Ttdfjt

Why is it "on the inside" and not "in the inside"?

How long does it take for electricity to be considered OFF by general appliances?

Problem with Eigenvectors

Should I intervene when a colleague in a different department makes students run laps as part of their grade?

How did the SysRq key get onto modern keyboards if it's rarely used?

Unknown indication below upper stave

How do I make my photos have more impact?

Why did I lose on time with 3 pawns vs Knight. Shouldn't it be a draw?

Convert graph format for Mathematica graph functions

A cubeful of three-dimensional devilry

Who said "one can be a powerful king with a very small sceptre"?

How to season a character?

How should I quote American English speakers in a British English essay?

Is there an antonym(a complementary antonym) for "spicy" or "hot" regarding food (I DO NOT mean "seasoned" but "hot")?

Why put copper in between battery contacts and clamps?

Shouldn't there be "us" instead of "our" in this sentence?

Why would anyone ever invest in a cash-only etf?

Is it okay for me to decline a project on ethical grounds?

How can Paypal know my card is being used in another account?

Desktop app status bar: Notification vs error message

How to improve king safety

To find islands of 1 and 0 in matrix

Exploiting the delay when a festival ticket is scanned

Rampant sharing of authorship among colleagues in the name of "collaboration". Is not taking part in it a death knell for a future in academia?

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.Field of characteristic p - perfect iff pth roots of elements are all in the field.A question regarding proving the fact that every finite field is perfectShow that if a field is perfect any irreducible polynomial is separable.Algebraic extension of perfect field in which every polynomial has a root is algebraically closedIf $Lmid K$ is a finite extension of fields then K is perfect iff L is perfect

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

19

2

$begingroup$

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 field-theory extension-field fake-proofs

share|cite|improve this question

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

$endgroup$

add a comment | 

19

2

$begingroup$

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 field-theory extension-field fake-proofs

share|cite|improve this question

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 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 field-theory extension-field fake-proofs

share|cite|improve this question

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

share|cite|improve this question

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

edited Jul 20 at 6:57

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

asked Jul 19 at 12:30

David C. UllrichDavid C. Ullrich

65.1k44 gold badges4444 silver badges101101 bronze badges

1 Answer
1

active

oldest

votes

28

$begingroup$

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

To understand how this might happen, suppose

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.
  

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

Then letting $p(x)=x^q-c^q$, we get $p'(x)=0$.

Claim $p$ is irreducible in $K[x]$.

To verify the irreducibility of $p$, suppose $fmid p$ in $K[x]$, where $f$ is a monic polynomial of degree $n$, with $0 < n < q$.

Since $fmid p$ in $K[x]$, we also have $fmid p$ in $E[x]$.

Since $q$ is prime and $textchar(K)=q$, we have the identity
$$x^q-c^q=(x-c)^q$$
hence, since $f$ is monic and $textdeg(f)=n$, it follows that $f=(x-c)^n$.

By the binomial theorem, the coefficient of the $x^n-1$ term of $f$ is $-nc$.

But then from $0 < n < q$ and $cnotin K$, we get $-ncnotin K$, contrary to $fin K[x]$.

Therefore $p$ is irreducible in $K[x]$, as claimed.

For an explicit example of such a polynomial $p$, let $t$ be an indeterminate, and let

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q$.
  

Then we have

• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $tin E$.$\[4pt]$
  


• 
  $t^qin K$, but $tnotin K$.
  

hence $p$ is irreducible in $K[x]$ and $p'=0$.

share|cite|improve this answer

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:45
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:49
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3297710%2fproof-that-every-field-is-perfect%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

28

$begingroup$

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

To understand how this might happen, suppose

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.
  

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

Then letting $p(x)=x^q-c^q$, we get $p'(x)=0$.

Claim $p$ is irreducible in $K[x]$.

To verify the irreducibility of $p$, suppose $fmid p$ in $K[x]$, where $f$ is a monic polynomial of degree $n$, with $0 < n < q$.

Since $fmid p$ in $K[x]$, we also have $fmid p$ in $E[x]$.

