Field with char p

  • #1
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Homework Statement:
Let ##F## be a field with characteristic ##p## and let ##f(x) = x^p - a \in F[x]##. Show ##f## is irreducible over ##F## or ##f## splits in ##F##.
Relevant Equations:
##F## has characteristic ##p## means that ##p## is the smallest positive integer such that for all ##x \in F## we have ##px = 0##.

##f(x)## splits in ##F## means ##f(x) = c_0(x-c_1)(x-c_2)\cdot\dots\cdot(x-c_n)## where ##c_i \in F##.

##f(x)## is irreducible over ##F## means if ##f(x) = g(x)h(x)## in ##F##, then ##g(x)## or ##h(x)## is a unit.
Suppose ##f## is reducible over ##F##. Then there exists ##g, h \in F## such that ##g, h## are not units and ##f = gh##. If there exists ##b \in F## such that ##b^p = a##, then ##(x - b)^p = x^p - b^p = x^p - a##, using the fact that ##F## has characteristic ##p##. So, if such a ##b \in F## exists, then ##f## splits in ##F##. But I don't think we can guarantee ##b## does exist. And I realize I didn't really use the assumption that ##f## is reducible. How to proceed?

Does ##f## being reducible over ##F## somehow imply ##a## has a pth root in ##F##?
 

Answers and Replies

  • #2
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If ##a## is a ##p##-th power, say ##a=c^p##, then ##f(x)=x^p-c^p=(x-c)^p## splits.

If ##a## is not a ##p##-th power, you can still factor ##f=(x-c)^p## in a splitting field for ##f##. Do you see why this implies that ##f## must be irreducible over ##F##?
 
  • #3
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If ##a## is a ##p##-th power, say ##a=c^p##, then ##f(x)=x^p-c^p=(x-c)^p## splits.

If ##a## is not a ##p##-th power, you can still factor ##f=(x-c)^p## in a splitting field for ##f##. Do you see why this implies that ##f## must be irreducible over ##F##?
Thanks for the response. I'm sorry I don't. The only thing I can think to do is try to show ##F[x]/(f)## is a field which would show ##f## is irreducible over ##F##. I can think of polynomials that are reducible over a field, but don't split, such as ##x^2 - 4## is reducible over ##\mathbb{Z}## but does not split in ##\mathbb{Z}##. So it has to do with ##f## having only one zero?

Am I just completely not understanding a definition here?
 
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  • #4
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If ##f=(x-c)^p##, what are the possibilities for how it can factor as ##f=gh##?

##x^2 - 4## is reducible over ##\mathbb{Z}## but does not split in ##\mathbb{Z}##
Besides ##\mathbb{Z}## not being a field, ##x^2-4=(x-2)(x+2)## splits. Something like ##(x^2+1)^2## is reducible but doesn't split (over ##\mathbb{Q}##).
 
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  • #5
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If ##f=(x-c)^p##, what are the possibilities for how it can factor as ##f=gh##?


Besides ##\mathbb{Z}## not being a field, ##x^2-4=(x-2)(x+2)## splits. Something like ##(x^2+1)^2## is reducible but doesn't split.
##f## can factor as ##f = (x-c)^m(x-c)^n## for integers ##m,n## where ##m + n = p## and I think we need to show that ##m = 0## or ##n = 0##. Assume by contradiction that ##f## is reducible over ##F##. Then ##f = (x-c)^m(x-c)^n## for some ##m, n \ge 2## (since ##c## is, by assumption, not an element of ##F##).






Also thanks for the correction, I remind myself now that ##\mathbb{Z}## is not a field because not everything has an inverse...my bad.
 
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  • #6
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Right, so you need to show that ##(x-c)^n## is not in ##F[x]## when ##0<n<p##, meaning that not all of its coefficients lie in ##F##. Can you find a coefficient that doesn't lie in ##F##?
 
  • #7
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Right, so you need to show that ##(x-c)^n## is not in ##F[x]## when ##0<n<p##, meaning that not all of its coefficients lie in ##F##. Can you find a coefficient that doesn't lie in ##F##?
It think it is ##c^n \not\in F[x]## and I show it by contradiction.

Proof: Assume ##f## does not split in ##F## and ##f## is reducible over ##F##. Then ##f = (x-c)^m(x-c)^n## for some integers ##m,n \ge 2## such that ##m + n = p##. This implies ##c^m, c^n \in F##. By previous chapter, ##p## is ##0## or prime. If ##p = 0##, then contradiction. Otherwise ##p## is prime. This implies ##\gcd(m, n) = 1##. Then there exists integers ##u, v## such that ##mu + nv = 1## i.e. ##(c^m)^u(c^n)^v = c^{mu + nv} = c^1 = c \in F##. But if ##c \in F##, then we have ##x - c \in F[x]## which would imply ##f## splits in ##F##. We have reached contradiction and can conclude ##(x-c)^m, (x-c)^n \not\in F[x]## i.e., ##f## is irreducible over ##F##. []
 
  • #8
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Right, this is fine. Even a little bit simpler is just to look at the coefficient ##-nc## for the ##x^{n-1}## term.
 
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  • #9
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Right, this is fine. Even a little bit simpler is just to look at the coefficient ##-nc## for the ##x^{n-1}## term.
I'm a fool for this, but how do we show ##-nc \in F## implies ##c \in F##, if I understand what you're saying? Does ##n## always have an inverse in a field?
 
  • #10
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Every nonzero element of a field has an inverse. Since ##0<n<p##, it is not a multiple of ##p##, so nonzero.
 
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  • #11
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Every nonzero element of a field has an inverse. Since ##0<n<p##, it is not a multiple of ##p##, so nonzero.
I understand now, thank you for your help on this thread.
 

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