebra.

3.2 Matrices and linear transformations

Our first step is to note that some of the familiar constructions from linear algebra of vector spaces over fields can actually be carried through to modules over rings. Let A be an integral domain, and M afreemoduleoffiniterankoverA n (so that M ⇠= A as A-modules, for some n). In pratice we will generally be taking A = Z,soM is the Z-span of some vectors in a vector space over Q,orwewillbetakingA = Q in which case M is just a vector space. An endomorphism of M is a function T : M M such that ! T (x + y)=T (x)+T (y)

43 and T (ax)=aT (x) for all x, y M and a A (i.e., T is a “linear transformation”). Given a 2 2 basis e1,...,en for M over A,wecanassociatetoT amatrix(aij)determined by n

T (ei)= aijej. j=1 X The determinant, trace,andcharacteristic polynomial of T will be the corresponding quantities associated to the matrix. Recall that the characteristic polynomial of an n n matrix B with entries in A is det(XI B) A[X], ⇥ n 2 where In is the n n identity matrix. Note that, a priori,thisdependson the choice of basis,⇥ and so really we should specify a basis when we speak of the trace, etc., of an endomorphism. On the other hand, the matrices of the same endomorphism relative to di↵erent bases are similar matrices, and so have the same trace, determinant, and characteristic polynomial. So, after a small amount of linear algebra, we can convince ourselves that these quantities do not depend on the particular basis used.

Definition 3.5. Let B be an integral domain, and let A be a subring such that B is a free A-module of finite rank. Then for ↵ B,wedefineTr (↵) 2 B/A to be the trace of the endomorphism T↵(x)=↵x.Similarly,wedefinethe norm NB/A(↵)tobethedeterminantofT↵,andthecharacteristicpolynomial of ↵ to be the characteristic polynomial of T↵. Note that all three of the defined quantities are independent of the choice of basis (i.e., the choice of representing matrix), by standard facts from linear algebra.

Example 3.6. Let A = R and B = C,whichisafreeA-module of rank 2, with basis 1,i .Thenif↵ = x + iy,, we have { } T (1) = x 1+y i ↵ · · T (i)= y 1+x i, ↵ · · and so the matrix for T↵ is x y . yx ✓ ◆

44 We obtain N (x + iy)=x2 + y2 = x + iy C/R | | TrC/R(x + iy)=2x =2Re(x + iy), and the characteristic polynomial for x + iy over R is

X2 2xX + x2 + y2. Example 3.7. The case we are usually interested in is that in which A = Z, and B is the ring of integers in a number field, or some other subring of a number field. For example, let K = Q(pD), where D 3(mod4)isa ⌘ square-free integer. We know that K = Z[pD], and a Z-basis for this is 1, pD . Now, for any ↵ = a + bpOD,wecanfindthematrixforx ↵x. In{ particular,} ↵1=↵ and ↵pD = bD + apD,sothematrixis 7!

abD . ba ✓ ◆ We then have

Tr K /Z(↵)=2a O 2 2 N K /Z(↵)=a b D. O Note that simple linear algebra shows the the characteristic polynomial of ↵ always has the form

n n 1 n X Tr (↵)X + +( 1) N (↵), B/A ··· B/A where n is the rank of B over A.Italsofollowsimmediatelyfromthe deﬁnition that NB/A(↵)=NB/A(↵)NB/A() and TrB/A(↵ + )=TrB/A(↵)+TrB/A(). If A = Q and B is a number ﬁeld, it’s fairly easy to describe the norm and trace and characteristic polynomial of an element.

Lemma 3.8. Let K/Q be a ﬁnite extension, and let L/K be a ﬁnite extension. Then for any ↵ K, 2

TrL/Q(↵)=[L : K]TrK/Q(↵),

45 and [L:K] NL/Q(↵)=NK/Q(↵) .

Similarly, if K/Q(↵) is the charactersitic polynomial for x ↵x over K, then 7! [L:Q] L/Q(↵)=K/Q(↵) .

