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Characterization of Hermitian Trace Forms
210, 690]696 Ž1998. JA987600
JOURNAL OF ALGEBRA ARTICLE NO.
Characterization of Hermitian Trace Forms Gregory Berhuy ´ Equipe de Mathematiques, Uni¨ ersite´ Besanc ¸on, Besanc¸on, France ´ Communicated by E¨ a Bayer-Fluckiger Received February 20, 1998
INTRODUCTION In w2x, Epkenhans obtained a characterization of quadratic forms that can be realized as trace forms of number fields Žsee Theorem 1 below.. The aim of this paper is to prove a similar result for hermitian trace forms.
DEFINITIONS AND NOTATION Let K be a field of characteristic different from 2, and let q be a quadratic form over K. If P is an ordering on K, then sign P q denotes the signature of q at the ordering P. A quadratic form q over K is called positi¨ e if sign P q G 0 for all orderings P of K. If K has no ordering, then all quadratic forms are positive. In this paper, the words ‘‘quadratic form’’ are reserved to mean ‘‘nondegenerate quadratic form.’’ If two forms q1 and q2 over K are isomorphic Žrespectively Witt equivalent., we write q1 , q2 Žrespectively q1 ; q2 .. If E is a finite separable extension of K, the trace form of ErK is the form E ª K, x ¬ TrEr K Ž x 2 .. Moreover, if s is a nontrivial involution of E over K, the hermitian trace form of ErK relati¨ e to s is the form E ª K, x ¬ TrEr K Ž xx s .. H will denote the hyperbolic plane cx 2 y cy 2 for any c g K *. Finally, we say that a finite field extension E of K has property Ž Q . if there exists a field extension E0 of K and a square-class a / "1 in E0U rE0U 2 such that E s E0 Ž'a .. 690 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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CHARACTERIZATION OF TRACE FORMS Žsee w2x. to In 1992, Epkenhans used results of Mestre and Kruskemper ¨ prove the following theorem: THEOREM 1. Let K be a number field. Let q be a quadratic form of dimension n o¨ er K. Then q is isomorphic to the trace form of a finite field extension F of K if and only if q is positi¨ e and any one of the following holds: 1. 2. 3. 4.
n s 1 and q , ²1:. n s 2, q , ²2, 2 D :, with some D g K * y K *2 . n s 3, q , ²1, 2, 2 D :, with some D g K *. n G 4.
CHARACTERIZATION OF HERMITIAN TRACE FORMS THEOREM 2. Let K be a number field. Let q be an e¨ en-dimensional quadratic form o¨ er K. Then q is isomorphic to a hermitian trace form of a finite field extension of K if and only if q is positi¨ e and 1. dim q s 2, q , ²2, y2 D :, with some D g K * y K * 2 . 2. dim q G 4. If LrK is a finite separable extension of K, l g L*, and P is an ordering of K, let EL r K Ž P . be the set of the orderings of L that extend P, and let HL r K Ž l. be the set of the orderings of L such that l is positive. Then sign P Tr L r K Ž² l:. s aEL r F Ž P . l H L r K Ž l. s aEL r F Ž P . l HL r K Žyl. Žsee w5, 3.4.5x.. An easy computation gives TrEr K Ž xx s . s TrE 0 r K Ž²2:. H yTrE 0 r K Ž²2 d :., where E0 is the fixed field of s and E s E0 Ž'd .. So we get sign P TrEr K Ž xx s . s aEE 0 r K Ž P . y aEE 0 r K Ž P . l HE 0 r K Ž d . q aEE 0 r K Ž P . l HE 0 r K Žyd . G 0, for all orderings P of K. Hence a hermitian trace form is positive. Moreover, the first statement of the theorem is easy and is left to the reader. The proof of the second statement is based on the following lemmas. LEMMA 1. If ErK has property Ž Q ., then the trace form of ErK is isometric to a hermitian trace form. Proof. By assumption, E s E0 Ž'a ., with a / "1 in E0U rE0U 2 . Let M s E0 Ž'ya .. We define s : M ª M by 'ya ¬ y 'ya , and s < E 0 s Id. It is a nontrivial involution, with fixed field E0 . We easily get TrEr K Ž²1:. s TrE 0 r K Ž²2:. H TrE 0 r K Ž²2 a:.. We also have TrM r K Ž xx s . s TrE 0 r K Ž²2:. H yTrE 0 r K Ž² y 2 a:.. So we get the desired isomorphism.
