the discriminant of a quadratic extension of an …...the discriminant of a quadratic extension of...

MATHEMATICS OF COMPUTATIONVOLUME 40. NUMBER 162APRIL 1983. PAGES 685-707

The Discriminant of a Quadratic Extension

of an Algebraic Field

By Theresa P. Vaughan

Abstract. Let F be an algebraic field, and K an extension of F of degree 2. We describe a

method for computing the relative discriminant D for K over F. We work out the details for

the case when F is quadratic and give tables which yield D very easily. We also apply the

method to one type of cubic field F, and give tables for it.

1. Introduction. Let F be an algebraic number field, and K an extension of F, of

degree 2 over F. We seek a " practical" method of computing the relative discrimi-

nant D of K over F.

Suppose K= F(fy), where y is an integer in F; we write (y) for the principal

ideal generated by y in the ring 5 of integers of F. Put D = 2kDx where Dx is odd. If

p is an odd rational prime and 5 is a prime ideal divisor of (p) in 5, we must

discover whether 5 divides (y) to an even or odd power. Then (Theorem 3.4) Dx is

the product of all norms of such ideals P, which divide (y) to an odd power. Thus,

the difficulty of finding D, is about the same as that of factoring (y).

The determination of the integer k is rather more complicated. In Section 4, we

show the following:

(a) There is a y, in 5 so that y/y, is a square in F and (y,) is not contained in the

square of any prime factor of (2) in 5;

(b) There is a ß in 5 so that ß2yx is congruent to a square modulo 4, and ß

satisfies certain minimality conditions:

(c) 2k divides the norm of ß2y, exactly (Theorem 4.6).

In actual computation, most of the action takes place in 5/(4). If [F: Q] = n,

then | 5/(4) | = 4", and clearly it is desirable to have as many restrictions as possible

on the nature of a suitable ß. We address this problem in Section 5. The behavior of

the squares in 5/(4) is of considerable interest, and we investigate this monoid in

Section 6.

In Section 7, we discuss the case of a quadratic field F = Q(fñ ). The only case

which is not almost trivial is that of n = 1 (mod 8). In Appendix 1, we give tables for

some of the arithmetic of 5/(4) in case n = 1 (mod 8); Table V of Appendix 1

summarizes the results for all n. With the aid of these tables, one can easily find the

relative discriminant; the only computational difficulty lies in finding and factoring

the norm of y. For specific examples, we work out relative discriminants for a list of

fields given by D. Shanks in [2].

Received July 21, 1982.

1980 Mathematics Subject Classification. Primary 12A05, 12A50.

©1983 American Mathematical Society

0025-5718/82/0OO0-0993/$06.25

685

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

686 THERESA P. VAUGHAN

In Section 8, we give a partial discussion of the case when F is a cubic field. The

monoid of squares in 5/(4) is completely determined (up to isomorphism) by the

factorization of (2) in 5; we give tables for these monoids in Appendix 2. We work

out the details for the field F = Q(a) where a is a root of x3 + 2.x2 + 1. (Here we

have (2) = PQ; this type is of moderate difficulty; the worst case is when (2) splits.)

Note that if g(x) E Z[x] and g(x) is irreducible and congruent to x3 + 2.x2 + 1

(mod4), then all the work done for Q(a) carries over, mutatis mutandis, to Q(ß)

where ß is a root of g(x).

In a later paper we hope to complete the work begun in Section 8, and give a

complete discussion of all the types of cubic fields.

2. Preliminaries. Let F = Q(a) be an extension of Q of degree n, where a is a root

of an irreducible monic polynomial with integer coefficients,

f(x) = a0 + axx + a2x2 + • • • +x".

The conjugates of a are the roots oif(x) in C; we denote these by a(,)(; = 1,2,...,«).

The trace and norm of a are defined by

Tr(a) = 2 «(,); N(a) = ft «(,>>;=i i=i

and onehasTr(a) = —an_x and N(a) = ( — l)"a0.

All of the following material may be found, in one form or another, in [1].

Let 5 be the ring of algebraic integers in F. Let & = {a,, a2,... ,a„} be a basis for

F over Q, with a, G 5 (/' = 1,2,...,«), and let A be the matrix whose (/', j) entry is

of.Then the discriminant of & is disc & — \ A \2. If 6£ is an integral basis for 5, then

disc 6? = disc 5, and otherwise, disc & = A:2 disc 5 where k E Z, k > 1.

Now let AT be a quadratic extension of F, and let 5 be the ring of integers in K.

Then for some y in F, with y not a square in F, we have K = F(Jy). Evidently, one

may assume without loss of generality, that y G 5.

Suppose that e = (ß + óVy)// G S, where ß, o G 5 and,/' G Z. Define the set %

by

® - {«i»«2>- ••-«„, eau ea2,... ,ean}.

2.1. Lemma. With all notation as above, if {ax,a2,...,an} is an integral basis for 5,

then

disc® = (disc5)2- (-) -N(S2y).

Proof. See [1, p. 43]. D

Finally, we use the following notation. If p, n E Z and p is prime, then p' \\ n

means that p' \ n and p'+ ' { n. If ß G 5, then (ß) is the principal ideal generated by

ß. If 5 is a prime ideal in 5, then P' || (/?) means that ß G P' and ß G P'+x.

3. A Reduction Process. Define the equivalence relation ~ on F* by y, ~ y2 if and

only if y, • y2 ' is a square in F. Let [y] denote the equivalence class of y.

3.1. Definition. Let y E F* and suppose p is a rational prime. Then there exists

some y, G [y] satisfying:

(a) y, G 5,


THE DISCRIMINANT OF A QUADRATIC EXTENSION 687

(b)p'\\N(yx)(t>0,tEZ),

(c) If y2 G [y] n 5 and if ps II N(y2), then t < s.

Such a y, is said to be reduced relative to p.

3.2. Lemma. Let y G 5, and let p be a rational prime, where [P¡: i = 1,2,... ,k} is

the set of the prime ideal divisors of(p)inR. Then for any i= 1,2,... ,k, y E Pf if

and only if there exists some k E Z with (k, p) = 1, and some 8 E R such that

5,. || (8), and Pj\(8) if i +j, such that

k2y = 82yx

for some y, in 5.

