Incompatibility between different p-adics?
Mike Oliver
they have a base-p expansion extending infinitely many
places to the left), and let Q_p be its quotient field
(whose elements have a natural base-p expansion with infinitely
many places to the left of the radix point, and finitely
many to the *right*).
How different are these structures for different values
of p? My guess is the following: (I'm not going to
call it a conjecture because it or its negation is undoubtedly
a standard result.)
Guess: if p and q are distinct primes, then the only ring
homomorphism from Z_p into Q_q is the constantly-zero function.
At the moment I have no idea how to prove it; in fact, I don't
even know how to prove that Q_p and Q_q aren't isomorphic.
(Well, OK, I do have *some* notion. I have a sketch in my
head of a proof that the theory of algebraically closed
fields of characteristic 0 is categorical in every infinite
cardinality (can anyone refute me on that?), so Q_p and Q_q
must not be algebraically closed, or they *would* be the
same thing. The most natural possibility is that you can
take p'th roots in Q_q but not in Q_p; that's the first
place I'd look if I wanted to spend a lot of time on this.)
Any help?
--
Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
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Mike Oliver
I, Mike Oliver, wrote:
> I have a sketch in my
> head of a proof that the theory of algebraically closed
> fields of characteristic 0 is categorical in every infinite
> cardinality (can anyone refute me on that?),
Oops, I can refute myself on that one. Q-bar, and the
algebraic closure of Q(pi).
Make it every *uncountable* cardinality.
Kurt Foster
. For a prime p, let Z_p be the ring of p-adic integers (i.e. they have a
. base-p expansion extending infinitely many places to the left), and let
. Q_p be its quotient field (whose elements have a natural base-p
. expansion with infinitely many places to the left of the radix point,
. and finitely many to the *right*).
. How different are these structures for different values of p?
.
Depends how much of the "structure" you're worried about. The ordinary
integers are a dense additive subgroup of Z_p for any given prime p.
Similarly for the ordinary nonzero rationals in the multiplicative group
of the nonzero elements of Q_p. For each p, the p-adic integers/rationals
are a completion of the ordinary integers/rationals in the p-adic
topology.
However, the topologies for different values of p are inequivalent, and
this is a well known basic result that should be covered in most any
introductory treatment of valuations or p-adic fields and rings. The
usual statement is to the effect that, given distinct primes p and q,
there is an x in Q such that |x|_p < 1 and |x|_q > 1. The powers of such
an x give a sequence converging to 0 in one topology, but to infinity in
the other.
Mike Oliver
Kurt Foster wrote:
>
> In <365A5F84...@math.ucla.edu>, Mike Oliver said:
> . How different are these structures for different values of p?
> Depends how much of the "structure" you're worried about.
Algebraic structure only.
> The ordinary
> integers are a dense additive subgroup of Z_p for any given prime p.
> Similarly for the ordinary nonzero rationals in the multiplicative
> group of the nonzero elements of Q_p. For each p, the p-adic
> integers/rationals
> are a completion of the ordinary integers/rationals in the p-adic
> topology.
Sure. But this doesn't seem in any obvious way to give nontrivial
homomorphisms from Z_p to Q_q for p <> q, which is what I asked
about.
Mike Oliver
I, Mike Oliver, wrote:
> Guess: if p and q are distinct primes, then the only ring
> homomorphism from Z_p into Q_q is the constantly-zero function.
>
> At the moment I have no idea how to prove it; in fact, I don't
> even know how to prove that Q_p and Q_q aren't isomorphic.
OK, I have a partial answer. I have it on Knuth's authority
(Seminumerical Algorithms, section 4.1, problem 31) that there
is a square root of -7 in the 2-adics; in fact, in the answers
to exercises, he gives an algorithm for computing
sqrt(n) in the 2-adics whenever n = 2^k (mod 2^{k+3}) for
some k.
However there can be no sqrt(-7) in the 5-adics, because
-7 is not a quadratic residue mod 5 (more computationally,
observe that -7 = ...444433. But if the 5-adic number x
ends in 0, then x^2 ends in 0; if x ends in 1 then x^2 ends in 1;
2 gives 4, 3 gives 4, 4 gives 1 -- so you can't get 3 as the
last digit).
