$mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$ if and only if $d$ divides $n$ [duplicate]
$begingroup$
This question already has an answer here:
Proving that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$.
2 answers
I am trying to solve the following exercise of Dummit and Foote Book(page # 551).
Let $a>1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$ if and only if $d$ divides $n$.
I did the first part and I know that for all $alphain mathbb{F}_{p^d}$, $alpha^{p^d}=alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
abstract-algebra field-theory finite-fields
asked Dec 5 '18 at 20:26
usmndjusmndj
484
$endgroup$
marked as duplicate by Jyrki Lahtonen abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 6 '18 at 10:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proving that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$.
2 answers
I am trying to solve the following exercise of Dummit and Foote Book(page # 551).
Let $a>1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$ if and only if $d$ divides $n$.
I did the first part and I know that for all $alphain mathbb{F}_{p^d}$, $alpha^{p^d}=alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
abstract-algebra field-theory finite-fields
asked Dec 5 '18 at 20:26
usmndjusmndj
484
$endgroup$
marked as duplicate by Jyrki Lahtonen abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 6 '18 at 10:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07
add a comment |
$begingroup$
This question already has an answer here:
Proving that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$.
2 answers
I am trying to solve the following exercise of Dummit and Foote Book(page # 551).
Let $a>1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$ if and only if $d$ divides $n$.
I did the first part and I know that for all $alphain mathbb{F}_{p^d}$, $alpha^{p^d}=alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
abstract-algebra field-theory finite-fields
asked Dec 5 '18 at 20:26
usmndjusmndj
484
$endgroup$
This question already has an answer here:
Proving that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$.
2 answers
I am trying to solve the following exercise of Dummit and Foote Book(page # 551).
Let $a>1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$ if and only if $d$ divides $n$.
I did the first part and I know that for all $alphain mathbb{F}_{p^d}$, $alpha^{p^d}=alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
This question already has an answer here:
Proving that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$.
2 answers
abstract-algebra field-theory finite-fields
abstract-algebra field-theory finite-fields
asked Dec 5 '18 at 20:26
usmndjusmndj
484
asked Dec 5 '18 at 20:26
usmndjusmndj
484
asked Dec 5 '18 at 20:26
usmndjusmndj
484
asked Dec 5 '18 at 20:26
usmndjusmndj
484
asked Dec 5 '18 at 20:26
usmndjusmndj
484
484
marked as duplicate by Jyrki Lahtonen abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 6 '18 at 10:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 6 '18 at 10:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07
add a comment |
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathbb F_{p^n}$ is the splitting field of $x^{p^n}-x$.
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
add a comment |
$begingroup$
If $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$, then $n=[mathbb{F}_{p^n}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}][mathbb{F}_{p^d}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}]d$, and so $d$ divides $n$.
If $d$ divides $n$, then $mathbb{F}_{p^d}^times$ is a subgroup of $mathbb{F}_{p^n}^times$ and so $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$. Here we use that $mathbb{F}_{p^n}^times$ is cyclic and that there is at most one finite field of a given size.
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$mathbb F_{p^n}$ is the splitting field of $x^{p^n}-x$.
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
add a comment |
$begingroup$
$mathbb F_{p^n}$ is the splitting field of $x^{p^n}-x$.
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
$endgroup$
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
add a comment |
$begingroup$
$mathbb F_{p^n}$ is the splitting field of $x^{p^n}-x$.
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
$endgroup$
$mathbb F_{p^n}$ is the splitting field of $x^{p^n}-x$.
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
answered Dec 5 '18 at 20:34
Chris CusterChris Custer
11.6k3824
11.6k3824
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
add a comment |
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
$begingroup$
Thanks, I got it.
$endgroup$
– usmndj
Dec 5 '18 at 21:32
add a comment |
$begingroup$
If $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$, then $n=[mathbb{F}_{p^n}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}][mathbb{F}_{p^d}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}]d$, and so $d$ divides $n$.
If $d$ divides $n$, then $mathbb{F}_{p^d}^times$ is a subgroup of $mathbb{F}_{p^n}^times$ and so $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$. Here we use that $mathbb{F}_{p^n}^times$ is cyclic and that there is at most one finite field of a given size.
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
$endgroup$
add a comment |
$begingroup$
If $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$, then $n=[mathbb{F}_{p^n}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}][mathbb{F}_{p^d}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}]d$, and so $d$ divides $n$.
If $d$ divides $n$, then $mathbb{F}_{p^d}^times$ is a subgroup of $mathbb{F}_{p^n}^times$ and so $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$. Here we use that $mathbb{F}_{p^n}^times$ is cyclic and that there is at most one finite field of a given size.
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
$endgroup$
add a comment |
$begingroup$
If $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$, then $n=[mathbb{F}_{p^n}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}][mathbb{F}_{p^d}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}]d$, and so $d$ divides $n$.
If $d$ divides $n$, then $mathbb{F}_{p^d}^times$ is a subgroup of $mathbb{F}_{p^n}^times$ and so $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$. Here we use that $mathbb{F}_{p^n}^times$ is cyclic and that there is at most one finite field of a given size.
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
$endgroup$
If $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$, then $n=[mathbb{F}_{p^n}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}][mathbb{F}_{p^d}:mathbb{F_p}]=[mathbb{F}_{p^n}:mathbb{F}_{p^d}]d$, and so $d$ divides $n$.
If $d$ divides $n$, then $mathbb{F}_{p^d}^times$ is a subgroup of $mathbb{F}_{p^n}^times$ and so $mathbb{F}_{p^d}subseteqmathbb{F}_{p^n}$. Here we use that $mathbb{F}_{p^n}^times$ is cyclic and that there is at most one finite field of a given size.
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
answered Dec 5 '18 at 23:02
lhflhf
164k10170395
164k10170395
add a comment |
add a comment |
$begingroup$
An alternative way of doing the last part is to note that if $mathbb{F}_{p^d}suseteq mathbb{F}_{p^n}$, then $mathbb{F}_{p^n}$ is a vector space over $mathbb{F}_{p^d}$. In particular, it must be of cardinality $(p^d)^k$, where $k$ is the dimension. So $p^n = (p^d)^k = p^{dk}$, so $n=dk$.
$endgroup$
– Arturo Magidin
Dec 5 '18 at 23:07