`ALGEBRAIC NUMBER THEORY - COURSE NOTESSTEVE DONNELLYHousekeeping (1) There'll be a final exam (3 hours or so) on the whole semester (representation theory and algebraic number theory) weighted towards algebraic number theory (2) There'll be homework assignments each week, due on Tuesdays by lunchtime. (3) The assignments will consist of core problems and more challenging `problems for discussion'; we'll schedule a couple of extra meetings to discuss them (4) I'll hand out complete lecture notes for each class. Useful books in the library On the course material: Stewart &amp; Tall: Algebraic Number Theory And Fermat's Last Theorem. Simple and elementary treatment of the most basic points. Janusz: Algebraic Number Fields. Well explained, taking a fairly algebraic point of view. Neukirch: Algebraic Number Theory. Editions in English and German. Thorough and meticulously written by a number theorist of the old German school. Swinnerton-Dyer: A Brief Guide to Algebraic Number Theory (QA241.S85). Concisely presented (possibly too terse in places), by a famous number theorist. Some nice books of wider or historical interest: Cox: Primes of the form x2 + ny 2 . Introduces several classical subjects beautifully, with the goal of motivating class field theory. Ribenboim, Paulo: Fermat's last theorem for amateurs. Ribenboim is a well known number theorist who loves to explain things in the most elegant, elementary way. Ribenboim, Paulo: 13 lectures on Fermat's last theorem. (We have several of Ribenboim's number theory books that you may enjoy browsing...)12STEVE DONNELLYLecture One: Introduction and examples Quadratic rings and fields. Let's start by considering the ring of Gaussian integers Z+Zi = {a+bi : a, b  Z}. We'll view it as a natural generalization of the ring Z, and investigate to what extent they are similar. The two rings are connected via the norm map N (a + bi) = (a + bi)(a - bi) = a2 + b2 = |a + bi|2 , which will often be useful. Recall that for an integral domain, Euclidean  Principal Ideal Domain  Unique Factorization Domain. Our domain Z + Zi is Euclidean with respect to the norm, since for any ,   Z + Zi we can find q, r  Z + Zi such that  = q + r with N (r) &lt; N () as follows. Notice that the ideal (Z + Zi) is a square lattice and r can be chosen inside a central square 1 1 {a + bi : -  a, b  } . 2 2 (Drawing a picture makes it very clear.) So Z + Zi is a PID and a UFD. The units in Z are the multiplicative group {1, -1}. What are the units in Z + Zi? Notice that the norm is a multiplicative map, meaning N () = N ()N () for all , . So the norm of a unit must be a unit, and one finds that the group of units is the group of 4th roots of unity, {i, -1, -i, 1}. Recall that in a UFD, irreducible elements generate prime ideals. Reason: suppose  is irreducible, and suppose ab is in the ideal (). Writing a and b as products of irreducibles, we have a1 . . . an  (). Since  is irreducible, we must have  = uai for one of the ai and some unit u. Thus ai  (), which means a or b is in (); this shows () is prime. The elements 1 + i, 3, 2 + i, 7, 11 and 3 + 2i are examples of prime Gaussian integers (generators of prime ideals in the PID Z + Zi). To see this, we could check that they are irreducible. For instance, if one could write  = 2 + i, it would follow that N ()N () = 5 and hence either  or  is a unit. Similarly, if one had a nontrivial decomposition  = 7, then N () and N () are ±7. But one can't write 7, or -7, in the form a2 + b2 . An odd prime p  Z can be written in this form iff p  1 mod 4. Hence, p remains prime in Z + Zi iff p  3 mod 4. Here's a more algebraic way of looking at it. An ideal () of Z+Zi is prime iff the &quot;residue ring&quot; Z + Zi/(Z + Zi) is a domain (equivalently, a field, since it is finite). Now, Z + Zi/(2 + i) is isomorphic to the field of 5 elements, denoted F5 = GF (5) = Z/(5), because Z  (2 + i) = 5Z and i  -2 mod 2 + i. Therefore 2 + i is a prime. To see that (7) is prime, note that (Z + Zi)/(7)  F7 [x]/(x2 + 1) = is a field because x2 + 1 is irreducible (has no root) over F7 . Indeed,ALGEBRAIC NUMBER THEORY3x2 + 1 has a