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Solutions to equations in the padic integers (rationals)

  1. Jul 24, 2010 #1
    I will post my solutions to a few problems i've considered here, I just need feedback from people who can tell me if my ideas are correct because i'm feeling shaky about the methods.

    Z7 is the dvr in the completion of the rationals w.r.t. the 7adic metric. Then for what integers a is there a solution to the polynomial 7x^2-a in Z7.

    If there was one then root(a/7) is in Z7 so it's square a/7 is too, thus |a/7| can be no bigger than one i.e. 7 divides a in the integers. We are looking at solutions of x^2-a/7 and these exist if and only if solutions exist modulo 7 (necessity is obvious and Hensel's lemma gives sufficiency). The integers a for which a/7 is a quadratic residue (mod7) are exactly those that are congruent to 0,7,14, or 28 modulo 49.

    Now I'm wondering about the same question for what rational 7adic numbers (aka Q7), a, does there exist a solution to the equation in Q7? I will post about this when I have an idea.
     
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  3. Jul 24, 2010 #2

    Hurkyl

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    Often, when asking for solutions to a polynomial, rational solutions are permitted. Are you sure the problem was asking for which a in Z7 there exists a solution for x in Z7, rather than for which a in Z7 there exists a solution for x in Q7?



    Anyways, let's assume you interpreted correctly. Then you correctly argue that 7 divides a.

    I don't believe your claim that Hensel's lemma gives sufficiency -- that lemma has a nonsingularity condition that you haven't checked. (i.e. a condition involving the derivative). You argue that 7 has a square root in Z7. What is its valuation?
     
  4. Jul 24, 2010 #3
    Notice that the roots of x^2-c/d are in 1-1correspondence with those of 7x^2-c/d because root(7) is in Q7 (by hensel's lemma since x^2-7 is reducible modulo 7). So we look at when solutions of the former exist.

    By gauss's lemmma (let a=c/d relatively prime) then a solution to dx^2-c=0 exists in Q7 iff it exists in Z7.

    Now dx^2-c has a root iff it does modulo (7) which means dx^2=c mod (7). A solution exists if and only if BOTH d and c are residues or BOTH nonresidues (product of a residue and nonresidue is a nonresidue and is both arent the solution exists b/c the reisdues are a subgroup of index two).

    So a solution exists if and only if c and d have the aformentioned property of both being nonresidues or both residues modulo 7.

    Is this correct?! the reasoning seems awkward.

    If it is however it will give me a method to (begin to) figure out the exact number of quadratic extensions of Q2.
     
  5. Jul 24, 2010 #4

    Hurkyl

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    No, Hensel's lemma doesn't say that.

    The function f(x) = x²-7 does indeed have the root 0 mod 7. However, this does not satisfy the conditions for Hensel's lemma to guarantee a root modulo 49. You don't have enough precision for Hensel lifting to work yet, and thus it doesn't guarantee a root in Z7 (or Q7).
     
  6. Jul 24, 2010 #5
    aha! you are correct indeed: in hensel's lemma the factors modulo p need to be relatively prime! which they are not in the case of x^2-7! ok back to square 1. I will think about this more.
     
  7. Jul 24, 2010 #6

    Hurkyl

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    Note there are more detailed versions of Hensel's lemma that deal with cases such as when f'(r) also vanishes modulo p.



    Anyways, I think the change of variable does help simplify things -- you can't set x = y/sqrt(7), but you can set x = y/7, and simplify to a monic integer polynomial.
     
  8. Jul 25, 2010 #7
    solutions (over Q7) to 7x^2-a (for the time being a is an integer) are in one to one correspondence with those of x^2-7a. Now the solutions of the latter will be in Z7 by Gauss's lemma.

    It is necessarily true that if a solution exists it will exist modulo 7^k for all k so let this k be precisely one larger than the largest power of 7 dividing 7a. Then 7a=7^{k-1}A where A is not divisible by 7. Then we must have that 7^{k-1}A is a residue modulo 7^k. Which occurs exactly when A and 7^{k-1} are residues (i.e. A is 1,2,4 modulo 7 and k-1 is even). I will try and prove this is sufficient. Let m be half of k-1.

    We change variables again to see the solutions of x^2-7^{k-1}A can be modified (take y to be x/7^m) to give solutions of y^2-A. Finally the latter has solutions by Hensel's lemma. We can work our way backwards to thus give solutions of 7x^2-a, in Q7. Now say x is a solution so that 7(x^2-a/7)=0 but then x^2-a/7 is reducible over Q7 and thus it is over Z7.

    So a must be an odd power of 7 times a nonzero residue modulo 7.

    whew
     
  9. Jul 25, 2010 #8
    for a in Q by basically the same reasoning we must have (if a=c/d) that cd is an odd (positive) power of 7 times a nonzero residue modulo 7. Since d^2=7^{2r}A^2 where d=7^mA then even if A is a nonresidue then it's square won't be and thus we have that a= an odd power of 7 (possibly negative) times a nonzero residue modulo 7.
     
  10. Jul 25, 2010 #9

    Hurkyl

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    This sounds correct.


    I confess I would have used unique factorization (i.e. the discrete valuation) to deal with 7. If v has a square root, then from its factorization v=u7n (where u is a unit; i.e. something nonzero modulo 7; i.e. something with absolute value 1), we see that u must be a square and that n must be even.

    Of course, this only works when considering m-th roots, and not with roots of general polynomials.
     
    Last edited: Jul 25, 2010
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