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Show that $a^p^n=amod p$

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1

1

$begingroup$

My book says that for elements $alpha$ in $mathbb F_p$, where $p$ is prime, it holds that
$$
alpha^p^n=alpha,
$$
because of Fermat's little theorem, which says that
$$
a^p=amod p.
$$
Of course it's clear that $alpha^p=alpha$, but I don't see why this would hold for arbitrary powers of $p$. We have
$$
alpha^p^n=alpha^n,
$$
but this isn't necessarily equal to $alpha$, is it?

For context: my book uses this argument, to show that the roots of $X^p^n-X$ in $overlineF_p$ form a field of $p^n$ elements, and they argue that $F_p$ is a subset of the set of roots of $X^p^n-X$.

finite-fields

share|cite|improve this question

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  $alpha^p^2 equiv alpha^pp equiv (alpha^p)^p equiv alpha^p equiv alpha pmodp$.
  $endgroup$
  – Hubble
  Mar 29 at 2:52
  

  
  
  
  

add a comment | 

1

1

$begingroup$

My book says that for elements $alpha$ in $mathbb F_p$, where $p$ is prime, it holds that
$$
alpha^p^n=alpha,
$$
because of Fermat's little theorem, which says that
$$
a^p=amod p.
$$
Of course it's clear that $alpha^p=alpha$, but I don't see why this would hold for arbitrary powers of $p$. We have
$$
alpha^p^n=alpha^n,
$$
but this isn't necessarily equal to $alpha$, is it?

For context: my book uses this argument, to show that the roots of $X^p^n-X$ in $overlineF_p$ form a field of $p^n$ elements, and they argue that $F_p$ is a subset of the set of roots of $X^p^n-X$.

finite-fields

share|cite|improve this question

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  $alpha^p^2 equiv alpha^pp equiv (alpha^p)^p equiv alpha^p equiv alpha pmodp$.
  $endgroup$
  – Hubble
  Mar 29 at 2:52
  

  
  
  
  

add a comment | 

$begingroup$

My book says that for elements $alpha$ in $mathbb F_p$, where $p$ is prime, it holds that
$$
alpha^p^n=alpha,
$$
because of Fermat's little theorem, which says that
$$
a^p=amod p.
$$
Of course it's clear that $alpha^p=alpha$, but I don't see why this would hold for arbitrary powers of $p$. We have
$$
alpha^p^n=alpha^n,
$$
but this isn't necessarily equal to $alpha$, is it?

For context: my book uses this argument, to show that the roots of $X^p^n-X$ in $overlineF_p$ form a field of $p^n$ elements, and they argue that $F_p$ is a subset of the set of roots of $X^p^n-X$.

finite-fields

share|cite|improve this question

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

$endgroup$

share|cite|improve this question

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

share|cite|improve this question

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

asked Mar 29 at 2:29

Sha VukliaSha Vuklia

1,4011719

2 Answers
2

active

oldest

votes

3

$begingroup$

Fixed points stay fixed under iteration, by an obvious induction:

if $,f(x) = x,$ then $, color#c00f^large n(x) = x,Rightarrow, f^large n+1(x) = f(color#c00f^large n(x)) = f(color#c00x) = x$

OP is the special case $, f(x) := x^large p,$ so $,f^large n(x) = x^large p^Large npmod p$

share|cite|improve this answer

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

1

$begingroup$

Observe that, in any ring $R$, for $alpha in R$ and $m in Bbb N$ we have

$alpha^m^2 = (alpha^m)^m; tag 1$

$alpha^m^3 = (alpha^m^2)^m; tag 2$

and in general, for $n in Bbb N$,

$alpha^m^n = alpha^m^n - 1m = (alpha^m^n - 1)^m. tag 3$

Now suppose that $R$ is such that

$alpha^m = alpha; tag 4$

then from (1) and (2),

$alpha^m^2 = (alpha^m)^m = alpha^m = alpha, tag 5$

$alpha^m^3 = (alpha^m^2)^m = alpha^m = alpha; tag 6$

thus if

$alpha^m^j = alpha, tag 7$

then

$alpha^m^j + 1 = alpha^m^jm = (alpha^m^j)^m = alpha^m = alpha, tag 7$

and by induction it may be concluded that

$alpha^m^n = alpha tag 8$

holds for all $n in Bbb N$.

Now taking

$m = p in Bbb P, tag 9$

and

$R = Bbb F_p tag10$

we have

$alpha^p = alpha; tag11$

thus it follows that

$alpha^p^n = alpha, ; forall n in Bbb N. tag12$

share|cite|improve this answer

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Simpler: fixed points stay fixed on iteration - see my answer.
  $endgroup$
  – Bill Dubuque
  Mar 30 at 2:11
  

  
  
  
  

add a comment | 

Your Answer

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3

$begingroup$

Fixed points stay fixed under iteration, by an obvious induction:

if $,f(x) = x,$ then $, color#c00f^large n(x) = x,Rightarrow, f^large n+1(x) = f(color#c00f^large n(x)) = f(color#c00x) = x$