Proof. Fix i, and put P = Pt. Choose 8 as described above. Then there exists e G 5

so that 8e = pk, where k E Z and (k, p) = 1 (e is in the conjugate of 5). Then, if

y G P2, we have

2 2e y=p -Yi

for some y, in 5, and hence, multiplying both sides by 82, we have k2y = 82 ■ y, as

required. The converse is obvious. D

3.3. Lemma. Let y and p be as in Lemma 3.2. Then y is reduced relative to p if and

only if y <2 Pf (i = 1,2,...,*).

Proof. Suppose first that y G P2 for some /'. Then by Lemma 3.2 we have

A:2y = ô2y, where (k, p) = 1, 5,11(0), and Pj\(8) if i¥>j. Then y~y,, and if

p" II N(y) andpv || N(yx), clearly v < u. Then y is not reduced relative top.

On the other hand, suppose y G Pf for any i, and y ~ y, where y, is reduced

relative top. By the above, we know y, G Pf for any i, and we also have a2y = ß2y,

for some a, ß G 5. Then Pfr+X ll(a2y) is possible if and only if 5,ll(y); since

<*2Y = ß2Yi, we have 5;H(y) if and only if 5,||(y,). Then 7Y(y) and A/(y,) are

divisible by exactly the same power of p, and so by definition, since y, is reduced

relative to p, so is y. □

If p is an odd prime, the situation is easily described.

3.4. Theorem. Let p be an odd prime and y E F, where y is not a square in F, and y

is reduced relative to p. Suppose that p' || disc 5 and ps \\ N(y). Then p2,+s || disc S.

Proof. Since p > 2, we need only show that if a, ß G 5, then (a + ßfy)/p G 5

unless (p)\(a) and (p)\(ß) (Lemma 1.1). Thus suppose that (a + ßfy) = pe for

some e G S. Then also (a — ßfy) = pex for some e, G S. Since 2a = p(e + e,) and

p is odd, then (p) \ (a). Then a2 — ß2y E (p2) implies that ß2y G (p2). But since y

is reduced relative top, (y) is not divisible by the square of any prime factor of (p).

Then (p)\(ß) also. The result follows from Lemma 1.1. D

4. The Case p = 2. Not surprisingly, this case requires special treatment, begin-

ning with another definition.

4.1. Definition. Let y G 5 be reduced relative to 2, and not square in F. Choose

ß G 5 as follows:

(i) For some a E R, a2 — ß2y = 0 (mod4).

(ii)2'H N(ß2y).



(iii) lie, 8 ER and if e2 - ¿52y = 0 (mod 4) and if 2s \\ N(82y), then t < s.

We say that such a ß2 is a match for y.

This section is devoted to proving that 2'II (disc5)/(disc5)2. We assume

throughout the rest of the paper that

(2) = PexF^ ■ ■ ■ Pf-

in 5, where the 5, are prime ideals.

4.2. Lemma, (a) Let a, ß E R. Then a2 = ß2 (mod 4) if and only ifa = ß (mod 2).

(b) Suppose Pf-1| (a) and Pf' II (ß) for i = 1,2,... ,r, and let k be a positive integer.

Suppose that 0 < a¡, bt < /ce,/or i = 1,2,...,r. Then a = ß (mod2*) implies a, = b¡

for i = l,...,r.

Proof, (a) Let a2 - ß2 = 4e for some e in 5. If Pf || (a - ß) and Pf II (a + ß),

then a + b > 2e¡ and hence (say) a > e¡. Since a — ß = a + ß (mod 2), then also

b > e,, and we have a = ß (mod 2). The converse is obvious.

(b) Put 8 = a — ß, and suppose that, for some i, ai < b¡ « ke¡. Then ô G 5,a' -

Pf'+X, and since a, < /ce,, then 5 G (2*), a contradiction. D

4.3. Lemma. Let a,ß,yE R. Then (a ± ß/y )/2 G S if and only if (a2 - ß2y)/4

G 5.

5roo/. First suppose that a2 — ß2y = 4e in 5. Then in 5, we have ß2y = (ßfy)2

= ß2, and then a2 — ß2 = 0 (mod4) in 5 implies a = ± ß, (mod2) in 5, by Lemma

4.2. That is, (a ± j8/y )/2 G S.Conversely, if (a ± ß/y )/2 G S, then the conjugate (a + ß/y )/2 is also in 5.

Then the product (a2 — ß2y)/4 is in S. But this product is in F, so it is in 5. D

4.4. Lemma. Let y be reduced relative to 2, and suppose for some a, ß in R we have

(a + ßfy )/2 in S. Then (a + ß/y )/4 is not /'n 5 H/j/ew a = 2a, and ß = 2ß, /o/-

some a,, ß, ('« 5.

Proof. Suppose (a + ß/y )/4 G 5. Then also the conjugate (a — ß/y )/4 G 5, so

a/2 G S. Since a/2 G F, this gives a = 2a, for some a, G 5. Suppose next that

ß/2 G 5. That is

(ß) = Pxh'P2b^--Pf-X (0<b(,i= \,2,...,r),

where X is an ideal of odd norm, and for at least one i, bi < e¡. Since y is reduced

relative to 2, we know that Pf' \\ (y), where g, = 0 or 1, for / = 1,2,... ,r. Then we

have Pfh'+g'\\(ß2y) where for at least one /', b¡ < e¡, and 2o, + g. < 2e,. Hence

(ß2y) ¿(4), that is ß2y/4 G 5. But now, we can write (2a, + ß/y )/2 G S, which

gives ß/y/2 G 5 and ß2y/4 G 5, a contradiction. Thus we must have ß = 2ßx.

□

4.5. Lemma. Let s E 5, and suppose that Pf' II (s) (/' = 1,... ,r). Choose ß, G P¡ so

that: Pi\\(ßi)andPj\(ßi)ifi ^fjori = l,2,...,r. T/ie«

(a) 77tere exists x in R with odd norm, such that

s=xßsx'ß^---ß^(mod4).

