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My proof goes as follows:


If $\alpha$ is a multiple root of $f=\newcommand{\Irr}{\mbox{Irr}(\alpha,K)}\Irr$ then $$ \begin{align} f&=(X-\alpha)^2g\\ f'&=2(X-\alpha)g+(X-\alpha)g' \end{align} $$ so $\alpha$ is a root of $f'$. Thus $$ f'\in(f)=\mbox{Ker}(ev_\alpha) $$ but the ideal $(f)$ only contains polynomials of degree greater than or equal to that of $f$ - except for the zero polynomial. Thus $f'=0$.

Suppose now instead that $f'=0$. In characteristic $0$ this means that $f$ is a constant polynomial so it is not irreducible. In characteristic $p$ it means that $f\in K[X^p]$ since all powers of $X$ that are not powers of $X^p$ have non-zero derivatives. Well, don't know where to go from there with the $\mbox{char}(K)=p$ case...

I wonder if there is an easier way to phrase this...

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I do not understand why would $f'\in (f)$ – Praphulla Koushik Jan 14 '14 at 12:51
@PraphullaKoushik: Because $f$ generates the ideal of all polynomials having $\alpha$ as a root. – String Jan 14 '14 at 12:54
Oh yes... I think that makes sense! – Praphulla Koushik Jan 14 '14 at 12:57
up vote 1 down vote accepted

You have $f(X)=g(X^p)$ for some $g\in K[X]$ of positive degree. Now look at this equation over an algebraic closure $\overline{K}$ of $K$. Then all the coefficients of $g$ are $p$-th powers, and we can write $f(X)=g(X^p)=\tilde{g}(X)^p$ for a polynomial $\tilde{g}\in \overline{K}[X]$ of positive degree. It follows that $f$ has repeated roots in $\overline{K}$.

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That was a clever argument! Thank you! – String Jan 14 '14 at 14:02
I find it rather strange that my textbook stated, that we should prove the $\Leftarrow$ direction as an exercise. I do not think, that I would have considered this myself... – String Jan 14 '14 at 14:04
Does this mean that roots of $f$ come with multiplicity $p$ or could it even be any multiple of $p$? – String Jan 14 '14 at 14:09
Dear @String, The multiplicities are at least $p$, but they could be larger powers of $p$ as well, because $g$ might also satisfy $g(X)=h(X^p)$ for some $h\in K[X]$, i.e., $g$ might not be separable. In general you can write $f(X)=f_{sep}(X^{p^f})$ for some (necessarily irreducible) separable $f_{sep}\in K[X]$ and some $f\geq 1$. – Keenan Kidwell Jan 14 '14 at 14:13
Ah, ok. I expected something like that, but it is nice to know! – String Jan 14 '14 at 14:15

Over $\mathbf{F}_p$, $f(x^p) = (f(x))^p$, so $f(x^p)$ can't be irreducible. This generalises over $\mathbf{F}_q$ for $q = p^k$, $f(x^p) = (g(x))^p$ where the coefficients of $g(x)$ are the inverse of those of $f(x)$ by the Frobenius map.

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I am having second thoughts... This seems to prove that there are NO $\alpha$'s that are inseparable in any fields since $f'=0$ does not correspond to an irreducible polynomial $f$ in any characteristic. – String Jan 14 '14 at 13:43
Also, the Frobenius map only fixes the coefficients if they are in $\mathbb F_p$ not if they are in a general characteristic $p$ fields, right? – String Jan 14 '14 at 13:44
Inseparable elements do not exist over any field of characteristic zero or any finite field. They can exist only over infinite fields of characteristic $>0$. And yes sorry $f(x^p) = (f(x))^p$ only over $\mathbf{F}_p$. But over $\mathbf{F}_{p^k}$ even if the coefficients are not fixed, you still have a factorisation of $f(x^p)$. – fkraiem Jan 14 '14 at 13:51
Thank you for clarifying that! I wonder how the inseparable elements come into play then in those infinite characteristic $p$ fields... – String Jan 14 '14 at 13:53
For an example, consider $L = \mathbf{F}_2(y)$ (the field of rational functions with coefficients in $\mathbf{F}_2$ in inteterminate $y$), which is infinite of characteristic $2$. Let $t = y^2$, and consider $K = \mathbf{F}_2(t)$. $L = K(y)$ is an algebraic extension of $K$, and the minimal polynomial of $y$ over $K$ is $x^2-t \in K[x]$. But in $L[x]$, $x^2-t = x^2-y^2 = (x-y)^2$, so $y$ is a multiple root. – fkraiem Jan 14 '14 at 14:06

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