# Question regarding separability of polynomials and perfect fields

I have this definition

A field $F$ is perfect if it has characteristic $0$ or if it has characteristic $p$ and $F^p = F$

From Wikipedia, I have this fact about separable polynomials:

Irreducible polynomials over perfect fields are separable.

I can show that if $p(x)$ is irreducible and $char(F) = 0$, then $p$ cannot have multiple zeros, i.e. it's separable. I can also show that, if $p(x)$ has multiple zeros, it has to be of the form $p(x) = q(x^p)$.

Now my question:

In the comment on Wikipedia, how does the second part follow? That is, that if $char(F)=p$ for $p$ prime and $f(x)$ is irreducible over $F$ then it cannot have multiple zeros?

Many thanks for your help.

• You know two things: p(x) = q(x^p) and every element of F is a pth power. Can you combine those two things? – Qiaochu Yuan Feb 15 '11 at 9:51

## 1 Answer

I'm expanding on this answer in response to a comment.

We'll use the following theorem: $f \in F[x]$ has a multiple root in some extension $E$ of $F$ if and only if $f$ and $f^\prime$ have a common factor $d(x) \in F[x]$ of positive degree.

First, let $char(F) = 0$ and let $a_0 + a_1 x + \dots + a_n x^nf \in F[x]$ be irreducible over $F$. We prove by contradiction that $f$ cannot have a multiple root. So, assume $f$ has a multiple root. Then $f$ and $f^\prime$ have a common factor $d(x) \in F[x]$ of positive degree. $f$ is irreducible so the only divisor of positive degree is $f$ itself, so that $f$ divides $f^\prime$. A polynomial cannot divide a polynomial of smaller degree hence $f^\prime = 0$. But this means that $a_1 = 2 a_2 = \dots = n a_n = 0$. Since $char(F) = 0$ we get $a_k = 0$ and therefore $f = a_0$ which is not irreducible.

If $F$ has characteristic $p$ then $(x + y)^p = x^p + y^p$. Then if $f(x) = g(x^p)$ with coefficients in $F$ and $F^p = F$ then $f(x) = a_n^px^{pn} + \dots + a_0^p = (a_nx^n \dots + a_0)^p = g(x)^p$ because all $a_i \in F$ can be written as powers of $p$ by assumption. But then by assumption $f$ is irreducible over $F$ therefore contradiction.

• @MattN.Hi. I am working on the same issue. Maybe you would help me with the math logic here. The lemma I was looking at stipulates that (as you mention) $f$ is irreducible. So I see the contradiction. What I am stuck on is, that being so, how do you get distinct roots in the case where $x$ -> $x^p$? Thanks. – user12802 May 29 '12 at 17:47
• @Andrew I can't be of help tonight so I posted your comment in chat. Let's hope someone can help. If not, I'll be back tomorrow. If it's urgent you might want to post a question. – Rudy the Reindeer May 29 '12 at 18:15
• @MattN.Thanks. I'll check or see you tomorrow. – user12802 May 29 '12 at 18:43
• @MattN.Found this googling. I think it will do it - have to work on understanding it. PERFECT FIELDS Characteristic 0 fields have a very handy feature ... – user12802 May 29 '12 at 19:21
• @Andrew The proof for a field $F$ of characteristic $p$ goes like this: assume there is an irreducible polynomial with multiple roots. Then we can produce a contradiction. Hence there cannot be any irreducible polynomials with multiple roots in a field of prime characteristic. – Rudy the Reindeer May 30 '12 at 12:06