ALGEBRAIC NUMBER THEORY - COURSE NOTES

ALGEBRAIC NUMBER THEORY - COURSE NOTES STEVE DONNELLY Housekeeping (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 On the course material: Useful books in the library Stewart & 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 (QA241S85) 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 x 2 + 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

2 STEVE DONNELLY Lecture 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) = a 2 + b 2 = 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 α, β Z + Zi we can find q, r Z + Zi such that α = qβ + r with N(r) < N(β) as follows Notice that the ideal (Z + Zi)β is a square lattice and r can be chosen inside a central square {aβ + biβ : 1 2 a, b 1 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(αβ) = N(α)N(β) for all α, β 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 α is irreducible, and suppose ab is in the ideal (α) Writing a and b as products of irreducibles, we have a 1 a n (α) Since α is irreducible, we must have α = ua i for one of the a i and some unit u Thus a i (α), which means a or b is in (α); this shows (α) 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 αβ = 2 + i, it would follow that N(α)N(β) = 5 and hence either α or β is a unit Similarly, if one had a nontrivial decomposition αβ = 7, then N(α) and N(β) are ±7 But one can t write 7, or 7, in the form a 2 + b 2 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 (α) of Z+Zi is prime iff the residue ring Z+Zi/α(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 F 5 = 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) = F 7 [x]/(x 2 + 1) is a field because x 2 + 1 is irreducible (has no root) over F 7 Indeed,

ALGEBRAIC NUMBER THEORY 3 x 2 + 1 has a root over F p 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 split 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 ζ 3 Its minimal polynomial (the monic polynomial in Q[x] of smallest possible degree for which ζ 3 is a zero) is x 2 + x + 1, so ζ3 2 = 1 ζ 3, and the ring generated by ζ 3 is Z + Zζ 3 Since ζ 3 = 1± 3, one finds that Z + 2 Z 3 Z + Zζ 3 Now we have two interesting rings inside Q + Qζ 3 So, which is the most natural one to take as the ring of integers, 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 non-ufd 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 + Zζ 3 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 ideal class group ) 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 > 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 > 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) = a 2 + db 2 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 x 2 + d has no root

4 STEVE DONNELLY ( in F p, equivalently d p ) = 1 For example, 3 does not remain prime 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 p ) = 1 for all primes less than d, implying that there are only finitely many d for which Z + Z d is a PID Cyclotomic rings and fields Let p be prime, and let ζ p be a primitive pth root of unity The minimal polynomial of ζ p is p 1 ( ) Φ p (x) := x p 1 + + x + 1 = x ζ k p, so Z+Zζ p + +Zζp p 2 is a ring which can be identified with Z[x]/(Φ p ) and is denoted Z[ζ p ] Similarly, Q[ζ p ] := Q[x]/(Φ p ) = Q + Qζ p + + Qζ p 2 p is a field (using that Φ p is irreducible) If you know some Galois theory, you ll realize that the extension of fields Q[ζ p ] Q is Galois and that Q[ζ p ] and Z[ζ p ] possess p 1 automorphisms sending ζ 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 σ k : Z[x] Z[x] that sends x to x k One checks that σ k maps the ideal Φ p Z[x] into itself by calculating σ(φ p ) = xpk 1 x k 1 (xp 1)Z[x] x k 1 k=1 = Φ p(x)z[x] x k 1 + + 1, and noting that Φ p (x) and x k 1 + +1 are relatively prime elements of Z[x] (meaning they generate the unit ideal) Therefore σ k gives a well defined ring homomorphism from Z[x]/(Φ p ) to itself Moreover, if kl 1 mod p then σ k and σ l are inverse to each other; in particular all the σ k are automorphisms of Z[ζ p ] (and they fix Z, as any automorphism must) They also induce automorphisms of Q[ζ p ] (fixing Q) We can deduce from this that Φ p is irreducible (which we haven t used in this paragraph) For let f be the minimal polynomial of ζ p, which is one of the irreducible factors of Φ p The coefficients of f, because they are rational, are fixed by each σ k Therefore each σ k (ζ p ) = ζ k p also satisfies f, which means f is all of Φ p Define a norm map N : Z[ζ p ] Z by p 1 N(α) = σ k (α) k=1 To see that the image really lies in Z, note that (1) for any α, N(α) is fixed by every σ k, and (2) the set of elements of Z[ζ p ] that are fixed by every σ k is Z