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LEMMA 2. Let K be a number field. If q , mH, m ) 1 odd, then q is isomorphic to a hermitian trace form of a field extension of K. Proof. The proof consists of four parts. First, we prove that a certain polynomial is irreducible in Cw X, Y x. Then, we construct a l algebraic over K such that l f K Ž l.* 2 . Next, we determine the trace form of the extension K Ž l.rK. Finally, we prove that TrK Ž l.r K Ž² l:. , TrK Ž l.r K Ž²1:.. From these four facts, established below, it follows easily that TrK Ž 'l .r K Ž xx s . is isometric to m copies of a hyperbolic plane. Before starting the proof, we recall some facts about Newton polygons. Let Ž L, ¨ . be a discretely valuated field such that ¨ Ž L. s Z j q`4 . Let P Ž X . s Ý nisk a i X i be a nonzero polynomial in Lw X x with a n / 0 and a k / 0. The Newton polygon of P is the convex envelope of Ž i, ¨ Ž a i .., i s k, . . . , n, such that a i / 04 . If none of the points of the Newton polygon is in Z 2 , then P is irreducible. Let i j be the integers such that the points Ž i j , x i j . of the Newton polygon are in Z 2 . Then the different sums of the numbers i jq1 y i j correspond to the degrees of the possible irreducible divisors of P. For more details on Newton polygons, see w3x, for example. In w3x, Gouvea ˆ considers Newton polygons of polynomials in C p w X x, but most of the results can be generalized to the case of polynomials with coefficients in any discretely valuated field. First, we show that RŽ X, Y . s Ž X 2 q Ž m y 1.Ž Y 2 y m m .. m y mŽ Y ym m .Ž X 2 q Ž my1.Ž Y 2 ym m .. my 1 y m m Ž my1. my 1 Ž Y 2 ym m . my 1 is an irreducible polynomial in Cw X, Y x. If we consider R in CŽ Y .w X x, the Newton polygon of R relative to ¨ Yy ' m is Ž x, m y 1 y ŽŽ m y m 1.r2 m. x ., x g w0, 2 m x4 . Now, we compute the intersection between Z 2 and the Newton polygon to know the possible decompositions of R. We have m y 1 y ŽŽ m y 1.r2 m. x g Z if and only if ŽŽ m y 1.r2. x ' 0 Žmod m.. Since m y 1 and m are relatively prime, Ž m y 1.r2 and m are relatively prime too, and the previous equation is equivalent to x g mZ. So we finally get x s 0, m, and 2 m. This implies that if R is reducible, then the degree of an irreducible factor of R is m. Suppose we are in this case. We have RŽyX, Y . s RŽ X, Y .. This implies that, if I Ž X, Y . is a monic irreducible factor of degree m in X, I ŽyX, Y . is also an irreducible factor of degree m in X dividing RŽ X, Y .. Furthermore, R is a monic polynomial in X, with coefficients in Cw Y x. So, R is irreducible in CŽ Y .w X x if and only if it is irreducible in Cw X, Y x. So we can suppose that I g Cw X, Y x. v
2
Now we show I ŽyX, Y . / cI Ž X, Y .. If I ŽyX, Y . s cI Ž X, Y ., we have I Ž X, Y . s c 2 I Ž X, Y ., so we have c s "1. We have I ŽyX, Y . / I Ž X, Y . since deg X Ž I . s m is odd. If I ŽyX, Y . s yI Ž X, Y ., we have I Ž0, Y . s 0,
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so RŽ0, Y . s 0, which is not the case. Finally, I ŽyX, Y . / cI Ž X, Y .. Comparing the leading coefficient relative to X, we get RŽ X, Y . s yI Ž X, Y . I ŽyX, Y .. This implies deg Y Ž I . s m. We have RŽ X, yY . s RŽ X, Y ., so I Ž X, yY . I ŽyX, yY . s I Ž X, Y . I ŽyX, Y .. Thus we have I Ž X, yY . s a I Ž X, Y . or b I ŽyX, Y .. As previously, we get a s "1 or b s "1, respectively. We have I Ž X, yY . / I Ž X, Y . since deg Y Ž I . is odd. If I Ž X, yY . s yI Ž X, Y ., we have I Ž X, 0. s 0, so RŽ X, 0. s 0, which is not the case. I Ž X, yY . / I ŽyX, Y ., since deg X Ž I . is odd. So I Ž X, yY . s yI ŽyX, Y .. This implies RŽ X, Y . s I Ž X, Y . I Ž X, yY .. Then we have RŽ X, 0. s I Ž X, 0. 2 in Cw X x. Since RŽ X, 0. s RŽ X, 0., we get I Ž X , 0 . s "I Ž X, 0., that is, I Ž X, 0. g Rw X x or I Ž X, 0. g iRw X x. Since I Ž X, Y . is monic relative to X, we get I Ž X, 0. g Rw X x. Thus we have RŽ X, 0. s I Ž X, 0. 2 in Rw X x. 2 Since RŽ Ž m y 1 . m m , 0. s ym m Ž m y 1. my1 - 0, we get a contradiction.
