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Topic: Algebraic Roots (Read 1743 times) |
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Sir Col
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Algebraic Roots
« on: Jan 27th, 2005, 3:49pm » |
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Let p1, p2, ... , pn [in] [bbq], and assume that each pk(1/k) is irrational.
(1) Prove that no irrational number is the root of a linear equation containing integral coefficients.
(2) Prove that x = p21/2 + p1 is the root of some quadratic with integral coefficients.
(3) What degree polynomial does x = p31/3 + p21/2 + p1 come from? Prove it!
(4) What if x is a series of irrational roots: x = pn1/n + pn-11/(n-1) + ... + p1?
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Icarus
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Re: Algebraic Roots
« Reply #1 on: Jan 31st, 2005, 5:10pm » |
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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.
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Barukh
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Re: Algebraic Roots
« Reply #2 on: Feb 1st, 2005, 12:08am » |
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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 Jan 27th, 2005, 3:49pm, Sir Col wrote:| Let p1, p2, ... , pn [in] [bbq], and assume that each pk(1/k) is irrational. |
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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, rk is a shortcut for pk(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, …, rk) has degree at most k! over [bbq], and this gives the upper bound for the answer to this question.
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Deedlit
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Re: Algebraic Roots
« Reply #3 on: Mar 31st, 2005, 5:36am » |
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The degree of the polynomial is in fact k!. You can establish this by taking field automorphisms that take pk1/k to pk1/k e2[pi]i a_k / k for all k, while fixing [bbr] - note that this works because all the pk's are relatively prime. Since [bbr] is fixed, so is the minimal polynomial, but the root
[sum]pk1/k
gets taken to
[sum] pk1/k e2[pi]i a_k / k
for any assignment of ak. 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!.
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Eigenray
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Re: Algebraic Roots
« Reply #4 on: Apr 8th, 2005, 1:38pm » |
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Relatively prime isn't a strong enough assumption. For example, 1+21/2 + 31/3 + 251/4 satisfies a polynomial of degree 12. Something like, each pk is divisible by some prime exactly once, which would ensure at least that [Q(pk1/k):Q]=k (why?). Heck, lets just take pk 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 pn1/n isn't in Q(p21/2,...,pn-11/(n-1)). Admittedly, it's "obvious", but it looks non-trivial.
Then the splitting field of
(x-2)(x2-3)(x3-5)(x4-7)...(xk-pk)
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 e2pi i/M -> e2pi 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] pk1/k e2 pi i a_k/k
actually does give k! distinct numbers. Again, "obvious," but why?
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| « Last Edit: Apr 8th, 2005, 1:41pm by Eigenray » |
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Deedlit
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Re: Algebraic Roots
« Reply #5 on: Apr 9th, 2005, 4:40pm » |
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on Apr 8th, 2005, 1:38pm, Eigenray wrote:
Now, Deedlit you seem to be assuming things that are as hard as what is being asked. Crucial is the fact that pn1/n isn't in Q(p21/2,...,pn-11/(n-1)). Admittedly, it's "obvious", but it looks non-trivial.
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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)...(xk-pk)
will be Galois of degree k!*phi(M), where M=lcm(1,2,...,k).
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Case in point: why is that again?
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Then all that's left is to demonstrate that
[sum] pk1/k e2 pi i a_k/k
actually does give k! distinct numbers. Again, "obvious," but why? |
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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? 
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Eigenray
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Re: Algebraic Roots
« Reply #6 on: Apr 9th, 2005, 6:13pm » |
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on Apr 8th, 2005, 1:38pm, Eigenray wrote:Then the splitting field of
(x-2)(x2-3)(x3-5)(x4-7)...(xk-pk)
will be Galois of degree k!*phi(M), where M=lcm(1,2,...,k). |
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Actually, I'm not completely sure how to prove that. Let an=pn1/n, Fn=Q(a1,...,an). Even assuming [Fn : Fn-1] > 1, just having [Fn : Q] = n isn't enough to have [Fn:Fn-1]=n. Now, the splitting field of the above polynomial with be Fk(z1,...,zk), where zn is a primitive n-th root of unity. But that's the same as Fk(zM), where M=lcm(1,...,k), and [Q(zM):Q]=phi(M). Obviously [Fk(zM):Fk]>1, as zM 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 zM over Fk, we'll still have that the maps an -> znran give k! distinct automorphisms.
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