Week 3 the Ring of Integers
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Week 3 The ring of integers 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<j+1. On the other hand, if a = mZ for some m =0,thenthereexistsan x N with x = y + mx ,forsomey N . Note6 that if z N , then 2 j+1 j+1 2 j 2 j+1 z = y + aj+1xj+1 for some aj+1 Z,byconstruction,andwenowknowthat m a ,sowritea = dm.Wethenhave2 | j+1 j+1 z dmx N , − 2 j showing that Nj + mxZ = Nj+1. On the other hand, since x1,...,xn are linearly independent, a x + + a x N implies a =0,andso 1 1 ··· j+1 j+1 2 j j+1 Nj mxZ = 0 . From this it follows that if y1,...,ys is a Z-basis for Nj, \ { } then y1,...,y2,x is a Z-basis for Nj+1,showingthatNj+1 is free of rank at most j +1. By induction, it follows that Nj is a free Z-module of rank at most j,and in particular N = Nn is a free Z-module of rank at most n. 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 corre- sponding 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.