'
RŽ X, Y . is irreducible in Cw X, Y x, so it is also irreducible in K w X, Y x. By Hilbert’s irreducibility theorem, there is a y g K such that RŽ X, y . is irreducible. Let Q g K w X x, such that QŽ X 2 . s RŽ X, y .. Then QŽ X 2 . is irreducible, so QŽ X . is also irreducible. Let a s y 2 y m m and let ly1 g C be a root of Q. We have v
Q Ž X . s Irr Ž ly1 , K . s Ž X q Ž m y 1 . a . yam Ž X q Ž m y 1 . a .
my 1
m
y m m Ž m y 1.
my1
a my1 .
We now prove that l is not a square in K Ž l.. Suppose that l s b 2 , b g K Ž l.. We have K Ž l. s K Ž b 2 . s K Ž b . s K Ž by 1 ., so degŽ Irr Ž by1 , K .. s m. We have QŽ by2 . s 0, so B Ž X . s Irr Ž by1 , K . is an irreducible factor of QŽ X 2 .. Since QŽ X 2 . is irreducible, we get QŽ X 2 . s B Ž X .. Since QŽ X 2 . and B Ž X . have different degrees we get a contradiction. Let u s amŽ m y 1. lrŽ1 q Ž m y 1. a l.. Then l s urŽ m y 1. aŽ m y u . and thus K Ž l. s K Ž u .. One checks that for P Ž X . [ X m q aX q Ž1 y m. a, we have P Ž u . s 0. Since deg P s m s w K Ž l. : K x, we have Irr Ž u , K . s P Ž X .. Furthermore, discŽ P . s Žy1. Ž my1.r2 Ž a q m m . s Žy1.Ž my1.r2 y 2 Žsee w6, Annex IIx.. By Serre Žsee w6, Annex II, Proposition 6x., we have TrK Ž u .r K Ž²1:. , ²1: H ŽŽ m y 1.r2.H, that is, TrK Ž l.r K Ž²1:. , ²1: H ŽŽ m y 1.r2.H. v
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694 v
1 Ý my is1
Recall that u m q au q Ž1 y m . a s 0. From the identity m my1yiu i s u Ž m my 1 y u my1 .rŽ m y u ., we get my1
Ý
m my 1yiu i s
m my 1u q au q Ž 1 y m . a myu
is1
.
Using this, a straightforward computation yields m
Ž a q m m . Ž m y 1. a
ž
Ž m y 1. a m
my1
q
Ý
m my 1yiu i s l .
is1
/
Using the relations Žrecalled in w1, Lemma VI.2.3x. TrK Ž u .r K Ž u i .
¡0
Ž 1 y m. a s~ m Ž m y 1. a Ž m y 1 . a2
¢
for 1 F i F m y 2 and m q 1 F i F 2 m y 3 for i s m y 1 for i s m for i s 2 m y 2,
we get TrK Ž u .r K Ž lu i . s 0 for 0 F i F m y 2 and TrK Ž u .r K Ž lu my 1 . s 1. Then 1, u , . . . , u Ž my3.r2 span a totally isotropic subspace of dimension Ž m y 1.r2 of Ž K Ž u ., TrK Ž u .r K Ž²1:... So, TrK Ž u .r K Ž² l:. , ²Žy1.Ž my1.r2 d : H ŽŽ m y 1.r2.H, where d s detŽTrK Ž u .r K Ž² l:.. in K *rK * 2 . Writing the matrix of TrK Ž u .r K Ž² l:. relative to the basis 1, u , . . . , u my 1 , we get d s Žy1.Ž my1.r2 , so TrK Ž l .r K Ž² l:. s TrK Ž u .r K Ž² l:. , ²1: H ŽŽ m y 1.r2.H. Thus TrK Ž l.r K Ž² l:. , TrK Ž l .r K Ž²1:.. Finally, let define s : K Ž'l . ª K Ž'l . by 'l ¬ y 'l and s < K Ž l. s Id. Since l f K Ž l.* 2 , we get TrK Ž 'l .r K Ž xx s . s TrK Ž l.r K Ž²2:. H yTrK Ž l .r K Ž²2 l:. , mH. LEMMA 3. Let F be a number field. If f is a positi¨ e quadratic form of dimension 4 o¨ er F, then f is isomorphic to a trace form of an extension ErK ha¨ ing property Ž Q .. In particular, f is isomorphic to a hermitian trace form. Proof. Epkenhans proved that every positive quadratic form of dimension 4 over a number field F is isomorphic to a trace form of a field extension F Ž u .rF, with Irr Ž u , F . s X 4 q a X 2 q b Žsee w2, Lemma 5x.. Let E0 s F Ž u 2 .. We have w F Ž u .: E0 x s 2, and w E0 : F x s 2. Now we show u 2 / y1 in E0UrE0U 2 . By way of contradiction, assume u 2 s yc 2 with some c g E0U . We have degŽ Irr Ž c, F .. F 2, since c g E0U and w E0 : F x s 2. On the other hand, degŽ Irr Ž c, F .. / 1; otherwise c g F, and u 2 g F, which implies w E0 : F x s 1. So degŽ Irr Ž c, F .. s 2. Since u 4 q au 2 q b s 0,