Thus there is no nontrivial ring homomorphism from Z_2 to Q_5.
(If 1 is not sent to 0, it must be sent to 1, which is
the only other solution of x^2=x; therefore the homomorphism
must fix the integers pointwise, so the image of -7 is -7).
So to answer the question in general, it may be enough to
generalize the algorithm that Knuth gives -- if it does
in fact generalize.
KRamsay
In article <365A5F84...@math.ucla.edu>,
|Guess: if p and q are distinct primes, then the only ring
|homomorphism from Z_p into Q_q is the constantly-zero function.
You call that a ring homomorphism? :-)
|At the moment I have no idea how to prove it; in fact, I don't
|even know how to prove that Q_p and Q_q aren't isomorphic.
The rational polynomials having roots in each are quite different. For
example (and various other proofs could no doubt be found), in order
for x^2-m=0 to have a p-adic root, where m is an integer, one must at
least have a solution to the congruence x^2-m=0 (mod p). On the other
hand, if p>2, p does not divide m, and x^2-m=0 (mod p), Hensel's lemma
ensures that x^2-m=0 has a root in Z_p. It also ensures that x^2-m=0
for integer m has a root in Z_2 when m=1 (mod 8). Thus for any
distinct primes p and q, there are m such that m has a p-adic integer
square root but no q-adic square root at all.
Sometimes people embed Q_p into a field called C_p which according to
the axiom of choice is isomorphic to the complex number field.
Keith Ramsay "Thou Shalt not hunt statistical significance with
kra...@aol.com a shotgun." --Michael Driscoll's 1st commandment
ilias kastanas 08-14-90
Mike Oliver <oli...@math.ucla.edu> wrote:
@
@
@I, Mike Oliver, wrote:
@> Guess: if p and q are distinct primes, then the only ring
@> homomorphism from Z_p into Q_q is the constantly-zero function.
@>
@> At the moment I have no idea how to prove it; in fact, I don't
@> even know how to prove that Q_p and Q_q aren't isomorphic.
If they were, Q_p would have two discrete valuations (the p-adic
and the carry-over of the q-adic by the isomorphism); that's impossible
by Hensel's lemma.
@OK, I have a partial answer. I have it on Knuth's authority
@(Seminumerical Algorithms, section 4.1, problem 31) that there
@is a square root of -7 in the 2-adics; in fact, in the answers
@to exercises, he gives an algorithm for computing
@sqrt(n) in the 2-adics whenever n = 2^k (mod 2^{k+3}) for
@some k.
@
@However there can be no sqrt(-7) in the 5-adics, because
@-7 is not a quadratic residue mod 5 (more computationally,
@observe that -7 = ...444433. But if the 5-adic number x
@ends in 0, then x^2 ends in 0; if x ends in 1 then x^2 ends in 1;
@2 gives 4, 3 gives 4, 4 gives 1 -- so you can't get 3 as the
@last digit).
@
@Thus there is no nontrivial ring homomorphism from Z_2 to Q_5.
@(If 1 is not sent to 0, it must be sent to 1, which is
@the only other solution of x^2=x; therefore the homomorphism
@must fix the integers pointwise, so the image of -7 is -7).
@
@So to answer the question in general, it may be enough to
@generalize the algorithm that Knuth gives -- if it does
@in fact generalize.
For any x in Z, not a multiple of p^2 (odd prime p)
if x is a multiple of p, then x is not a square in Q_p; other-
wise, x is a square in Q_p if and only if x is a quadratic
residue mod p.
Ilias
Gerry Myerson
a square root in exactly one of the two rings Z_p & Z_q.
This should not be hard to prove from quadratic reciprocity (see any
good Number Theory textbook), though it may be necessary to split the
proof into several cases. E.g., -1 has a square root if and only if
p is congruent to 1 mod 4.
Original message appended.
Gerry Myerson (ge...@mpce.mq.edu.au)
In article <365B0D21...@math.ucla.edu>, oli...@math.ucla.edu wrote:
=> I, Mike Oliver, wrote:
=> > Guess: if p and q are distinct primes, then the only ring
=> > homomorphism from Z_p into Q_q is the constantly-zero function.
