RELATIVE RAMIFICATION AND INERTIA DEGREES.
IAN KIMING
Let K be an algebraic number eld with ring of integers O, and let p be a prime
ideal of O and Kp the completion of K w.r.t. p . Let us choose a prime element
2 p, i.e., an element with p () = 1.
The prime ideal p divides some prime number p. Let e and f denote the ramication index and the inertia degree of p, respectively. I.e., we have
p (p) = e and #(O=p) = pf :
So, e 1 p is an extension of the exponential valuation p on Q attached to the
prime number p, and as we discussed earlier, we can in an obvious fashion view Qp
as sitting inside Kp . It is clear that the eld extension Kp =Qp is a nite extension
(if is a primitive element for the extension K=Q then also for Kp =Qp ).
We will prove the following statement which is a `local' version of equation (3.5.3)
of [1] (for number elds).
Proposition 1. In the above situation we have:
ef = [Kp : Qp ] :
Before the proof we need a number of simple remarks.
First notice that for m 2 N the ideal (pOp )m = pm Op is a principal ideal
generated by m : For clearly m 2 pm Op since 2 p. On the other hand, if
2 pm Op we see that p () m (use that p (x1 + : : : + xs ) minfp (xi )g). Then
p ( m ) 0 so that m 2 Op , hence 2 m Op . So pm Op = m Op , and
we also see that pm Op consists precisely of the elements of Kp that have valuation
m.
In a similar fashion we prove the following:
Op . Then a = pm Op where
m := minfp () j 2 ag :
In particular, a is a principal ideal, and Op is a principal ideal domain.
Proof. Notice that m < 1 since a 6= 0. If 2 a then p () m so by the above
we have 2 pm Op . Suppose that 2 pm Op . Then p ( ) m. Choose 2 a
with p () = m. Then 6= 0 and p ( 1 ) 0 so that 1 2 Op . We deduce
2 Op a.
Proposition 2. Let a be a non-zero ideal of
Corollary 1. We have
pOp = pe Op :
Proof. Immediate from the previous proposition as p (p) = e.
1
2
IAN KIMING
Applying Proposition 2 to the case K = Q we have that Zp is a principal ideal
domain. Clearly Zp Op . In particular we can view Op as a Zp -module. As
such Op is torsion-free as a Zp -module: This follows because Kp is a eld. We
then claim that Op is nitely generated as Zp -module: If it were not there would
be no bound on sizes of nite sets of Zp -linearly independent elements of Op ; but
any nite number of Zp -linearly independent elements of Op are also Qp -linearly
independent (Qp is the quotient eld of Zp ). Since Kp has nite dimension as
vector space over Qp we reach a contradiction, and the claim follows.
So Op is a torsion-free, nitely generated module over the principal ideal domain
Zp . By the structure theory of torsion-free, nitely generated modules over a
principal ideal domain (accept this, or consult for instance Appendix A.4 of [1]) we
have
()
Op = Zp 1 : : : Zp r
with certain elements 1 ; : : : ; r
()
2 Op . We claim that
r = [Kp : Qp ] :
For rst of all we clearly have r [Kp : Qp ] by the above independence argument.
On the other hand, Kp has a Qp basis 1 ; : : : ; s where s := [Kp : Qp ]. Now, if
t is suciently large we have p (pt i ) 0 for all i, i.e., pt i 2 Op for all i. But
then pt 1 ; : : : ; pt s are Zp -linearly independent elements of Op . By the theory of
torsion-free, nitely generated modules over a principal ideal domain this implies
s r, and we are done.
Proof of Proposition 1. Look at the quotient Op =pOp : By () above we have
Op =pOp = (Zp =pZp )r = (Fp )r
where r = [Kp : Qp ] by () above. In particular,
(y)
#(Op =pOp ) = p[Kp :Qp ] :
On the other hand, by Corollary 1 we have pOp = pe Op so that
Op =pOp = Op =pe Op :
Now notice that we have pm 1 Op =pm Op = Op =pOp (as abelian groups) for any
m 2 N. This can be seen because pt Op = t Op : Then map x 7! 1 m x then gives
the desired isomorphism.
Consequently, we easily deduce by induction on v u that we have
#(pu Op =pv Op ) = #(Op =pOp )v
u
for v u. In particular,
(z)
#(Op =pOp ) = #(Op =pe Op ) = #(Op =pOp )e = #(O=p)e = pef
where we used the isomorphism Op =pOp = O=p discussed earlier.
