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I am reading separable extension. I don't quite understand why $\text{char} 0$ fields must be perfect. If $F$ is a $\text{char} 0$ field, $f(x)$ is an irreducible polynomial in $F[x]$. Then why $f$ must be separable? Assume $f$ has multiple roots, say $r$ is a multiple root of $f(x)$. Then $r$ is also a root of $f'(x)$. $r$ is in the splitting field of $f(x)$ over $F$, denote it as $E$. Then why $f$ must divide $f'$? I know $f$ is irreducible but I don't know how the argument goes on. Thanks for any help!

cali
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3 Answers3

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Suppose $f$ is an irreducible polynomial and $\xi$ is a root of $f$. If $g$ is any other polynomial such that $g(\xi)=0$, then $f\mid g$. Indeed, if we call $h$ the greatest common divisor of $f$ and $g$, then $h$ is not $1$, since it has $\xi$ a a root. As $h$ is then a non-contant polynomial which divides $f$, it must be a scalar multiple of $f$ itself. Since $h$ divides also $g$, we see that $f$ divides $g$.

Mariano Suárez-Álvarez
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    Oh right. f(x) is irreducible so it is the minimal polynomial of its roots. Thanks! – cali May 03 '15 at 02:24
  • @Mariano CAn you please tell how you argument proves that polynomial will always be separable in char0 field.? According to definition of separability, I have to prove that roots are distinct in any algebraic closure of K. But i am not able to prove it with the help of your argument. Do you mind elaborating? –  May 16 '21 at 04:53
  • @No-One Let $g = f'$. Clearly $g$ is non-zero as we are working in characteristic $0$. But $f$ is a degree $n$ polynomial dividing $g$, which by definition of the derivative is a degree $n-1$ polynomial, which is impossible. Therefore, $f$ and $g$ couldn't have shared a root to begin with. – Arun Bharadwaj Jun 23 '21 at 04:16
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    Hello! Where did you use that the characteristic is $0$? – dahemar Feb 05 '23 at 18:52
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We relied on the assumption that the field $K$ has characteristic 0 to conclude that $f'$ has degree $n-1$.

mahsa yg
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We can also prove this in another way. Let us assume that $f$ and $f'$ have a common root. By the assumption that $f$ is irreducible, we can conclude that $f$ is the minimal polynomial of that root. Therefore, we have $deg(f) \leq deg(f')$, which is contradiction . Note that this contradiction arises because the field $K$ has characteristic zero, so if $f$ has degree $n$, then $f'$ has degree $n-1$.

mahsa yg
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