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Math Help - Very tough polynomial question.

  1. #1
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    Very tough polynomial question.

    Let f(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n be a polynomial with integer coefficients and assume that for a given prime integer p that:

    1. p does not divide a_0,

    2. p divides a_1,a_2,...,a_n

    3. p^2 does not divide a_{n-1}
    Prove that either  f(x) has a rational root, or  f(x) is irreducible over the filed Q

    What is the meaning of the last sentence? Do I have to show that there are two cases: case1 irreducible, case2 has rational root?

    I assume that I should consider two cases. Case1 is easy, it occurs when p^2 does not divide  a_{n}. In this case Eisenstein's criterion gives answer. But what about case2 when p^2 divides  a_{n} ?

    I managed to show the following in case2:

    f(x)=x(a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1})+a_n where a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1} is irreducible according to Eisenstein's criterion. I have a feeling that f(x) has a rational root, but do not know how to show it...
    Last edited by mr fantastic; December 1st 2008 at 02:27 PM. Reason: removed initial comment about having posted this question elsewhere.
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  2. #2
    Senior Member vincisonfire's Avatar
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    First you should have learned that roots in  \mathbb Q *are of the form  \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0} If not tell me I'll show you.
    Notice that it could be the case that none of them works. In this case the polynomial is irreducible over  \mathbb Q .
    Else, let *c_0 be a root found using the criterion  \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0} .
    Then it is possible to rewrite the polynomial as  g(x)=(x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1}) . You can now simply factor the second polynomial with the criterion  \frac{\pm \text{ factors of }b_{n-1}}{ \text{ factors of }b_0} .
    At this point you can use the Eisenstein's criterion.
     h(x)=b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1}
    Even though you have divided these coefficients by some number, you can again put it in the form of a polynomial with integer coefficient by multiplying the whole thing by the least common multiple of the denominator.
    You still have that some  p|b_i but  p^2 \text{ doesn't divide } b_{n-1} that is some multiple of   a_{n-1} .
    Therefore, this h(x) is irreducible and you thus are either one or no root.
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  3. #3
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    =vincisonfire;229638]First you should have learned that roots in  \mathbb Q *are of the form  \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}
    I know that and I tried but did not find very helpful.

    You did not use a_n so you continued my case2 ?
    Last edited by andreas; December 1st 2008 at 03:46 AM.
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  4. #4
    Senior Member vincisonfire's Avatar
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    The three conditions you post add up all together.
    You have only one case with three conditions. Not three with one condition.
    What you did in your "case 2" doesn't work.
    Any polynomial of the form  ax^2 +bx is reducible. But you can't divide the polynomial in the way you did. You can get a factor out but not a term.
    Remark that I used the three conditions in my answer. I'm pretty sure I'm right even though my explanation can seem a little rusty.
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  5. #5
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    You still have that some  p|b_i but  p^2 \text{ doesn't divide } b_{n-1} that is some multiple of   a_{n-1} .
    Therefore, this h(x) is irreducible and you thus are either one or no root.

    How do you prove this?
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  6. #6
    Senior Member vincisonfire's Avatar
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    When you make the division, you'll find that  b_{n-1} = \frac{a_n}{c_0} because  a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n = (x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1})
    You thus have an integer that p does not divide, divided by an integer that p does divide.
    Since p divides all  a_i expcept  a_0 , when you multiply everything by the least common multiple you are sure you won't have a  p^2 factor in the factorization of  b_{n-1} .
    Although, you are sure that you will have p factor in some  a_i since  p doesn't divide  a_0
    Last edited by vincisonfire; December 1st 2008 at 08:12 AM.
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