OP is the special case $, f(x) := x^large p,$ so $,f^large n(x) = x^large p^Large npmod p$

share|cite|improve this answer

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

3

$begingroup$

Fixed points stay fixed under iteration, by an obvious induction:

if $,f(x) = x,$ then $, color#c00f^large n(x) = x,Rightarrow, f^large n+1(x) = f(color#c00f^large n(x)) = f(color#c00x) = x$

OP is the special case $, f(x) := x^large p,$ so $,f^large n(x) = x^large p^Large npmod p$

share|cite|improve this answer

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

$begingroup$

Fixed points stay fixed under iteration, by an obvious induction:

if $,f(x) = x,$ then $, color#c00f^large n(x) = x,Rightarrow, f^large n+1(x) = f(color#c00f^large n(x)) = f(color#c00x) = x$

OP is the special case $, f(x) := x^large p,$ so $,f^large n(x) = x^large p^Large npmod p$

share|cite|improve this answer

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

share|cite|improve this answer

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

answered Mar 30 at 2:09

Bill DubuqueBill Dubuque

213k29196654

1

$begingroup$

Observe that, in any ring $R$, for $alpha in R$ and $m in Bbb N$ we have

$alpha^m^2 = (alpha^m)^m; tag 1$

$alpha^m^3 = (alpha^m^2)^m; tag 2$

and in general, for $n in Bbb N$,

$alpha^m^n = alpha^m^n - 1m = (alpha^m^n - 1)^m. tag 3$

Now suppose that $R$ is such that

$alpha^m = alpha; tag 4$

then from (1) and (2),

$alpha^m^2 = (alpha^m)^m = alpha^m = alpha, tag 5$

$alpha^m^3 = (alpha^m^2)^m = alpha^m = alpha; tag 6$

thus if

$alpha^m^j = alpha, tag 7$

then

$alpha^m^j + 1 = alpha^m^jm = (alpha^m^j)^m = alpha^m = alpha, tag 7$

and by induction it may be concluded that

$alpha^m^n = alpha tag 8$

holds for all $n in Bbb N$.

Now taking

$m = p in Bbb P, tag 9$

and

$R = Bbb F_p tag10$

we have

$alpha^p = alpha; tag11$

thus it follows that

$alpha^p^n = alpha, ; forall n in Bbb N. tag12$

share|cite|improve this answer

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Simpler: fixed points stay fixed on iteration - see my answer.
  $endgroup$
  – Bill Dubuque
  Mar 30 at 2:11
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that, in any ring $R$, for $alpha in R$ and $m in Bbb N$ we have

$alpha^m^2 = (alpha^m)^m; tag 1$

$alpha^m^3 = (alpha^m^2)^m; tag 2$

and in general, for $n in Bbb N$,

$alpha^m^n = alpha^m^n - 1m = (alpha^m^n - 1)^m. tag 3$

Now suppose that $R$ is such that

$alpha^m = alpha; tag 4$

then from (1) and (2),

$alpha^m^2 = (alpha^m)^m = alpha^m = alpha, tag 5$

$alpha^m^3 = (alpha^m^2)^m = alpha^m = alpha; tag 6$

thus if

$alpha^m^j = alpha, tag 7$

then

$alpha^m^j + 1 = alpha^m^jm = (alpha^m^j)^m = alpha^m = alpha, tag 7$

and by induction it may be concluded that

$alpha^m^n = alpha tag 8$

holds for all $n in Bbb N$.

Now taking

$m = p in Bbb P, tag 9$

and

$R = Bbb F_p tag10$

we have

$alpha^p = alpha; tag11$

thus it follows that

$alpha^p^n = alpha, ; forall n in Bbb N. tag12$

share|cite|improve this answer

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Simpler: fixed points stay fixed on iteration - see my answer.
  $endgroup$
  – Bill Dubuque
  Mar 30 at 2:11
  

  
  
  
  

add a comment | 

$begingroup$

Observe that, in any ring $R$, for $alpha in R$ and $m in Bbb N$ we have

$alpha^m^2 = (alpha^m)^m; tag 1$

$alpha^m^3 = (alpha^m^2)^m; tag 2$

and in general, for $n in Bbb N$,

$alpha^m^n = alpha^m^n - 1m = (alpha^m^n - 1)^m. tag 3$

Now suppose that $R$ is such that

$alpha^m = alpha; tag 4$

then from (1) and (2),

$alpha^m^2 = (alpha^m)^m = alpha^m = alpha, tag 5$

$alpha^m^3 = (alpha^m^2)^m = alpha^m = alpha; tag 6$

thus if

$alpha^m^j = alpha, tag 7$

then

$alpha^m^j + 1 = alpha^m^jm = (alpha^m^j)^m = alpha^m = alpha, tag 7$

and by induction it may be concluded that

$alpha^m^n = alpha tag 8$

holds for all $n in Bbb N$.

Now taking

$m = p in Bbb P, tag 9$

and

$R = Bbb F_p tag10$

we have

$alpha^p = alpha; tag11$

thus it follows that

$alpha^p^n = alpha, ; forall n in Bbb N. tag12$

share|cite|improve this answer

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

$endgroup$

share|cite|improve this answer

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167

answered Mar 29 at 3:56

Robert LewisRobert Lewis

48.6k23167