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If we are working in a field of characteristic $p>0$, $K$, and we have $f$ irreducible and non-separable over $K$. If we have $L/K$ be the splitting field of $f$ and $r$ a root of $f$, is it true that $f$ splits into $(x-r)^d$.

I get the impression that this is true but cannot figure out how I would go about proving it.

Thank you for any help.

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up vote 4 down vote accepted

This is not always the case. Assume that $\alpha$ is a root of a separable irreducible polynomial $f(x)\in K[x]$ of degree $\ge2$, and further assume that $\alpha$ is not a $p^{th}$ power in $K[\alpha]$. Then the minimal polynomial of $r=\alpha^{1/p}$ over $K$ will be $f(x^p)$. Its zeros are the $p^{th}$ roots of those of $f(x)$, so they are not all equal.

For a concrete example let $K=\mathbb{F}_2(t)$, $t$ transcendental, and let $f(x)=x^2+x+t$. The zeros of $f(x)$ are $\alpha$ and $\alpha+1$. Techniques of valuations show that they both have a simple pole in the unique point lying above the infinite point of $K$, so they cannot be squares in $K[\alpha]$. Therefore the polynomial $p(x)=x^4+x^2+t$ is irreducible in $K[x]$ and it factors in its splitting field $K[\sqrt{\alpha}]$ as follows $$ p(x)=(x^2-\alpha)(x^2-\alpha-1)=(x-\sqrt{\alpha})^2(x-\sqrt{\alpha}-1)^2. $$

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@PeteL.Clark: You are right, need to rephrase... – Jyrki Lahtonen Feb 4 '13 at 7:38

The conclusion holds iff $f$ is purely inseparable, in fact almost by definition.

For an explicit counterexample, let $k$ be a field of positive characteristic $p$, let $\ell$ be a prime different from $p$, take $K = k(t)$ and $f(x) = x^{\ell p} - t$.

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Small $k$'s are integers, not fields. – Git Gud Feb 4 '13 at 7:37
@GitGud It's common in commutative algebra to let $k$ denote a field so that the extension of fields $k \subset K \subset \bar{k}$ is as confusing as possible to say. – JSchlather Feb 4 '13 at 7:41
@JacobSchlather Didn't know that, sounds funny, though. It was just a joke anyway. – Git Gud Feb 4 '13 at 7:42

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