root over Fp iff the multiplicative group {1, 2, . . . , p - 1} has order divisible by 4, iff p  1 mod 4. Thus we see again that an = odd prime p remains prime in Z + Zi iff p  3 mod 4. Note that the odd primes which don't remain prime all &quot;split&quot; as products of two distinct primes (p) = (a + bi)(a - bi). What happens to the prime 2? Well, 2 = -i(1 + i)2 where -i is a unit and 1 + i is a prime. Thus the ideal (2) = (1 + i)2 decomposes as the square of a prime ideal. This phenomenon, where a Z-prime is contained in a higher power of a prime ideal, is called ramification. As already noted, 2 is the only prime that ramifies in Z + Zi. We will see later that in general the set of ramified primes is finite, and how to determine them. Recall the construction known as the field of fractions of a domain. The field of fractions of Z is Q. The field of fractions of Z+Zi is Q+Qi (because you can always rationalize the denominator).  Let's look now at the ring Z + Z -3. One might also be interested in the ring generated by a primitive cube root of unity 3 . Its minimal polynomial (the monic polynomial in Q[x] of smallest possible degree 2 for which 3 is a zero) is x2 + x + 1, so 3 = -1 - 3 , and the ring  -1± -3 generated by 3 is Z + Z3 . Since 3 = , one finds that Z + 2  Z -3  Z + Z3 . Now we have two interesting rings inside Q + Q3 . So, which is the most natural one to take as &quot;the ring of integers&quot;, in analogy with Z inside Q? The largest possible ring (if it exists)? Notice that all the elements we've discussed have minimal polynomials with integer coeffecients. Maybe we could take the set of all such elements ... but is that a ring?  Next let's consider the ring Z+Z -5, which is the textbook example of a &quot;non-UFD&quot; because   3 × 2 = (1 + -5) × (1 - -5) , where all four elements are irreducible since there are no elements of  norm 2 or 3. Another example: Z + Z -3 is not a UFD, even though Z + Z3 is a UFD and even a PID. The issue of whether a given ring is a PID, or to what extent it fails to be (as measured by the &quot;ideal class group&quot;) raises many open questions in number theory. One of my favourite ironies is that we have a precise conjecture for the proportion  of primes p &gt; 0, p  3 mod 4 such that the real quadratic field Z+Z p is a PID (approx 77% of them, counting primes in increasing order);  however we can't even prove that Z + Z p is a PID for infinitely many primes. For imaginary quadratic fields, in contrast, it is known that  the only values of d &gt; 0 for which Z + Z -d is a PID are 1 and 2. It is straightforward to see that, for many values of d, it is not a PID,  by observing that the norm N (a + b -d) = a2 + db2 takes no prime  values less than d. That means that if Z + Z -d were a PID, then all  the primes less than d would have to remain prime in Z + Z -d. But, using reasoning we used before, p remains prime iff x2 + d has no root4STEVE DONNELLYin Fp , equivalently -d = -1. For example, 3 does not remain prime p whenever -d  1 mod 3; so in these cases there must be some other  prime ideal of Z + Z -d containing 3, which cannot be principal. A heuristic calculation suggests that beyond some finite bound, no values of d will satisfy -d = -1 for all primes less than d, implying that p  there are only finitely many d for which Z + Z -d is a PID. Cyclotomic rings and fields. Let p be prime, and let p be a primitive pth root of unity. The minimal polynomial of p isp-1p (x) := xp-1 + · · · + x + 1 =k=1k x - p ,p-2 so Z + Zp + · · · + Zp is a ring which can be identified with Z[x]/(p ) and is denoted Z[p ]. Similarly, Q[p ] := Q[x]/(p )  Q + Qp + · · · + = p-2 Qp is a field (using that p is irreducible). If you know some Galois theory, you'll realize that the extension of fields Q[p ]  Q is Galois and that Q[p ] and Z[p ] possess p - 1 automorphisms sending p to the other primitive pth roots of unity. We'll construct these automorphisms by hand, without Galois theory. For 1  k  p - 1, consider the ring homomorphism k : Z[x]  Z[x] that sends x to xk . One checks that k maps the ideal p Z[x] into itself by