(b) There exists s' in R such that s = s' (mod 2), and y in 5 with odd norm, such that

s'= yßpßp • ■ ■ ß?',

where u¡ = min(s,, e¡) for i = \,2,...,r.



Proof, (a) For each ß,, we can find 5, G 5 so that ß, • Ô, = 2m ¡, where m¡ is an odd

rational integer. Since m,= ± 1 (mod 4), (a) follows from successive applications of

the process described in Lemma 3.2. To see (b), use the expression from (a). If

si < e„ then w, = st: if s¡ > e„ e, = ß?< + 2 has 5/' || (e,), 5y | (e,) if ; #y, and by (a)

we can write e, = ßfxt (mod 4), where x¡ is some integer of odd norm. Substituting e,

in the expression from (a), for all /' for which e, < s,, we get (b).

At last, we are in a position to prove:

4.6. Theorem. Suppose that y is reduced relative to 2, that y is not square in F, and

that ß2 is a match for y, with 2' \\ N(ß2y). Then 2' || (disc S)/(disc 5)2.

Proof. There is some a in 5 so that a2 — ß2y = 0 (mod 4), so by Lemma 4.3,

(a + ß/y )/2 is in S. Using the notation of Lemma 1.1 with / = 2, we have

disc "3d = (disc 5)2 N(ß2y). The result will follow if we can show that, for all u, s in

5, if X = (u + s(a + ß/y )/2)/2 is in S, then u = 2«, and s = 2s, in 5. Thus

suppose that X E S. By Lemma 4.5(b) it is clear we may assume that, if for any

i = 1,2,..., r we have Pf || ( ß), then y =s e,, or if Pf || (s) thenj < e,. Since

X = ((2u + sa) + sßyC )/4 G S,

it follows from Lemma 4.4 that sa = 2a, and sß = 2ß, for some a,, ß, in 5.

Suppose that Pf' || (s) and Pf- \\ (ß), with s, < e,, £>, < e„ for /' = 1,2,- • -,r. Then

sß = 2ß, implies that o, + s, s* e, (/' = 1,2,... ,r); or

s, = e,-o, + w, (0<w,=£o,).

Sowehave^'IKß,).

Next, from Lemma 4.3, we get

(w + a,)2-ß2y = 0(mod4);

say 2V || N(ßfy). If any of the wt were less than b¡, we would have v < t, contradict-

ing the choice of ß. Thus every w¡ = b¡, and then every s, = e,, so that s = 2s, for

some s, in 5. Now also we have 2w/4 in S, that is, u = 2ux, and this completes the

proof. □

5. Some Refinements. As it stands, Theorem 4.6 does not look very useful, since

the squares in 5 fall in 2" congruence classes (mod 4) and 2" is rather a large

number. In this section we give some conditions that a and ß must satisfy in order

that a2 - ß2y = 0 (mod4).

5.1. Theorem. Suppose that y is reduced relative to 2, and that a2 — ß2y = 0

(mod 4), and that we have

Pf-\\(a),Pf'\\(ß),Pf'\\(y) (i=l,2,...,r).

(a) We may assume that 0 =£ a¡, bt■ < e,.

(b) For i = 1,2,... ,r, a¡ = b¡, and if g, = 1, then a, = b¡ = e,.

Proof. Part (a) follows from Lemma 4.5(b), and (b) from (a) and Lemma 4.2(b).

□

5.2. Lemma. Suppose that a, ß are in 5 and that P,.a'||(a) and 5,a'|l(ß) for

i = 1,2,... ,r. Then there exists some x E R with odd norm, such that a2 = x2 ■ ß2

(mod 4).


690 THERESA P VAUGHAN

Proof. Lemma 4.2(a) and Lemma 4.5(b). D

5.3. Corollary. Let a, ß, y be as usual. Then a2 — ß2y = 0 (mod 4) if and only if

there exist x, y in 5, both having odd norm, such that

(i)ß2(x2-y) = 0(mod4),

(ii)a2(l -j2y)EE0(mod4). U

5.4. Theorem. Let y be reduced relative to 2, and suppose that ß2 is a match for y,

with a2 - ß2y = 0 (mod4). Suppose that a, and ß, in R also satisfy a2 — ß2y = 0

(mod 4). Finally suppose that

Pf'Hß) and Pf'\\(ßx) (i=l,2.r).

Then for i = 1,2,... ,r, 0 < b,, < c¡.

Proof. We use Theorem 4.6. The basis % defined there is not necessarily an

integral basis for S, but we do have that (disc<ä>)/(discS) is an odd integer. Then,

putting e = (a + ß/y )/2, every integer in 5 can be written in the form (u + ve)/m,

where u, v are in 5 and m is an odd rational integer.

By Lemma 4.3, we have (a, + ß,/y )/2 in S, and thus, for some u, v in 5 and odd

m in Z, we can write

u + ve = mi a, + ß,fy 1/2,

and hence mo, = 2u + va and mßx = uß. Since m is odd, we are done. D

5.5. Corollary, (i) Lef y be reduced relative to 2 and suppose that a2 - ß2y = 0

(mod 4) for some a, ß in 5. Then ß = 0 (mod 2) unless there exists x in R with odd

norm so that

x2 - y = 0 (mod P2a<P2a> ■ ■ ■ Pfa-)

for some integers a, satisfying 0 < a,- < e, (/' = l,...,r).

(ii) // the condition of (i) is satisfied, and if ß2 is a match for y, with Pf' II (ß,)

(i = 1,2.r), thenQ*zbi<ei-ai(i= 1,2,..., r). D

Remarks. We shall see later that the number N of squares of odd norm,

incongruent modulo 4, varies inversely with the number of prime factors of (2) in 5.

The amount of work we have to do to find a match for some y depends on N

(Corollary 5.3) and on the number of factors of (2) (Corollary 5.5). The most

manageable cases are those with N small and r close to n, or with (y) having many

of the prime factors of (2) as divisors (Theorem 5.1).

