THE DIFFERENT IDEAL KEITH CONRAD. Introduction The discriminant of a number field K tells us which primes p in Z ramify in O K : the prime factors of the discriminant. However, the way we have seen how to compute the discriminant doesn t address the following themes: (a) determine which prime ideals in O K ramify (that is, which p in O K have e(p p) > rather than which p have e(p p) > for some p), (b) determine the multiplicity of a prime in the discriminant. (We only know the multiplicity is positive for the ramified primes.) Example.. Let K = Q(α), where α 3 α = 0. The polynomial T 3 T has discriminant 3, which is squarefree, so O K = Z[α] and we can detect how any prime p factors in O K by seeing how T 3 T factors in F p [T ]. Since disc(o K ) = 3, only the prime 3 ramifies. Since T 3 T (T 3)(T 0) mod 3, (3) = pq. One prime over 3 has multiplicity and the other has multiplicity. The discriminant tells us some prime over 3 ramifies, but not which ones ramify. Only q does. The discriminant of K is, by definition, the determinant of the matrix (Tr K/Q (e i e j )), where e,..., e n is any Z-basis of O K. By a finer analysis of the trace, we will construct an ideal in O K which is divisible precisely by the ramified primes in O K. This ideal is called the different ideal. (It is related to differentiation, hence the name I think.) In the case of Example., for instance, we will see that the different ideal is q, so the different singles out the particular prime over 3 that ramifies. While the discriminant lies downstairs in Z, the different lies upstairs in O K. The ideal norm of the different is the absolute value of the discriminant, and this connection between the different and discriminant will tell us something about the multiplicity of primes in the discriminant. So the different ideal gives answers to both (a) and (b) above (only a partial answer in the case of (b)). The main idea needed to construct the different ideal is to do something in number fields that is analogous to the classical notion of a dual lattice in Euclidean space. We will start off in Section describing dual lattices in R n and some of their basic properties. Armed with that intuition, we will meet the analogous construction in number fields in Section 3 and then construct the different ideal in Section 4.. The Z-dual of a lattice in R n In R n, the standard dot product gives a notion of orthogonal complement: when V R n is a subspace, we set V = {w R n : w V } = {w R n : w V = 0}. Then R n = V V, V = V, and V V V V.
KEITH CONRAD A lattice in R n is, by definition, the Z-span of a basis of R n. The standard lattice is Z n, for instance. There is a concept for lattices in R n somewhat like the orthogonal complement of a subspace. Definition.. For a lattice L R n its Z-dual is L = {w R n : w L Z}. This Z-dual of a lattice is not an orthogonal complement. The condition for a vector to lie in the Z-dual of L is to have integral dot product against all elements of L, not to have dot product 0 against all elements of L. Some similarities with properties of orthogonal complements will be seen in Corollary.7. If e,..., e n is a Z-basis of L, then to have w v Z for all v L it suffices to check w e i Z for i =,..., n since every element of L is a Z-linear combination of the e i s. Here are three examples of dual lattices in R. Example.. Let L = Z = Z ( ( 0) + Z 0 ). For w R, w L if and only if w ( 0) Z and w (0 ( ) Z. Writing w = a b), the two dot products are a and b, so L = Z = L. The lattice Z is self-dual. Example.3. Let L = Z ( ( 0) + Z ( ). Then a b) is in L when a Z and a + b Z, which is equivalent to a Z and b (/)Z, so L = Z ( ) ( 0 + Z 0 /). Example.4. Let L = Z ( ) + Z ( 3). To say ( a b) L is equivalent to a + b Z and a + 3b Z. The system of equations a + b = x a + 3b = y is equivalent to so Thus L = Z ( 3/5 /5 a = 3 5 x 5 y b = 5 x + 5 y, ( ) ( ) ( a 3/5 /5 = x + y b /5 /5 ) ( + Z /5 ). /5 Theorem.5. If L = n i= Ze i is a lattice in R n then its Z-dual is L = n i= Ze i, where {e i } is the dual basis to {e i} relative to the dot product on R n : e i e j = δ ij. In particular, L is a lattice. Proof. For w R n, write it in the dual basis {e,..., e n} as w = n i= c ie i. Then w e i = c i, so to say w L is equivalent to the coefficients c i being integers. Therefore L is the Z- span of the e i s. Topologically, the lattices in R n are the subgroups Λ such that Λ is discrete and R n /Λ is compact. The Z-dual is the dual space of L as a Z-module: every Z-linear map L Z has the form v w v for a unique w L. ).
