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This is from Gallian's Contemporary Abstract Algebra: enter image description here

I don't understand the step in the red box: why can we assume that the common factor has a zero? Isn't it possible that for example $f(x)$ and $f'(x)$ has a common factor $x^2+1$ in ${\mathbb R}[x]$?

  • Yes, it is possible, but even for those zeros (non-real zeros) the argument carries over. If you want to avoid working with complex numbers you can always factor out, instead of $(x-a)$, the factor $x^2+1$ from $f$ and from $f'$. – OR. Aug 04 '13 at 13:56

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The root is allowed by the phrasing of the theorem to be in some extension $E$.

Small edit: Your question brings to mind one of my favorite and somewhat surprising exercises. The gcd of $f(x)$ and $g(x)\in F[x]$ is the same as their gcd in $E[x]$ for any extension $E$ of $F$.

Ted Shifrin
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    This is certainly the right answer. Still, to my taste the text should have said "let $a$ be a zero of the common factor in some extension field $E$". This is obviously intended but is not what was said, hence the confusion. – Pete L. Clark Aug 04 '13 at 13:58
  • See also this answer, which explains why gcds in PID/Bezout domains persist in extension rings. – Bill Dubuque Jun 18 '14 at 00:37
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While it is correct that the common factor $g$ of $f$ and $f'$ may not have a zero in the field $F$, the theorem is about multiple zeroes "in some extension $E$". As soon as the common factor $g$ has positive degree, it has a zero $a$ at least in the algebraic closure $\overline F$ of $F$. This is in general an infinite extension, but $a$ is also a root of $g$ (and hence a common root of $f$ and $f'$ and ultimately a multiple root of $f$) in the finite extension $E=F(a)$ of $F$. Alternatively, if $h$ is an irreducible factor of $g$, let $E=F[x]/(h)$; there $h$ (and hence $g$) has a root. So the important thing here is that the claim to be shown allows us to switch to an extension of $F$.

Hagen von Eitzen
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