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Tag 09H0

Chapter 9: Fields > Section 9.12: Separable extensions

Lemma 9.12.1. Let $F$ be a field. Let $P \in F[x]$ be an irreducible polynomial over $F$. Let $P' = \text{d}P/\text{d}x$ be the derivative of $P$ with respect to $x$. Then one of the following two cases happens

  1. $P$ and $P'$ are relatively prime, or
  2. $P'$ is the zero polynomial.

Then second case can only happen if $F$ has characteristic $p > 0$. In this case $P(x) = Q(x^q)$ where $q = p^f$ is a power of $p$ and $Q \in F[x]$ is an irreducible polynomial such that $Q$ and $Q'$ are relatively prime.

Proof. Note that $P'$ has degree $< \deg(P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg(P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2).

Assume we are in case (2) and $P = a_d x^d + \ldots + a_0$. Then $P' = da_d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_d, \ldots, a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_i = 0$ whenever $p \not | i$. In other words, $P(x) = P_1(x^p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. $\square$

    The code snippet corresponding to this tag is a part of the file fields.tex and is located in lines 1101–1114 (see updates for more information).

    \begin{lemma}
    \label{lemma-irreducible-polynomials}
    Let $F$ be a field. Let $P \in F[x]$ be an irreducible polynomial over $F$.
    Let $P' = \text{d}P/\text{d}x$ be the derivative of $P$ with respect
    to $x$. Then one of the following two cases happens
    \begin{enumerate}
    \item $P$ and $P'$ are relatively prime, or
    \item $P'$ is the zero polynomial.
    \end{enumerate}
    Then second case can only happen if $F$ has characteristic $p > 0$.
    In this case $P(x) = Q(x^q)$ where $q = p^f$ is a power of $p$ and
    $Q \in F[x]$ is an irreducible polynomial such that $Q$ and $Q'$
    are relatively prime.
    \end{lemma}
    
    \begin{proof}
    Note that $P'$ has degree $< \deg(P)$. Hence if $P$ and $P'$ are not relatively
    prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg(P)$
    contradicting the irreducibility of $P$. This proves we have the dichotomy
    between (1) and (2).
    
    \medskip\noindent
    Assume we are in case (2) and $P = a_d x^d + \ldots + a_0$. Then
    $P' = da_d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see
    that this forces $a_d, \ldots, a_1 = 0$ which would mean $P$ is constant
    a contradiction. Thus we conclude that the characteristic $p$ is positive.
    In this case the condition $P' = 0$ forces $a_i = 0$ whenever $p \not | i$.
    In other words, $P(x) = P_1(x^p)$ for some nonconstant polynomial $P_1$.
    Clearly, $P_1$ is irreducible as well. By induction on the degree we
    see that $P_1(x) = Q(x^q)$ as in the statement of the lemma, hence
    $P(x) = Q(x^{pq})$ and the lemma is proved.
    \end{proof}

    Comments (1)

    Comment #2780 by Pieter Belmans (site) on August 20, 2017 a 10:12 am UTC

    It's not very important, but it would be better to write \not\mid compared to \not|: compare $\not\mid$ to $\not|$.

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