# Formal derivatives over finite fields.

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?

-
I am not sure what the first sentence of your question has to do with the rest of it. – Qiaochu Yuan May 30 '12 at 21:34
The formal derivative obeys the usual rule for the derivative of a product. So if a polynomial has a repeated root (possibly in an extension field), then it has a repeated irreducible factor $m(x)$ = the minimal polynomial of the said root. If $f(x)=m(x)^2g(x)$, then $f'(x)=2m(x)m'(x)g(x)+m(x)^2g'(x)$, and hence $m(x)$ is a common factor. This would show in the calculation of $\gcd(f(x),f'(x))$ by other means, e.g. Euclid's algorithm. If that was tl;dr; the the quick answer is: YES. Your logic is valid :-) – Jyrki Lahtonen May 31 '12 at 5:04
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. – Jyrki Lahtonen Jun 2 '12 at 7:40

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)