6. The Square-Classes (mod 4) in 5. Our purpose here is to find out more about

the nature of the squares (mod 4) in 5, with particular attention to those of odd

norm. A square-class S is defined by

S = G(x2) = {a E R: a = x2 (mod4)}.

Let M be the set of all square-classes in 5. The obvious multiplication, ß(x2)Q(y2)

= Q(x2y2), makes M a monoid, with identity 6(1); by Lemma 4.2 we have

| M | = 2". Many of the results of previous sections can be restated as properties of

M and we list these without proof.



Let U be the set of invertible elements in M. That is,

U= {6(jc2):/V(.x)isodd}.

Evidently, U is a group.

Choose ß, G P¡ so that P¡ || (ß,) and 5,1 (ß,) if i #/ Then each product of the

form

o=][ß!' (0</,<eí)i=i

gives rise to a square class 6(ô2), and these classes are distinct. Let D be the set of

all these classes. Then

6.1. Theorem. Let 6 be any square-class. Then there exist x, 8 in R so that

6(x2) E U and 6(52) ED and 0, = 6(x2)6(S2). (We shall say that 6 is associated

with the r-tuple (tx, t2,.. .,tr).)

We define two sets for every square-class in M:

U(6) = {e(x2): #(x) is odd and 6(x2)6 = 6},

5(6) = {6(x2)6:/V(x)isodd}.

6.2. Lemma. Suppose that 6, and 62 in M are associated to the same r-tuple

(tx,...,tr).ThenexEF(e2).

6.3. Theorem. Let y be reduced relative to 2, and suppose ß2 is a match for y. Then

there exists a in 5 such that a2 - ß2y = 0 (mod4), and 6(a2) G 5(6(ß2)).

6.4. Theorem, (a) For every 6 in M, U(Q) is a subgroup of U. (b) If6x G 5(6),

then 1/(6,) = U(6). (c)IfQx is associated to (tx, t2,...,tr) and&2 to (sx,...,sr), and

iftxsx + t2s2 + • ■ ■ +trsr = 0, then U(6X) n U(62) = (6(1)}.

Proof. Parts (a) and (b) follow directly. To see (c), choose a2 G 6, and a\ E <22,

and suppose Q(x2) E U(QX) n U(Q2). Then x2a2 E 6, and x2a\ E 62, so we have

x2a2 = a2 (mod4), x2a\ = a\ (mod4),

«2(1 - x2) = 0 (mod4), a\(l - x2)=0 (mod4),

(1 - x2) = 0 mod n pf~'' and (1 - x2) = 0 mod n Pf'~s' ■

Since Ir,s, = 0, it follows that (1 - x2) = 0 (mod Pfe') for all i=l,2,...,r. Then

(1 - x2) = 0 (mod4) and 6(x2) = 6(1).

6.5. Theorem. If\ U\= m,and\ U(Q)\- k,then\F(Q)\= m/k. We also have

2|5(6)|=2«e

where the sum is taken over 6 such that the sets F(G) are mutually disjoint.

The cardinality of U is the number of units in the ring 5/(2) (Lemma 4.2); in case

5 s Z/(f(x)) (more or less), this number was given by Dedekind. We provide here

the slight generalization to any 5.



6.6. Definition. If / is an ideal in 5, put || /1| = | R/I | , and let <j>(I) be the number

of units in R/I.

6.7. Theorem. </>: / — <>(/) is a multiplicative function from the set of ideals in 5 to

Z; that is, if Ix and I2 are relatively prime, then <¡>(IXI2) = <¡>(Ix)<f>(I2).

Proof. We can write

where the 5, are prime ideals in R. Then

k

R/I at 0 R/Pfi=l

and u is a unit in R/I if and only if, under the isomorphism above, U corresponds to

some (w,, u2,...,uk) where w,is a unit in R/Pf' (i = l,2,...,/c). D

6.8. Theorem. // 5 is a prime ideal in 5, and r is a positive integer, then

Proof. The chain R D P D P2 D ■■■ projects naturally to 5 = R/Pr, giving the

chain R D P D P2 D • ■ • D Pr~] D P' = {0} (where 5 is the unique maximal ideal

of 5 ). Then we have

5 = \R/P\ ■ \P/P2\ ■ \P2/P3\ ■■■= \\P(,

P = \P/P2\ ■ \P2/P3\ ■■■ = \\P\\r~ '.

Since P is the unique maximal ideal of 5, then « is a unit in 5 if and only if

u E R - P. Thus the number of units is || 51|r — || 51|r" \ as required. D

6.9. Corollary. Let (2) = Px< ■ • ■ Pf-, where 5, is a prime ideal of degree f¡

(i = 1,... ,r). Then

|l/| = «((2)) = 2»n(l-2-'').1 = 1

7. Quadratic Fields. We use the same notation as before, and in addition

m = (disc S )/ (disc 5 )2.

We have K = F(fy), where y is a nonsquare in 5; we wish to compute m. In

practice, one is often faced with a y which is not reduced relative to one or more

primes; to find the power of prime p dividing m, we need to know something about

reduced forms of y, relative to p. It is not necessary, in general, to find such a form

explicitly, as our examples show. Indeed, for odd primes, all we need is the prime

ideal factorization of the principal ideal (y) (Lemma 3.3; Theorem 3.4). If p = 2,

however, it is also necessary to work with the arithmetic of 5/(4). In this section, we

work out the details for the case of a quadratic field F = Q(fZ), together with an

assortment of specific examples borrowed from a paper of Shanks [2]. The results are

summarized in Table V of Appendix 1.

Let 5 = Q(fZ), where Z is a squarefree integer. An integral basis for 5 is (1, w},

where co = fZ if Z = 2,3 (mod4) and w = (1 + fZ)/2 if Z = 1 (mod4). We shall

denote a + bu by (a, b).



