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
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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,
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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).
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688 THERESA P. VAUGHAN
(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.
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 689
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).
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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.
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 691
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.
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692 THERESA P. VAUGHAN
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).
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 693
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.
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694 THERESA P. VAUGHAN
(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.
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 695
K
H
+
2 l
hK
i
vs fs<N —
+ Xr*l OS<S iCM
K+IN
OS
+ 5<
IEx
IE>
^
un
SO — _*
*L+ rn
& x
C: X "".
+
b s u
Ol
i
X
£X— 04
S X— r»
X
 ̂:¿sxtN (N
Ex a
3OsOs
SO x
S x— rs
+ ■* ^x
£: X ~
bs
X■»
!
X
I
X
I
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696 THERESA P. VAUGHAN
\
+ X _- tx OSO i _~(N I w
3 X
+ <»
■*
+ -£ xCI
3
+SO 00
oo ^
t S
I+OS
«i
ios-m »
f- ' t»)Tj- (N w
f> 2!
3 £
os rsi ^
3— <N
±xin rsj
X
I ■*
<n
Îm
Î
1
c<s rs|
X
I
X
!
3
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 697
— ** C_TO (N
P I ci
3 x
Z> xo «
- I O
m3 ^
en xOO ON
Tx
S x
+ 2r- s
Os Tf
3
en
+ 3soX
+ X
be
^
^
I
^
^
^
a
^r- m
+ -
E X
^
+ -os X
+ X -
+Ex
+
Ex
Xr-I
XIX
I
î
O
r
Xtx
I
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698 THERESA P. VAUGHAN
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.
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 699
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
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700 THERESA P. VAUGHAN
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.
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 701
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
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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
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 703
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)
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704 THERESA P. VAUGHAN
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
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THE DISCRIMINANT OF A QUADRATIC EXTENSION 705
(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
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706 THERESA P. VAUGHAN
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
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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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