# Proving Existence of b in F for Field with Char p and Reducible f

• fishturtle1
In summary, we have shown that if ##f## is reducible over ##F##, then it splits in ##F##. Furthermore, if ##a## is a ##p##-th power, then ##f## splits in ##F##.
fishturtle1
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##?

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##?

Infrared said:
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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If ##f=(x-c)^p##, what are the possibilities for how it can factor as ##f=gh##?

fishturtle1 said:
##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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fishturtle1
Infrared said:
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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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##?

Infrared said:
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##. []

Right, this is fine. Even a little bit simpler is just to look at the coefficient ##-nc## for the ##x^{n-1}## term.

fishturtle1
Infrared said:
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?

Every nonzero element of a field has an inverse. Since ##0<n<p##, it is not a multiple of ##p##, so nonzero.

fishturtle1
Infrared said:
Every nonzero element of a field has an inverse. Since ##0<n<p##, it is not a multiple of ##p##, so nonzero.

## 1. What is the significance of proving the existence of b in a field with characteristic p and reducible polynomial f?

The existence of b in a field with characteristic p and reducible polynomial f is significant because it allows us to understand the structure and properties of the field. It also helps us to solve equations and perform operations in the field.

## 2. How is the existence of b proven in a field with characteristic p and reducible polynomial f?

The existence of b is proven by showing that the polynomial f has a root in the field. This can be done by using techniques such as the Euclidean algorithm or the factor theorem to find a root of f.

## 3. Can the existence of b be proven in any field with characteristic p and reducible polynomial f?

Yes, the existence of b can be proven in any field with characteristic p and reducible polynomial f. However, the method of proof may vary depending on the specific field and polynomial.

## 4. What are the implications of not being able to prove the existence of b in a field with characteristic p and reducible polynomial f?

If the existence of b cannot be proven in a field with characteristic p and reducible polynomial f, it means that the field is not algebraically closed. This can limit the types of equations that can be solved and the operations that can be performed in the field.

## 5. How does the existence of b relate to the concept of roots of polynomials?

The existence of b is closely related to the concept of roots of polynomials. In fact, the existence of b is equivalent to the existence of a root of the polynomial f in the field. This means that if b exists, then f has a root in the field, and vice versa.

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