THE DIFFERENT IDEAL 3 Example.6. Let L = Z ( ( 0) + Z ( ), as in Example.3. The dual basis to ( 0) and ) is ( ) ( / and 0 ) /, so by Theorem.5, L = Z ( ) ( / + Z 0 /). If we do a change of basis, replacing ( ) ( / with ) ( / + 0 ( /) = ) ( 0 and keeping 0 /) then we recover the spanning set for L in Example.3. Corollary.7. For lattices in R n, the following properties hold: () L = L, () L L L L, (3) (L + L ) = L L, (4) (L L ) = L + L. Proof. (): Theorem.5 and double duality of vector spaces tells us L = L since the dual basis of a dual basis is the original basis, whose Z-span is the original lattice. (): It is easy to see from the definitions that if L L then L L. Applying this to the inclusion L L gives us L L, so L L. (3): If v (L + L ) then v has integral dot product with any vector in L + L, and hence with any vector in L and any vector in L. Thus v L and v L, so v L L. We have shown (L + L ) L L. The reverse inclusion is just as easy to check. (4): Rather than directly verify that (L L ) = L + L, we will check the two sides are equal by checking their dual lattices are equal. That means, by double duality, we want to check L L = (L + L ). From (3), (L + L ) = L L = L L, so we are done. One property of orthogonal complements that is not shared by dual lattices is V V = R n. The lattices L and L are not complementary in the direct sum sense. Both a lattice and its dual lattice have rank n. 3. Lattices in Number Fields The ideas about lattices in R n will not literally be used, but they are the motivation for what we do now in number fields. Replace R n and its dot product (v, w) v w with a number field K and the operation (x, y) Tr K/Q (xy) on it, which we ll call the trace product. 3 The trace product has values in Q. Instead of being concerned with vectors having a dot product in Z, we will look at algebraic numbers having a trace product in Z. Definition 3.. In a number field K of degree n, a lattice in K is the Z-span of a Q-basis of K. Examples of lattices in K include O K, fractional ideals, and orders. Definition 3.. Let L be a lattice in K. Its dual lattice is L = {α K : Tr K/Q (αl) Z}. As with lattices in Euclidean space, where one can check membership in a dual lattice by checking the dot products with a basis of the lattice are all in Z, to check α L it suffices to check its trace products with a basis of L are all in Z: Tr K/Q (αe i ) Z for some Z-basis e,..., e n of L. 3 The usual term is trace pairing, not trace product. But since we re trying to emphasize the similarity to the dot product on R n, the term trace product seems helpful.