CHARACTERIZATION OF HERMITIAN TRACE FORMS
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we have c 4 y a c 2 q b s 0. Then Irr Ž c, F . divides X 4 y a X 2 q b . We have X 4 y a X 2 q b s Ž X 2 y Ž a q 'D .r2.Ž X 2 y Ž a y 'D .r2., with D s a 2 y 4b . Since this polynomial has an irreducible factor of degree 2, this implies Irr Ž c, F . s X 2 y Ž a " 'D .r2, and we get 'D g F, since Irr Ž c, F . g F w X x. Since X 4 q a X 2 q b s Ž X 2 y Žya q 'D .r2.Ž X 2 q Ž a q 'D .r2., this implies Irr Ž u , F . is reducible over F, and we get a contradiction. Thus, F Ž u . has property ŽQ., and we can apply Lemma 1. Before proving the theorem, we recall the following lemma, proved by Žsee w4, Lemma 2x.: Kruskemper ¨ LEMMA 4. Let q be a positi¨ e form o¨ er K with dim q s kr, k G 1, r G 3. Then there exists an extension LrK of degree k and a positi¨ e form r o¨ er L with dim r s r such that TrL r K Ž r . , q. In w4x, Kruskemper also proved the following results Žsee w4, proof of ¨ Proposition 2x.. LEMMA 5.
Let m ) 1 odd.
1. If y1 f K *2 , there exists a field extension LrK of degree m, and an a g L* y L* 2 such that Ža. The Witt class of q y TrL r K Ž²2:. in W Ž K . is represented by a quadratic form r of dimension m. Žb. r , TrL r K Ž²2 a:.. Žc. q , TrLŽ 'a .r K Ž²1:.. 2. If y1 g K *2 , there exists a field extension LrK and a b g L* y L* 2 such that q , TrLŽ 'b .r K Ž²1:. Ž see w4, proof of Proposition 2x.. Proof of Theorem 2. Let q be a positive quadratic form of dimension m, with m ) 1. We discuss three cases: m ) 1 odd, m s 2, m ) 2 even. m ) 1 odd. If q ; 0, apply Lemma 2. Now, assume that q ¤ 0. If y1 g K * 2 , let LrK, r , and a be as in the first statement of Lemma 5. Let us show that a / y1 in L*rL* 2 . By way of contradiction, assume a s y1. We have r , TrL r K Ž² y 2:., so r ; yTrL r K Ž²2:.. Since r ; q y TrL r K Ž²2:., this implies q ; 0 in W Ž K ., and we get a contradiction. So, LŽ'a . has property ŽQ., and we can apply Lemma 1. v
If y1 g K * 2 , let LrK and b be as in the second statement of Lemma 5. Since y1 g K *2 : L* 2 , we get b / y1 in L*rL* 2 . Thus, we can conclude as before. m s 2, so dim q s 4. Apply Lemma 3 with F s K. m ) 2 even. Apply Lemma 4, with k s mr2 and r s 4. There exists an extension LrK of degree mr2 and a positive form r over L with v v
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dim r s 4 such that TrL r K Ž r . , q. Now, apply Lemma 3 with F s L and f s r . There exists an extension E of L having property ŽQ. such that r , TrEr LŽ²1:.. Then we have TrL r K ŽTrEr LŽ²1:.. , q; that is, TrEr K Ž²1:. , q. Now, apply Lemma 1.
ACKNOWLEDGMENTS I thank Eva Bayer-Fluckiger, Vincent Fleckinger, Detlev Hoffmann, and Robert Perlis for fruitful conversations.
REFERENCES 1. P. E. Conner and R. Perlis, A survey of trace forms of algebraic number fields, World Scientific, Singapore, 1984. 2. M. Epkenhans, On trace forms of algebraic number fields, Arch. Math. 60 Ž6. Ž1993., 527]529. 3. F. Q. Gouvea, ˆ ‘‘ p-adic Numbers: An Introduction,’’ pp. 199]215, Universitext, SpringerVerlag, BerlinrNew York, 1997. 4. M. Kruskemper, Algebraic number field extensions with prescribed trace form, J. Number ¨ Theory 40 Ž1992., 120]124. 5. W. Scharlau, ‘‘Quadratic and Hermitian Forms,’’ Springer-Verlag, HeidelbergrNew York, 1985. 6. J.-P. Serre, L’invariant de Witt de la forme TrŽ x 2 ., Comment. Math. Hel¨ . 59 Ž1984..