calculating(p ) =xpk - 1 (xp - 1)Z[x] p (x)Z[x]  = k-1 , k -1 k -1 x x x + ··· + 1and noting that p (x) and xk-1 +· · ·+1 are relatively prime elements of Z[x] (meaning they generate the unit ideal). Therefore k gives a well defined ring homomorphism from Z[x]/(p ) to itself. Moreover, if kl  1 mod p then k and l are inverse to each other; in particular all the k are automorphisms of Z[p ] (and they fix Z, as any automorphism must). They also induce automorphisms of Q[p ] (fixing Q). We can deduce from this that p is irreducible (which we haven't used in this paragraph). For let f be the minimal polynomial of p , which is one of the irreducible factors of p . The coefficients of f , because they are k rational, are fixed by each k . Therefore each k (p ) = p also satisfies f , which means f is all of p . Define a norm map N : Z[p ]  Z byp-1N () =k=1k () .To see that the image really lies in Z, note that (1) for any , N () is fixed by every k , and (2) the set of elements of Z[p ] that are fixed by every k is Z.ALGEBRAIC NUMBER THEORY5The argument given above is more conceptual than the usual proof that p (x) is irreducible, in which one proves (equivalently) that the polynomialp-1 k x - (1 - p ) = p (1 - x) k=1is irreducible. The constant coefficient is p (1 - 0) = p (1) = p, and it's an elementary exercise to show the other coefficients (except the leading coefficient) are also divisible by p; hence the polynomial is irreducible according to Eisenstein's criterion.1 This argument draws attention to two noteworthy features of Z[p ]. The first is that 1 - p , because it has norm p, generates a prime ideal with residue field Fp k (exercise: prove this carefully). Moreover for any k, 1 - p = (1 - k-1 k p )(1 + · · · + p )  (1 - p )Z[p ] and reversing the roles of p and p k k we get 1-p  (1-p )Z[p ]. Hence the ideals (1-p ) for 1  k  p-1 are all the same ideal, and so the ideal (p) = (1 - p )p-1 (we say (p) is &quot;totally ramified&quot;). We'll see later that p is the only ramified prime. This explains why the coefficients of p (1 - x) are divisible by p: they are in the ideal generated by the roots, which is (1 - p ), and they are integers. The second interesting feature, which follows from what we've k-1 said above, is that 1 + · · · + p is a unit for 1  k &lt; p. Note that each such unit is the product of a root of unity in Z[p ] and a unit of the &quot;real subring&quot; Z[p ]  R, and that the same group UR is generated using only 2  k  p-1 . UR turns out to be free abelian of rank p-1 -1. 2 2 By a standard theorem that determines how many units there are in rings of &quot;algebraic integers&quot; (which we won't quite have time to prove in this course), UR has finite index inside the group of all units of Z[p ] (in particular, up to finite index all units are real). Further, a result special to cyclotomic rings states that the group generated UR and the obvious roots of unity equals the whole group of units of Z[p ] exactly when Z[p ] is a PID.Homework: Read through these notes, and make sure you can give solid proofs of all assertions beginning with &quot;Note that ... &quot;, as well as any other assertions for which an argument is given (but not given in full detail). Ask next time or by email if there are any details that you can't clarify. (These examples will guide us as we develop the subject abstractly, starting next time.)says that if a monic polynomial has all coefficients divisible by a prime p, and the constant coefficient not divisible by p2 , then it is irreducible.1Eisenstein's criterion6STEVE DONNELLYLecture Two: Number fields and rings of integers A number field is a field obtained from Q by adjoining a root of an irreducible polynomial. Let's review the construction known as &quot;adjoining the root of a polynomial&quot;. Proposition 2.1. . Let F be any field. For an irreducible element f  F [x], the ring F [x]/(f ) is a field. Proof: Exercise. As a consequence of the proposition, given any ring R  F containing the coefficients of f , the ring R[x]/(f ) is an integral domain, because you can define an injective ring homomorphism R[x]/(f )  F [x]/(f ). Notation. Q[] and Z[]. When  is a root of f in some given ring R (such as C), then Z[] denotes the smallest subring of R that contains Z and . Otherwise, when  does not explicitly denote an element of a previously defined ring, then Z[] is simply a name for the ring Z[x]/(f ), and  is a name for the image of x in Z[x]/(f ). We don't automatically identify Z[x]/(f ) with a subring of R or C. The same comments apply to Q[]. When Q[] is a field we also write Q(). Example. Adjoining the square root of an integer d  Z.   