7.1. Case 1. Let Z = 2 (mod4). An integral basis for 5 is {l,fZ }, and (2) = P2.

If y = a + bfZ, we can write y = 2*(c + dfZ) where either c or d is odd;

y, = c + d/Z is reduced relative to 2, and y ~ y,. We have | U | = 2 and the "odd

squares" are (1,0) and (3,2) (mod 4). If c is odd and d is even, but y = x2 (mod4),

then (c, d) - (1,0) G (2) = P2 and a match, ß2, will have 221| ß2. Then 4 || m. If

c, d are odd, the only match will have 24 || ß2 and then 24 || m. If c is even, d odd,

again the only match has 24 II ß2; since 2 || N(c + d{Z ) we have 251| m.

The case Z = 3 (mod 4) is similar to the above; in this case the "odd squares" are

(1,0) and (3,0).

7.2. Case 2. Let Z = 5 (mod8). Then w = (1 + fZ)/2, and (2) = 5. We can write

y = a + bu = 2k(c + du) = 2*y,,

and if k is even, y ~ y,, while if k is odd, then y ~ 2y,. If Z = 8/ + 5, the odd

squares (mod 4) are: (1,0), (2j + 1,1), (2j + 2,3). Then a match for the reduced

form of y is 1 if it is congruent to one of the odd squares (mod 4) and is 4 otherwise.

7.3. Case 3. Let Z s 1 (mod8), so u = (1 + fZ)/2. Now (2) = PQ, and the

situation is more complicated accordingly. The tables of Appendix 1 give enough of

the arithmetic of 5/(4) for our purposes; note that (writing Z = 8j + 1) there are

different tables for y even and y odd. Tables la and lb give some of the multiplica-

tion of 5/(4). Table II gives the norm modulo 4, for y z 0 (mod 2). If y is already

reduced then these tables allow the determination of the power of 2 dividing m; one

need only decide if 1 — y is in P2, Q2, (4), or none of these (Corollary 5.3). The

results are in Table V.

The reduction process is more involved, and for this we need Tables III and IV.

Suppose y = n + mu has even norm and y G (2); say y G 5. Choose ß G Q so that

N(ß) = 2b, b = 1 (mod4). Write N(n + mu) = Vx, x odd. Then ß(n + mu) =

2(a + bu); Table III gives the values of (a, b). Note that we need to know whether

x is congruent to 1 or 3 mod 4 to take care of the case when y is even. For Table IV,

we have X = 2(n + mu) where N(n + mu) = 2x, x odd. Choose ß as for Table III;

then Xß2 = 4(a + bu). In Table III, if; > 4, and if ß2(n + mu) = 4(u + vu), then

(n, m) = (u,v) (mod 2), so that in fact this process is reasonably short.

7.4. Examples. As an illustration, we find the value of m for some quartic fields

discussed by Daniel Shanks in [2]. The situation is this: Let T = X + YfZ and

t = T+ )/T2 - 1 , where Shanks' requirements are that Z,4X, and 4(X2 - Y2Z)

are integers, and | X — YfZ | < 1. Then t is a root of the reciprocal polynomial:

f(y) =y4- 4A>3 + (2 + 4(A-2 - Y2Z))y2 - 4Xy + 1.

Dr. Shanks has shown that

disc(/) = ((47)2z)2/(l)/(-l)

(personal communication), but it is not very easy to find the value of m from this; in

practice disc/ seems to have a lot of extraneous factors. It is easier to tackle

y = T2 - 1 directly. Below, we give the values of m for most of the examples listed

in [2]. We do two of these in detail, and for the rest we indicate the main steps.



(A) Let A = (13 + /Ï93 )/2. Then Z = 8 • 24 + 1, and y = 24 is even. We have

A=6 + uandA2- l=(A- l)(A + 1) = (5 + u)(l + u). Mod 4, this is A2 - 1

= (1,1) • (3,1) = (3,1) (Table la). One computes N(A - 1) = -18, N(A + 1) = 8,

so N(A2 - 1) = 24(-9); for our tables,; = 4 and x = -9 = 3 (mod4). Then four

multiplications by a suitable ß yield the sequence

ß1 ß ß

(3,1) -(3,1)^(1, l)-(3,2).

We use Table III; observe that it is not necessary to carry out any actual calcula-

tions, nor to know anything more about ß than that it exists. So y = A2 — 1 ~ y,

where y, = (3,2) (mod 4). From Table V, 4 || m; alternatively, from Table la we see

that (3,2) • (0,1) = (0,1) = (2, l)2 and a match for y, is ß2 = (2,1)2 (mod4), with

2 || N(ß). N(yx) is odd, so 4 || 7V(ß2y,) and 4 \\m. Finally, 3 is not ramified in F, and

A2 — 1 £ 0 (mod 3) so we have y ~ y2 where 3 f N(y2); hence 31 m. The factor - 1

is not square, so m = —4.

(B) Let 5 = (25 + fWf )/4 = (12 + u)/2. We have 697 = 8.87 + 1; y is odd.Since 52 - 1 ~ 452 - 4, we use 25 - 2 = 10 + u = (2,1) and 25 + 2 = 14 + u

= (2,1); N(2B - 2) = -26 and 7V(25 + 2) = 36; 7V(452 - 4) = 28(-9); j = 8,

x = -9 = 3 (mod4). We have (2,1) ■ (2,1) = (2,1), and the sequence is

(2,1) -(2,1)^(0,3) -(3,2)/36 ß ß

and as before, 4 || m, 3 \ m, and m = — 4.

The remaining examples from [2] are given below in tabular form.

8. Cubic Fields. We use the notation of the previous sections, where now 5 is a

cubic extension of Q. Then | 5/(4) | = 64, | 5/(2) | = 8, and there are eight equiva-

lence classes of squares (mod 4). The structure of the monoid M is determined, up to

isomorphism, by the factorization of (2) in F. (Since 5 is a cubic field, this is an easy

consequence of the results of Section 6; it is also easy, if tedious, to show this

directly.) When (2) = 5, a prime in F, then M — {0} is a cyclic group of order 7; for

all other cases, we give the tables for M in Appendix 2.