JOURNAL OF ALGEBRA ARTICLE NO.
Characterization of Hermitian Trace Forms Gregory Berhuy ´ Equipe de Mathematiques, Uni¨ ersite´ Besanc ¸on, Besanc¸on, France ´ Communicated by E¨ a Bayer-Fluckiger Received February 20, 1998
INTRODUCTION In w2x, Epkenhans obtained a characterization of quadratic forms that can be realized as trace forms of number fields Žsee Theorem 1 below.. The aim of this paper is to prove a similar result for hermitian trace forms.
DEFINITIONS AND NOTATION Let K be a field of characteristic different from 2, and let q be a quadratic form over K. If P is an ordering on K, then sign P q denotes the signature of q at the ordering P. A quadratic form q over K is called positi¨ e if sign P q G 0 for all orderings P of K. If K has no ordering, then all quadratic forms are positive. In this paper, the words ‘‘quadratic form’’ are reserved to mean ‘‘nondegenerate quadratic form.’’ If two forms q1 and q2 over K are isomorphic Žrespectively Witt equivalent., we write q1 , q2 Žrespectively q1 ; q2 .. If E is a finite separable extension of K, the trace form of ErK is the form E ª K, x ¬ TrEr K Ž x 2 .. Moreover, if s is a nontrivial involution of E over K, the hermitian trace form of ErK relati¨ e to s is the form E ª K, x ¬ TrEr K Ž xx s .. H will denote the hyperbolic plane cx 2 y cy 2 for any c g K *. Finally, we say that a finite field extension E of K has property Ž Q . if there exists a field extension E0 of K and a square-class a / "1 in E0U rE0U 2 such that E s E0 Ž'a .. 690 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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CHARACTERIZATION OF TRACE FORMS Žsee w2x. to In 1992, Epkenhans used results of Mestre and Kruskemper ¨ prove the following theorem: THEOREM 1. Let K be a number field. Let q be a quadratic form of dimension n o¨ er K. Then q is isomorphic to the trace form of a finite field extension F of K if and only if q is positi¨ e and any one of the following holds: 1. 2. 3. 4.
n s 1 and q , ²1:. n s 2, q , ²2, 2 D :, with some D g K * y K *2 . n s 3, q , ²1, 2, 2 D :, with some D g K *. n G 4.
CHARACTERIZATION OF HERMITIAN TRACE FORMS THEOREM 2. Let K be a number field. Let q be an e¨ en-dimensional quadratic form o¨ er K. Then q is isomorphic to a hermitian trace form of a finite field extension of K if and only if q is positi¨ e and 1. dim q s 2, q , ²2, y2 D :, with some D g K * y K * 2 . 2. dim q G 4. If LrK is a finite separable extension of K, l g L*, and P is an ordering of K, let EL r K Ž P . be the set of the orderings of L that extend P, and let HL r K Ž l. be the set of the orderings of L such that l is positive. Then sign P Tr L r K Ž² l:. s aEL r F Ž P . l H L r K Ž l. s aEL r F Ž P . l HL r K Žyl. Žsee w5, 3.4.5x.. An easy computation gives TrEr K Ž xx s . s TrE 0 r K Ž²2:. H yTrE 0 r K Ž²2 d :., where E0 is the fixed field of s and E s E0 Ž'd .. So we get sign P TrEr K Ž xx s . s aEE 0 r K Ž P . y aEE 0 r K Ž P . l HE 0 r K Ž d . q aEE 0 r K Ž P . l HE 0 r K Žyd . G 0, for all orderings P of K. Hence a hermitian trace form is positive. Moreover, the first statement of the theorem is easy and is left to the reader. The proof of the second statement is based on the following lemmas. LEMMA 1. If ErK has property Ž Q ., then the trace form of ErK is isometric to a hermitian trace form. Proof. By assumption, E s E0 Ž'a ., with a / "1 in E0U rE0U 2 . Let M s E0 Ž'ya .. We define s : M ª M by 'ya ¬ y 'ya , and s < E 0 s Id. It is a nontrivial involution, with fixed field E0 . We easily get TrEr K Ž²1:. s TrE 0 r K Ž²2:. H TrE 0 r K Ž²2 a:.. We also have TrM r K Ž xx s . s TrE 0 r K Ž²2:. H yTrE 0 r K Ž² y 2 a:.. So we get the desired isomorphism.
