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Distinct zeros of irreducible polynomial

  1. Apr 19, 2013 #1
    This claim is supposed to be true. Assume that [itex]p\in\mathbb{F}[X][/itex] is an irreducible polynomial over a field [itex]\mathbb{F}\subset\mathbb{C}[/itex]. Also assume that

    [tex]
    p(X)=(X-z_1)\cdots (X-z_N)
    [/tex]

    holds with some [itex]z_1,\ldots, z_N\in\mathbb{C}[/itex]. Now all [itex]z_1,\ldots, z_N[/itex] are distinct.

    Why is this claim true?

    For example, if [itex]z_1=z_2[/itex], then [itex](X-z_1)^2[/itex] divides [itex]p[/itex], but I see no reason to assume that [itex](X-z_1)^2\in\mathbb{F}[X][/itex], so the claim remains a mystery to me.
     
  2. jcsd
  3. Apr 19, 2013 #2
    This is because every field F of characteristic zero is a perfect field. This means that every irreducible polynomial over F is separable ( splits in the splitting field with distinct linear factors ).

    A polynomial p(x) has multiple roots in the splitting field, iff the polynomials p(x) and Dp(x) [its formal derivative ] have a root in common. In other words, if c is the root of p(x), the minimal polynomial of c over F will divide both p(x) and Dp(x). So, if p(x) has multiple roots in the splitting field, p(x) and Dp(x) are not co-prime.

    Assume p(x) in F[x] is irreducible, and has degree n. Then, Dp(x) is degree n - 1; however, p(x) is irreducible, so its only factors in a factorization are p( x ) and constants. So, since Dp(x) is degree n -1 < n, its non-constant factors must be degree d: 1 <= d <= n - 1. So, p(x ) and Dp(x) do not have any common non-unit factors, and hence the polynomials are co-prime. So, p(x) is separable
     
  4. Apr 19, 2013 #3
    Thank you for the good answer, but I did not understand the purpose of every part of it. Is there something wrong with proof like this:

    Assume that [itex]p\in\mathbb{F}[X][/itex] has a zero [itex]x[/itex] of higher degree than one. Now [itex]Dp\in\mathbb{F}[X][/itex] has the same zero. Let [itex]m[/itex] be the minimial polynomial of [itex]x[/itex] in [itex]\mathbb{F}[/itex]. Now [itex]m|p[/itex] and [itex]m|Dp[/itex], so in particular [itex]m|p[/itex] with such [itex]m[/itex] that its degree is less than the degree of [itex]p[/itex], which is a contradiction.
     
  5. Apr 20, 2013 #4
    There isn't anything wrong with saying it like that. That's the idea: if p is irreducible over F, then p is a minimal polynomial ( once you scale it to be monic ), of some x which is a root in the splitting field. If x is also a root of Dp, then the minimal polynomial of x has to divide Dp. As Dp has a smaller degree, this is impossible.
    This is exactly what was going on in the last paragraph. Actually it's a proof (although it's pretty trivial ) of the obvious fact that since Dp has smaller degree, the minimal polynomial cannot divide Dp. the irreducible factors of Dp must have degree <= n - 1. But the minimal polynomial must be degree n, and if the min polynomial divided Dp, it would appear in the decomposition of Dp into irreducibles
     
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