4 KEITH CONRAD Example 3.3. Let K = Q(i) and L = Z[i] = Z + Zi. For a + bi Q(i), a + bi L when Tr Q(i)/Q (a + bi) Z and Tr Q(i)/Q ((a + bi)i) Z. This is equivalent to a Z and b Z, so Z[i] = Z + Zi = Z[i]. Taking L = ( + i)z[i] = Z( + i) + Z( + i), calculations like those in Example.4 (but using trace products instead of dot products) lead to ( L = Z 0 i ) ( +Z 5 5 i ) = Z i 0 0 +Z i = i 0 0 (Z Zi) = ( + i) Z[i]. This calculation shows a relation between the dual lattice and the inverse ideal. Writing a for L = ( + i), an ideal, the calculation of L says a = a. Theorem 3.4. For a number field K and a lattice L K with Z-basis e,..., e n, L = n i= Ze i, where {e i } is the dual basis to {e i} relative to the trace product on K/Q. In particular, L is a lattice. Proof. This is virtually identical to Theorem.5, except for the use of the trace product in place of the dot product. Corollary 3.5. For lattices in K, the following properties hold: () L = L, () L L L L, (3) (L + L ) = L L, (4) (L L ) = L + L, (5) (αl) = α L for α K. Proof. The first four properties are proved in the same way as the proof of Corollary.7, and the last one is left to the reader. Example 3.6. For any quadratic field K = Q( d), with a squarefree integer d, we use Theorem 3.4 to compute L for the lattices Z + Z d and Z + Z + d. (The second lattice isn t a ring unless d mod 4, but it is a lattice for all d, so we can speak of its dual lattice in all cases.) The dual basis of {, d} for K/Q relative to the trace product on K is {, }, so d (Z + Z d) = Z + Z d = d (Z d + Z) = d (Z + Z d). The dual basis of {, + d } relative to the trace product on K is { + a calculation yields (Z + Z + d ) = d (Z + d + Z) = d (Z + Z + d ). d, d d d}, and The next theorem tells us the dual basis of a power basis of K (a basis consisting of powers of a single element). In terms of dual lattices, a tight connection between dual lattices and differentiation is revealed. Theorem 3.7. Let K = Q(α) and let f(t ) be the minimal polynomial of α in Q[T ]. Write f(t ) = (T α)(c 0 (α) + c (α)t + + c n (α)t n ), c i (α) K. The dual basis to {, α,..., α n } relative to the trace product is { c 0(α) f (α), c (α) f (α),..., c n (α) f (α) }. In particular, if K = Q(α) and α O K then (Z + Zα + + Zα n ) = f (α) (Z + Zα + + Zαn ).
THE DIFFERENT IDEAL 5 Proof. Let α,..., α n be the Q-conjugates of α in a splitting field, with α = α. A beautiful polynomial identity of Euler says n f(t ) f =. (α i ) T α i i= Indeed, both sides are polynomials of degree less than n that are equal at n values. By the same argument, n αi k f(t ) f = T k (α i ) T α i i= for 0 k n. Comparing coefficients of like powers of T on both sides, n αi k f (α i ) c j(α i ) = δ jk. i= The left side is Tr K/Q (α k (c j (α)/f (α))), so {c j (α)/f (α)} is the dual basis to {α j } and To show (Z + Zα + + Zα n ) = f (α) (Zc 0(α) + Zc (α) + + Zc n (α)). Zc 0 (α) + Zc (α) + + Zc n (α) = Z + Zα + + Zα n when α O K, we find a formula for c j (α), the coefficient of T j in f(t )/(T α). Letting f(t ) = a 0 + a T + + a n T n + a n T n Z[T ], where a n =, f(t ) T α f(t ) f(α) = T α n T i α i = a i T α = = i= n i= n j=0 α i j T j a i i j=0 n i=j+ a i α i j T j, so c j (α) = n i=j+ a iα i j, whose top term is α n j (since a n = ). A transition matrix from, α,..., α n to c n (α),..., c (α), c 0 (α) is triangular with integral entries and s on the main diagonal, so it is invertible over Z and shows the Z-span of the two sets is the same. Example 3.8. Returning to Example 3.6, we recompute (Z + Zα) for α = d or + d, with f(t ) = T d or T T + d 4, respectively. In the first case f (α) = d, and in the second case f (α) = + d = d. The formula (Z + Zα) = f (α)(z + Zα) from Theorem 3.7 recovers the formulas for the dual lattices in Example 3.6. The most interesting lattice in K is O K. What can we say about O K = {α K : Tr K/Q (αo K ) Z}?