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LEMMA 2. Let K be a number field. If q , mH, m ) 1 odd, then q is isomorphic to a hermitian trace form of a field extension of K. Proof. The proof consists of four parts. First, we prove that a certain polynomial is irreducible in Cw X, Y x. Then, we construct a l algebraic over K such that l f K Ž l.* 2 . Next, we determine the trace form of the extension K Ž l.rK. Finally, we prove that TrK Ž l.r K Ž² l:. , TrK Ž l.r K Ž²1:.. From these four facts, established below, it follows easily that TrK Ž 'l .r K Ž xx s . is isometric to m copies of a hyperbolic plane. Before starting the proof, we recall some facts about Newton polygons. Let Ž L, ¨ . be a discretely valuated field such that ¨ Ž L. s Z j q`4 . Let P Ž X . s Ý nisk a i X i be a nonzero polynomial in Lw X x with a n / 0 and a k / 0. The Newton polygon of P is the convex envelope of Ž i, ¨ Ž a i .., i s k, . . . , n, such that a i / 04 . If none of the points of the Newton polygon is in Z 2 , then P is irreducible. Let i j be the integers such that the points Ž i j , x i j . of the Newton polygon are in Z 2 . Then the different sums of the numbers i jq1 y i j correspond to the degrees of the possible irreducible divisors of P. For more details on Newton polygons, see w3x, for example. In w3x, Gouvea ˆ considers Newton polygons of polynomials in C p w X x, but most of the results can be generalized to the case of polynomials with coefficients in any discretely valuated field. First, we show that RŽ X, Y . s Ž X 2 q Ž m y 1.Ž Y 2 y m m .. m y mŽ Y ym m .Ž X 2 q Ž my1.Ž Y 2 ym m .. my 1 y m m Ž my1. my 1 Ž Y 2 ym m . my 1 is an irreducible polynomial in Cw X, Y x. If we consider R in CŽ Y .w X x, the Newton polygon of R relative to ¨ Yy ' m is Ž x, m y 1 y ŽŽ m y m 1.r2 m. x ., x g w0, 2 m x4 . Now, we compute the intersection between Z 2 and the Newton polygon to know the possible decompositions of R. We have m y 1 y ŽŽ m y 1.r2 m. x g Z if and only if ŽŽ m y 1.r2. x ' 0 Žmod m.. Since m y 1 and m are relatively prime, Ž m y 1.r2 and m are relatively prime too, and the previous equation is equivalent to x g mZ. So we finally get x s 0, m, and 2 m. This implies that if R is reducible, then the degree of an irreducible factor of R is m. Suppose we are in this case. We have RŽyX, Y . s RŽ X, Y .. This implies that, if I Ž X, Y . is a monic irreducible factor of degree m in X, I ŽyX, Y . is also an irreducible factor of degree m in X dividing RŽ X, Y .. Furthermore, R is a monic polynomial in X, with coefficients in Cw Y x. So, R is irreducible in CŽ Y .w X x if and only if it is irreducible in Cw X, Y x. So we can suppose that I g Cw X, Y x. v
2
Now we show I ŽyX, Y . / cI Ž X, Y .. If I ŽyX, Y . s cI Ž X, Y ., we have I Ž X, Y . s c 2 I Ž X, Y ., so we have c s "1. We have I ŽyX, Y . / I Ž X, Y . since deg X Ž I . s m is odd. If I ŽyX, Y . s yI Ž X, Y ., we have I Ž0, Y . s 0,
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CHARACTERIZATION OF HERMITIAN TRACE FORMS
so RŽ0, Y . s 0, which is not the case. Finally, I ŽyX, Y . / cI Ž X, Y .. Comparing the leading coefficient relative to X, we get RŽ X, Y . s yI Ž X, Y . I ŽyX, Y .. This implies deg Y Ž I . s m. We have RŽ X, yY . s RŽ X, Y ., so I Ž X, yY . I ŽyX, yY . s I Ž X, Y . I ŽyX, Y .. Thus we have I Ž X, yY . s a I Ž X, Y . or b I ŽyX, Y .. As previously, we get a s "1 or b s "1, respectively. We have I Ž X, yY . / I Ž X, Y . since deg Y Ž I . is odd. If I Ž X, yY . s yI Ž X, Y ., we have I Ž X, 0. s 0, so RŽ X, 0. s 0, which is not the case. I Ž X, yY . / I ŽyX, Y ., since deg X Ž I . is odd. So I Ž X, yY . s yI ŽyX, Y .. This implies RŽ X, Y . s I Ž X, Y . I Ž X, yY .. Then we have RŽ X, 0. s I Ž X, 0. 2 in Cw X x. Since RŽ X, 0. s RŽ X, 0., we get I Ž X , 0 . s "I Ž X, 0., that is, I Ž X, 0. g Rw X x or I Ž X, 0. g iRw X x. Since I Ž X, Y . is monic relative to X, we get I Ž X, 0. g Rw X x. Thus we have RŽ X, 0. s I Ž X, 0. 2 in Rw X x. 2 Since RŽ Ž m y 1 . m m , 0. s ym m Ž m y 1. my1 - 0, we get a contradiction.