Comparing (y) and (z) the proposition follows.
RELATIVE RAMIFICATION AND INERTIA DEGREES.
3
Now let us look at a relative situation: Suppose that L is a nite extension of
K , OL the ring of integers of L, and P a prime ideal of OL that divides the given
prime p of K :
Q K L
[
[
[
OK OL
[
[
[
p p P
Z
Considering the completions we have then a tower of eld extensions
Qp Kp LP :
One denes the relative ramication index of LP over Kp (or sometimes one
says just of P over p), denoted by e(LP =Kp ) (or just as e(P=p)), as the exponent
of P in the prime factorization (in OL ) of the ideal p OL . Thus, the number e that
we had attached to K in the beginning is in fact the relative ramication index of
Kp over Qp and could also be denoted e(Kp =Qp ).
Similarly, one has the notion of relative inertia degree of LP over Kp (or of P
over p): There is a natural homomorphism OK ! OL =P; the kernel of this is some
ideal a of OK ; then OK =a is a subring of the nite eld OL =P; so a is a prime ideal
of OK and hence also a maximal ideal; we have that a is contained in P; but p is
also a maximal ideal of OK contained in P; if we had a 6= p then a + p = OK , and
we would deduce the contradiction 1 2 P; so, a = p, and we have an injection of
nite elds
OK =p ,! OL =P :
The above relative inertia degree, denoted by f (LP =Kp ) (or just as f (P=p)), is
dened as the degree of this extension of nite elds, i.e., as the dimension of
OL =P as vector space over OK =p.
We say that LP =Kp is unramied (or that P is unramied over p) if
e(LP =Kp ) = 1 :
We say that LP =Kp is totally ramied (or that P is totally ramied over p) if
f (LP =Kp ) = 1 :
By the next proposition we will then have:
LP =Kp unramied , e(LP =Kp ) = 1 , f (LP =Kp ) = [LP : Kp ] ;
and
LP =Kp totally ramied , f (LP =Kp ) = 1 , e(LP =Kp ) = [LP : Kp ] :
We have now the following easy consequence of the denitions and results above.
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IAN KIMING
Proposition 3. We have
e(LP =Qp ) = e(LP =Kp ) e(Kp =Qp ) ;
f (LP =Qp ) = f (LP =Kp ) f (Kp =Qp ) ;
and
[LP : Kp ] = e(LP =Kp ) f (LP =Kp ) :
Proof. By denition the exponent of p in the prime factorization of pOK is e(Kp =Qp ).
Similarly, the exponent of P in the prime factorization of pOL is e(LP =Kp ). So,
the exponent of P in the prime factorization of pOL is e(Kp =Qp ) e(LP =Kp ); but
by denition this exponent is e(LP =Qp ).
The second claim follows from denitions thus:
pf (LP =Qp ) = #(Z=p)f (LP =Qp ) = #(OL =P) = #(OK =p)f (LP =Kp )
= #((Z=p)f (Kp =Qp ) )f (LP =Qp ) = pf (LP =Kp )f (Kp =Qp ) :
Now, the third claim follows from the 2 rst and Proposition 1:
[L : Q ] e(LP =Qp ) f (LP =Qp )
[LP : Kp ] = P p =
= e(LP =Kp ) f (LP =Kp ) :
[Kp : Qp ]
e(Kp =Qp ) f (Kp =Qp )
Finally, we can easily derive from the last proposition the analogous statement
in a truly relative situation: Suppose further that M is a nite extension of L, and
e is a prime of M dividing P. Then we have:
that P
e(MPe =Kp ) = e(MPe =LP ) e(LP =Kp ) ;
f (MPe =Kp ) = f (MPe =LP ) f (LP =Kp ) ;
and of course the following statement that is already in the proposition:
[MP
e =Kp ) :
e =Kp ) f (MP
e : Kp ] = e(MP
The proofs of the rst 2 statements are easy consequences of Proposition 3 and
are left to the reader.
References
[1] H. Koch: `Number Theory. Algebraic Numbers and Functions'. Graduate Studies in Mathematics
24,
AMS 2000.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen , Denmark.
E-mail address :
[email protected]