Proof. Let x1,x2,...,xn be a basis for K over Q. We have seen that, if y1,...,ym is a basis for L/K,thenthexiyj form a basis for L over Q. Now, both Tr and N are deﬁned in terms of the matrix giving x ↵x K/Q K/Q 7! relative to the basis x1,...,xn. In other words, if M is this matrix, then

x1 x1 x2 x2 ↵ 0 . 1 = M 0 . 1 . . . B C B C B xn C B xn C B C B C @ A @ A For any yj, we then have

x1yj x1yj x2yj x2yj ↵ 0 . 1 = M 0 . 1 . . . B C B C B xnyj C B xnyj C B C B C @ A @ A If we write yj for the vector x1yj,x2yj,...,xnyj,thenthisshowsthat

y M 0 0 y 1 ··· 1 y2 0 M 0 y2 ↵ 0 . 1 = 0 . . ··· . 1 0 . 1 . . . . . B C B C B C B ym C B 00 M C B ym C B C B ··· C B C @ A @ A @ A as block matrices. In other words, the matrix representing x ↵x relative to 7! the basis x1y1,x2y1,...,xnym is and m m block-diagonal matrix, with Mson the diagonal. Such a matrix has trace⇥m Tr(M)anddeterminantdet(M)m. Since m =[L : K], this proves the claim. The claim for the characteristic polynomial also follows from the structure of characteristic polynomials of blcok-diagonal matrices.

46 Theorem 3.9. Let K be a number ﬁeld of degree n, let ↵ K, and let 2 1,...,n be the distinct embeddings of K into C. Then for any ↵ K, 2 N (↵)= (↵) (↵) (↵), K/Q 1 2 ··· n Tr (↵)= (↵)+ (↵)+ + (↵), K/Q 1 2 ··· n and the characteristic polynomial for ↵ is

(X (↵))(X (↵)) (X (↵)). 1 2 ··· n

Note, if K = Q(↵), then 1(↵),...,n(↵)arejusttheconjugatesof↵. It turns out that in general, even if K = Q(↵), the list 1(↵),...,n(↵)is 6 just the conjugates of ↵ each repeated [K : Q(↵)] times (check this as an exercise!).

Proof. We’ll ﬁrst show this in the case K = Q(↵). It suces to show that the characteristic polynomial is the minimal polynomial, since the characteristic polynomial has the form

n n 1 n X Tr (↵)X + +( 1) N (↵), K/Q ··· K/Q while the minimal polynomial is (writing ↵i for i(↵))

n n 1 n (X ↵ ) (X ↵ )=X (↵ + + ↵ )X + +( 1) (↵ ↵ ). 1 ··· n 1 ··· n ··· 1 ··· n Now, the minimal polynomial of ↵ over Q has degree n and is monic, and the same is true of the characteristic polynomial of ↵ over Q.Sowejustneedto show that ↵ is a root of the characteristic polynomial, which just amounts to showing that the matrix ↵I T is not invertible (here we’re using T to n ↵ ↵ denote the matrix for T↵). Let n n 1 f(X)=X + an 1X + + a1X + a0 ··· be the minimal polynomial of ↵ over Q.Wecanuse

n 1 e ,e ,...,e = 1,↵,...,↵ { 1 2 n} { } as a basis for K over Q,andthen

i 1 i T (e )=↵ ↵ = ↵ = e i · i+1

47 for 0 i

So the matrix for T↵ is

000 0 a0 100··· 0 a 0 ··· 1 1 010 0 a2 . B . . . ··· . . C B . . . . . C B C B 000 1 an 1 C B ··· C @ A Thus the matrix ↵I T is n ↵ ↵ 00 0 a0 1 ↵ 0 ··· 0 a 0 ··· 1 1 0 1 ↵ 0 a2 . B . . . ··· . . C B . . . . . C B C B 000 1 ↵ + an 1 C B ··· C @ A Adding ↵ times the second row to the ﬁrst row, we obtain

2 0 ↵ 0 0 a0 + a1↵ 1 ↵ 0 ··· 0 a 0 ··· 1 1 0 1 ↵ 0 a2 . B . . . ··· . . C B . . . . . C B C B 000 1 ↵ + an 1 C B ··· C @ A Adding ↵2 times the third row to the ﬁrst row, we obtain