'
RŽ X, Y . is irreducible in Cw X, Y x, so it is also irreducible in K w X, Y x. By Hilbert’s irreducibility theorem, there is a y g K such that RŽ X, y . is irreducible. Let Q g K w X x, such that QŽ X 2 . s RŽ X, y .. Then QŽ X 2 . is irreducible, so QŽ X . is also irreducible. Let a s y 2 y m m and let ly1 g C be a root of Q. We have v
Q Ž X . s Irr Ž ly1 , K . s Ž X q Ž m y 1 . a . yam Ž X q Ž m y 1 . a .
my 1
m
y m m Ž m y 1.
my1
a my1 .
We now prove that l is not a square in K Ž l.. Suppose that l s b 2 , b g K Ž l.. We have K Ž l. s K Ž b 2 . s K Ž b . s K Ž by 1 ., so degŽ Irr Ž by1 , K .. s m. We have QŽ by2 . s 0, so B Ž X . s Irr Ž by1 , K . is an irreducible factor of QŽ X 2 .. Since QŽ X 2 . is irreducible, we get QŽ X 2 . s B Ž X .. Since QŽ X 2 . and B Ž X . have different degrees we get a contradiction. Let u s amŽ m y 1. lrŽ1 q Ž m y 1. a l.. Then l s urŽ m y 1. aŽ m y u . and thus K Ž l. s K Ž u .. One checks that for P Ž X . [ X m q aX q Ž1 y m. a, we have P Ž u . s 0. Since deg P s m s w K Ž l. : K x, we have Irr Ž u , K . s P Ž X .. Furthermore, discŽ P . s Žy1. Ž my1.r2 Ž a q m m . s Žy1.Ž my1.r2 y 2 Žsee w6, Annex IIx.. By Serre Žsee w6, Annex II, Proposition 6x., we have TrK Ž u .r K Ž²1:. , ²1: H ŽŽ m y 1.r2.H, that is, TrK Ž l.r K Ž²1:. , ²1: H ŽŽ m y 1.r2.H. v
GREGORY BERHUY ´
694 v
1 Ý my is1
Recall that u m q au q Ž1 y m . a s 0. From the identity m my1yiu i s u Ž m my 1 y u my1 .rŽ m y u ., we get my1
Ý
m my 1yiu i s
m my 1u q au q Ž 1 y m . a myu
is1
.
Using this, a straightforward computation yields m
Ž a q m m . Ž m y 1. a
ž
Ž m y 1. a m
my1
q
Ý
m my 1yiu i s l .
is1
/
Using the relations Žrecalled in w1, Lemma VI.2.3x. TrK Ž u .r K Ž u i .
¡0
Ž 1 y m. a s~ m Ž m y 1. a Ž m y 1 . a2
¢
for 1 F i F m y 2 and m q 1 F i F 2 m y 3 for i s m y 1 for i s m for i s 2 m y 2,
we get TrK Ž u .r K Ž lu i . s 0 for 0 F i F m y 2 and TrK Ž u .r K Ž lu my 1 . s 1. Then 1, u , . . . , u Ž my3.r2 span a totally isotropic subspace of dimension Ž m y 1.r2 of Ž K Ž u ., TrK Ž u .r K Ž²1:... So, TrK Ž u .r K Ž² l:. , ²Žy1.Ž my1.r2 d : H ŽŽ m y 1.r2.H, where d s detŽTrK Ž u .r K Ž² l:.. in K *rK * 2 . Writing the matrix of TrK Ž u .r K Ž² l:. relative to the basis 1, u , . . . , u my 1 , we get d s Žy1.Ž my1.r2 , so TrK Ž l .r K Ž² l:. s TrK Ž u .r K Ž² l:. , ²1: H ŽŽ m y 1.r2.H. Thus TrK Ž l.r K Ž² l:. , TrK Ž l .r K Ž²1:.. Finally, let define s : K Ž'l . ª K Ž'l . by 'l ¬ y 'l and s < K Ž l. s Id. Since l f K Ž l.* 2 , we get TrK Ž 'l .r K Ž xx s . s TrK Ž l.r K Ž²2:. H yTrK Ž l .r K Ž²2 l:. , mH. LEMMA 3. Let F be a number field. If f is a positi¨ e quadratic form of dimension 4 o¨ er F, then f is isomorphic to a trace form of an extension ErK ha¨ ing property Ž Q .. In particular, f is isomorphic to a hermitian trace form. Proof. Epkenhans proved that every positive quadratic form of dimension 4 over a number field F is isomorphic to a trace form of a field extension F Ž u .rF, with Irr Ž u , F . s X 4 q a X 2 q b Žsee w2, Lemma 5x.. Let E0 s F Ž u 2 .. We have w F Ž u .: E0 x s 2, and w E0 : F x s 2. Now we show u 2 / y1 in E0UrE0U 2 . By way of contradiction, assume u 2 s yc 2 with some c g E0U . We have degŽ Irr Ž c, F .. F 2, since c g E0U and w E0 : F x s 2. On the other hand, degŽ Irr Ž c, F .. / 1; otherwise c g F, and u 2 g F, which implies w E0 : F x s 1. So degŽ Irr Ž c, F .. s 2. Since u 4 q au 2 q b s 0,
