wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Algebraic Roots
        
(Message started by: Sir Col on Jan 27^th, 2005, 3:49pm)

Title: Algebraic Roots
Post by Sir Col on Jan 27^th, 2005, 3:49pm

Let p₁, p₂, ... , p_n [in] [bbq], and assume that each p_k^(1/k) is irrational.

(1) Prove that no irrational number is the root of a linear equation containing integral coefficients.

(2) Prove that x = p₂^1/2 + p₁ is the root of some quadratic with integral coefficients.

(3) What degree polynomial does x = p₃^1/3 + p₂^1/2 + p₁ come from? Prove it!

(4) What if x is a series of irrational roots: x = p_n^1/n + p_n-1^1/(n-1) + ... + p₁?

Title: Re: Algebraic Roots
Post by Icarus on Jan 31^st, 2005, 5:10pm

Since no one has replied to this one in 4 days, I thought I would get the ball rolling by proving those extremely difficult first two problems. ::)

(1) if ax+b = 0, then x = -b/a, which is rational when a, b are rational.

(2) (x - p₁)² = p₂. Simplify and multiply by the lcd of p₂ and p₁².

Title: Re: Algebraic Roots
Post by Barukh on Feb 1^st, 2005, 12:08am

I have been thinking about this problem in the last 4 days, but did not come to any final solution. Here’s what I know so far.

on 01/27/05 at 15:49:54, Sir Col wrote:

Let p₁, p₂, ... , p_n [in] [bbq], and assume that each p_k^(1/k) is irrational.

To avoid a contradiction, we should assume k > 1.

First of all, it is easy to see that if [alpha] is an algebraic number of degree n (i.e. there is a polynomial f(x) with coefficients in [bbq] of degree n s.t. f([alpha]) = 0 and no polynomial with smaller degree having this property), and q [in] [bbq], then [alpha]+q is again algebraic with degree n. (I leave this as an exercise). So, we may always omit the number p₁. In what follows, r_k is a shortcut for p_k^(1/k).

(3)
It is not difficult to construct a polynomial of degree 6 with [alpha] = r₃ + r₂ being its root. Here’s one way:

([alpha]-r₂)³ = r₃³ = p₃. Expanding the binom and rearranging, gives:
[alpha]³ + 3[alpha]p₂ – p₃ = (3[alpha]² + p₂)r₂. Squaring both sides and rearranging, finally gives:
[alpha]⁶ – 3p₂[alpha]⁴ – 2p₃[alpha]³ + 3 p₂²[alpha]² – 6p₂p₃[alpha] + (p₃² - p₂³) = 0

The hard part is to show that this polynomial is irreducible, and so [alpha] is algebraic with degree 6 (which I believe it is). Using the theory of Extension Fields from Abstract Algebra, one can show that the degree of [alpha] should divide 6: this is because the extension field [bbq]([alpha]) is a subfield of the extension field [bbq](r₂, r₃), and the latter has degree 6. So, [alpha] may be of degree 2, 3, 6.

(4)
The extension field [bbq](r₂, r₃, …, r_k) has degree at most k! over [bbq], and this gives the upper bound for the answer to this question.

Title: Re: Algebraic Roots
Post by Deedlit on Mar 31^st, 2005, 5:36am

The degree of the polynomial is in fact k!. You can establish this by taking field automorphisms that take p_k^1/k to p_k^1/k e^{2[pi]i a_k / k} for all k, while fixing [bbr] - note that this works because all the p_k's are relatively prime. Since [bbr] is fixed, so is the minimal polynomial, but the root

[sum]p_k^1/k

gets taken to

[sum] p_k^1/k e^{2[pi]i a_k / k}

for any assignment of a_k. This generates k! different roots of the minimal polynomial, so it must be at least degree k!, and Barukh has already shown it is at most degree k!.

Title: Re: Algebraic Roots
Post by Eigenray on Apr 8^th, 2005, 1:38pm

Relatively prime isn't a strong enough assumption. For example, 1+2^1/2 + 3^1/3 + 25^1/4 satisfies a polynomial of degree 12. Something like, each p_k is divisible by some prime exactly once, which would ensure at least that [Q(p_k^1/k):Q]=k (why?). Heck, lets just take p_k to be the k-th prime.

Now, Deedlit you seem to be assuming things that are as hard as what is being asked. Crucial is the fact that p_n^1/n isn't in Q(p₂^1/2,...,p_n-1^1/(n-1)). Admittedly, it's "obvious", but it looks non-trivial.

Then the splitting field of
(x-2)(x²-3)(x³-5)(x⁴-7)...(x^k-p_k)
will be Galois of degree k!*phi(M), where M=lcm(1,2,...,k). Now, as the maps you list (together with the phi(M) choices e^{2pi i/M} -> e^{2pi im/M}) are the only possible candidates for automorphisms, they will in fact all be automorphisms.

Then all that's left is to demonstrate that
[sum] p_k^1/k e^{2 pi i a_k/k}
actually does give k! distinct numbers. Again, "obvious," but why?

Title: Re: Algebraic Roots
Post by Deedlit on Apr 9^th, 2005, 4:40pm

on 04/08/05 at 13:38:47, Eigenray wrote:

Now, Deedlit you seem to be assuming things that are as hard as what is being asked. Crucial is the fact that p_n^1/n isn't in Q(p₂^1/2,...,p_n-1^1/(n-1)). Admittedly, it's "obvious", but it looks non-trivial.

Yeah, I was just saying what I knew, without being able to prove anything. I think I've forgotten most of Galois theory.

Quote:

Then the splitting field of
(x-2)(x²-3)(x³-5)(x⁴-7)...(x^k-p_k)
will be Galois of degree k!*phi(M), where M=lcm(1,2,...,k).

Case in point: why is that again?

Quote:

Then all that's left is to demonstrate that
[sum] p_k^1/k e^{2 pi i a_k/k}
actually does give k! distinct numbers. Again, "obvious," but why?

Well, I think there's some result about the logarithms of integers being algebraically indepedent if they are rationally independent; how about that for assuming something harder than we are trying to prove? :P

Title: Re: Algebraic Roots
Post by Eigenray on Apr 9^th, 2005, 6:13pm

on 04/08/05 at 13:38:47, Eigenray wrote:

Then the splitting field of
(x-2)(x²-3)(x³-5)(x⁴-7)...(x^k-p_k)
will be Galois of degree k!*phi(M), where M=lcm(1,2,...,k).

Actually, I'm not completely sure how to prove that. Let a_n=p_n^1/n, F_n=Q(a₁,...,a_n). Even assuming [F_n : F_n-1] > 1, just having [F_n : Q] = n isn't enough to have [F_n:F_n-1]=n. Now, the splitting field of the above polynomial with be F_k(z₁,...,z_k), where z_n is a primitive n-th root of unity. But that's the same as F_k(z_M), where M=lcm(1,...,k), and [Q(z_M):Q]=phi(M). Obviously [F_k(z_M):F_k]>1, as z_M isn't real, but again, what I said doesn't quite follow from that. But I don't think the actual value matters quite so much there: whatever the degree of z_M over F_k, we'll still have that the maps a_n -> z_n^ra_n give k! distinct automorphisms.