6 KEITH CONRAD First of all, O K is not the elements of K with integral trace. It is smaller than that. To lie in O K, the trace product with all elements of O K must lie in Z. This includes the condition the trace of the element is in Z only as a special case (taking the trace product with ). Since algebraic integers have integral trace, O K O K, so O K always contains O K. We saw earlier that Z[i] = Z[i], for instance. The next theorem says the dual lattice of O K is a fractional ideal that controls the dual lattice of any fractional ideal. Theorem 3.9. For any fractional ideal a in K, a is a fractional ideal and a = a O K. This formula explains a = a from Corollary 3.5 in the special case of fractional ideals. Proof. By definition, a = {α K : Tr K/Q (αa) Z}. First we check a is a fractional ideal. We know it is finitely generated as a Z-module (any dual lattice is a lattice), so the key point is that it is preserved by multiplication by O K. For α a and x O K, xα a since, for any β a, Tr K/Q ((xα)β) = Tr K/Q (α(xβ)) Z, as xβ a and α a. To show a = a O K, pick α a. For any β a, Tr K/Q (αβo K ) Z since βo K a. Therefore αβ O K. Letting β vary in a we get αa O K, so α a O K. Thus a a O K. The reverse inclusion is left to the reader. Theorem 3.0. The dual lattice O K is the largest fractional ideal in K whose elements all have trace in Z. Proof. For a fractional ideal a, a = ao K. Thus Tr K/Q (a) Z if and only if Tr K/Q (ao K ) Z, which is equivalent to a O K. This theorem isn t saying O K is the set of all elements in K with integral trace. It s the largest fractional ideal whose elements have integral trace. The set of all elements with integral trace is an additive group, but it is not a fractional ideal. Example 3.. Since Z[i] = Z[i], any fractional ideal whose elements have integral trace is inside Z[i]. All the elements of Q(i) with integral trace are Z + Qi, which isn t a fractional ideal at all. 4. The Different Ideal The construction of O K provides us with an interesting canonical fractional ideal in K other than O K. Because O K O K, the inverse of O K is a fractional ideal inside O K, hence is an integral ideal. Definition 4.. The different ideal of K is D K = (O K) = {x K : xo K O K }. Example 4.. Since Z[i] = Z[i] by Example 3.3, D Q(i) = Z[i]. Theorem 4.3. If O K = Z[α] then D K = (f (α)), where α has minimal polynomial f(t ) Z[T ]. Proof. Use Theorem 3.7. Example 4.4. For a quadratic field K = Q( d) with squarefree d Z, O K is Z[ d] or Z[( + d)/], depending on d mod 4. Using Example 3.8, { ( d), if d mod 4, (4.) D Q( d) = ( d), if d mod 4.
THE DIFFERENT IDEAL 7 Remark 4.5. It is not true in general that O K = Z[α] for some α, so the different can t be calculated as a principal ideal (f (α)) all the time. However, the different does divide ideals of this type. Specifically, for any α O K, D K (f α(α)) where f α (T ) is the characteristic polynomial of α in Q[T ]. This is automatic if Q(α) K since f α(α) = 0, as f α (T ) is a proper power of the minimal polynomial of α. If Q(α) = K and α O K, then Z[α] is a lattice in K and the inclusion Z[α] O K implies O K Z[α], which is equivalent to D K f α(α) Z[α], so D K f α(α) O K. Inverting ideals, (f α(α)) D K, so D K (f α(α)). If we let α vary then D K divides the ideal generated by f α(α) as α runs over O K. It can be shown that D K is equal to the ideal generated by all f α(α) for α O K. This is in marked contrast to the situation for discriminants: disc(o K /Z) does not always equal the greatest common divisor of all disc(f α (T )) as α ranges over algebraic integers in K. Theorem 4.6. For any number field K, N(D K ) = disc(k). Proof. Let e,..., e n be a Z-basis for O K, so O K = n i= Ze i. Then D K The norm of an ideal is its index in O K, so N(D K ) = [O K : D K ] = 4 [D K : O K] = [O K : O K ]. = O K = n i= Ze i. To compute the index [O K : O K], recall that for finite free Z-modules M M of equal rank, [M : M ] = det(a) where A is a matrix expressing a Z-basis of M in terms of a Z-basis of M. For our application, with M = O K and M = O K, let s write the e i s in terms of the e i s: if e j = n i= a ije i, the meaning of being dual bases relative to the trace product is that a ij = Tr K/Q (e j e i ) = Tr K/Q (e i e j ). Therefore (a ij ) = (Tr K/Q (e i e j )). The determinant of this matrix, by definition, is