CHARACTERIZATION OF HERMITIAN TRACE FORMS
695
we have c 4 y a c 2 q b s 0. Then Irr Ž c, F . divides X 4 y a X 2 q b . We have X 4 y a X 2 q b s Ž X 2 y Ž a q 'D .r2.Ž X 2 y Ž a y 'D .r2., with D s a 2 y 4b . Since this polynomial has an irreducible factor of degree 2, this implies Irr Ž c, F . s X 2 y Ž a " 'D .r2, and we get 'D g F, since Irr Ž c, F . g F w X x. Since X 4 q a X 2 q b s Ž X 2 y Žya q 'D .r2.Ž X 2 q Ž a q 'D .r2., this implies Irr Ž u , F . is reducible over F, and we get a contradiction. Thus, F Ž u . has property ŽQ., and we can apply Lemma 1. Before proving the theorem, we recall the following lemma, proved by Žsee w4, Lemma 2x.: Kruskemper ¨ LEMMA 4. Let q be a positi¨ e form o¨ er K with dim q s kr, k G 1, r G 3. Then there exists an extension LrK of degree k and a positi¨ e form r o¨ er L with dim r s r such that TrL r K Ž r . , q. In w4x, Kruskemper also proved the following results Žsee w4, proof of ¨ Proposition 2x.. LEMMA 5.
Let m ) 1 odd.
1. If y1 f K *2 , there exists a field extension LrK of degree m, and an a g L* y L* 2 such that Ža. The Witt class of q y TrL r K Ž²2:. in W Ž K . is represented by a quadratic form r of dimension m. Žb. r , TrL r K Ž²2 a:.. Žc. q , TrLŽ 'a .r K Ž²1:.. 2. If y1 g K *2 , there exists a field extension LrK and a b g L* y L* 2 such that q , TrLŽ 'b .r K Ž²1:. Ž see w4, proof of Proposition 2x.. Proof of Theorem 2. Let q be a positive quadratic form of dimension m, with m ) 1. We discuss three cases: m ) 1 odd, m s 2, m ) 2 even. m ) 1 odd. If q ; 0, apply Lemma 2. Now, assume that q ¤ 0. If y1 g K * 2 , let LrK, r , and a be as in the first statement of Lemma 5. Let us show that a / y1 in L*rL* 2 . By way of contradiction, assume a s y1. We have r , TrL r K Ž² y 2:., so r ; yTrL r K Ž²2:.. Since r ; q y TrL r K Ž²2:., this implies q ; 0 in W Ž K ., and we get a contradiction. So, LŽ'a . has property ŽQ., and we can apply Lemma 1. v
If y1 g K * 2 , let LrK and b be as in the second statement of Lemma 5. Since y1 g K *2 : L* 2 , we get b / y1 in L*rL* 2 . Thus, we can conclude as before. m s 2, so dim q s 4. Apply Lemma 3 with F s K. m ) 2 even. Apply Lemma 4, with k s mr2 and r s 4. There exists an extension LrK of degree mr2 and a positive form r over L with v v
696
GREGORY BERHUY ´
dim r s 4 such that TrL r K Ž r . , q. Now, apply Lemma 3 with F s L and f s r . There exists an extension E of L having property ŽQ. such that r , TrEr LŽ²1:.. Then we have TrL r K ŽTrEr LŽ²1:.. , q; that is, TrEr K Ž²1:. , q. Now, apply Lemma 1.
ACKNOWLEDGMENTS I thank Eva Bayer-Fluckiger, Vincent Fleckinger, Detlev Hoffmann, and Robert Perlis for fruitful conversations.
REFERENCES 1. P. E. Conner and R. Perlis, A survey of trace forms of algebraic number fields, World Scientific, Singapore, 1984. 2. M. Epkenhans, On trace forms of algebraic number fields, Arch. Math. 60 Ž6. Ž1993., 527]529. 3. F. Q. Gouvea, ˆ ‘‘ p-adic Numbers: An Introduction,’’ pp. 199]215, Universitext, SpringerVerlag, BerlinrNew York, 1997. 4. M. Kruskemper, Algebraic number field extensions with prescribed trace form, J. Number ¨ Theory 40 Ž1992., 120]124. 5. W. Scharlau, ‘‘Quadratic and Hermitian Forms,’’ Springer-Verlag, HeidelbergrNew York, 1985. 6. J.-P. Serre, L’invariant de Witt de la forme TrŽ x 2 ., Comment. Math. Hel¨ . 59 Ž1984..