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The problem is that you can have $p'=0$, so $\gcd(p,p')=p$, which doesn't imply that $p$ is reducible.

To understand how this might happen, suppose

$q$ is prime. $\\[4pt]$

is prime. $\text{char}(K)=q$ .

and suppose $c\in E$ is such that $c^q\in K$, but $c

ot\in K$.

Then letting $p(x)=x^q-c^q$, we get $p'(x)=0$.

Claim $p$ is irreducible in $K[x]$.

To verify the irreducibility of $p$, suppose $f\mid p$ in $K[x]$, where $f$ is a monic polynomial of degree $n$, with $0 < n < q$.

Since $f\mid p$ in $K[x]$, we also have $f\mid p$ in $E[x]$.

Since $q$ is prime and $\text{char}(K)=q$, we have the identity $$x^q-c^q=(x-c)^q$$ hence, since $f$ is monic and $\text{deg}(f)=n$, it follows that $f=(x-c)^n$.

By the binomial theorem, the coefficient of the $x^{n-1}$ term of $f$ is $-nc$.

But then from $0 < n < q$ and $c

ot\in K$, we get $-nc

ot\in K$, contrary to $f\in K[x]$.

Therefore $p$ is irreducible in $K[x]$, as claimed.

For an explicit example of such a polynomial $p$, let $t$ be an indeterminate, and let

$K=F_q(t^q)$ . $\\[4pt]$

. $E=F_q(t)$ . $\\[4pt]$

. $p(x)=x^q-t^q$ .

Then we have

$\text{char}(K)=q$ . $\\[4pt]$

. $t\in E$ . $\\[4pt]$

. $t^q\in K$ , but $t

ot\in K$ .

hence $p$ is irreducible in $K[x]$ and $p'=0$.