Q[ d] := Q[x]/(x2 - d), Z[ d] := Z[x]/(x2 - d) Definition. An algebraic number field is a field of the form Q[x]/(f ) for some irreducible element f  Q[x]. Recall that given an extension of fields K  L, also denoted L/K, L is a vector space over K and the dimension is called the degree of L over K, deg(L/K). Proposition 2.2. (&quot;Theorem of the primitive element.&quot;) A number field is the same thing as a finite extension of Q. Proof. Clearly every element in a finite extension K/Q is algebraic (satisfies a polynomial in Q[x]), so K = Q(1 , 2 , . . . , k ). It suffices to show that the field generated by any two elements ,   K can be generated by a single element (and then apply induction). We'll use the fact from Galois theory that a finite extension K/Q has only finitely many intermediate subfields Q  F  K. Consider the subfields Q( + q) for q  Q. Since there are only finitely many subfields, we must have Q( + q1 ) = Q( + q2 ) for some q1 = q2 . But then that field must contain  and , so Q(, ) = Q( + q1 ) is generated by a single element. In particular, a finite extension is algebraic, so every element satisfies a polynomial with coefficients in the ground field.ALGEBRAIC NUMBER THEORY7Definition. For an element  in a number field K, the minimal polynomial of  is the monic polynomial in Q[x] satisfied by  of smallest possible degree. The minimal polynomial is uniquely determined: it is determined up to a unit as a generator of the ideal in Q[x] of polynomials satisfied by , and there is a unique monic generator since the units in Q[x] are Q× . Check that the minimal polynomial of any element is irreducible. In these notes K will always denote a number field. Algebraic integers. Inside any number field, we wish to define a &quot;ring of integers&quot;, generalising Z inside Q. Definition. Let K be a number field. An element   K is an algebraic integer if its minimal polynomial has coefficients in Z. Example. The golden number 1+2 minimal polynomial is x2 - x - 1. 5is an algebraic integer, since itsThe next lemma tells us that a number is an algebraic integer iff it satisfies some monic polynomial in Z[x] (the polynomial doesn't need to be irreducible). Lemma 2.3 (Gauss' Lemma). Let f  Z[x] be monic and suppose f = gh for monic g, h  Q[x]. Then g, h  Z[x]. Proof. See any general purpose algebra book. We want to show that sums and products of algebraic integers are again algebraic integers. It is too awkward to work with the minimal polynomial of a sum or product, so we take a more abstract approach, and use the &quot;finiteness criterion&quot; for integrality given in the next lemma. The intuition here is that if  is not an integer, n will have larger and larger denominators as n grows, and n won't be an integer combination of smaller powers of . In other words the ring Z[] generated by  will not be finitely generated as a Z-module. For instance 1 a Z = { n : a  Z, n  0} 2 21 whereas the Z-module generated by any finite subset of Z[ 2 ] has bounded denominators.Lemma 2.4.  