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\author{Jason Wojciechowski}
\title{Notebook Problem Fraleigh 9.4.18}

\newtheorem{thm}{Theorem}

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\begin{thm}
  Every irreducible polynomial over a field $F$ of characteristic 0 is
  seperable.
\end{thm}

\begin{proof}
  The irreducible polynomial $f$ for an arbitrary $\alpha \in \bar F$
  over some field $F$ is also the polynomial of minimum degree for
  $\alpha$, so $f'$, the derivative of $f$, which has degree one less
  than $f$, does not have $\alpha$ as a zero unless $f'$ is the zero
  polynomial.  However, there are only two ways $f'$ could be the zero
  polynomial: if $f$ was a constant polynomial or if the coefficients
  of $f'$ were all zeros. We know $f$ is non-constant because constant
  polynomials are not irreducible.  We know the coefficients of $f'$
  are not zero because they are all of the form $a_i(i \cdot 1)$.
  There is at least one $a_i$ which is non-zero, or else we would have
  the zero polynomial, and $i \cdot 1$ is not zero because $F$ has
  characteristic $0$.
  
  Thus, by exercise $17$, $\alpha$ is a zero of multiplicty one, so
  all of $f$'s zeros are of multiplicity one, so $f$ is seperable.
\end{proof}

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