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I am slightly confused about what formal derivatives over finite fields mean.

Example 1: Consider $f(x)=x^3-2\in \mathbb{F}_7[x]$.

By checking each element of $\mathbb{F}_7$ we easily see that this is irreducible. What about separable? Can we look at the formal derivative $f’(x)=3x^2$ which has a double zero at $0$ and hence gcd$(f,f’)=1$ and so $f(x)$ is separable?

Example 2: Consider $f(x)=x^p-x+1\in\mathbb{F}_p[x]$.

Observe that if $\alpha$ is a root then $\alpha +a$ is a root for any $a\in\mathbb{F}_p$ (because $(\alpha +a)^p=\alpha^p+a^p$). Hence has $0$ is not a root, $f$ is irreducible.

Now $f$ is separable because we have $p$ distinct roots. Can we show this also by the formal derivative? What is the formal derivative here?

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    Wait a second!? In your second example you can conclude that $f$ has no multiple roots as in DonAntonio's good answer. But you cannot conclude that $f$ is irreducible!! It is easy to find reducible polynomials without roots in the coefficient field: e.g. $(x^2-2)(x^2-3)$ over rationals. The irreducibility of this polynomial is discussed e.g. [here.](http://math.stackexchange.com/q/81583/11619) – 2012-06-02

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Yes to your first question, and $(x^p-x+1)'=px^{p-1}-1=-1\pmod p$ so the pol. is separable (as is any irreducible pol. over any field of characteristic zero or over any finite field)