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On page 549 in Dummit and Foote is a proposition which states "Every irreducible polynomial over a finite field is separable. A polynomial in $\mathbb{F}_p[x]$ is separable if and only if it is the product of distinct irreducible polynomials in $\mathbb{F}_p[x]$." However, we know that if the derivative of a polynomial $f(x)$ is $0$, then every root is a multiple root, and so $f(x)$ is not separable. Does this mean, then, that any polynomial in $\mathbb{F}_p$ with derivative $0$ is reducible? The reason I ask is because on the next page, the authors state "If $p(x)$ is an irreducible polynomial which is not separable..." But doesn't the proposition above preclude this?

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  • $\begingroup$ I don't have Dummit and Foote's book, but I suppose the last sentence either refers to the case where the field is not necessarily finite, or the proposition is not yet proved. $\endgroup$ – Bernard Feb 21 '18 at 16:44
  • $\begingroup$ It says "over a field of characteristic p" $\endgroup$ – ponchan Feb 21 '18 at 16:47
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    $\begingroup$ @ponchan There are infinite fields of characteristic $p$. For instance, $\mathbb{F}_p(t)$, the field of rational functions. $\endgroup$ – Viktor Vaughn Feb 21 '18 at 16:53
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Over $K=\Bbb{F}_p$ it does hold that if, for some $f(x)\in K[x]$, we have $f'(x)=0$, then $f$ is reducible. The reason is the following.

If $f'(x)=0$ this means that all the terms in $f(x)$ have degrees that are multiples of $p$. In other words, $$ f(x)=\sum_{i=0}^na_ix^{pi} $$ for some natural number $n$ and some coefficients $a_i\in\Bbb{F}_p$. Two key results then come to the fore:

  • In a commutative ring $R$ of characteristic $p$ we have the formula (also known as Freshman's dream) for all $a,b\in R$: $$(a+b)^p=a^p+b^p.$$
  • For all $a\in\Bbb{F}_p$ we have $a^p=a$ (Little Fermat).

Put together these imply that the above polynomial $f(x)$ $$ f(x)=\sum_{i=0}^na_ix^{pi}=\sum_{i=0}^n(a_ix^{i})^p=\left(\sum_{i=0}^na_ix^i\right)^p $$ is actually the $p$th power of a lower degree polynomial, hence reducible.


The same result holds for other fields $K$ of characteristic $p$ as long as all the elements of $K$ are $p$th powers of some element of $K$ (above it would have been enough to have $a_i=b_i^p$ for some $b_i\in K$). Such fields are called perfect, and whenever $K$ is a perfect field we see that irreducible polynomials over $K$ are necessarily separable.

Therefore we can use as the field $K$ any finite field. This is because the Frobenius automorphism $z\mapsto z^p$ is an injective endomorphism of $K$ (trivial kernel). When $K$ is finite "injectivity $\implies$ surjectivity" and we are done.

It doesn't work as nicely for all fields of characteristic $p$. The textbook counterexample is $K=\Bbb{F}_p(t)$. The polynomial $m(T)=T^p-t$ is irreducible (Eisenstein), but it is not separable. It has a single zero $t^{1/p}$ of multiplicity $p$ in an extension field of $K$.

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  • $\begingroup$ I can't understand how the first sentence matches the reasoning that follows. The reasoning basically says that if $f'(x)=0$ then $f$ is reducible, and yet "Over $K=\Bbb{F}_p$ it does hold that if, for some $f(x)\in K[x]$, we have $f'(x)=0$, then $f$ is irreducible."? Did you mean "reducible" maybe? $\endgroup$ – polettix Apr 14 '19 at 17:37
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    $\begingroup$ @polettix Yes. There was a typo.Thanks. $\endgroup$ – Jyrki Lahtonen Apr 14 '19 at 18:40

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