Proving Existence of b in F for Field with Char p and Reducible 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 $F$?

 

[Infrared]

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 $F$?

 

[fishturtle1]

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?

 

[Infrared]

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 $\mathbb{Q}$).

 

[fishturtle1]

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.

 

[Infrared]

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 $F$?

 

[fishturtle1]

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 $F$. []

 

[Infrared]

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

 

[fishturtle1]

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?

 

[Infrared]

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

 

[fishturtle1]

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.