$K$ finite field, $P'(X) = 0$ then P is reductible

138 Views Asked by At

In this exercise I've shown first that if we have $P'(X)=0$ then the $Car K$ different $ 0$. Now they are specifing that if$ K$ is a finite field than $P$ is reductible. I've been searching for hours, tried to write the derivative of $P$ but I get nothing intersting.

Any Hint ?

Thanks for reading,

1

There are 1 best solutions below

1
On BEST ANSWER

Call $p$ the characteristic of $K$. Consider the linear map $$\partial : K[X] \longrightarrow K[X]$$ defined by $\partial (P(X))=P'(X)$.

It is defined on the basis $\{ X^n \}_{n \ge 0}$ by the formula $\partial (X^n)=nX^{n-1}$.

From this definition you can see that the kernel of this map is $$\ker \partial = \mathrm{span} \{ X^{np} \}_{n \ge 0}$$ in other words a ploynomial has zero derivative if and only if it has form $$P(X) = \sum_{k=0}^n a_k X^{kp}$$ But now, all these polynomials have the nice property that they can be written as $$\sum_{k=0}^n a_k X^{kp} = \left( \sum_{k=0}^n a_k X^{k} \right)^p$$ this is a consequence of the fact that in $K[X]$ the magic formula $$(A+B)^p=A^p+B^p$$ holds.