3 2 00↵ 0 a0 + a1↵ + a2↵ 1 ↵ 0 ··· 0 a 0 ··· 1 1 0 1 ↵ 0 a2 . B . . . ··· . . C B . . . . . C B C B 000 1 ↵ + an 1 C B ··· C @ A Proceeding in this fashion, we see that ↵In T↵ is row-equivalent to a matrix whose first row is ( 000 0 f(↵) ). ··· 48 So det(↵I T ) = 0, and hence ↵ is a root of the characteristic polynomial n ↵ of ↵.ThisprovesthetheoremwhenK = Q(↵). Now for the general case. First note that if ↵ K,and ,..., : 2 1 n K C are the n =[K : Q]distinctembeddingsofK,then1,...,n all ! restrict to embeddings of Q(↵) C,thoughtheirrestrictionsmightnotbe ! distinct (in fact, will not be unless K = Q(↵)). So, 1(↵),...,n(↵)areall conjugates of ↵.Fromthepreviouscaseandthepreviouslemma,theproof in this case amounts to showing that each conjugate is repeated the same number of times in the list above. We will sketch this argument. In general, suppose that ↵ and ↵0 C are conjugate over Q,sothereisanisomorphism 2 : K = Q(↵) K0 = Q(↵0). We can assume that our extension L/K is primitive, so L != K(), and let g K[X] be the minimal polynomial of 2 . Note that K[X] ⇠= K0[X] by applying to coecients, so we may speak about (g)(X) K0[X], which is a monic, irreducible polynomial of degree deg(g)=[L : K2]. For any root 0 C of (g), we then have a string of isomorphisms 2 L = K() K[X]/gK[X] K0[X]/( (g)K0[X]) K0(0) C. $ $ $ ✓ In other words, each root of (g) defines a distinct embedding of L into C,whichtakes↵ to ↵0.Expandingonthat,eachconjugate↵0 of ↵ defines [L : K]embeddings : L C satisfying (↵)=↵0.Thisisatotalof ! [L : K][K : Q]=[L : Q]embeddings,sothisisalloftheembeddingsofL into C.Inotherwords,giventheusuallist1,...,n : L C of embeddings, ! i sends ↵ to each of its conjugates exactly [L : K]times,provingthetheorem in the general case.

Note, though, that if ↵ K,thenthematrixforT↵ is just ↵I[L:K],and so 2 [L:K] NL/K (↵)=↵ and TrL/K (↵)=[L : K]↵. So, combining this with the above lemma, we have for any ↵ L, 2 [L:K(↵)] NL/K (↵)=NK(↵)/K (↵) and TrL/K (↵)=[L : K(↵)] TrK(↵)/K (↵).

49 If f(X)istheminimalpolynomialfor↵ over K,itturnsoutthatwealso ﬁnd that characteristic polynomial for ↵ L is 2 f(X)[L:K(↵)].

So, while these quantities depend on the ﬁeld L as well as ↵,theydosoina fairly predictable way. Lemma 3.10. Let K be a number ﬁeld, and let ↵ . Then Tr (↵), N (↵) 2OK K/Q K/Q 2 Z. Also, ↵ ⇥ if and only if NK/ (↵)= 1. 2OK Q ±

Proof. By the comments above, TrK/Q(↵)=[K : Q(↵)] TrQ(↵)/Q(↵), and [K:Q(↵)] NK/Q(↵)=NQ(↵)/Q(↵) ,soitsucestoconsiderthecaseK = Q(↵). In this case, the minimal polynomial of ↵ is exactly

n n 1 X Tr (↵)X + N (↵). K/Q ···± K/Q Since ↵ K ,weknowthatallofthecoecientsofthisareinZ, and so in 2O particular TrK/Q(↵), NK/Q(↵) Z. For the last claim, note that2 if ↵ =1for↵, ,then 2OK