disc(k), so N(D K ) = disc(k). Theorem 4.6 tells us that in the inclusions of lattices D K O K O K each successive inclusion has index disc(k). In particular, O K is strictly larger than O K if and only if disc(k) >. That inequality holds for all K Q (Minkowski s theorem), so O K is not its own dual lattice when K Q. The different ideal could be considered a measure of how much O K fails to be self-dual as a lattice in K. Example 4.4 and Theorem 4.6 recover the formulas for (the absolute value of) the discriminant of a quadratic field from the norm of the different of a quadratic field: (4.) disc(q( { 4 d, if d mod 4, d)) = N(D Q( d) ) = d, if d mod 4. Theorem 4.6 suggests, from experience with the discriminant, there should be a relation between the different and ramified primes. We will show the prime ideal factors of the different are the ramified primes in K. Lemma 4.7. For a nonzero ideal a in O K, a D K if and only if Tr K/Q (a ) Z. Proof. Since divisibility is the same as containment for integral ideals, a D K if and only if a D K = (O K ), which is equivalent to O K a. By Theorem 3.0, this last containment is equivalent to Tr K/Q (a ) Z. 4 For fractional ideals a, b, and c, with a b, ac/bc = a/b as OK-modules. When c is a nonzero ideal, using a = c and b = O K gives us O K/c = c /O K. Thus [O K : c] = [c : O K].
8 KEITH CONRAD Here is the central theorem about the different ideal. It not only tells us the prime ideal factors of the different, but also the multiplicities in most cases. Theorem 4.8 (Dedekind). The prime ideal factors of D K are the primes in K that ramify over Q. More precisely, for any prime ideal p in O K lying over a prime number p, with ramification index e = e(p p), the exact power of p in D K is p e if e 0 mod p, and p e D K if p e. When p e, the theorem does not tell us the exact multiplicity of p in D K, but only says the multiplicity is at least e. Proof. If we grant that p e D K, any ramified prime in K divides D K. If p is unramified over Q, so e =, then the last part of the theorem says p doesn t divide D K since p doesn t divide. Therefore it remains to check the two divisibility relations. To show p e D K, write (p) = p e a. Since p e (p), p a. To say p e D K is equivalent, by Lemma 4.7, to saying Tr K/Q (p (e ) ) Z. Since p (e ) = p a, Tr K/Q(p (e ) ) Z if and only if Tr K/Q (a) pz, which is the same as Tr K/Q (α) 0 mod p for all α a. This congruence is what we will actually show. For α a, Tr K/Q (α) = Tr OK /Z(α) and Tr OK /Z(α) mod p = Tr (OK /(p))/f p (α). This last trace is the trace of multiplication by α on O K /(p) as an F p -linear map, and α is a general element of a/(p). (The quotient a/(p) makes sense since a (p) by the definition of a.) Since a is divisible by every prime ideal factor of (p) (including p), a high power of a is divisible by (p). Therefore a high power of α is 0 in O K /(p), which means multiplication by α on O K /(p) is nilpotent as an F p -linear map. Nilpotent linear maps have trace 0, so Tr (OK /(p))/f p (α) = 0. That completes the proof that p e D K. Now we want to show p e D K if and only if p e. Write (p) = p e b, so b is not divisible by p. To say p e D K is equivalent to Tr OK /Z(b) pz by Lemma 4.7, which is equivalent to (4.3) Tr (OK /(p))/f p (β) = 0 for all β b. We will break up O K /(p) into a product of two rings and analyze the trace separately on both. Since p e and b are relatively prime, O K /(p) = O K /p e O K /b as rings by the natural map, so (4.4) Tr (OK /(p))/f p (x) = Tr (OK /p e )/F p (x) + Tr (OK /b)/f p (x) for all x O K, where x on the left is x mod (p) and x on the right is x mod p e and x mod b. (Both O K /b and O K /p e contain F p since p e and b both divide (p), so p is 0 in both rings.) If x b, then x = 0 in O K /b, so Tr (OK /(p))/f p (x) = Tr (OK /p e )/F p (x). For any y O K, there is an x O K satisfying x y mod p e and x 0 mod b, so Tr (OK /p e )/F p (y) = Tr (OK /p e )/F p (x) = Tr (OK /(p))/f p (x). Therefore proving (4.3) is equivalent to proving (4.5) Tr (OK /p e )/F p (y) = 0 for all y O K. Unlike (4.3), which is quantified over the ideal b, (4.3) runs over O K. We will show (4.5) happens if and only if p e. To study the trace down to F p of y on O K /p e, we will filter O K /p e by subspaces made from powers of p: O K /p e p/p e p /p e p e /p e p e /p e = {0}.
