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On Residually Reducible Representations on Local Rings
Journal of Algebra 212, 738]742 Ž1999. Article ID jabr.1998.7635, available online at http:rrwww.idealibrary.com on
On Residually Reducible Representations on Local Rings* E. Urban CNRS-UMR 7539 Uni¨ ersite´ de Paris-Nord, Paris, France and UCLA Department of Mathematics, Los Angeles, California 90095-1555 E-mail: [email protected] Communicated by A. W. Goldie Received February 27, 1998
1. THE THEOREM Let A be a local Artinian ring with maximal ideal M and residual field k . Let R be an A-algebra. We mean by an n-dimensional A-representation of R any homomorphism of A-algebra r from R to MnŽ A .. For such a representation, we denote by r the residual representation with values in MnŽ k .. It is proven by Carayol that such a representation, if r is absolutely irreducible, is completely determined by its trace. In the reducible case, one cannot generalize this theorem without assuming more hypothesis Žsee the counterexample given in w1x.. The purpose of this note is to give a generalization in the reducible case. It is the best generalization one can get Žsee remarks.. THEOREM.
Let r 1 , r 2 , and r be three A-representations. We assume:
Ži.
r and r 1 [ r 2 ha¨ e the same characteristic polynomials
Žii.
r 1 and r 2 are absolutely irreducible
Žiii.
r1 \ r2
Živ.
r is indecomposable and the subrepresentation of r is isomorphic
to r 1.
* This work was partially supported by an NSF grant. 738 0021-8693r99 $30.00 Copyright Q 1999 by Academic Press All rights of reproduction in any form reserved.
739
ON RESIDUALLY REDUCIBLE REPRESENTATIONS
Then there exists g g GLnŽ A . such that
rŽ r. s g
ž
r 1Ž r .
w
0
r2 Ž r .
/
gy1 ,
for all r g R. Remarks. Ža. In fact, one can get a more general result Žby the same proof. replacing in Ži. r 1 [ r 2 by r 1 [ ??? [ r t and the other points by appropriate properties for the r i ’s and r . Žb. One cannot remove assumption Živ.. Indeed consider the algebra, Rs
a c
½ž
b g M2 Ž k w t x r Ž t 2 . . ; b, c g t k w t x r Ž t 2 . , d
5
/
with A s k w t xrŽ t 2 .. Let x 1 and x 2 be the character of R defined by
x1
žž
a c
b d
//
s a and
x2
žž
a c
b d
//
s d,
and let r be the standard representation of R in M2 Ž k w t xrŽ t 2 ... Then r and x 1 [ x 2 have the same characteristic polynomial although it is clear that r contains neither x 1 nor x 2 . Žc. One cannot remove assumption Žii.. Indeed consider the algebra, Rs
½ž
a bt 2
b g M2 Ž k w t x r Ž t 3 . . ; a, b g k w t x r Ž t 3 . , a
5
/
with A s k w t xrŽ t 3 . and r the standard representation of M2 Ž k w t xrŽ t 3 ... If we set
x1
žž
a bt 2
b a
//
s a q bt and
x2
žž
a bt 2
b a
//
R in
s a y bt,
then r and x 1 [ x 2 have the same characteristic polynomial, moreover r is indecomposable. However r does not verify the conclusion of the theorem. Žd. This result is still true if A is Henselian and Hausdorff. Že. A variant of this theorem is used in w2x to produce nontrivial elements in Ext 1R Ž r 1 , r 2 ..
740
E. URBAN
2. PROOF OF THE THEOREM First, we assume that k contains at least n s dim r elements. From Ži., Žii., and Žiii. and the Brauer]Nesbitt theorem, the image of the semisimplification of r is isomorphic to Mn1Ž k . = Mn 2Ž k . where n i s dim r i . Thus there exists an element r 0 g R such that the characteristic polynomial of r Ž r 0 . has exactly n different roots in i . Therefore r Ž r 0 . has n eigenvalues which are different mod M . We denote them by a 1 , . . . , a n . Then it is easy to see that we can assume that
r Ž r 0 . s Diag Ž a 1 , . . . , a n . , Žby changing r by g. pgy1 for g g GLnŽ A . if necessary .. Moreover by Ži., we can also assume that r 1Ž r 0 . s DiagŽ a 1 , . . . , a n1 . and r 2 Ž r 0 . s DiagŽ a n1q1 , . . . , a n .. Now for all r g R, we set by blocks,
rŽ r. s
ž
Ar Cr
Br , Dr
/
with A r g Mn1Ž A ., Br g Mn1 , n 2Ž A ., Cr g Mn 2 , n1Ž A . and Dr g Mn 2Ž A .. Because r Ž r 0k . is diagonal, we have for all r g R and all integers k, tr Ž r Ž rr 0k . . s tr A r Diag Ž a 1 , . . . , a n1 .