is an algebraic integer iff Z[] is finitely generated as a Z-module. Proof: Exercise. Corollary 2.5. The set of algebraic integers inside a number field K forms a ring, which we denote OK .8STEVE DONNELLYProof. We must show the set is closed under addition and multiplication. Suppose  and  are integers in K, so Z[] and Z[] are finitely generated Z-modules, say with generators {i : i  I} and {j : j  J} respectively. Then the ring Z[, ] generated by  and , is generated as a Z-module by the finite set {i j : (i, j)  I × J}. Recall that submodules of finitely generated Z-modules are themselves finitely generated. Since Z[, ] contains Z[ + ] and Z[], they are also finitely generated Z-modules, which shows that  +  and  are algebraic integers. Corollary 2.6. Let Q denote an algebraic closure of Q. The set of algebraic integers in Q forms a ring. Note that every element of Q is algebraic (satisfies a polynomial in Q[x]), so Q is the union of the number fields it contains. Proof of the corollary. For any algebraic integers  and  in Q, the field Q[, ] which they generate is a number field by Prop 2.2. Hence  +  and  are algebraic integers by the previous corollary. Proposition 2.7. For any number field K, OK is finitely generated as a Z-module. Hence OK = Z1 +· · ·+Zn for some algebraic integers i . Proof: Exercise. Real and complex embeddings, trace and norm. Let K = Q[x]/(f ) be a number field, with f monic. Note that since f is irreducible, it has distinct roots (over any field containing Q). By the fundamental theorem of algebra, f = (x - 1 ) . . . (x - r )(x - 1 )(x - 1 ) . . . (x - s )(x - s ) where 1 . . . r are the real roots and i , i are the conjugate pairs of complex roots. So we get r embeddings (injective homomorphisms of fields) K  R, sending x to i , and additionally s pairs of embeddings K  C. They are well defined homomorphisms from Q[x]/(f ) because the image of f is f (i ) = 0. Note that any nonzero homomorphism of fields is injective. Conversely there are no other embeddings K  C, since any homomorphism from Q[x] to any ring R that sends f to 0 must send x to a root of f in R. Definition. Trace and norm. For   K let T rK/Q () :=:K C() , () .:K CNK/Q () :=Note that for ,   K, T rK/Q ( + ) = T rK/Q () + T rK/Q (), and NK/Q ( + ) = NK/Q ()NK/Q () .ALGEBRAIC NUMBER THEORY9Proposition 2.8. Suppose K = Q() has degree d, and let the minimal polynomial of  be xd + a1 xd-1 + · · · + ad . Then T rK/Q () = -a1 and NK/Q () = (-1)d ad . Proof: By definition. In particular, the trace and norm lie in Q in this situation. We now look at the situation where Q() is a proper subfield of K. In particular for a  Q, it's clear that T rK/Q (a) = da and NK/Q (a) = ad , where d = deg(K/Q). Proposition 2.9. Suppose   K. Let M = Q(), so Q  M  K, and let d = deg(M/K). Then T rK/Q () = dT rM/Q () and NK/Q () = NM/Q ()d Proof. By Prop 2.2, K = Q() for some beta. Now let m,M  M [x] be the minimal polynomial of  over M , which has degree d. The embedddings K  C can be described as follows. For each embedding  : M  C, the polynomial  m,M  C[x] obtained by applying  to the coefficients of m,M is irreducible over (M ), since  is an isomorphism of fields M  (M ). The embeddings K  C extending  must = send  to a root of m,M , so there are precisely d such embeddings. The proposition follows. Corollary 2.10. For any   K, T rK/Q () and NK/Q () belong to Q. If   OK , they belong to Z. Proof. Use the previous two propositions. For the second statement, since  is integral, each () is integral, and together they generate a number field L  C. Since the algebric integers in L form a ring, the trace and norm are algebraic integers lying in Q, and so they belong to Z. Homework Set One. Core Problems. Worth 40 points. Due Tuesday April 19 by 2pm. (1) Prove Prop 2.1. (2) For any integral element  in a number field K = Q[x]/(f ) and any prime number p, show that the &quot;residue ring&quot; Z[]/(p) is  isomorphic to Fp [x]/(f ). With K = Q[ 3 2] give an example of  and p for which Z[]/(p) is a field; give another example for which it is not a field and express it as a direct sum of fields. (3) Prove Lemma 2.4 and Prop 2.7.  (4) Find the ring of integers in Q[ 2d] for an odd integer d, and in Q[x]/(x3 - x  1). - (5) Let K = Q( m, n) where m, n are coprime, odd integers. What are the subfields of K? For which values of m, n is OK =   Z[ m, n]? For which values does OK have odd discriminant?10STEVE DONNELLY(6) The theory we're studying started its life around the 1840s with the work of Kummer, in the hope of proving Fermat's Last Theorem. For an odd prime p, suppose there are coprime x, y, z  Z satisfyingp-1 z p = xp + y p = (x + y)(x + p y) . . . (x + p y) .Show that the factors on the right are &quot;nearly&quot; coprime to eachother in Z[p ], by identifying the possibilities for the ideal generated by two of them. Deduce that if Z[p ] is a UFD, each factor is &quot;nearly&quot; a pth power. Problems for discussion (in 2 weeks time). Worth 20 points, no deadline for solutions. What's expected here? At a minimum (for 10 points), you should think about some of the problems, try out several ideas/examples, and come to our discussion session ready to explain where you got stuck. (1) (15 points.) Let K = Q(p ) for an odd prime p. Recall that (1 - p )p-1 Z[p ] = pZ[p ]. (a) Show that the discriminant of Z[p ] is a power of p. (b) Show that (1 - p )OK is a prime ideal. (c) Show that OK = Z[p ]. (Hint: Think in Z[1 - p ] instead. Suppose  = a0 +a1 (1-p )+· · ·+ap-1 (1-p )p-1 is integral, and consider powers of 1 - p .) You can do (b) and (c) in either order. (2) (25 points.) Investigate whether there exists a number field K for which OK is not equal to Z[] for any   K. Representations and integrality. Proposition 2.11. Let G be a finite group and let  be the character associated to a representation of G over C. Then for each g  G, (g) is an algebraic integer. Proof. Let n be the order of g. The representation associates to g a matrix M which must satisfy M d = I; in particular all eigenvalues of M must satisfy  d = 1. By Gauss' Lemma 2.3, they are algebraic integers in C. The set of all algebraic integers in C forms a ring. Therefore (g), which is the sum of the eigenvalues, is also an algebraic integer. The following proposition was stated without proof by Michael, and used in showing that the dimension of an irreducible representation over C divides the group order. Proposition 2.12. Suppose u = g u(g)g  C[G], where the coefficients u(g) are algebraic integers. Then u is integral over Z, in other words u satisfies a monic polynomial in Z[x].ALGEBRAIC NUMBER THEORY11Proof. We can work in OK [G], where K  C is a number field containing u(g) for all g. As a Z-module, OK [G] is generated by the finite set {i g : 1  i  n, g  G} where 1 , . . . , n is an integral basis of OK . Let M be the Z-submodule of OK [G] generated by the powers of u. Since any submodule of a finitely generated Z-module is also finitely generated, M must be generated by {1, u, u2 , . . . , uk } for some k. But uk+1 is also in M , so uk+1 = a0 + a1 u + . . . ak uk with ai  Z, showing that u satisfies a monic polynomial in Z[x]. Lecture Three: Discriminants We would like to be able to easily determine the ring of integers in a number field. The discriminant will be a helpful tool. Definition. Let K be a number field of degree n, and let 1 , . . . n denote the embeddings K  C. We define the discriminant of a basis 1 , . . . n of K (as a vector space over Q) to be 1 (1 ) . . . 1 (n ) . . . . disc(1 , . . . n ) := . . n (1 ) . . . n (n )2= det T rK/Q (i j )1i,jnThe two formulas for the discriminant are equal because both of them are equal to 1 (1 ) . . . . . . 1 (n ) . . . n (1 ) 1 (1 ) . . . . . . . . . n (n ) n (1 ) . . . 1 (n ) . . . n (n )Where does this crazy determinant come from? The ith column is the image of i under the map (1 , . . . , n ) : K  Cn . This map is injective (indeed each embedding individually is injective). We will show later that disc(1 , . . . n ) is nonzero, which means that the images of 1 , . . . , n under this map are C-linearly independent. It's unnecessary to use all the embeddings when some of them occur in conjugate pairs: one from each pair is enough. So it's nicer instead to map the K into Rr × Cs  Rn using all r real embeddings of K and one = embedding chosen from each pair of complex conjugate embeddings. The images of 1 , . . . , n are linearly independent over R. (Otherwise, if there were real numbers rj such that r1 i (1 ) + · · · + rn i (n ) for all the chosen embeddings i , then the same relation would hold for the other embeddings too, simply by taking conjugates. But this would contradict the fact that disc(1 , . . . n ) is nonzero.) Therefore the image is a lattice in Rn (in other words, a copy of Zn  Rn that