THE DIFFERENT IDEAL 9 Each power p i is an ideal, so multiplication by y is a well-defined linear operator on each p i /p e (0 i e). We now appeal to a result from linear algebra: if V is a finite-dimensional vector space over a field F, A: V V is a linear operator and W is a subspace of V such that A(W ) W, then Tr(A: V V ) = Tr(A: V/W V/W ) + Tr(A: W W ). (This identity is proved by a matrix calculation, using a basis of V/W lifted to V together with a basis of W to form a basis of V. The matrix for A on V in this basis is block triangular.) Taking F = F p, V = O K /p e, W = p/p e, and A to be multiplication by y, Tr (OK /p e )/F p (y) = Tr(m y : O K /p e O K /p e ) = Tr(m y : O K /p O K /p) + Tr(m y : p/p e p/p e ). In a similar way, using multiplication by y on the vector space p i /p e and subspace p i+ /p e, where 0 i e, Tr(m y : p i /p e p i /p e ) = Tr(m y : p i /p i+ p i /p i+ ) + Tr(m y : p i+ /p e p i+ /p e ). Using this recursively for i = 0,,..., e, we get e (4.6) Tr (OK /p e )/F p (y) = Tr(m y : p i /p i+ p i /p i+ ). i=0 The traces in the sum take values in F p. We will show the traces are all equal. Choosing π p p, (π i ) is divisible by p i but not by p i+, so p i = (π i ) + p i+. Therefore O K /p = p i /p i+ as O K -modules by x mod p π i x mod p i+. This O K -module isomorphism commutes with multiplication by y on both sides, so Thus (4.6) becomes Tr(m y : p i /p i+ p i /p i+ ) = Tr(m y : O K /p O K /p). (4.7) Tr (OK /p e )/F p (y) = e Tr (OK /p)/f p (y) for any y O K, so (4.5) is true if and only if e Tr (OK /p)/f p (y) = 0 for all y O K /p. Since O K /p is a finite field, the trace map from O K /p to F p is not identically 0, so (4.5) holds if and only if e = 0 in F p, i.e., p e. Up to now the fact that the prime numbers dividing disc(k) are the ramified primes in K hasn t been used (outside of Example.). Theorem 4.8 leads to a new proof of that result. Corollary 4.9. The prime factors of disc(k) are the primes in Q that ramify in K. Proof. Since disc(k) = N(D K ), if p is a prime dividing disc(k) then D K must have a prime ideal factor whose norm is a power of p, which means (p) is divisible by a prime factor of D K, so p ramifies in K. Conversely, if p ramifies in K then D K is divisible by a prime ideal factor of (p), so N(D K ) is divisible by p. Corollary 4.0. Write po K = p e peg g and f i = f(p i p). If no e i is a multiple of p then the multiplicity of p in disc(k) is (e )f + + (e g )f g = n (f + + f g ). If some e i is a multiple of p then the multiplicity of p in disc(k) is larger than this amount.
