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Proof that every field is perfect???

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Proof that every field is perfect???

Perfect closure is perfectEvery algebraic extension of a perfect field is separable and perfecttwo Non isomorphic root field-extension of the field.Show that an field extension is algebraic (normal).If $E/F$ is a finite extension and $E$ is algebraically closed, then $F$ is perfect.Irreducible polynomial is always separable in char 0 fieldPerfect field of characteristic $p$Every finite field is perfectLet $F$ be a field, and let $f(x)in F[x]$ be a polynomial of prime degree. Suppose for every field extension $K$ of $F$If $Lmid K$ is a finite extension of fields then K is perfect iff L is perfect

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5

$begingroup$

The following must be wrong, since it shows that every field is perfect, which I gather is not so. But I can't find the error:

Suppose $E/K$ is a field extension and $pin K[x]$ is irreducible (in $K[x]$). Then every root of $p$ in $E$ is simple.

Proof: Suppose otoh that $lambdain E$ and $(x-lambda)^2mid p(x)$. Then $(x-lambda)mid p'$, so $gcd_E(p,p')ne1$. But the euclidean algorithm shows that $gcd_K(p,p')=gcd_E(p,p')$, hence $p$ is not irreducible.

abstract-algebra

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edited 8 hours ago

Bernard

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asked 9 hours ago

David C. UllrichDavid C. Ullrich

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$endgroup$

add a comment | 

5

$begingroup$

The following must be wrong, since it shows that every field is perfect, which I gather is not so. But I can't find the error:

Suppose $E/K$ is a field extension and $pin K[x]$ is irreducible (in $K[x]$). Then every root of $p$ in $E$ is simple.

Proof: Suppose otoh that $lambdain E$ and $(x-lambda)^2mid p(x)$. Then $(x-lambda)mid p'$, so $gcd_E(p,p')ne1$. But the euclidean algorithm shows that $gcd_K(p,p')=gcd_E(p,p')$, hence $p$ is not irreducible.

abstract-algebra

share|cite|improve this question

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

asked 9 hours ago

David C. UllrichDavid C. Ullrich

64.7k44 gold badges4444 silver badges9999 bronze badges

$endgroup$

add a comment | 

$begingroup$

The following must be wrong, since it shows that every field is perfect, which I gather is not so. But I can't find the error:

Suppose $E/K$ is a field extension and $pin K[x]$ is irreducible (in $K[x]$). Then every root of $p$ in $E$ is simple.

Proof: Suppose otoh that $lambdain E$ and $(x-lambda)^2mid p(x)$. Then $(x-lambda)mid p'$, so $gcd_E(p,p')ne1$. But the euclidean algorithm shows that $gcd_K(p,p')=gcd_E(p,p')$, hence $p$ is not irreducible.

abstract-algebra

share|cite|improve this question

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

asked 9 hours ago

David C. UllrichDavid C. Ullrich

64.7k44 gold badges4444 silver badges9999 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

asked 9 hours ago

David C. UllrichDavid C. Ullrich

64.7k44 gold badges4444 silver badges9999 bronze badges

share|cite|improve this question

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

asked 9 hours ago

David C. UllrichDavid C. Ullrich

64.7k44 gold badges4444 silver badges9999 bronze badges

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

edited 8 hours ago

Bernard

130k77 gold badges4343 silver badges121121 bronze badges

asked 9 hours ago

David C. UllrichDavid C. Ullrich

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asked 9 hours ago

David C. UllrichDavid C. Ullrich

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1 Answer
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13

$begingroup$

The problem is that you can have $p'=0$, so $gcd(p,p')=p$, which doesn't imply that $p$ is reducible.

Explicitly, if

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $cin E$ is such that $c^qin K$, but $cnotin K$.
  

then letting $p(x)=x^q-c^q$, we get

• 
  $p(x)=(x-c)^q$ which is irreducible in $K[x]$.$\[4pt]$
  


• 
  $p'(x)=0$.
  

As an example, if $t$ is an indeterminate, and

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q=(x-t)^q$.
  

then $p(x)$ is irreducible in $K[x]$ and $p'(x)=0$.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

quasiquasi

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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!... My first reaction was ok, we make an exception for $deg(p)=0$. But no, we can have $deg(p)>0$ and $p'=0$.
  $endgroup$
  – David C. Ullrich
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. This is obviously impossible in fields with characteristic zero. But if the characteristic is positive then it is very easy to find such a polynomial.
  $endgroup$
  – Mark
  9 hours ago
  

  
  
  
  

add a comment | 

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13

$begingroup$

The problem is that you can have $p'=0$, so $gcd(p,p')=p$, which doesn't imply that $p$ is reducible.

Explicitly, if

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $cin E$ is such that $c^qin K$, but $cnotin K$.
  

then letting $p(x)=x^q-c^q$, we get

• 
  $p(x)=(x-c)^q$ which is irreducible in $K[x]$.$\[4pt]$
  


• 
  $p'(x)=0$.
  

As an example, if $t$ is an indeterminate, and

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q=(x-t)^q$.
  

then $p(x)$ is irreducible in $K[x]$ and $p'(x)=0$.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!... My first reaction was ok, we make an exception for $deg(p)=0$. But no, we can have $deg(p)>0$ and $p'=0$.
  $endgroup$
  – David C. Ullrich
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. This is obviously impossible in fields with characteristic zero. But if the characteristic is positive then it is very easy to find such a polynomial.
  $endgroup$
  – Mark
  9 hours ago
  

  
  
  
  

add a comment | 

13

$begingroup$

The problem is that you can have $p'=0$, so $gcd(p,p')=p$, which doesn't imply that $p$ is reducible.

Explicitly, if

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $cin E$ is such that $c^qin K$, but $cnotin K$.
  

then letting $p(x)=x^q-c^q$, we get

• 
  $p(x)=(x-c)^q$ which is irreducible in $K[x]$.$\[4pt]$
  


• 
  $p'(x)=0$.
  

As an example, if $t$ is an indeterminate, and

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q=(x-t)^q$.
  

then $p(x)$ is irreducible in $K[x]$ and $p'(x)=0$.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!... My first reaction was ok, we make an exception for $deg(p)=0$. But no, we can have $deg(p)>0$ and $p'=0$.
  $endgroup$
  – David C. Ullrich
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. This is obviously impossible in fields with characteristic zero. But if the characteristic is positive then it is very easy to find such a polynomial.
  $endgroup$
  – Mark
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The problem is that you can have $p'=0$, so $gcd(p,p')=p$, which doesn't imply that $p$ is reducible.

Explicitly, if

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $cin E$ is such that $c^qin K$, but $cnotin K$.
  

then letting $p(x)=x^q-c^q$, we get

• 
  $p(x)=(x-c)^q$ which is irreducible in $K[x]$.$\[4pt]$
  


• 
  $p'(x)=0$.
  

As an example, if $t$ is an indeterminate, and

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q=(x-t)^q$.
  

then $p(x)$ is irreducible in $K[x]$ and $p'(x)=0$.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges

answered 9 hours ago

quasiquasi

38.3k22 gold badges2727 silver badges6868 bronze badges