ALGEBRAIC NUMBER THEORY 5 The argument given above is more conceptual than the usual proof that Φ p (x) is irreducible, in which one proves (equivalently) that the polynomial p 1 ( x (1 ζ k p ) ) = Φ p (1 x) k=1 is irreducible The constant coefficient is Φ p (1 0) = Φ 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[ζ p ] The first is that 1 ζ p, because it has norm p, generates a prime ideal with residue field F p (exercise: prove this carefully) Moreover for any k, 1 ζp k = (1 ζ p )(1 + + ζp k 1 ) (1 ζ p )Z[ζ p ] and reversing the roles of ζ p and ζp k we get 1 ζ p (1 ζp k )Z[ζ p ] Hence the ideals (1 ζp k ) for 1 k p 1 are all the same ideal, and so the ideal (p) = (1 ζ p ) p 1 (we say (p) is totally ramified ) We ll see later that p is the only ramified prime This explains why the coefficients of Φ p (1 x) are divisible by p: they are in the ideal generated by the roots, which is (1 ζ p ), and they are integers The second interesting feature, which follows from what we ve said above, is that 1 + + ζp k 1 is a unit for 1 k < p Note that each such unit is the product of a root of unity in Z[ζ p ] and a unit of the real subring Z[ζ p ] R, and that the same group U R is generated using only 2 k p 1 U 2 R turns out to be free abelian of rank p 1 1 2 By a standard theorem that determines how many units there are in rings of algebraic integers (which we won t quite have time to prove in this course), U R has finite index inside the group of all units of Z[ζ p ] (in particular, up to finite index all units are real) Further, a result special to cyclotomic rings states that the group generated U R and the obvious roots of unity equals the whole group of units of Z[ζ p ] exactly when Z[ζ p ] is a PID Homework: Read through these notes, and make sure you can give solid proofs of all assertions beginning with Note that, 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) 1 Eisenstein s criterion says that if a monic polynomial has all coefficients divisible by a prime p, and the constant coefficient not divisible by p 2, then it is irreducible

6 STEVE DONNELLY Lecture 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 adjoining the root of a polynomial Proposition 21 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[α] and Z[α] When α is a root of f in some given ring R (such as C), then Z[α] denotes the smallest subring of R that contains Z and α Otherwise, when α does not explicitly denote an element of a previously defined ring, then Z[α] is simply a name for the ring Z[x]/(f), and α 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[α] When Q[α] is a field we also write Q(α) Example Adjoining the square root of an integer d Z Q[ d] := Q[x]/(x 2 d), Z[ d] := Z[x]/(x 2 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 22 ( Theorem of the primitive element ) 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(α 1, α 2,, α k ) It suffices to show that the field generated by any two elements α, β 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(α + qβ) for q Q Since there are only finitely many subfields, we must have Q(α + q 1 β) = Q(α + q 2 β) for some q 1 q 2 But then that field must contain α and β, so Q(α, β) = Q(α + q 1 β) 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 THEORY 7 Definition For an element β in a number field K, the minimal polynomial of β is the monic polynomial in Q[x] satisfied by β 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 β, 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 ring of integers, generalising Z inside Q Definition Let K be a number field An element α K is an algebraic integer if its minimal polynomial has coefficients in Z Example The golden number 1+ 5 2 is an algebraic integer, since its minimal polynomial is x 2 x 1 The 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 23 (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 finiteness criterion for integrality given in the next lemma The intuition here is that if α is not an integer, α n will have larger and larger denominators as n grows, and α n won t be an integer combination of smaller powers of α In other words the ring Z[α] generated by α will not be finitely generated as a Z-module For instance [ ] 1 Z = { a : a Z, n 0} 2 2n whereas the Z-module generated by any finite subset of Z[ 1 ] has bounded 2 denominators Lemma 24 α is an algebraic integer iff Z[α] is finitely generated as a Z-module Proof: Exercise Corollary 25 The set of algebraic integers inside a number field K forms a ring, which we denote O K

8 STEVE DONNELLY Proof We must show the set is closed under addition and multiplication Suppose α and β are integers in K, so Z[α] and Z[β] are finitely generated Z-modules, say with generators {α i : i I} and {β j : j J} respectively Then the ring Z[α, β] generated by α and β, is generated as a Z-module by the finite set {α i β j : (i, j) I J} Recall that submodules of finitely generated Z-modules are themselves finitely generated Since Z[α, β] contains Z[α+β] and Z[αβ], they are also finitely generated Z-modules, which shows that α + β and αβ are algebraic integers Corollary 26 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 α and β in Q, the field Q[α, β] which they generate is a number field by Prop 22 Hence α + β and αβ are algebraic integers by the previous corollary Proposition 27 For any number field K, O K is finitely generated as a Z-module Hence O K = Zα 1 + +Zα n for some algebraic integers α 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 α 1 ) (x α r )(x β 1 )(x β 1 ) (x β s )(x β s ) where α 1 α r are the real roots and β i, β i are the conjugate pairs of complex roots So we get r embeddings (injective homomorphisms of fields) K R, sending x to α 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(α 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 α K let T r K/Q (α) := σ(α), N K/Q (α) := σ:k C σ:k C σ(α) Note that for α, β K, T r K/Q (α + β) = T r K/Q (α) + T r K/Q (β), N K/Q (α + β) = N K/Q (α)n K/Q (β) and