spans Rn ). For any generating set b1 , . . . , bn of a lattice in Rn , the associated &quot;fundamental domain&quot; is the set {1 b1 + · · · + n bn : 0  i  1}  Rn . Its volume, up to sign, is the determinant of the matrix of basis vectors bi . In our situation, when all the embeddings are real, the first matrix12STEVE DONNELLYappearing in the definition above is the matrix of basis vectors. In general there is a slight change of basis involved, and one has |disc(1 , . . . n )| = 22s V ol(Z1 + · · · + Zn )2 . Using the second formula in the definition of the discriminant, we see that the discriminant of any basis 1 , . . . n of K lies in Q, and the discriminant of an integral basic (a basis consisting of algebraic integers) lies in Z. This means that the discriminant gives us information about the ring of integers, as we will see next. First we prove a fact about Z-modules which is useful in many situations. Lemma 3.13. Suppose  is a homomorphism of Z-modules from Zn to itself, given by a matrix M . Then | det M | equals the index [Zn : (Zn )]. Proof. View Zn  Rn . Let H be the &quot;basic hypercube&quot; in Rn , in other words H = {a1 e1 + · · · + an en  Rn : 0  ai &lt; 1} where the ei 's denote the standard basis of Rn . Now,  : Zn  Zn extends to a linear ~ transformation  : Rn  Rn , also given by M . Since (Zn ) has finite n ~ index in Z , (Zn ) spans Rn , which implies that  is surjective. Note n n ~ that H is a set of coset representatives for R /Z , and hence (H) is a set of coset representatives for Rn /(Zn ). Since Zn /(Zn ) has order ~ m, this implies V ol((H)) = mV ol(H) = m. On the other hand, ~ (H) = {a1 (e1 ) + · · · + an (en )  Rn : 0  ai &lt; 1}, where the vectors ~ (ei ) are the columns of M , so V ol((H)) = | det(M )|. Proposition 3.14. Suppose L = Z1 + · · · + Zn is a submodule of index m in L = Z1 + · · · + Zn . Then disc(1 , . . . n ) = m2 disc(1 , . . . , n ) . Proof. Let  : L  L be the map sending i to i for each i. Thus (L ) = L, which has index m in L . Therefore using the previous lemma, the associated matrix M  Mn (Z) has determinant m. Since [1 , . . . , n ]M = [1 , . . . , n ] and M has integer entries, we have     1 (1 ) . . . 1 (n ) 1 (1 ) . . . 1 (n ) .  . M =  .  . . . . . . . . . n (1 ) . . . n (n ) n (1 ) . . . n (n ) and the proposition follows. Recall Prop 2.7, that OK = Z1 + · · · + Zn for some integral basis. Corollary 3.15. Suppose 1 , . . . n is an integral basis with discriminant D. Let d be the largest integer with d2 | D. Then n 1 OK  Z + · · · + Z . d d Proof. Using Prop 3.14, the order of OK /Z1 + · · · + Zn divides d. Hence dOK  Z1 + · · · + Zn .ALGEBRAIC NUMBER THEORY13 Example. Rings of  integers in K = Q( d)  odd d. for We calculate disc(1, d) = 4d. Therefore Z[ d] is either equal to OK  or is a submodule of index 2 in OK , and OK  1 Z[ d]. Considering 2   the quotient of abelian groups 1 Z[ d]/Z[ d]  Z/2 × Z/2, we see =  2 1 there are three possibilities for OK . Clearly 2 and 2d are not integral.  On the other hand we calculate that the minimal polynomial of 1+2 d  is x2 - x - d-1 . Therefore OK equals Z[ 1+2 d ] for d  1 mod 4, and 4  Z[ d] for d  3 mod 4. Definitions. The discriminant of   K is disc(1, , . . . , deg(K/Q)-1 ). The discriminant of K, or of OK is the discriminant of an integral basis that generates OK as a Z-module. Proposition 3.16. The discriminant of an irreducible polynomial equals the discriminant of one of its roots. Proof. Suppose f (x) = (x - 1 ) . . . (x - n ) is irreducible. Then  n-1 2 1 1 . . . 1 .  = . . . disc(1 ) = det  . . (i - j )2 = disc(f ) . . . n-1 1i&lt;jn 1 n . . . n using the &quot;Vandemonde determinant&quot;. Corollary 3.17. The discriminant of any basis of K over Q is nonzero. Proof. The discriminant of a basis of the form 1, , . . . deg(K/Q)-1 is nonzero by the previous proposition, since an irreducible polynomial over Q has distinct roots. The discriminants of any two bases differ by a nonzero square factor, as a consequence of Prop 3.14.Office 126, Research 1, phone 3183 E-mail address: [email protected]`

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