The following must be wrong, since it shows that every field is perfect, which I gather is not so. But I can't find the error:
Suppose $E/K$ is a field extension and $p\in K[x]$ is irreducible (in $K[x]$). Then every root of $p$ in $E$ is simple.
Proof: Suppose OTOH that $\lambda\in E$ and $(x-\lambda)^2\mid p(x)$. Then $(x-\lambda)\mid p'$, so $\gcd_E(p,p')\ne1$. But the euclidean algorithm shows that $\gcd_K(p,p')=\gcd_E(p,p')$, hence $p$ is not irreducible.
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
and suppose $c\in E$ is such that $c^q\in K$, but $c\not\in K$.
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 $c\not\in K$, we get $-nc\not\in K$, 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
Then we have
hence $p$ is irreducible in $K[x]$ and $p'=0$.
