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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?

Please help.

Michael Hardy
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John Michael
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3 Answers3

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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).

J. W. Tanner
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Hagen von Eitzen
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    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
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    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
  • Generally risky to reference comments, particularly comments to other answers. – Thomas Andrews Aug 26 '21 at 22:36
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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.

Andreas Caranti
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  • Can you give/link such an example? – Berci May 19 '13 at 10:21
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    @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
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    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
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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 $\mathbb{F}_p(t^{1/p})$.

Robin Balean
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    Could you please explain why $x^p-t$ is irreducible ? – Friedrich Jan 13 '18 at 20:49
  • @Friedrich, using Gauss lemma, $x^p-t$ is irreducible in $\Bbb{F}_p(t)[x]$ if and only if it is irreducible in $\Bbb{F}_p[t][x]=\Bbb{F}_p[t,x]$. The latter is clear, since $x^p-t$ has degree $1$ in $t$. – rmdmc89 Oct 09 '20 at 14:10