NK/Q(↵)NK/Q()=NK/Q(1) = 1 and NK/ (↵), NK/ () Z. In other words, NK/ must take units to units. Q Q 2 Q But we know that Z⇥ = 1 . {± } For the other direction of the last claim, suppose that ↵ K and NK/Q = 1. Then if f is the characteristic polynomial of ↵,wehave2Of(0) = 1. In ± ± other words, there are integers c1,...,cd 1 Z such that 2 d d 1 ↵ + cd 1↵ + + c1↵ 1=0. ··· ± But then d 2 ↵ (cd 1↵ + + c1 )=1, ⌥ ··· d 2 and since (cd 1↵ + + c1) K ,thisshowsthat ↵ is a unit. ⌥ ··· 2O We can use the above fact to confirm some of our previous claims about unit groups. For example, if x + iy Z[i]isaunit,then 2 x2 + y2 =N (x + iy)= 1. Q(i)/Q ± The only solutions are (x, y)=( 1, 0), (0, 1), corresponding to the four units 1and i. ± ± ± ± 50 Similarly, the units in Q(⇣) with ⇣2 +⇣ +1 = 0 mustbeoftheform a+b⇣ with a b 1 N + p 3 = (a2 +3b2)= 1, K/Q 2 2 4 ± ✓ ◆ with a and b either both even or both odd. There are only finitely many solutions, and we can check them all. In fact, this theorem generalizes quite nicely to Q(pD)withD<0. Theorem 3.11. Let K = Q(pD), where D<0 is a square-free integer, and let denote the ring of integers in K (that is, the integral closure of OK Z K). Then ⇥ = 1 unless D = 1 or D = 3. (In the cases ✓ OK {± } D = 1 and D = 3, we just checked that ⇥ is isomorphic to either Z/4Z OK or Z/6Z, respectively.) Proof. By our characterization of integers in quadratic number fields, we have Z[pD]ifD 2, 3(mod4) K = ⌘ O Z[ 1 + 1 pD]ifD 1(mod4). ( 2 2 ⌘ We’ll look at the case D 2, 3 (mod 4) first. In this case, ↵ has the ⌘ 2OK form ↵ = a + bpD,fora, b Z,and 2 N (↵)=a2 Db2. K/Q Notice that, since D<0, it’s not possible that a2 Db2 = 1. Thus, if ↵ is aunit,wehave a2 Db2 =1. Now, if D< 1, then either b =0or Db2 > 1. In the latter case, it’s impossible that a2 Db2 =1,andsofor D< 1theonlysolutionsto a2 Db2 =1are(a, b)=( 1, 0). This means that the only units are 1+0 pD = 1. ± ± ± a b p In the case D 1(mod4)theintegersinK have the form 2 + 2 D, where a and b have⌘ the same parity. This gives a unit if and only if

a2 Db2 =4, where a and b have the same parity. An analysis just as in the case above shows that the only solutions are (a, b)=( 2, 0) except in the case D = 3, where we also have the solutions ( 1, ±1) (for all four combinations of sign). ± ±

51 It turns out that the only number ﬁelds K such that ⇥ is ﬁnite are OK K = Q,andK = Q(pD)forD<0.

3.3 A bilinear form

Recall that a bilinear form on a vector space V over a ﬁeld F is a function f : V V F which is linear in each coordinate, in other words ⇥ ! f(x + y, z)=f(x, z)+f(y, z)

f(cx, y)=cf(x, y) f(x, y + z)=f(x, y)+f(x, z) f(x, cy)=cf(x, y) for all x, y, z V and c F .Suchaformisnon-degenerate if and only if f(x, y)=0forall2 y implies2 x =0,andf(x, y)=0forallx implies y =0.If f is non-degenerate and e1,...,en is an F -basis for V ,thenwecanrepresent f by a non-singular matrix A,whoseijth entry is f(ei,ej). In particular, if x, y F n are the coordinates of x, y V relative to this basis, then 2 2 f(x, y)=xT Ay.

Lemma 3.12. Let K be a number ﬁeld. Then the function

(↵,) Tr (↵) 7! K/Q is a non-degenerate binary linear form on K (as a vector space over Q). Proof. From the deﬁnition, it is fairly easy to check that this function is bilinear over Q.ThisjustfollowsfromthefactthatTrK/Q is a Q-linear transformation. The tricky thing is to show that this pairing is non-degenerate.