0 KEITH CONRAD Proof. The multiplicity of p in disc(k) is determined by the multiplicities of the p i s in D K, since the norm of the different is the discriminant (in absolute value). By Theorem 4.8, p e p eg g is a factor of D K. Applying the ideal norm, p (e )f + +(e g )f g is a factor of disc(k). If no e i is a multiple of p then p i appears in D K with multiplicity e i, so this power of p is the full power of p in disc(k). If some e i is a multiple of p then p i appears in D K with multiplicity larger than e i, so the p-multiplicity of disc(k) is larger than what we just computed. Example 4.. In Q( 0), the prime numbers that ramify are and 5: () = p and (5) = p 5. By Example 4.4, the different of Q( 0)/Q is D = ( 0) = p 3 p 5. The multiplicity of p 5 in D is e(p 5 5) =, and the multiplicity of p is at least e(p ) =, which is consistent with Theorem 4.8 since e(p 5 5) is not a multiple of 5 and e(p ) is a multiple of. Notice the p -multiplicity in D is in fact greater than e(p ). Example 4.. Let K = Q(α), where α 3 α = 0. Since O K = Z[α] and disc(k) = 3, D K has norm 3. Therefore D K is a prime ideal. Since (3) = pq and q must divide D K by Theorem 4.8, D K = q. The multiplicity of q in the different is = e(q 3), as expected. Example 4.3. Let K = Q( 3 ), so Z[ 3 ] O K. We will use Theorem 4.8 to prove O K = Z[ 3 ]. Since disc(z[ 3 ]) = [O K : Z[ 3 ]] disc(o K ) and disc(z[ 3 ]) = 08 = 4 7, it suffices to show disc(o K ) = 08. The only prime factors of 08 are and 3, so the only primes that can ramify in K are and 3. The polynomials T 3 and (T ) 3 = T 3 3T + 3T 3 are Eisenstein at and 3 with roots generating K, so both and 3 are totally ramified in O K : () = p 3 and (3) = q 3. By Theorem 4.8, D K is divisible by p q 3. Taking norms, disc(o K ) is divisible by 3 3 = 08. We already knew disc(o K ) is a factor of 08, so disc(o K ) = 08 and O K = Z[ 3 ] (and D K = p q 3 ). Example 4.4. If p has exactly one prime lying over it, say (p) = p e, then D K is divisible by p e, so disc(k) is divisible by N(p) e = p f(e ). If e 0 mod p then p f(e ) is the exact power of p in disc(k). Example 4.5. Let K = Q( 3 75). Using our knowledge of the multiplicity of primes in the discriminant, based on the connection between the different and discriminant, we will determine O K and then prove O K does not have a power basis. View K as a subfield of R, so there is no ambiguity about the meaning of cube roots. Set α = 3 75. Since 75 = 5 7, K also contains 3 5 7 = 35/α. Set β = 3 5 7. The minimal polynomials for α and β over Q are T 3 5 7 and T 3 5 7, which are Eisenstein at 5 and 7, so 5 and 7 are both totally ramified in O K : (5) = p 3 and (7) = q 3. Therefore D K is divisible by p 3 q 3 = p q, so disc(o K ) is divisible by N(p q ) = 5 7. Moreover, (T + ) 3 75 = T 3 + 3T + 3T 74 is Eisenstein at 3, so 3 is totally ramified in K: (3) = p 3 3. Since e(p 3 3) is a multiple of 3, disc(o K ) is divisible by 3 3, not just 3, so disc(o K ) is a multiple of 3 3 5 7. The lattice Z + Zα + Zβ lies inside O K and is a ring: (4.8) α = 5β, β = 7α, αβ = 35.
