Separability in a finite field












0












$begingroup$


Let $F$ be a finite field. Then any $f in F[x]$ is separable.



How do you tackle this?



I thought of letting $f in F[x]$ be irreducible, and let $alpha$ be a root of $f$. Towards a contradiction, assume that $alpha$ is a repeated root. Then $alpha$ is also a root of the formal derivative of $f$ ; $D(f)$.



Since $f$ is irreducible, it must be the minimal polynomial of $alpha$ over $F$. But $alpha$ is a root of $D(f)$ which has degree less than $f$. I think this is a contradiction, but I’m not sure?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 17:23










  • $begingroup$
    Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
    $endgroup$
    – reuns
    Jan 18 at 17:30










  • $begingroup$
    @reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
    $endgroup$
    – the man
    Jan 18 at 18:38












  • $begingroup$
    @theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 18:52










  • $begingroup$
    @ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
    $endgroup$
    – reuns
    Jan 18 at 19:05


















0












$begingroup$


Let $F$ be a finite field. Then any $f in F[x]$ is separable.



How do you tackle this?



I thought of letting $f in F[x]$ be irreducible, and let $alpha$ be a root of $f$. Towards a contradiction, assume that $alpha$ is a repeated root. Then $alpha$ is also a root of the formal derivative of $f$ ; $D(f)$.



Since $f$ is irreducible, it must be the minimal polynomial of $alpha$ over $F$. But $alpha$ is a root of $D(f)$ which has degree less than $f$. I think this is a contradiction, but I’m not sure?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 17:23










  • $begingroup$
    Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
    $endgroup$
    – reuns
    Jan 18 at 17:30










  • $begingroup$
    @reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
    $endgroup$
    – the man
    Jan 18 at 18:38












  • $begingroup$
    @theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 18:52










  • $begingroup$
    @ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
    $endgroup$
    – reuns
    Jan 18 at 19:05
















0












0








0





$begingroup$


Let $F$ be a finite field. Then any $f in F[x]$ is separable.



How do you tackle this?



I thought of letting $f in F[x]$ be irreducible, and let $alpha$ be a root of $f$. Towards a contradiction, assume that $alpha$ is a repeated root. Then $alpha$ is also a root of the formal derivative of $f$ ; $D(f)$.



Since $f$ is irreducible, it must be the minimal polynomial of $alpha$ over $F$. But $alpha$ is a root of $D(f)$ which has degree less than $f$. I think this is a contradiction, but I’m not sure?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$




Let $F$ be a finite field. Then any $f in F[x]$ is separable.



How do you tackle this?



I thought of letting $f in F[x]$ be irreducible, and let $alpha$ be a root of $f$. Towards a contradiction, assume that $alpha$ is a repeated root. Then $alpha$ is also a root of the formal derivative of $f$ ; $D(f)$.



Since $f$ is irreducible, it must be the minimal polynomial of $alpha$ over $F$. But $alpha$ is a root of $D(f)$ which has degree less than $f$. I think this is a contradiction, but I’m not sure?






abstract-algebra galois-theory finite-fields




abstract-algebra galois-theory finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 17:17









the manthe man

836716




836716












  • $begingroup$
    A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 17:23










  • $begingroup$
    Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
    $endgroup$
    – reuns
    Jan 18 at 17:30










  • $begingroup$
    @reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
    $endgroup$
    – the man
    Jan 18 at 18:38












  • $begingroup$
    @theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 18:52










  • $begingroup$
    @ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
    $endgroup$
    – reuns
    Jan 18 at 19:05




















  • $begingroup$
    A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 17:23










  • $begingroup$
    Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
    $endgroup$
    – reuns
    Jan 18 at 17:30










  • $begingroup$
    @reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
    $endgroup$
    – the man
    Jan 18 at 18:38












  • $begingroup$
    @theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
    $endgroup$
    – ÍgjøgnumMeg
    Jan 18 at 18:52










  • $begingroup$
    @ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
    $endgroup$
    – reuns
    Jan 18 at 19:05


















$begingroup$
A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
$endgroup$
– ÍgjøgnumMeg
Jan 18 at 17:23




$begingroup$
A polynomial is separable iff it is coprime to its formal derivative. If your field has characteristic $p$ and $f(X) = Q(X^p)$ then $p mid f^prime$ so $f^prime = 0$ in this field.. so $(f, f^prime) = f$, and $f$ is inseparable.
$endgroup$
– ÍgjøgnumMeg
Jan 18 at 17:23












$begingroup$
Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
$endgroup$
– reuns
Jan 18 at 17:30




$begingroup$
Let $F = mathbb{F}_p(t^p),K = mathbb{F}_p(t)$ then $K/F$ is not separable because $f(X) = X^p-t^p = (X-t)^pin F[x]$ is irreducible. $f'(X)= p X^{p-1} =0 in F[X]$. $t$ is a root of $f$ as well as of $f'$. If $f' ne 0$ then yes it would contradict $f$ is the minimal polynomial. With $F = mathbb{F}_p$ : $f$ the minimal polynomial of $alpha$ and $f$ not separable. Thus $f' = 0$ so $f(X) = g(X^p) = g(X)^p$. And $g(alpha)^p = 0$ implies $g(alpha) = 0$ (why ?) contradicting the minimality of $f$.
$endgroup$
– reuns
Jan 18 at 17:30












$begingroup$
@reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
$endgroup$
– the man
Jan 18 at 18:38






$begingroup$
@reuns Ok, thank you. So is my proof incorrect or did I just need to explain the contradiction for it to be correct?
$endgroup$
– the man
Jan 18 at 18:38














$begingroup$
@theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
$endgroup$
– ÍgjøgnumMeg
Jan 18 at 18:52




$begingroup$
@theman the statement isn't true.. you're saying any $f in F[X]$ with $F$ a finite field is separable?
$endgroup$
– ÍgjøgnumMeg
Jan 18 at 18:52












$begingroup$
@ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
$endgroup$
– reuns
Jan 18 at 19:05






$begingroup$
@ theman What do you not understand in what we said ? $f$ is the minimal polynomial of $alpha$ and $alpha$ is a double root implies $f' = 0$, because this is the only way for $f'(alpha) = 0$ not to contradict the minimality of $f$. I proposed a way to finish the proof from there.
$endgroup$
– reuns
Jan 18 at 19:05












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