ž
k
/ q tr ž D DiagŽ a r
n 1 q1 , . . . ,
an .
k
/,
and by the assumption Ži., we get also, tr Ž r Ž rr 0k . . s tr Ž r 1 Ž rr 0k . . q tr Ž r 2 Ž rr 0k . . s tr r 1 Ž r . Diag Ž a 1 , . . . , a n1 .
ž
k
/
q tr r 2 Ž r . Diag Ž a n1q1 , . . . , a n .
ž
k
/.
Noting that the A-submodule generated by DiagŽ a 1 , . . . , a n . k is exactly equal to the submodule of diagonal matrices, we get from the two previous equalities, ; r g R,
tr Ž A r . s tr Ž r 1 Ž r . .
and tr Ž Dr . s tr Ž r 2 Ž r . . . Ž Tr1.
If we knew that r ¬ A r was a representation, we could easily get our result. But we cannot get it directly and we have to take another way. From ŽTr1., we get ; r , s g R,
tr Ž A r s . s tr Ž A sr . ,
ON RESIDUALLY REDUCIBLE REPRESENTATIONS
741
and writing r Ž rs . s r Ž r . r Ž s ., we have A r s s A r A s q Br C s and because trŽ A r A s . s trŽ A s A r . we get ; r , s g R,
tr Ž Br C s . s tr Ž Bs Cr . .
Ž Tr2.
We are going to use that relation and the indecomposability of r to prove that Cr s 0 for all r g R. For that purpose, we prove by induction on i that Cr g M i MnŽ A . for all positive integers i. The first step of the induction is given by Živ. and the choice of the basis for which we give the expression of r ; we assume now that our assumption is true for the order i. Step 1. We prove that Cr g M iq1 MnŽ A . for all r g Ker r . Indeed, on the one hand, one can easily see from the assumption Žii., Žiii., and Živ. that the map, R ª Mn 1 , n 2Ž k . , r ¬ Br mod M is surjective. On the other hand, if r g Ker r we have by ŽTr2. that trŽ Bs Cr . g M iq1 MnŽ A . for all s g R. By the surjectivity noted earlier and the nondegenerescence of the trace, this implies that Cr g M iq1 MnŽ A . as claimed. Consider now the map of k-vector spaces, Fr
M iq1 , R ª Mn 2 , n1Ž A . m M irM r ¬ Cr mod M iq1 . By Step 1, there exists a map Fr such that the following diagram commutes
,
Fr
6
Fr
Tn1 , n 2Ž k . 6
r
6
R
Mn 2 , n 1 M
iq1
M rM
i
where Tn1 , n 2Ž k . is the subalgebra of MnŽ k . constituted by matrices whose left bottom block n 2 = n1 is null. Then we know that Im r s Tn1 , n 2Ž k ..
742
E. URBAN
Step 2. In order to finish the induction, we have to prove that Fr is null. From the formula Cr s s Cr A s q Dr C s , we get Fr
A 0
žž
B D
/ž
A9 0
B9 D9
//
s Fr
A 0
žž
B D
//
A9 q DFr
žž
A9 0
B9 D9
//
.
From the remark that all diagonal matrices are linearly generated by the power of DiagŽ a 1 , . . . , a n . we see that the image by Fr of any diagonal matrices of Tn1 , n 2 is zero. By the previous formula, this implies that Fr is trivial. Indeed we have Fr
žž
A 0
0 0
//
s Fr
žž
1 n1
0
0
0
A 0
0 0
s Fr
/ž // žž
1 n1
0
0
0
//
A s 0,
similarly we have that Fr
žž
0 0
0 D
//
s 0,
B 0
and Fr
žž
0 0
B 0
//
s Fr
žž
s 0.Fr
1 n1
0
0
0
0 0
B 0
q Fr
/ž // žž // žž 0 0
1 n1
0
0
0
//
.0 s 0.
This finishes the proof by induction. Finally, A is Artinian, thus Cr s 0 for all r g R and therefore r ¬ A r is a representation Žisomorphic to r 1 by ŽTr1. and w1x.. So we get our theorem. If k does not contain n elements, we can find a faithfully flat extension A 9 of A whose residue field k 9 does. Therefore we have an exact sequence, 0 ª r1
mA
A9 ª r
mA
A9 ª r 2
mA
A 9 ª 0,
and our result follows by a simple descent argument.
REFERENCES 1. H. Carayol, Formes modulaires et representations galoisiennes ` a valeurs dans un anneau ´ local complet, Contemp. Math. 165 Ž1994., 213]235. 2. E. Urban, Selmer groups and the Eisenstein ]Klingen ideal, Oct. 1997, preprint.