ALGEBRAIC NUMBER THEORY 9 Proposition 28 Suppose K = Q(α) has degree d, and let the minimal polynomial of α be x d + a 1 x d 1 + + a d Then T r K/Q (α) = a 1 and N K/Q (α) = ( 1) d a d Proof: By definition In particular, the trace and norm lie in Q in this situation We now look at the situation where Q(α) is a proper subfield of K In particular for a Q, it s clear that T r K/Q (a) = da and N K/Q (a) = a d, where d = deg(k/q) Proposition 29 Suppose α K Let M = Q(α), so Q M K, and let d = deg(m/k) Then T r K/Q (α) = dt r M/Q (α) and N K/Q (α) = N M/Q (α) d Proof By Prop 22, K = Q(β) for some beta Now let m β,m M[x] be the minimal polynomial of β over M, which has degree d The embedddings K C can be described as follows For each embedding σ : M C, the polynomial σ m β,m C[x] obtained by applying σ to the coefficients of m β,m is irreducible over σ(m), since σ is an isomorphism of fields M = σ(m) The embeddings K C extending σ must send β to a root of m β,m, so there are precisely d such embeddings The proposition follows Corollary 210 For any α K, T r K/Q (α) and N K/Q (α) belong to Q If α O K, they belong to Z Proof Use the previous two propositions For the second statement, since α is integral, each σ(α) 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 21 (2) For any integral element α in a number field K = Q[x]/(f) and any prime number p, show that the residue ring Z[α]/(p) is isomorphic to F p [x]/(f) With K = Q[ 3 2] give an example of α and p for which Z[α]/(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 24 and Prop 27 (4) Find the ring of integers in Q[ 2d] for an odd integer d, and in Q[x]/(x 3 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 O K = Z[ m, n]? For which values does O K have odd discriminant?

10 STEVE 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 satisfying z p = x p + y p = (x + y)(x + ζ p y) (x + ζ p 1 p y) Show that the factors on the right are nearly coprime to eachother in Z[ζ p ], by identifying the possibilities for the ideal generated by two of them Deduce that if Z[ζ p ] is a UFD, each factor is nearly 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(ζ p ) for an odd prime p Recall that (1 ζ p ) p 1 Z[ζ p ] = pz[ζ p ] (a) Show that the discriminant of Z[ζ p ] is a power of p (b) Show that (1 ζ p )O K is a prime ideal (c) Show that O K = Z[ζ p ] (Hint: Think in Z[1 ζ p ] instead Suppose α = a 0 +a 1 (1 ζ p )+ +a p 1 (1 ζ p ) p 1 is integral, and consider powers of 1 ζ p ) You can do (b) and (c) in either order (2) (25 points) Investigate whether there exists a number field K for which O K is not equal to Z[α] for any α K Representations and integrality Proposition 211 Let G be a finite group and let χ be the character associated to a representation of G over C Then for each g G, χ(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 ω d = 1 By Gauss Lemma 23, they are algebraic integers in C The set of all algebraic integers in C forms a ring Therefore χ(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 212 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 THEORY 11 Proof We can work in O K [G], where K C is a number field containing u(g) for all g As a Z-module, O K [G] is generated by the finite set {α i g : 1 i n, g G} where α 1,, α n is an integral basis of O K Let M be the Z-submodule of O K [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, u 2,, u k } for some k But u k+1 is also in M, so u k+1 = a 0 + a 1 u + a k u k with a i 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 σ 1, σ n denote the embeddings K C We define the discriminant of a basis α 1, α n of K (as a vector space over Q) to be 2 σ 1 (α 1 ) σ 1 (α n ) disc(α 1, α n ) := = det ( T r K/Q (α i α j ) ) 1 i,j n σ n (α 1 ) σ n (α n ) The two formulas for the discriminant are equal because both of them are equal to σ 1 (α 1 ) σ n (α 1 ) σ 1 (α 1 ) σ 1 (α n ) σ 1 (α n ) σ n (α n ) σ n (α 1 ) σ n (α n ) Where does this crazy determinant come from? The ith column is the image of α i under the map (σ 1,, σ n ) : K C n This map is injective (indeed each embedding individually is injective) We will show later that disc(α 1, α n ) is nonzero, which means that the images of α 1,, α 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 R r C s = R n using all r real embeddings of K and one embedding chosen from each pair of complex conjugate embeddings The images of α 1,, α n are linearly independent over R (Otherwise, if there were real numbers r j such that r 1 σ i (α 1 ) + + r n σ i (α n ) for all the chosen embeddings σ i, then the same relation would hold for the other embeddings too, simply by taking conjugates But this would contradict the fact that disc(α 1, α n ) is nonzero) Therefore the image is a lattice in R n (in other words, a copy of Z n R n that spans R n ) For any generating set b 1,, b n of a lattice in R n, the associated fundamental domain is the set {β 1 b 1 + +β n b n : 0 β i 1} R n Its volume, up to sign, is the determinant of the matrix of basis vectors b i In our situation, when all the embeddings are real, the first matrix