In other words, if TrK/Q(↵)=0forall K,then↵ =0.Thisisnot obvious, since there can certainly be non-zero2 elements of K with trace zero.

Let x1,...,xn be a basis for K over Q.WeneedonlycomputeTrK/Q(xixj) for every pair xi, xj,sincethesearetheentriesofthematrixwhichwewant to show is non-singular. By the calculation above, we have

Tr (x x )= (x x )+ (x x )+ + (x x ). K/Q i j 1 i j 2 i j ··· n i j

52 Since the k are ring homomorphisms, we have k(xixj)=k(xi)k(xj)for all k,andso

Tr (x x )= (x ) (x )+ (x ) (x )+ + (x ) (x ) K/Q i j 1 i 1 j 2 i 2 j ··· n i n j 1(xj) 2(xj) = 1(xi) 2(xi) n(xi) 0 . 1 . ··· . B C B n(xj) C B C @ A In other words, if N is the matrix whose ijth entry is i(xj), then we have shown that M = N T N. It follows that

det(M) = det(N T N) = det(N T )det(N)=det(N)2 =0. 6

Note that, since M is a matrix with entries in Q,thisconﬁrmsourcon- jecture that the square of the determinant of the matrix i(xj)wasarational number! In fact, we’d like to do one better: we would like to show that it is an integer.

3.4 K is a free Z-module O Note that there is something incomplete about our work above. Note that if we have a Z-basis for K /Z,thenwecandeﬁneTr K /Z.Ifwehaveabasis O O for K/Q,thenwecandeﬁneTrL/Q. In some of the examples above, we do in fact have a Z-basis for K ,andinadditionitturnsouttoalsobeaQ-basis O for K. So the trace functions turn out to be the same (well, technically TrK/Q is an extension of Tr K /Z,sinceithasalargerdomain). O So two questions come up naturally:

1. Is there always a Z-basis for K ?Inotherwords,areringsofintegers O in number ﬁelds always free Z-modules (of ﬁnite type)?

2. Is a Z-basis for K always also a Q-basis for K? O

53 It turns out that the second question is easier to answer than the ﬁrst, but they are closely related. Our next lemma states that, just how every rational number is an integer divided by and integer, every algebraic number is an algebraic integer divided by an integer. Lemma 3.13. Let K be a number ﬁeld, and let ↵ K. Then there is a positive integer d such that d↵ . 2 2OK Proof. Let

n n 1 f(X)=X + an 1X + + a1X + a0 Q[X] ··· 2 be the minimal polynomial of ↵.Thenforanyintegerd, we have

n n n 1 n 1 n 0=d f(↵)=(d↵) + dan 1(d↵) + + d a1(d↵)+d a0. ··· 2 n But of course we can choose d 1sothatdan 1,d an 2,... a a0 are all integers. Then we have d↵ as the root of a monic polynomial with integer coecients. Lemma 3.14. For any vector space V over a ﬁeld F , any non-degenerate pairing ( , ):V V F , and any basis x ,...,x of V over F , there is a · · ⇥ ! 1 n basis y1,...yn of V over F such that

(xi,yj)=ij.

Proof. Identifying elements of V with their coordinates relative to x1,..,xn, we can take V = F n,andsothepairingisgivenby

(↵,)=↵T A for some non-singular matrix A,whoseijth entry is (ei,ej)(thestandard n 1 basis vectors in F ). Since A is non-signular, it has an inverse A .Let 1 n yi = A ei.Theny1,...,yn forms a basis for F ,andwecancompute

T T 1 T (xi,yj)=xi Ayj = xi AA xj = xi xj = ij.

Theorem 3.15. Let K be a number ﬁeld of degree n. Then is a free OK Z-module of rank n.