On Residually Reducible Representations on Local Rings* E. Urban CNRS-UMR 7539 Uni¨ ersite´ de Paris-Nord, Paris, France and UCLA Department of Mathematics, Los Angeles, California 90095-1555 E-mail: [email protected] Communicated by A. W. Goldie Received February 27, 1998
1. THE THEOREM Let A be a local Artinian ring with maximal ideal M and residual field k . Let R be an A-algebra. We mean by an n-dimensional A-representation of R any homomorphism of A-algebra r from R to MnŽ A .. For such a representation, we denote by r the residual representation with values in MnŽ k .. It is proven by Carayol that such a representation, if r is absolutely irreducible, is completely determined by its trace. In the reducible case, one cannot generalize this theorem without assuming more hypothesis Žsee the counterexample given in w1x.. The purpose of this note is to give a generalization in the reducible case. It is the best generalization one can get Žsee remarks.. THEOREM.
Let r 1 , r 2 , and r be three A-representations. We assume:
Ži.
r and r 1 [ r 2 ha¨ e the same characteristic polynomials
Žii.
r 1 and r 2 are absolutely irreducible
Žiii.
r1 \ r2
Živ.
r is indecomposable and the subrepresentation of r is isomorphic
to r 1.
* This work was partially supported by an NSF grant. 738 0021-8693r99 $30.00 Copyright Q 1999 by Academic Press All rights of reproduction in any form reserved.
739
ON RESIDUALLY REDUCIBLE REPRESENTATIONS
Then there exists g g GLnŽ A . such that
rŽ r. s g
ž
r 1Ž r .
w
0
r2 Ž r .
/
gy1 ,
for all r g R. Remarks. Ža. In fact, one can get a more general result Žby the same proof. replacing in Ži. r 1 [ r 2 by r 1 [ ??? [ r t and the other points by appropriate properties for the r i ’s and r . Žb. One cannot remove assumption Živ.. Indeed consider the algebra, Rs
a c
½ž
b g M2 Ž k w t x r Ž t 2 . . ; b, c g t k w t x r Ž t 2 . , d
5
/
with A s k w t xrŽ t 2 .. Let x 1 and x 2 be the character of R defined by
x1
žž
a c
b d
//
s a and
x2
žž
a c
b d
//
s d,
and let r be the standard representation of R in M2 Ž k w t xrŽ t 2 ... Then r and x 1 [ x 2 have the same characteristic polynomial although it is clear that r contains neither x 1 nor x 2 . Žc. One cannot remove assumption Žii.. Indeed consider the algebra, Rs
½ž
a bt 2
b g M2 Ž k w t x r Ž t 3 . . ; a, b g k w t x r Ž t 3 . , a
5
/
with A s k w t xrŽ t 3 . and r the standard representation of M2 Ž k w t xrŽ t 3 ... If we set
x1
žž
a bt 2
b a
//
s a q bt and
x2
žž
a bt 2
b a
//
R in
s a y bt,
then r and x 1 [ x 2 have the same characteristic polynomial, moreover r is indecomposable. However r does not verify the conclusion of the theorem. Žd. This result is still true if A is Henselian and Hausdorff. Že. A variant of this theorem is used in w2x to produce nontrivial elements in Ext 1R Ž r 1 , r 2 ..
740
E. URBAN
2. PROOF OF THE THEOREM First, we assume that k contains at least n s dim r elements. From Ži., Žii., and Žiii. and the Brauer]Nesbitt theorem, the image of the semisimplification of r is isomorphic to Mn1Ž k . = Mn 2Ž k . where n i s dim r i . Thus there exists an element r 0 g R such that the characteristic polynomial of r Ž r 0 . has exactly n different roots in i . Therefore r Ž r 0 . has n eigenvalues which are different mod M . We denote them by a 1 , . . . , a n . Then it is easy to see that we can assume that
r Ž r 0 . s Diag Ž a 1 , . . . , a n . , Žby changing r by g. pgy1 for g g GLnŽ A . if necessary .. Moreover by Ži., we can also assume that r 1Ž r 0 . s DiagŽ a 1 , . . . , a n1 . and r 2 Ž r 0 . s DiagŽ a n1q1 , . . . , a n .. Now for all r g R, we set by blocks,
rŽ r. s
ž
Ar Cr
Br , Dr
/
with A r g Mn1Ž A ., Br g Mn1 , n 2Ž A ., Cr g Mn 2 , n1Ž A . and Dr g Mn 2Ž A .. Because r Ž r 0k . is diagonal, we have for all r g R and all integers k, tr Ž r Ž rr 0k . . s tr A r Diag Ž a 1 , . . . , a n1 .
