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Let $F$ be of characteristic $p$. A polynomial $f(x)\in F[x]$ is called a $p$-polynomial if it has the form $x^{p^m}+a_1x^{p^{m-1}}+\cdots+a_mx$. Show that a monic polynomial of positive degree is a $p$-polynomial iff its roots form a finite subgroup of the additive group of the splitting field, and every root has the same multiplicity $p^e$. [Basic Algebra I, by Nathan Jacobson. (3 pp. 234.)]

I think I proved the $\Longleftarrow$ direction. For $\implies$, assuming $f(x)$ is a $p$-polynomial, it is easy to show its roots form an additive subgroup, by using the fact that $p$-powers distribute over $\pm$. I'm having a hard time showing that every root has the same multiplicity $p^e$.

I have two cases:

  1. If $a_m\neq 0$, then $f'(x)=a_m\neq 0$, so $f$ is separable, and has no repeated roots. So every root has multiplicity $1=p^0$.

  2. If $a_m= 0$, by Cocopuff's suggestion, I can write $f(x)=g(x^{p^e})$. Normally, if $g(x)$ is separable, I can show that every root of $f(x)$ has multiplicity $p^e$, but here the $p$-polynomial $f(x)$ is reducible, so I can't tell if $g(x)$ is irreducible to apply the derivative criterion. How can I get around this? (Answered in comments.)

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    $\begingroup$ Write $f$ in the form $f = g^{p^e}$ where $g$ is separable, recalling that $p$-powers respect $+$ and $\cdot$ $\endgroup$ – Cocopuffs Mar 27 '13 at 20:51
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    $\begingroup$ @Cocopuffs You mean $f(x)=g(x^{p^e})$ for some $g(x)$? $\endgroup$ – Chelsea Dirks Mar 27 '13 at 21:00
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    $\begingroup$ @Cocopuffs What if the coefficients of $f$ are not necessarily all $p$-powers? $\endgroup$ – Chelsea Dirks Mar 27 '13 at 21:02
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    $\begingroup$ $g$ might have coefficients from the splitting field, in general $\endgroup$ – Cocopuffs Mar 27 '13 at 21:06
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    $\begingroup$ A polynomial is said to be separable when it is coprime to its derivative. Now, that is clearly the case when the derivative is a nonzero constant ;) $\endgroup$ – darij grinberg Mar 28 '13 at 5:58
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I am going to take as an assumption that your statement, that when $a_m\neq 0$, all roots have multiplicity $ 1 $, is true. It certainly sounds correct, but I'm no expert on that part.

However, you can easily show the case for $a_m=0$ from that point, as follows:

Suppose that the smallest index that arises is $p^e$. That is,

$$ f(x) = \sum_{i=0}^{m-e} a_ix^{p^{m-i}} $$ where $a_0=1$ and $a_e\neq 0$. Now, let $y=x^{p^e}$. This allows us to write

$$ f(x) = \sum_{i=0}^{m-e} a_i y^{p^{m-i}/p^e} = \sum_{i=0}^{m-e} a_i y^{p^{m-e-i}/p^e} $$ Let $M=m-e$, so we have

$$ f(x) = \sum_{i=0}^M a_i y^{p^{M-i}} = g(y) $$ Now, $g(y)$ has, from the assumed statement, $p^M$ roots of multiplicity $ 1 $.

Now is where we bring in the characteristic being $p$, which means that $(x+y)^p = x^p+y^p$. So we seek solutions to $x^p=a$. If $x_1^p=x_2^p=a$, then $x_1^p-x_2^p = (x_1-x_2)^p=0$, and so $x_1=x_2$. So $x^p-a=0$ has one root of multiplicity $p$, and by iterating this process, we find that $x^{p^e}-a=0$ has one root of multiplicity $p^e$.

But our roots of $g(y)=f(x)$ take the form $y=x^{p^e}=a$, so each root of $g(y)$ is associated uniquely with a root of $f(x)$, and has multiplicity $p^e$.

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    $\begingroup$ The reason why all roots have multiplicity $ 1 $ when $ a_m\ne 0 $ is $ \gcd(f(x), f'(x))=1 $ iff. $ f(x) $ has no multiple roots. $\endgroup$ – Bach Feb 5 '19 at 14:27

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