THE DIFFERENT IDEAL Using these equations and the formulas Tr K/Q (α) = 0 and Tr K/Q (β) = 0, the matrix of trace products for the Q-basis {, α, β} of K is 3 0 0 0 0 35 3, 0 35 3 0 whose determinant is 3 3 5 7. Therefore 3 3 5 7 = disc(z + Zα + Zβ) = [O K : (Z + Zα + Zβ)] disc(o K ). Since we know disc(o K ) is a multiple of 3 3 5 7, the index [O K : (Z + Zα + Zβ)] must be, so O K = Z + Zα + Zβ and thus K has discriminant 3 3 5 7 = 33075. We will now show in two ways that O K does not have a power basis. Method. For each γ O K Z we will show disc(z[γ]) disc(o K ) From the computation of O K, we can write γ = a + bα + cβ with integers a, b, and c, where b and c are not both 0. Using (4.8), the matrix for multiplication by γ relative to the ordered Z-basis, α, β is (4.9) [m γ ] = a 35c 35b b a 7c c 5b a Computing powers of this matrix in PARI we get Tr K/Q (γ) = 3a, Tr K/Q (γ ) = 3a + 0bc, Tr K/Q (γ 3 ) = 3a 3 + 55b 3 + 735c 3 + 630abc, and so Tr K/Q (γ 4 ) = 3a 4 + 00ab 3 + 940ac 3 + 050b c + 60a bc, disc(z[γ]) = det Tr K/Q () Tr K/Q (γ) Tr K/Q (γ ) Tr K/Q (γ) Tr K/Q (γ ) Tr K/Q (γ 3 ) Tr K/Q (γ ) Tr K/Q (γ 3 ) Tr K/Q (γ 4 ) = 3 3 5 7 (5b 3 7c 3 ). Since 3 3 5 7 is the discriminant of K, [O K : Z[γ]] = 5b 3 7c 3. Therefore this index is if and only if 5b 3 7c 3 = ±, which implies 5b 3 ± mod 7, but 5 mod 7 is not a cube. We have a contradiction. Method. We will show the different ideal D K is not principal, so O K can t have the form Z[γ] by Theorem 4.3. We already saw that disc(k) = 3 3 5 7 and (3) = p 3 3, (5) = p3, and (7) = q 3. Since N(D K ) = disc(k) = 3 3 5 7, D K = p 3 3 p q. Since (β) 3 = (45) = (5)(7) = p 3 q 6 we get (β) = pq, so D K = (3)(β)p (also D K = (3)(α)q). We will prove p is not principal, so D K is not principal. The ideal norm of p is 5, so to show p is not principal it suffices (and actually is equivalent) to show that no γ in O K has norm ±5. Write γ = a + bα + cβ with a, b, c Z. Using (4.9), N K/Q (γ) = det(m γ ) = a 3 + 75b 3 + 45c 3 05abc. Set this equal to ±5 and reduce modulo 7: the coefficients 75, 45, and 05 are multiples of 7, so ±5 = a 3 mod 7. This has no solution for a in Z/(7), so we have a contradiction.
KEITH CONRAD The reasoning in Example 4.5 can be generalized: for distinct primes p and q not equal to 3 such that p q mod 3 and p q mod 9, and p is not a cube in Z/(q), the cubic field Q( 3 p q) has a different ideal that is not principal. Examples besides (p, q) = (5, 7) are (p, q) = (, 7), (7, 7), and (, 3); that last example is Q( 3 5). I learned about this type of construction from [, pp. 460 46] and [3, Exer. 0B, pp. 0-0]. Theorem 4.8 gives us an exact formula for the multiplicity of p in D K if e(p p) 0 mod p, but not if e(p p) 0 mod p. Ramification when e(p p) 0 mod p is generally easier to study than when e(p p) 0 mod p. For this reason, when e(p p) 0 mod p we say p is tamely ramified over p, and when e(p p) 0 mod p we say p is wildly ramified over p. If every prime over p in K is tamely ramified then we say p is tamely ramified in K. Any unramified prime is tamely ramified, but ramified primes can also be tamely ramified. For example, is tamely ramified in Q( 3 ) (() = p 3 ), but is wildly ramified in Q(γ), where γ 3 + γ + 4 = 0 (() = pq ). Upper and lower bounds for the multiplicity of a nonzero prime ideal p in D K are e ord p (D K ) e + e ord p (e), where p p and e = e(p p). The lower bound was proved by Dedekind (Theorem 4.8), who conjectured the upper bound, which was later proved by Hensel as one of the first applications of p-adic fields to number theory. In the ideal class group of K, Hecke proved the ideal class of D K is always a square. For a proof in the number field and function field settings, see []. References [] J. V. Armitage, On a theorem of Hecke in number fields and function fields, Invent. Math. (967), 38 46. [] H. Hasse, Number Theory, Springer-Verlag, 980. [3] P. Samuel, Algebraic Theory of Numbers, Dover, 008.