# Roots of a $p$-polynomial all have same multiplicity $p^e$.

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.)

• Write $f$ in the form $f = g^{p^e}$ where $g$ is separable, recalling that $p$-powers respect $+$ and $\cdot$ – Cocopuffs Mar 27 '13 at 20:51
• @Cocopuffs You mean $f(x)=g(x^{p^e})$ for some $g(x)$? – Chelsea Dirks Mar 27 '13 at 21:00
• @Cocopuffs What if the coefficients of $f$ are not necessarily all $p$-powers? – Chelsea Dirks Mar 27 '13 at 21:02
• $g$ might have coefficients from the splitting field, in general – Cocopuffs Mar 27 '13 at 21:06
• 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 ;) – darij grinberg Mar 28 '13 at 5:58

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 $$p^e$$.

• 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. – Bach Feb 5 '19 at 14:27