We seek the power of 2 dividing m = (discS)/(disc5)2. The situation is fore-

shadowed, to some degree, by the quadratic case: if y is already reduced relative to

2, it is comparatively simple to discover this power of 2, while if y is not reduced,

some sort of reduction process is needed. If (2) has only one prime factor, reduction

is a simple matter; if (2) has two prime factors there are some manageable

difficulties. If (2) has three distinct factors, we have found (so far) only a partial

solution to the reduction problem. We shall work out one comparatively simple case,

with (2) = PQ. The necessary tables are given in Appendix 3. The methods used for

the quadratic case are not sufficient here, doubtless reflecting the fact that the two

factors of (2) are not of the same degree; nevertheless there is considerable

similarity.



K

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en

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The correspondence g(a)

ß=(a,b,c)

Let f(x) = x3 + 2x2 + 1. Then disc/= —59, and (where a is a root of f(x)),

{I, a, a2} is an integral basis for 5. We denote a + ba + ca2 by (a, b, c). The

multiplication in 5 (or in F) is given by (a, b, c) ■ (r, s, t) = (x, y, z) where:

x = ar — cs + ( — b + 2c)t,

y = br + as — ct,

z = cr+ (b- 2c)s + (a - 2b + 4c)t.

Let C be the companion matrix for/(x),

0 0 -1C= 1 0 0

.0 1 -2.

g(C) is an isomorphism of F with Q[C]. We have

a —c —b + 2c

5= ¿7 a -c

_c b — 2c a — 2b + 4c.

We shall have some use for the adjoint of 5; say adj 5 *-* (u, v, w), where

u = a2 - 2ab + be + 4ac + 2c2,

v = -c2 - ab + 2o2 - 4bc,

w = b2 — 2bc — ac.

Recall that \B\=N(ß); Tr(5) = Tr(ß); |adj5| = |5|2, B(ad] B) = N(ß)I. We

investigate only the power of 2 dividing w. Since/(x) = x3 + 1 (mod 2), then in 5,

(2) = PQ, where we choose 5 = (a + 1,2) = (a + 1) and Q = (a2 + a + 1,2) =

(a2 + a + 1).

The congruence classes modulo 2, are grouped as follows:

(2) (000)

5

Q(1)

If (a, b, c) is given modulo 2, then (a, o, c)2

The square table is:

(a,b,c) 000 100 010 001 110 101 011 111

(x,y,z) 000 100 001 230 121 332 031 113

In the first column of Table I of Appendix 3, we give a list of all (a, b, c) z 0

(mod 2), with the left-most entry 1. (All entries are given modulo 4.) The second

column gives (a, b, c) ■ (001) and the third gives (a, b, c) ■ (230). Thus the elements in

any row (in the first three columns) are equivalent mod ~ via multiplicaton by unit

squares. Every (a, b, c) is either congruent (mod 4) to one of these, or to a multiple

thereof by 0, 2, or 3. Thus we restrict our attention to the twelve entries of the first

column.

We need to know which of these are reduced relative to two. If ß, = ß2 (mod 4)

then N(ßx) = N(ß2) (mod4), so the entries of the fourth column tell that all the

units are reduced, and also (110) and (132). One checks that (111), (131), (133) are

all reduced, and (113) is not.

(110),(101),(011)

(111)

(100), (010), (001)

(x, y, z) is determined modulo 4.



Next, which y satisfy: y z 0 (2), ß2y is a square (mod 4), where y and ß are both

reduced? We first choose y by: ß G 5, 1 - y G g2, and then ß G Q, 1 - y G P2.

We require y reduced, and not a square (mod 4), so this gives:

(i)ß G 5, y G {(213), (322)},(ii) ß G Q, y = (u, v, w) with u + v + w = 1 (mod 4) or y G {(122), (Olü), (320),

(131)}.For every possible y listed above, we also include every y' ~ y from Table I; for

example, with (213) we also have (110) and (303). This set of y is a complete set

satisfying the stated requirements (Corollary 5.3).

From Theorem 5.1, if y is reduced and y = 0 (mod 2), then a match for y is

ß2 = 22. The only way such a y can be reduced is if it has the form 2(x, y, z) where

N(x, y, z) is odd. We have considered all possibilities and the results are listed in

Table V.

There remains the problem of finding a reduced form y' (mod 4) for some given y

which is not reduced relative to 2. We shall assume first that yzO (mod 2). In Table

II, we give the result (x, y, z) of multiplying y G 5 by (111) and dividing through by

two. Each y gives rise to two possible (x, y, z); where y is already reduced, one of

them has norm congruent to 1 (mod 4) and the other to 3 (mod 4). Where y is not

reduced, one of the (x, y, z) is reduced, and the other is not. Now if N(y) = 2'x,

and if y ~ y' where y' is reduced, then N(y') = 2 (mod 4) iij is odd, and N(y') = x

(mod 4) if j is even. Thus we can "chase the table" to a unique result, for y G 5,

y s 0 (mod 2). (We give an example later.)

Unfortunately, for y G Q, the "table approach" does not work unless 7V(y) = 4'x,

x odd, y odd. Of course, we can construct a table, using the multiplier (110) (Table

III) but the result is four possibilities for (x, y, z); in casej is even, we have found

no simple way to distinguish these in general. We get around the problem by using

the adjoint (described earlier). This works equally well whether y is even or odd; here

we assume y even. Let y = (u, v, w) so that

y «-» G = u

v — 2w

— v + 2w

u — 2v + 4w

We have N(y) = 4jx and y G Q. It is shown in [3] that the Smith form S of G has

the form

1 0 0

0 Vy 0

0 0 Vz

where y \ z and yz = x (indeed, every y G Q, y z 0 (mod 2), must have such a Smith

form and conversely). The Smith form Sx of adj G is then

Vy 0 00 z 0

0 0 2'x



That is, adj G = 2'B, where 5 is an integral and 5 *£ 0 (mod 2). We have 5 «-» (a, b, c)

and (a, o, c) G 5, (a, b, c) z (000) (mod 2). We have

(Müw)-27(aoc) = 47x:,

(u, v, w)- (xa, xb, xc) = 2'x2.

Since j is even, we have (u, v, w) ~ (xa, xb, xc); since x is odd, this is well-de-

termined modulo 4. Finally, since (xa, xb, xc) E 5, we can use Table II.