ž
k
/ q tr ž D DiagŽ a r
n 1 q1 , . . . ,
an .
k
/,
and by the assumption Ži., we get also, tr Ž r Ž rr 0k . . s tr Ž r 1 Ž rr 0k . . q tr Ž r 2 Ž rr 0k . . s tr r 1 Ž r . Diag Ž a 1 , . . . , a n1 .
ž
k
/
q tr r 2 Ž r . Diag Ž a n1q1 , . . . , a n .
ž
k
/.
Noting that the A-submodule generated by DiagŽ a 1 , . . . , a n . k is exactly equal to the submodule of diagonal matrices, we get from the two previous equalities, ; r g R,
tr Ž A r . s tr Ž r 1 Ž r . .
and tr Ž Dr . s tr Ž r 2 Ž r . . . Ž Tr1.
If we knew that r ¬ A r was a representation, we could easily get our result. But we cannot get it directly and we have to take another way. From ŽTr1., we get ; r , s g R,
tr Ž A r s . s tr Ž A sr . ,
ON RESIDUALLY REDUCIBLE REPRESENTATIONS
741
and writing r Ž rs . s r Ž r . r Ž s ., we have A r s s A r A s q Br C s and because trŽ A r A s . s trŽ A s A r . we get ; r , s g R,
tr Ž Br C s . s tr Ž Bs Cr . .
Ž Tr2.
We are going to use that relation and the indecomposability of r to prove that Cr s 0 for all r g R. For that purpose, we prove by induction on i that Cr g M i MnŽ A . for all positive integers i. The first step of the induction is given by Živ. and the choice of the basis for which we give the expression of r ; we assume now that our assumption is true for the order i. Step 1. We prove that Cr g M iq1 MnŽ A . for all r g Ker r . Indeed, on the one hand, one can easily see from the assumption Žii., Žiii., and Živ. that the map, R ª Mn 1 , n 2Ž k . , r ¬ Br mod M is surjective. On the other hand, if r g Ker r we have by ŽTr2. that trŽ Bs Cr . g M iq1 MnŽ A . for all s g R. By the surjectivity noted earlier and the nondegenerescence of the trace, this implies that Cr g M iq1 MnŽ A . as claimed. Consider now the map of k-vector spaces, Fr
M iq1 , R ª Mn 2 , n1Ž A . m M irM r ¬ Cr mod M iq1 . By Step 1, there exists a map Fr such that the following diagram commutes
,
Fr
6
Fr
Tn1 , n 2Ž k . 6
r
6
R
Mn 2 , n 1 M
iq1
M rM
i
where Tn1 , n 2Ž k . is the subalgebra of MnŽ k . constituted by matrices whose left bottom block n 2 = n1 is null. Then we know that Im r s Tn1 , n 2Ž k ..
742
E. URBAN
Step 2. In order to finish the induction, we have to prove that Fr is null. From the formula Cr s s Cr A s q Dr C s , we get Fr
A 0
žž
B D
/ž
A9 0
B9 D9
//
s Fr
A 0
žž
B D
//
A9 q DFr
žž
A9 0
B9 D9
//
.
From the remark that all diagonal matrices are linearly generated by the power of DiagŽ a 1 , . . . , a n . we see that the image by Fr of any diagonal matrices of Tn1 , n 2 is zero. By the previous formula, this implies that Fr is trivial. Indeed we have Fr
žž
A 0
0 0
//
s Fr
žž
1 n1
0
0
0
A 0
0 0
s Fr
/ž // žž
1 n1
0
0
0
//
A s 0,
similarly we have that Fr
žž
0 0
0 D
//
s 0,
B 0
and Fr
žž
0 0
B 0
//
s Fr
žž
s 0.Fr
1 n1
0
0
0
0 0
B 0
q Fr
/ž // žž // žž 0 0
1 n1
0
0
0
//
.0 s 0.
This finishes the proof by induction. Finally, A is Artinian, thus Cr s 0 for all r g R and therefore r ¬ A r is a representation Žisomorphic to r 1 by ŽTr1. and w1x.. So we get our theorem. If k does not contain n elements, we can find a faithfully flat extension A 9 of A whose residue field k 9 does. Therefore we have an exact sequence, 0 ª r1
mA
A9 ª r
mA
A9 ª r 2
mA
A 9 ª 0,
and our result follows by a simple descent argument.
REFERENCES 1. H. Carayol, Formes modulaires et representations galoisiennes ` a valeurs dans un anneau ´ local complet, Contemp. Math. 165 Ž1994., 213]235. 2. E. Urban, Selmer groups and the Eisenstein ]Klingen ideal, Oct. 1997, preprint.