# Irreducible polynomials have distinct roots?

I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.

Theorem 7.3 page 27 seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me.

Does the result hold for $\Bbb F_q$ where q is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots?

Consider a field $$F$$ of characteristic $$p$$. A polynomial has multiple roots only if it has a root in common with its (formal) derivative; that is, the multiple roots of $$f$$ are the roots of $$\gcd(f,f')$$. Since $$f$$ is irreducible, multiple roots can occur only if the $$\gcd$$ is $$f$$ itself, that is $$f'$$ is a multiple of $$f$$. And that is only possible if $$f'=0$$, that is, all monomials in $$f$$ have degree a multiple of $$p$$, so $$f(x)=g(x^p)$$ for some polynomial $$g$$.

If $$F$$ is finite, then $$\phi\colon a\mapsto a^p$$ is an automorphism of $$F$$ (and also of the splitting field $$E$$ of our polynomial), and there exists $$h(x)$$ such that $$\phi(h)=g$$. Then for $$\alpha\in E$$ with $$f(\alpha)=0$$ also $$h(\alpha)=0$$ (because $$\phi(h(\alpha))=\phi(h)(\phi(\alpha))=g(\alpha^p)=f(\alpha)=0$$). Since $$h$$ is of smaller degree than $$f$$, $$f$$ is not irreducible.

As this proof shows, one has to look for cases where $$\phi$$ is not an automorphism to find a counterexample (such as in Andreas Carantis comment).

• Sorry to ask, but why must there exist a h(x) such that phi(h) = g? I know phi is an automorphism of F but h(x) is in F[x]. – John Michael May 19 '13 at 11:33
• If $g(X)=\sum a_kX^k$, find $b_k\in F$ with $\phi(b_k)=a_k$ and let $h(X)=\sum b_kX^k$. (In other words, $\phi$ extends to an automorphism of $F[X]$ by coefficientwise application). – Hagen von Eitzen May 19 '13 at 11:35
• phi(h) means raising the whole polynomial to the pth power right? Or it means raising only each coefficient? eg. phi(2x + 3x^2) = (2^p)(x^p) + (3^p)(x^2p) or phi(2x + 3x^2) = (2^p)(x) + (3^p)(x^2) ? – John Michael May 19 '13 at 11:36

The result holds over any finite field. One way of seeing this is that if $h(x) \in \Bbb{F}_{q}[x]$ (where $q$ is a power of the prime $p$) is irreducible over $\Bbb{F}_{q}$, and $\alpha$ is one of its roots, then $\alpha$ is also algebraic over $\Bbb{F}_{p}$. If $f(x) \in \Bbb{F}_{p}[x]$ is the minimal polynomial of $\alpha$ over $\Bbb{F}_{p}$, then $h(x)$ divides $f(x)$, and you know that the latter has distinct roots.

There are examples, though, of irreducible polynomials of degree $> 1$ over an infinite field of positive characteristic which have only one root.

• Can you give/link such an example? – Berci May 19 '13 at 10:21
• @Berci, will try and post a reference. However the standard example is the polynomial $f(x) = x^p - z^ p \in F[x]$, where $z$ is an indeterminate over $\Bbb{F}_{p}$, and $F = \Bbb{F}_{p}(z^p) \subseteq \Bbb{F}_{p}(z) = E$. Clearly $f(x)$ has the only root $z$ in its splitting field $E$, and $f(x)$ can be readily shown to be irreducible in $F[x]$. – Andreas Caranti May 19 '13 at 10:24
• Thank you! I forgot that we can have an infinite field with positive characteristic. – John Michael May 19 '13 at 10:53
• @JohnMichael, you're welcome! – Andreas Caranti May 19 '13 at 10:54

As pointed out, to get an example of an irreducible polynomial with multiple roots, it cannot be finite. Here is an example where this happens in an infinite field of characteristic $p$.

Consider the irreducible polynomial $f(x)=x^p-t\in \mathbb{F}_p(t)[x]$ in the extension field $\mathbb{F}_p(t^{1/p})$ of the field $\mathbb{F}_p(t)$ of rational functions over $\mathbb{F}_p$.

Notice that $(x - t^{1/p})^p = x^p + (-1)^p t = x^p - t$, where the first equality follows from the binomial theorem and the second is obviously true for all odd $p$, and when $p=2$ we have $t=-t$ so it holds there as well.

Thus $f$ has one root of multiplicity $p$ in $\mathbb{F}_p(t^{1/p})$.

• Could you please explain why $x^p-t$ is irreducible ? – Friedrich Jan 13 '18 at 20:49