Now suppose y = 2y,, where y,£0 (mod2), and N(yx) = 2'x (x odd). If y, G 5

and t is even, or if y, G Q and t = 0 (mod 4), then we have y ~ 2y' where N(y') is

odd, and 26 || m from Table V. We now suppose that either y, G 5 and t is odd, or

y, G gandí = 2 (mod 4).

If y, G 5, we use Table II to find y2:

y2 = Y,-ß'/2'.

(Since t is odd, we do not have y, ~ y2; note that N(y2) will be odd.) Now compute

y3, using Table IV:

Y3=y2X (111) (mod4).

Then 2y, X ß'+ '/2'+ ' ~ y3, where y3 is reduced, and we use Table V to find m.

If y, G Q, we use the procedure described previously to find a y' ~ y, with

y' G 5, and then proceed as above.

Example (a). Let y = 5 + 7a + 4a2. Then y = (130) (mod 4), y G 5, y is not

reduced. We find N(y) = 388 = 4 • 97, so we know: y-(lll)/2 is reduced and

y • (111)2/4 must have odd norm, congruent to 1 (mod 4). This gives the sequence

130 ̂ 101-3- (110) -3- (010).

Since (030) is not in Table V, a match for this is ß2 = 22. Since (030) has odd norm,

we have 26 || m.

Example (b). Let y = 9 + 5a + 3a2. Then N(y) = 4s, and we can use Table III.

The sequence is (each arrow represents a single application of the multiplier (110))

113 -> 331 -> 113 - 331 = 3 • (113) -» 333.

We use here the fact that, in Table III, the entries (111), (133), (313) are all

equivalent mod ~ . Then from Table V, 26 11 m.

Example (c). Let y = 5 + 5a + 3a2: N(y) = 24 ■ 17. We have

5 -3 15 5 -33-17

adj G8 5 1

-11 8 5-5 -1 10

Then (5,5,3) • (8, - 11, - 5) = 4 • 17; since 17 = 1 (mod 4),

(5,5,3)-17- (8, -11, -5) =(8, -11,-5) =(0,1,3) (mod4).

Since | adj G | = | G \2 = 28 • 172, we know that A/(8, - 11, -5) = 4 ■ 172. Then from

Table II,

013 ~ 112 -+ 321 ~ 3 • (132) -+ 3 • (032).

Then y ~ y', where y' = (012) (mod 4). From Table V, 22 || m.



Example (d). Let y,

have the sequence:

8 + 3a + a2, y = 2y,. We find N(yx) = 25 X 13, and we

(031) ~ 3 X (112) - 3 X (130) - 3 X (112)-* 3 X (130) -» (303) - (232) ~ (120).

Here, y, = (031) and y2 = (232). Then y3 = (111) X (120) = (333). From Table V,

26||m.

Appendix 1

Let Z — iy + I, and u = (1 + fZ)/2. We denote n + mu by (n, m); in all these

tables, n, m are reduced modulo 4. 5 is the set of integers in Q(fZ). Tables la and

lb give the multiplication in 5/(4), for the two cases y even and y odd, for values of

(n, m) s (0,0) (mod2). Table II gives N(n + mu), reduced modulo 4. For Table

III: given (n, m) z (0,0) (mod2), where N(n + mu) is even. Choose ß in 5 so that

N(ß) = 2o, b=l (mod4) and ß-(n + mu) = 2(a + bu). Write N(n + mu) =

2'x, x odd. Then Table III gives the values of (a, b) modulo 4. For Table IV, we

have X = 2(n + mu) where N(n + mu) = 2x, x odd. Choose ß as for Table III;

X- ß2 = 4(a + bu); we give the values of (a, b) (mod 4).

Table la (y even)

10 20 30 01

10

20

30

01

11

21

31

02

12

22

32

03

13

23

33

10 I 20

00

30

20

10

01

02

03

01

11

11

22

33

02

13

2L

21

02

23

03

20

01

31

31

22

13

00

31

22

13

02 12

02

00

02

02

00

02

00

00

12

20

32

03

11

23

31

02

10

22

22

00

22

00

22

00

22

00

22

00

32 03

32

20

12

01

33

21

13

02

30

22

10

03

02

01

03

02

01

00

02

01

00

03

01

13

13

22

31

00

13

22

31

00

13

22

31

00

13

23 33

23

02

2!

01

20

03

22

02

21

00

23

03

22

01

33

22

11

02

31

20

13

00

33

22

11

02

31

20

13


702 THERESA P VAUGHAN

Table lb ( y odd)

10 20 30 01 11 21 31 02 12 22 32_03 13 23 33

10 10 20 30 01 11 21 31 02 12 22 32 03 13 23 33

20

30

01

11

21

31

02

12

22

32

03

13

23

33

00 20 02 22 02 22 00 20 00 20 02 22 02 22

10 03 33 23 13 02 32 22 12 01 31 21 11

21 22 23 20 02 03 00 01 23 20 21 22

33 00 11 00 11 22 33 22 33 00 11

21 02 02 23 00 21 21 02 23 00

33 00 31 22 13 20 11 02 33

00 02 00 02 02 00 02 00

10 22 30 01 13 21 33

00 22 00 22 00 22

10 03 31 23 11

21 20 23 22

33 02 11

21 00

33

Table II

Z = 8 y + 1

N(n + mu) = r (mod4), (n, m) reduced mod4, not both even.

