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Polynomials and function space over fields

  1. Nov 23, 2007 #1
    Hi,

    Can someone explain why the following is true? It seems to be an "accepted fact" everywhere I search, and I can't tell why.

    Let F be a field. Let E be the function from F[x] to F^F, where F[x] is the set of all polynomials over F, and F^F is the set of all functions from F to F.
    Then E is 1-to-1 if and only if F is infinite, and E is onto if and only if F is finite.

    Why is this true?

    For the first part, I can see that 1-to-1 would mean that the kernel of E is 0. So if F is finite, there should be some non-zero polynomial in F[x] which maps to the zero function in F^F.

    For the second part, I can see that if F is finite, then F^F is finite. F[x] is of course infinite, and so it makes sense that E is onto, but can someone give a more conceptual reason why? Or is that the essence of the reasoning?

    Thank you!

    Kriti
     
  2. jcsd
  3. Nov 24, 2007 #2

    matt grime

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    What do you mean by 'the' function? You simply mean the inclusion right? Well, for the first case, if I take a polynomial and consider it as a function, how can it be the zero function (i.e. send every element of F to zero)? Only if it was the zero polynomial.

    The second is also easy. Forget fields. Suppose I tell you, over the reals, that f(0)=1, f(1)=10 and f(3)=4, write down a polynomial through the points (0,1), (1,10) amd (3,4), you'd be able to do that easily. Well, the finite field question you ask is no harder.
     
  4. Nov 24, 2007 #3
    Yes, by "the" function, I meant the natural function mapping each polynomial in F[x] to it's associated function in F^F.

    For the first case, what you mentioned is true. I had figured that much myself, but that proves that F is infinite => ker E = 0 <=> E is 1-to-1. How do we go in the opposite direction? I.e., why is it that ker E = 0 => F is infinite?

    Similarly, for the second part, what you said was the basic concept I understood. But why does this require a finite field then? Given any function in F^F, couldn't I find some polynomial that could interpolate it?
     
  5. Nov 24, 2007 #4

    matt grime

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    If a field is finite it has a characteristic. It is easy to show that (x+1)^p=x^p+1 when p is the characteristic of a finite field. This is more than enough of a prod to help you with the problems you're having. If that isn't enough, then does the word Frobenious help?
     
  6. Nov 24, 2007 #5
    Here is a little something to add to what matt-grime was lecturing about.

    Theorem: Let [tex]|F|= \infty[/tex] and [tex]f(x)[/tex] be a polynomial if [tex]f(\alpha)=0[/tex] for every [tex]\alpha \in F[/tex] then [tex]f(x)=0[/tex].

    Proof: Let [tex]\deg f(x) = n[/tex] (assuming [tex]f(x)\not = 0[/tex]) then [tex]f(x)[/tex] can have at most [tex]n[/tex] zeros. But it clearly does not for any [tex]\alpha \in F[/tex] is a zero. So [tex]f(x)[/tex] must be the trivial polynomial.

    Note, if [tex]F[/tex] is finite and [tex]\deg f(x) > |F|[/tex] then the same conclusion can be drawn. But it need not to be always true. Consider [tex]F[/tex] a field of order 3. And [tex]f(x)=x^3-x[/tex] it maps all elements into zero.
     
  7. Nov 26, 2007 #6

    mathwonk

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    the proof for surjectivity matt gave also proves non injectivity in the finite case, [besides that it is obvious, since the domain is infinite and the target is finite].
     
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