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Abstract algebra Polynomials and Prime

  1. Feb 15, 2016 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    Let g(x) ∈ ℤ[x] have degree at least 2, and let p be a prime number such that:
    (i) the leading coefficient of g(x) is not divisible by p.
    (ii) every other coefficient of g(x) is divisible by p.
    (iii) the constant term of g(x) is not divisible by p^2.

    a) Show that if a ∈ ℤ such that [a]_p ≠ [0]_p, then [g(a)]_p ≠ [0]_p.

    b) Show that if b ∉ ℤ such that _p = [0]_p, then [g(b)]_p^2 ≠ [0]_p^2.

    c) Conclude that g(x) has no roots in ℤ

    2. Relevant equations


    3. The attempt at a solution

    Well, the problem I am having here is understanding the question at all.

    So g(x) exists in Z[x].

    What does Z[x] exactly mean?

    It has a degree of at least 2, and we have p as a prime number.

    We know the leading coefficient of g(x) is NOT divisible by p.

    So this means we have C*x^n+...
    Where c is not divisible by p.

    Every other coefficient of g(x) is divisible by p.
    So we have C*x^n + g*x^(n-1)+....
    where g IS divisible by p.

    The constant term fo g(x) is not divisible by p^2. So we now have
    C*x^n + g*x^(n-1)+....+ z
    where z is not divisible by p*p.

    So we look at question a. If a exists in the integers such that [a]_p cannot equal [0]_p then [g(a)]p cannot equal [0]p.
    So we have
    C*a^2+g*a+z as an example polynomial.

    and here is where I start to break down.
    Can't decide what to do from here.
     
  2. jcsd
  3. Feb 15, 2016 #2

    RJLiberator

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    Gold Member

    Any idea on this? :)
     
  4. Feb 15, 2016 #3

    RJLiberator

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    Gold Member

    Perhaps we can take a step back and instead just look at part a

    a) Show that if a exists in the integers such that [a]_p cannot equal [0]_p then [g(a)]_p cannot equal [0]_p.

    Logically, this seems consistent and almost obvious.

    Let's try this, using the division algorithm:

    We take the first term of some polynomial
    Q*a^n = dp+r

    Q is NOT divisible by p by (i).
    So if we assume, for a contradiction, that r = 0.
    Then Q*a^n/p cannot equal the integer d because
    a) Q is not divisible by p
    b) [a]_p =/= [0]_p

    Therefore, [g(a)]_p =/= [0]_p.
     
  5. Feb 15, 2016 #4

    fresh_42

    Staff: Mentor

    I read it this way:
    should be
    a) Show that if ##a∈ℤ: a ≠ 0 \mod{p} ⇒ g(a) ≠ 0 \mod{p}##

    and
    should be
    b) Show that if ##b ∉ ℤ: b = 0\mod{p} ⇒ g(b) ≠ 0 \mod{p^2}##

    That is my decoding of your input. Unfortunately ##b ∉ ℤ## generates two major problems:
    (i) Where is ##b## supposed to be in? ##ℚ##?, ##ℝ##? or ##ℂ##?
    (ii) If ##b ∉ ℤ## then ##b \mod{p}## doesn't make any sense.

    For the rest: I would write ##g(x) = a_n x^n + a_{n-1} p^{r_{n-1}} x^{n-1} +... + a_{1} p^{r_{1}} x + a_0 p ##
    where none of the ##a_i ∈ ℤ ## is divisible by ##p##, ## n ∈ℕ, n>1## and ##r_j ∈ ℕ, r_j>0##.

    That should make things easier to talk about. However, please resolve the ##b## mystery!
     
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