|
|
In article <11219718.1162304105704.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
James <james545@xxxxxxxxx> wrote:
>If Q is the field of rational numbers, and v_p is the p-adic
>valuation with respect to the positive prime integer p, then what is
>the valuation ring? That is, what is the ring {x in Q such that v_p
>(x) >= 0 }?
What is the definition?
For the rational numbers, if x = r/s with r and s integers, we have
v_p(x) = v_p(r) - v_p(s).
Now, some authors define it differently, but I suspect that the
definition of v_p that you have is:
For an integer s, the "usual" definition is that if you factor s into
primes
s= p_1^{a_1} * ... * p_t^{a_t}
where p_i are pairwise distinct primes, and a_i are positive integers,
then v_p(s)=0 if p is different from each p_i, and v_p(s) = a_i if
p=p_i for some i.
So... under what conditions will v_p(x) >= 0? If you assume r and s
are relatively prime integers, then at least one of them will have
p-adic valuation equal to 0; which means that v_p(x) >=0 if and only
if v_p(s)=0, which occurs if and only if gcd(r,s)=1.
So:
{ x in Q : v_p(x)>=0} = { x in Q : x = r/s, gcd(r,s)=gcd(s,p)=1}
i.e., the rational numbers that can be written as a fraction with
denominator prime to p.
>I don't think it's Z, (the inegers),
You are right.
>but my book makes me believe it is.
You are misreading the book.
>My book says : "In the case of
>the field of rational numbers Q and the p-adic valuation v_p with its
>completion Q_p, the numbers 0,1, ..., p-1 form a system of
>representatives R for the residue class field Z / pZ of the
>valuation, ...".
This is not the valuation ring. He is talking about the residue class
field.
Remember that the valuation ring will have a unique maximal ideal,
M = {x in Q : v_p(x)>=1}.
In this case, as you can see, it will consist of all rational numbers
which can be written as a fraction with denominator prime to p, and
numerator divisible by p.
Now, if R_p is the valuation ring, what is R_p/M (the residue class
field)?
Let x and y be elements of R_p, x=r/s, y = t/v, with
gcd(r,s)=gcd(t,v)=1, so that gcd(p,sv)=1. Then x = y (mod M) if and
only if x-y in M, if and only if (r/s)-(t/v) in M, if and only if
(vr-st)/sv in M
if and only if p|vr-st.
Now, trivially, 0, 1, ..., p-1 are all distinct elements of R_p, and
they are not congruent modulo M. I claim that each element of R_p is
congruent to one and only one element of {0,1,..,p-1}.
For, let x be an element of R_p. Write x = r/s with s prime to p, r an
integer. Then r is congruent to one and only one of {0,1,...,p-1}
modulo p in the integers, and s is congruent to one and only one of
{1,...,p-1} modulo p in the integers. Say r = i (mod p), and s = j
(mod p). Let k be the unique integer in {1,...,p-1} such that jk=1
(mod p). Then
x = ik (mod M)
For r/s is congruent to ik/1 (mod M) if and only if (r/s) - ik/1 is in
M, which happens if and only if r - sik = 0 (mod p). Since s = j (mod
p) and jk = 1 (mod p), this is equivalent to r - i = 0 (mod p), which
is equivalent to r=i (mod p), which is true. So x = ik (mod M). Now,
ik is congruent to one and only one integer in {0,1,...,p-1} modulo p,
so x is congruent to one and only one of {0,1,...,p-1} mod M.
Which means that {0,1,...,p-1} is a complete set of representatives
for the residue class field R_p/M.
Now, in fact, it is easy to verify that R_p/M is going to be
isomorphic to Z/pZ. The author probably already did so: in general,
for the q-adic valuation of Q, the residue class field will always be
isomorphic to Z/qZ by the argument above.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
|
|