Since $q$ is prime and $textchar(K)=q$, we have the identity
$$x^q-c^q=(x-c)^q$$
hence, since $f$ is monic and $textdeg(f)=n$, it follows that $f=(x-c)^n$.

By the binomial theorem, the coefficient of the $x^n-1$ term of $f$ is $-nc$.

But then from $0 < n < q$ and $cnotin K$, we get $-ncnotin K$, contrary to $fin K[x]$.

Therefore $p$ is irreducible in $K[x]$, as claimed.

For an explicit example of such a polynomial $p$, let $t$ be an indeterminate, and let

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q$.
  

Then we have

• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $tin E$.$\[4pt]$
  


• 
  $t^qin K$, but $tnotin K$.
  

hence $p$ is irreducible in $K[x]$ and $p'=0$.

share|cite|improve this answer

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:45
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:49
  

  
  
  
  

add a comment | 

28

$begingroup$

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

To understand how this might happen, suppose

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.
  

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

Then letting $p(x)=x^q-c^q$, we get $p'(x)=0$.

Claim $p$ is irreducible in $K[x]$.

To verify the irreducibility of $p$, suppose $fmid p$ in $K[x]$, where $f$ is a monic polynomial of degree $n$, with $0 < n < q$.

Since $fmid p$ in $K[x]$, we also have $fmid p$ in $E[x]$.

Since $q$ is prime and $textchar(K)=q$, we have the identity
$$x^q-c^q=(x-c)^q$$
hence, since $f$ is monic and $textdeg(f)=n$, it follows that $f=(x-c)^n$.

By the binomial theorem, the coefficient of the $x^n-1$ term of $f$ is $-nc$.

But then from $0 < n < q$ and $cnotin K$, we get $-ncnotin K$, contrary to $fin K[x]$.

Therefore $p$ is irreducible in $K[x]$, as claimed.

For an explicit example of such a polynomial $p$, let $t$ be an indeterminate, and let

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q$.
  

Then we have

• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $tin E$.$\[4pt]$
  


• 
  $t^qin K$, but $tnotin K$.
  

hence $p$ is irreducible in $K[x]$ and $p'=0$.

share|cite|improve this answer

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:45
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 12:49
  

  
  
  
  

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.

To understand how this might happen, suppose

• 
  $q$ is prime.$\[4pt]$
  


• 
  $textchar(K)=q$.
  

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

Then letting $p(x)=x^q-c^q$, we get $p'(x)=0$.

Claim $p$ is irreducible in $K[x]$.

To verify the irreducibility of $p$, suppose $fmid p$ in $K[x]$, where $f$ is a monic polynomial of degree $n$, with $0 < n < q$.

Since $fmid p$ in $K[x]$, we also have $fmid p$ in $E[x]$.

Since $q$ is prime and $textchar(K)=q$, we have the identity
$$x^q-c^q=(x-c)^q$$
hence, since $f$ is monic and $textdeg(f)=n$, it follows that $f=(x-c)^n$.

By the binomial theorem, the coefficient of the $x^n-1$ term of $f$ is $-nc$.

But then from $0 < n < q$ and $cnotin K$, we get $-ncnotin K$, contrary to $fin K[x]$.

Therefore $p$ is irreducible in $K[x]$, as claimed.

For an explicit example of such a polynomial $p$, let $t$ be an indeterminate, and let

• 
  $K=F_q(t^q)$.$\[4pt]$
  


• 
  $E=F_q(t)$.$\[4pt]$
  


• 
  $p(x)=x^q-t^q$.
  

Then we have

• 
  $textchar(K)=q$.$\[4pt]$
  


• 
  $tin E$.$\[4pt]$
  


• 
  $t^qin K$, but $tnotin K$.
  

hence $p$ is irreducible in $K[x]$ and $p'=0$.

share|cite|improve this answer

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

edited Jul 20 at 6:58

TheSimpliFire

14.6k66 gold badges2929 silver badges7272 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges

answered Jul 19 at 12:34

quasiquasi

39.1k33 gold badges2828 silver badges6868 bronze badges