54 Proof. Let x1,...,xn be a basis for K over Q.Sinced1x1,...,dnxn will also be a basis, for any non-zero integers d1,..,dn,wecanassumewithoutlossof generality that x1,...,xn K . Now let y1,...,yn be a dual basis with respect to the pairing 2O (↵,) Tr (↵). 7! K/Q For any ,wecanwrite 2OK = q y + + q y , 1 1 ··· n n with qi Q. Now we compute the coecients using the pairing. By the bilnearity,2 we have

TrK/Q(xi)= qj TrK/Q(xiyj)=qi. j=1 X In other words, n

= TrK/Q TrK/Q(xi)yi. i=1 X But xi, K ,soTrK/ Z.Thismeansthat is in the Z-module 2O Q 2 generated by y1,...,yn,themodule

q1y1 + + qnyn : qi Z K. { ··· 2 }✓

Moreover, y1,...,yn is a Z-basis for this module, since any non-trivial linear relation amongst the yj over Z would also be a non-trivial linear relation over Q. Since was arbitrary, we have contained in this submodule, 2OK OK which is a free Z-module of rank n.Sobyalemmaabove, K is a free O Z-module of rank at most n. Now, suppose that z1,...,zm is a Z-basis for K ,withm n.Forany O  ↵ K there is a non-zero d Z such that d↵ K .Sowecanwrited↵ 2 2 2O as a Z-linear combination of the zi.Dividingbyd,wecanthenwrite↵ as a Q-linear combination of the zi,sothezi span K as a vector space over Q. This means that m n,soinfactm = n.

Lemma 3.16. Let M be the matrix whose ijth entry is TrK/Q(xixj), for any basis x1,...,xn of K over Q. Then det(M) Z. 2

55 Proof. We know that det(M)isindependentoftheparticularchoiceofbasis, so we may choose our own basis. By a lemma above, we may do so with xi K for all i.Thenxixj K for all i, j,andsoTrK/Q(xixj) Z for all i and2Oj.SoM is a matrix with2O integer entries (relative to this basis),2 and hence has integer determinant.

3.5 is Noetherian OK Theorem 3.17. If K is a number field, then every ideal of is finitely OK generated. Moreover, any non-empty collection of ideals in K has a maximal element (i.e., is Noetherian). O OK Proof. This is actually fairly easy, given the above. We know that is a OK finitely generated Z-module of rank at most n.Butanyideala K is a subgroup under addition, and is closed under multiplication by elements✓O of Z K ,somustbeaZ-submodule of K .Itfollowsthata has a Z-basis ✓O O of size at most n,sayx ,...,x a such that 1 m 2

a = a1x1 + + amxm : ai Z . { ··· 2 } Note that a a x + + a x : a a, ✓{ 1 1 ··· m m i 2OK }✓ because Z K and because a is closed under addition and multiplication by elements✓ ofO . So in fact OK a = x + + x , 1OK ··· mOK aﬁnitelygeneratedideal. We now prove that K is Noetherian, relying heavily on the structure of ﬁnitely-generated freeO abelian groups. Note that it is sucient to prove that any sequence of ideals I1 I2 is eventually constant, since this is equivalent to every non-emtpy✓ collection✓··· having a maximal element. Now, let x1,...,xm be a Z-basis for K .EachidealIn is a Z-submodule of K ,and from our investigation of theO structure of these above, we know thatO each has the form

Im =SpanZ(a1,mx1,...,an,mxm), for some integers a1,m,...,an,m (some of which might be zero). Note that the inclusion I I forces a to be a divisor of a ,unlessa =0. m ✓ m+1 i,m+1 i,m i,m 56 As m increases, the number of ai,m which are zero cannot increase, so it is eventually constant. Without loss of generality, then, let’s assume that this number is constant. In fact, to simplify the notation, we might as well assume that none of the ai,m are zero. So now, every ai,m for every m,isa divisor of ai,1,foreachi.Sinceai,1 has only finitely many divisors, there are only finitely many choices for ai,m.Sinceai,m+1 ai,m for all i and all m,we have only finitely many choices for the ideal |

Im =SpanZ(a1,mx1,...,an,mxm) as m . So it must eventually be constant. !1 Note that the proof actually gives us something more: it tells us that every ideal of K is generated by at most [K : Q]elements.SoforK = Q, this says thatO every ideal has a single generator, which we already know. For K a quadratic ﬁeld, it does not say that, but it says that every ideal is generated by at most two elements of K .Inotherwords,everyidealis the gcd of two principal ideals. This seemsO like “close to” being a PID, but unfortunately it turns out that it isn’t really a useful way of being almost a PID.