# Proof that every field is perfect?

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 $$p'=0$$.

• Oh!... My first reaction was ok, we make an exception for $\deg(p)=0$. But no, we can have $\deg(p)>0$ and $p'=0$. – David C. Ullrich Jul 19 at 12:45
• 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. – Mark Jul 19 at 12:49