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I am going over my lecture notes in my Field theory class and I saw this following statement without a proof: if $\mathrm{char}(F)=p>0$ then if $g(x)\in F[x]$ is irreducible then $g(x)$ have multiple roots iff $g'(x)=0$.

I believe I can prove that if $g(x)$ have multiple roots then $g'(x)=0$ but I am not sure and I am unable to prove that the converse is also true.

My reasoning is as follows: $g(x)$ have multiple roots imply there is an extension $K/F$ and $\alpha\in K$ s.t $g(\alpha)=g'(\alpha)=0$ (since if the multiplicity of $\alpha$ in $g(x)$ is $m>1$ (since it is not a simple root) then the multiplicity of $\alpha$ in $g'(x)$ is at least $m-1\gt 0$). Since $g(x)$ is irreducible and we may assume WLOG that $g(x)$ is monic then it follows that the minimal polynomial of $\alpha$ over $F$ is $g(x)$ , but $g'(\alpha)=0$ and if $g'\neq0$ then $\deg(g')<\deg(g)$ and this is a contradiction.

Is my argument correct, and how can I prove the converse ? help is appreciated!

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Note that $\deg(g')$ could be $0$ in principle. Although you know that $\deg(g)\geq 2$, you could have something like $g(x) = x^p-2x$ or something like that, where the derivative is constant even though $g(x)$ is not linear. – Arturo Magidin Jul 17 '12 at 22:48
up vote 5 down vote accepted

Your argument for necessity is correct.

For sufficiency, if $g'(x)=0$ and $g(x)$ is irreducible, then it is not constant (constant polynomials are units, hence not irreducible by definition).

If we write $g(x) = a_nx^n+ a_{n-1}x^{n-1}+\cdots + a_0$, with $n\gt 0$ and $a_n\neq 0$, then we conclude that for every $i$ such that $a_i\neq 0$, we must have $ia_i=0$; therefore, $i$ is a multiple of $p$. Thus, $g(x)$ is a polynomial in $x^p$.

Therefore we have that $g(x) = a_0 + a_1x^p + \cdots +a_kx^{kp}$. Passing to an extension of $F$ where each coefficient is a $p$th power, if necessary, we can write $g(x)$ as $$\begin{align*} g(x) &= a_0 + a_1x^p + \cdots + a_kx^{kp}\\ &= r_0^p + r_1^px^p + \cdots + r_k^px^{pk} \\ &= (r_0 + r_1x + r_2x^2 + \cdots + r_kx^k)^p \end{align*}$$ hence $g(x) = h(x)^p$ for some $h(x)$, and hence $g(x)$ must have repeated roots.

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Thank you very much! – Belgi Jul 17 '12 at 22:47
One can also show that all the roots have the same multiplicity (a power of $p$), I think. – Dylan Moreland Jul 17 '12 at 22:50

Your proof looks good! I think you could prove the other direction by working more explicitly with the derivative, as follows.

Let $\alpha$ be a root of $g$ lying in some extension of $K/F$. In $K[x]$ one can write $g(x) = (x - \alpha)^mh(x)$ where $m \geq 1$ and $h(\alpha) \neq 0$. Old rules from calculus apply to give \[ g'(x) = m(x - \alpha)^{m - 1}h(x) + (x - \alpha)^mh'(x). \] Now the contrapositive follows: if $f$ has only simple roots then $m = 1$ and $g'(\alpha) = h(\alpha)$ is non-zero, so $g' \neq 0$.

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@Belgi I was just editing — I might have overreached with regard to the direction you proved. – Dylan Moreland Jul 17 '12 at 23:15
I thought I saw that you wrote that this shows both directions, my bad – Belgi Jul 17 '12 at 23:16
@Belgi Nono, my fault entirely. I still think it's a little unclear... – Dylan Moreland Jul 17 '12 at 23:17

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