=> >
=> > At the moment I have no idea how to prove it; in fact, I don't
=> > even know how to prove that Q_p and Q_q aren't isomorphic.
=>
=> OK, I have a partial answer. I have it on Knuth's authority
=> (Seminumerical Algorithms, section 4.1, problem 31) that there
=> is a square root of -7 in the 2-adics; in fact, in the answers
=> to exercises, he gives an algorithm for computing
=> sqrt(n) in the 2-adics whenever n = 2^k (mod 2^{k+3}) for
=> some k.
=>
=> However there can be no sqrt(-7) in the 5-adics, because
=> -7 is not a quadratic residue mod 5 (more computationally,
=> observe that -7 = ...444433. But if the 5-adic number x
=> ends in 0, then x^2 ends in 0; if x ends in 1 then x^2 ends in 1;
=> 2 gives 4, 3 gives 4, 4 gives 1 -- so you can't get 3 as the
=> last digit).
=>
=> Thus there is no nontrivial ring homomorphism from Z_2 to Q_5.
=> (If 1 is not sent to 0, it must be sent to 1, which is
=> the only other solution of x^2=x; therefore the homomorphism
=> must fix the integers pointwise, so the image of -7 is -7).
=>
=> So to answer the question in general, it may be enough to
=> generalize the algorithm that Knuth gives -- if it does
=> in fact generalize.
=>
=> --
=> Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
=>
=> Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
=> 1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9
Mike Oliver
>
> In article <365A5F84...@math.ucla.edu>,
> Mike Oliver <oli...@math.ucla.edu> writes:
>
You're saying it's not? Next you'll be telling me that {0} isn't
a field, either.
On reflection, I think {0} is my favorite field; it has so
many marvelous properties. I'll list just a few:
0) The zero element is invertible.
0) The field is algebraically closed -- and the computational
algorithm for finding a root for a given polynomial has this,
shall I say, pristine purity about it.
0) In spite of being algebraically closed, the field is also
rigid--find me another field like that!
0) It's a great structure for nihilists.
Moreover, a number of recurring sci.math threads can be resolved,
or at least unified, by considering this field:
0) Here, one really *does* equal zero -- so all those "proofs"
were right!
0) 1/0 and 0/0 are both perfectly well defined.
0) The field has characteristic 1, which should put to rest
once and for all the question, "is 1 prime?".
(I note in passing that the theory of algebraically closed fields
of characteristic 1 is *extremely* categorical--any two models
are isomorphic *whether*or*not* they have the same cardinality.)
I modestly propose that any of the threads referenced above
should now go under one of the following two subjects:
{0} is too a field, you doofus!
or
It is not either a field, stupid!
--
"Nihilists! You can say what you want about the tenets of National
Socialism, but at least it's an ethos!" -- John Goodman's character in
"The Big Lebowski"
T.Ward
>
>At the moment I have no idea how to prove it; in fact, I don't
>
I think you can argue as follows:
if f is a ring isomorphism (of rings with units) between them,
then it must send the maximal subrings to each other and must
send p to p. However one of the maximal subrings is invariant
under multiplication by p and the other is not.
Tom Ward
UEA
KRamsay
In article <365BC26E...@math.ucla.edu>,
|On reflection, I think {0} is my favorite field; it has so
|many marvelous properties.
Very soothing, isn't it. It fits the Buddha's remark, "void is form,
form is void".
|I'll list just a few:
|
|0) The zero element is invertible.
That would answer the periodic concerns over why we don't permit
division by zero.
|0) The field is algebraically closed
...
0) {0} has a decidable theory so Goedel incompleteness isn't a worry.
0) Both 0^0=1 and 0^0=0 are correct, which puts that argument to rest.
0) No more pesky inequalities to deal with.
0) 0 can always be integrated in closed form.
0) You won't get the wrong answer in a multiple choice test about {0}.
0) Every element of {0} is a palindrome.
0) {0} is Noetherian: only one ideal.
0) {0} is Artinian: only one ideal.
0) {0} is a P.I.D.: only one ideal, generated by 0.
0) It is easy to apply the Euclidean algorithm in {0}.
0) The spectrum of {0} was our favorite scheme, in algebraic geometry
class: no points, and just one section of the structure sheaf, 0,
which is defined globally.