(n, m) r (y even) r (y odd)

(1,1)(3,3)

(1.3)

(3,1)

(2,1)

(2,3)

(0,1)

(0,3)

(1,0)

(3,0)

(1,2)

(3,2)

2

2

0

0

2

2

0

0

1

1

3

3

0

0

2

2

0

0

2

2

1

1

3

3



Table III

;>i

y even

7=1

(n,m) J ;>2 (n,m) X =1(4) X = 3(4)

(1,3)

(3,1)

(0,1)

(0,3)

(3,3)

(1,1)

(2,3)

(2,1)

(3,1)

(1,3)

(0,1)

(0,3)

(1,1)

(3,3)

(2,1)

(2,3)

(3,0)

(1,0)

(3,0)

(1,0)

(3,2)

(1,2)

(1,2)

(3,2)

;>i

y odd

7=1

(n,m) 7 = 2 >2 (n,m) X =1(4) A- =3(4)

(1,1)

(3,3)

(2,1)

(2,3)

(3,1)

(1,3)

(0,3)(0,1)

(3,3)

(1,1)

(2,1)

(2,3)

(1,3)

(3,1)

(0,1)

(0,3)

(1,0)

(3,0)

(3,0)

(L0)

0,2)(3,2)

(1,2)

(3,2)

Table IV

For these tables: U = 2(n + mu) where N(n + mu) = 2x, x

odd. Let ß = u + 2 if y is even, n and m odd; ß = u + 1 if y

is even, « even, m odd; ß = « if y is odd, « and m odd; and

ß =u — 1 if >> is odd, n even, m odd. Then(ß2/4)£/ = a + ou.

j even

x = 1 (mod 4)

l(n, m)

(LI)a 3)(2,1)

(2,3)

(a, ¿7)

(2,3)

(2,1)

(3,3)(LI)

x = 3 (mod 4)

(«, m)

(1,1)

(3,3)

(2,1)

(2,3)

(a,b)

(2,1)

(2,3)

(1,1)

(3,3)

y odd

x = 1 (mod 4)

(n,m)

(1,3)

(3,1)

(0,1)

(0,3)

(a,b)

(0,1)

(0,3)

(1,3)

(3,1)

x = 3 (mod 4)

(n, m)

(1,3)

(3,1)

(0,1)(0,3)

(a,b)

(0,3)

(0,1)

(3,1)

(L3)



n

odd

odd

odd

even

m

even

even

odd

odd

Table V

(a) Z = 2 (mod 4)

n + m = 1 (mod 4)

n + m = 3 (mod 4)

Exact power of 2

dividing disc S

26

282io

2"

(b) Z = 3(mod4)

Exact power of 2

n m dividing disc S

odd 4j 24

odd 4j + 2 26

even odd 28

odd odd 29

(c) Z = 5 (mod 16)

«

4fc+ 1

4k + 1

4/: + 2

all others with n, m not

both even

2A:

m

4;

4j+l

47 + 3

2 y _/', /: not both even

Exact power of 2

dividing disc S

21 disc S

24

26

(d) Z = 13 (mod 16)

n

4k + 1

4/: + 3

4*

all others with «, m not

both even

2*

m

47

4y+ 1

4;+ 3

2y /, k not both even

Exact power of 2

dividing disc S

21disc S

24

26



(e) Z = 8y + 1

4k + 1 4j

4k+ 3 4j + 2

4k + 1 4^ + 2

4* + 3 4y

2A: 4j + I (k-y odd)

2/r+l 4y + 3 (Jfc-y odd)

2k 47+I (k-y even)

2A:+1 4y +3 (Â:-^ even)

4* + 2 4j

Exact power of 2

dividing disc S

2 \ disc S22

22

24

25

25

23

23

26

Appendix 2

Table I

For this table, (2) = PXP2P3 inR,U= {e},a~ Px2, b

f~P2,c~P22P32,g~P2P2,d. P2P2.

P2

e

a

b

c

d

d

0

0

/d

c

c

g

8

0

0

0

0

d

f

d

f00

8

Table II

For this table, (2) = PQ in 5 where Q has degree 2.

Then U= {e, a, b} ;/, c, d ~ 52 and g ~ Q2.

f 8

e

a

b

c

~d

fg0

d

c

ff

fd

c

c

~d

f

gg

0

0

g

0

ooo

oooo



Table III

For this table, (2) = 5g2in5; U = {e,g},a~Q2,b

c,d~P2.f~P2Q2.

d f g

Q\

d

f0

./'0

0

/

a

bd

0

0

0

0

d

fg0

/0

c

/

e

0

0

0

0

Table IV

For this table, (2) = P3 in R;U = {e, c, d, g) ; a, f~ P2 and ¿7 ~ 54.

/ 0

da

b

fb

0

8fb

0

0

0

0

d

Ifh

c

a

d

0

0

0

0

Appendix 3

Table I

The first column is ß = (abc); the second column is ß • (010)2,

the third is ß-(010)4, the last column gives N(ß) modulo 4.

All entries are given modulo 4.

ß

100

102

120122

110

112

130

132111

131

113

ßa2

001

021

201

221

303

323

103

123

133

333

113

ßa4

230

030

232

032

213

013

211

Oil

313

311

113

N(ß)

1

1

1

1

2

0

0

2

0

0

0


THE DISCRIMINANT OF A QUADRATIC EXTENSION

Table II

ß(lll)/2 = (xyz)

ß110

112

130

132

xyz

010,232

103,321

323,101

012,230

Table III

ß-(110)/2 = (xyz)

ß xyz

111 010,032,212,230

131 021,003,223,201

113 111,133,313,331

X 110 112

Table IV

130 132 111 131 113

102

120

122

312

132

330

310

130

332

332112

310

330110

312

333

333

111

313

313131

331

331113

Table V

We list those y (modulo 4) such that (a) y is reduced relative

to 2, and (b) y has a match ß2 with ß z 0 (mod 2). For each y

we give the corresponding ß, and the power of 2 dividing

N(ß2y).

100,001,230

213,110,303

322,223,012

131,333,311

122,221,032,

320,203,212,

010,023,302

ß

1

(HO)

(111)

(111)

N(ß2y)

odd

23

22

26

If y is reduced and not listed above, then a match for y is ß2 = 22.

Department of Mathematics

University of North Carolina at Greensboro

Greensboro, North Carolina 27412

1. Daniel A. Marcus, Number Fields, Springer-Verlag, New York, 1977.

2. Daniel Shanks, "Dihedral quartic approximations and series for n'\ J. Number Theory, v. 14,

1982, pp. 397-423.

3. Theresa P. Vaughan, On Computing the Discriminant of an Algebraic Number Field. (Preprint.)


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