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Polynomial over finite field

  1. May 27, 2008 #1
    Hello,

    I've got a quite simple question, but I don't get it:

    Say we've got a finite field [tex]\mathbb{F}_q[/tex] and a polynomial [tex]f \in \mathbb{F}_q[X][/tex]. Let [tex]v[/tex] denote the number of distinct values of [tex]f[/tex].

    Then, i hope, it should be possible to proof, that [tex]\deg f \geq 1 \Rightarrow v \geq \frac{q}{\deg f}[/tex].

    I'd appreciate any suggestions.

    Greetings,
    korollar
     
  2. jcsd
  3. May 27, 2008 #2

    Hurkyl

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    Isn't this just a straightforward counting argument?

    Further hint: (how many inputs does it have?)
     
    Last edited: May 27, 2008
  4. May 29, 2008 #3
    Sorry, I just don't see it.

    I have q inputs. I can try out some polynomials and fields of small degree. But doesn't it demand number theoretical arguments do show this estimation?
     
  5. May 29, 2008 #4
    Ah, it's too easy. Each value can at most [tex]\deg f[/tex] times assumed by f. So [tex]v \geq \frac{q}{\deg f}[/tex].

    But how can I show, that even [tex]v \geq \frac{q-1}{\deg f} + 1[/tex] is true?
     
  6. May 30, 2008 #5
    Hint: Show that for [tex]\deg f \neq 1, q[/tex], [tex]\frac{q}{\deg f}[/tex] has a remainder. Then do [tex]\deg f = 1, q[/tex] as special cases.

    EDIT: Also note that the inequality, as stated, isn't true: consider [tex]f = x^{q+1}-x[/tex], which has only one value, 0, yet [tex]1 \not\geq 1+\frac{q-1}{q+1}[/tex]. You need a floor around your fraction.
     
    Last edited: May 30, 2008
  7. Jun 1, 2008 #6
    Sorry, I forgot to write that I only look at polynomials of degree [tex]< q[/tex].

    But I still need help.

    Say [tex]\frac{q}{\deg f} = n \cdot \deg f + r[/tex] where [tex]r < \deg f[/tex]. But in general [tex]r + \frac{\deg f -1}{\deg f} > 1[/tex]. So I can't tell why the estimation is correct?
     
  8. Jun 1, 2008 #7
    You're right, I was assuming you meant to include a floor on the fraction.

    I'll need to think more about this; it's not obvious to me.
     
  9. Jun 2, 2008 #8
    Wan et al's "Value Sets over Finite Fields" give this result as their Corollary 2.4; their paper may be worth a look.

    If there's a simple counting argument, I don't see it.
     
  10. Jun 2, 2008 #9
    I'd love to have a look at them. You don't accidentally have it at hand?

    However - I found this inequation in a paper of Gomez-Calderon, where the notation is the following:

    [tex]\left[ \frac{q-1}{d} \right] + 1 \leq |V_f|[/tex] (d: degree of f, V_f: value set of f)

    And it says: [tex][x][/tex] denotes the greatest integer [tex]\leq x[/tex] - which makes it trivial.

    In my source there were no brackets. I assume the notation above is the correct one.
     
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