12 STEVE DONNELLY appearing in the definition above is the matrix of basis vectors general there is a slight change of basis involved, and one has disc(α 1, α n ) = 2 2s V ol(zα 1 + + Zα n ) 2 Using the second formula in the definition of the discriminant, we see that the discriminant of any basis α 1, α 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 313 Suppose φ is a homomorphism of Z-modules from Z n to itself, given by a matrix M Then det M equals the index [Z n : φ(z n )] Proof View Z n R n Let H be the basic hypercube in R n, in other words H = {a 1 e 1 + + a n e n R n : 0 a i < 1} where the e i s denote the standard basis of R n Now, φ : Z n Z n extends to a linear transformation φ : R n R n, also given by M Since φ(z n ) has finite index in Z n, φ(z n ) spans R n, which implies that φ is surjective Note that H is a set of coset representatives for R n /Z n, and hence φ(h) is a set of coset representatives for R n /φ(z n ) Since Z n /φ(z n ) has order m, this implies V ol( φ(h)) = mv ol(h) = m On the other hand, φ(h) = {a 1 φ(e 1 ) + + a n φ(e n ) R n : 0 a i < 1}, where the vectors φ(e i ) are the columns of M, so V ol( φ(h)) = det(m) Proposition 314 Suppose L = Zα 1 + + Zα n is a submodule of index m in L = Zα 1 + + Zα n Then disc(α 1, α n ) = m 2 disc(α 1,, α n) Proof Let φ : L L be the map sending α i to α i for each i Thus φ(l ) = L, which has index m in L Therefore using the previous lemma, the associated matrix M M n (Z) has determinant m Since [α 1,, α n]m = [α 1,, α n ] and M has integer entries, we have σ 1 (α 1) σ 1 (α n) σ 1 (α 1 ) σ 1 (α n ) M = σ n (α 1) σ n (α n) σ n (α 1 ) σ n (α n ) and the proposition follows Recall Prop 27, that O K = Zα 1 + + Zα n for some integral basis Corollary 315 Suppose α 1, α n is an integral basis with discriminant D Let d be the largest integer with d 2 D Then O K Z α 1 d + + Zα n d Proof Using Prop 314, the order of O K /Zα 1 + + Zα n divides d Hence do K Zα 1 + + Zα n In

ALGEBRAIC NUMBER THEORY 13 Example Rings of integers in K = Q( d) for odd d We calculate disc(1, d) = 4d Therefore Z[ d] is either equal to O K or is a submodule of index 2 in O K, and O K 1 2 Z[ d] Considering the quotient of abelian groups 1 2 Z[ d]/z[ d] = Z/2 Z/2, we see there are three possibilities for O K Clearly 1 and d are not integral 2 2 On the other hand we calculate that the minimal polynomial of 1+ d 2 is x 2 x d 1 Therefore O 4 K equals Z[ 1+ d] for d 1 mod 4, and 2 Z[ d] for d 3 mod 4 Definitions The discriminant of α K is disc(1, α,, α deg(k/q) 1 ) The discriminant of K, or of O K is the discriminant of an integral basis that generates O K as a Z-module Proposition 316 The discriminant of an irreducible polynomial equals the discriminant of one of its roots Proof Suppose f(x) = (x α 1 ) (x α n ) is irreducible Then 1 α 1 α n 1 2 1 disc(α 1 ) = det = (α i α j ) 2 = disc(f) 1 α n αn n 1 1 i<j n using the Vandemonde determinant Corollary 317 The discriminant of any basis of K over Q is nonzero Proof The discriminant of a basis of the form 1, α, α 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 314 Office 126, Research 1, phone 3183 E-mail address: sdonnelly@iu-bremende