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

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

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

  • $K=F_q(t^q)$.$\\[4pt]$
  • $E=F_q(t)$.$\\[4pt]$
  • $p(x)=x^q-t^q$.

Then we have

  • $\text{char}(K)=q$.$\\[4pt]$
  • $t\in E$.$\\[4pt]$
  • $t^q\in K$, but $t\not\in K$.

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

  • 5
    $\begingroup$ Oh!... My first reaction was ok, we make an exception for $\deg(p)=0$. But no, we can have $\deg(p)>0$ and $p'=0$. $\endgroup$ – David C. Ullrich Jul 19 at 12:45
  • 4
    $\begingroup$ Yes. This is obviously impossible in fields with characteristic zero. But if the characteristic is positive then it is very easy to find such a polynomial. $\endgroup$ – Mark Jul 19 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.