If $f (x)$ is an irreducible polynomial over a perfect field $F$, then $f (x)$ has no multiple zeros$(1)$.

But I have also $2$ theorems, which are contradictory to the statement above

$\textbf{Theorem1}$: Let $f(x)$ be an irreducible polynomial over a field $F$. If $F$ has characteristic $p$, then $f(x)$ has a multiple zero only if it is of the form $f(x)=g(x^p)$ for some $g(x)$ in $F[x]$.

for example $f(x)=x^{4p}+3x^{2p}+x^p+1$, then $g(x)=x^4+3x^2+x+1$

$\textbf{Theorem2}$: Every finite field is perfect.

So every field of characteristic $p$ is finite, hence perfect, combined with $(1)$, Theorem $1$ is wrong, or not ?

  • 2
    $\begingroup$ Not every field of characteristic $p$ is finite. $\endgroup$ – Matt Samuel Jan 27 '15 at 13:19
  • $\begingroup$ There must be something wrong with theorem 1. Irreducible polynomials with multiple roots? $\endgroup$ – Git Gud Jan 27 '15 at 13:21
  • $\begingroup$ @Git Gud maybe roots are in some extension of $F$ $\endgroup$ – implicit lee Jan 27 '15 at 13:22
  • $\begingroup$ @GitGud these exist. $\endgroup$ – Matt Samuel Jan 27 '15 at 13:22
  • $\begingroup$ I am not sure what your example is meant to illustrate. In any case, $f$ is not irreducible; indeed, $g(x^p) = g(x)^p$ (in this case). $\endgroup$ – quid Jan 27 '15 at 13:33

$\newcommand{\F}{\mathbf{F}}$I believe you need to consider the following classical example.

Let $x, y$ be independent indeterminates over the field $\F$ with $p$ elements, $p$ a prime. Consider the fields $E = \F(y)$ of rational functions over $\F$ in the indeterminate $y$, and its subfield $K = \F(y^{p})$.

Then one can prove that the polynomial $$ f(x) = x^{p} - y^{p} \in K[x] $$ is irreducible in $K[x]$. And clearly $f(x) = g(x^{p})$, where $g(x) = x - y^{p}$.

Note that $f(x) = (x-y)^{p}$ has the single root $y$ in $E$, with multiplicity $p$.

The latter fact allows for an easy proof of the irreducibility of $f(x)$ in $K[x]$. In fact, if $h(x) \in K[x]$ is a monic, proper, non-constant divisor of $f(x)$, then since $K[x]$ is a UFD $h(x) = (x - y)^{k}$ for some $0 < k < p$. The coefficient of $x^{k-1}$ in $h(x)$ is $- k y$. Thus $-k y \in K$, and thus $y \in K$, which is easily seen to be a contradiction.

  • $\begingroup$ Thanks for the example, The problem was that I assumed :$char(p)\implies$finite. $\endgroup$ – implicit lee Jan 27 '15 at 14:09
  • $\begingroup$ @implicitlee, you're welcome. $\endgroup$ – Andreas Caranti Jan 27 '15 at 14:18

There is no problem. There is just no irreducible polynomial over a finite field of the form $g(x^p)$.


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