Mixing
Order Of The Intersection of Two Subfields.
$begingroup$
Let $E$ and $F$ be subfields of $GF(p^n)$. If $|E| = p^r$ and $|F| = p^s$, what is the order of $Ecap F$?
I read a corollary that "A finite field of order $p^n$ contains a unique subfield of order p^m for each $m$ | $n$ and no other subfields.
If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $0$? So in this case, the order of $E cap F$ = $0$?
abstract-algebra field-theory finite-fields
edited Feb 2 at 3:44
Pradeep Bihani
599
asked Apr 6 '14 at 5:43
JamesJames
303
$endgroup$
add a comment |
$begingroup$
Let $E$ and $F$ be subfields of $GF(p^n)$. If $|E| = p^r$ and $|F| = p^s$, what is the order of $Ecap F$?
I read a corollary that "A finite field of order $p^n$ contains a unique subfield of order p^m for each $m$ | $n$ and no other subfields.
If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $0$? So in this case, the order of $E cap F$ = $0$?
abstract-algebra field-theory finite-fields
edited Feb 2 at 3:44
Pradeep Bihani
599
asked Apr 6 '14 at 5:43
JamesJames
303
$endgroup$
2
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
1
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25
add a comment |
$begingroup$
Let $E$ and $F$ be subfields of $GF(p^n)$. If $|E| = p^r$ and $|F| = p^s$, what is the order of $Ecap F$?
I read a corollary that "A finite field of order $p^n$ contains a unique subfield of order p^m for each $m$ | $n$ and no other subfields.
If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $0$? So in this case, the order of $E cap F$ = $0$?
abstract-algebra field-theory finite-fields
edited Feb 2 at 3:44
Pradeep Bihani
599
asked Apr 6 '14 at 5:43
JamesJames
303
$endgroup$
Let $E$ and $F$ be subfields of $GF(p^n)$. If $|E| = p^r$ and $|F| = p^s$, what is the order of $Ecap F$?
I read a corollary that "A finite field of order $p^n$ contains a unique subfield of order p^m for each $m$ | $n$ and no other subfields.
If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $0$? So in this case, the order of $E cap F$ = $0$?
abstract-algebra field-theory finite-fields
abstract-algebra field-theory finite-fields
edited Feb 2 at 3:44
Pradeep Bihani
599
asked Apr 6 '14 at 5:43
JamesJames
303
edited Feb 2 at 3:44
Pradeep Bihani
599
asked Apr 6 '14 at 5:43
JamesJames
303
edited Feb 2 at 3:44
Pradeep Bihani
599
edited Feb 2 at 3:44
Pradeep Bihani
599
edited Feb 2 at 3:44
Pradeep Bihani
599
599
asked Apr 6 '14 at 5:43
JamesJames
303
asked Apr 6 '14 at 5:43
JamesJames
303
asked Apr 6 '14 at 5:43
JamesJames
303
303
2
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
1
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25
add a comment |
2
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
1
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25
2
2
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
1
1
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Extended hint:
So if $tmid r$, and $tmid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $Ecap F$.
In view of this the intersection $Ecap F$ is a copy of $GF(p^ell)$, where $ell=gcd(r,s)$.
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
add a comment |
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$begingroup$
Extended hint:
So if $tmid r$, and $tmid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $Ecap F$.
In view of this the intersection $Ecap F$ is a copy of $GF(p^ell)$, where $ell=gcd(r,s)$.
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
add a comment |
$begingroup$
Extended hint:
So if $tmid r$, and $tmid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $Ecap F$.
In view of this the intersection $Ecap F$ is a copy of $GF(p^ell)$, where $ell=gcd(r,s)$.
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
add a comment |
$begingroup$
Extended hint:
So if $tmid r$, and $tmid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $Ecap F$.
In view of this the intersection $Ecap F$ is a copy of $GF(p^ell)$, where $ell=gcd(r,s)$.
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
Extended hint:
So if $tmid r$, and $tmid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $Ecap F$.
In view of this the intersection $Ecap F$ is a copy of $GF(p^ell)$, where $ell=gcd(r,s)$.
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
edited Apr 6 '14 at 8:43
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
answered Apr 6 '14 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172390
110k13172390
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
add a comment |
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E cap F|$ $geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E cap F| = p^t + p$ in this case?
$endgroup$
– James
Apr 6 '14 at 6:43
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
@James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield?
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 6:51
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$?
$endgroup$
– James
Apr 6 '14 at 7:08
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Correct, @James!
$endgroup$
– Jyrki Lahtonen
Apr 6 '14 at 7:41
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
$begingroup$
Thank you for the hints, helped me out. :)
$endgroup$
– James
Apr 6 '14 at 15:07
add a comment |
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2
$begingroup$
The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 in E, F$!
$endgroup$
– Robert Lewis
Apr 6 '14 at 6:04
$begingroup$
Very good point, thank you for the insight.
$endgroup$
– James
Apr 6 '14 at 6:33
1
$begingroup$
If you think about that corollary, you should be able to answer your question.
$endgroup$
– Gerry Myerson
Apr 6 '14 at 6:35
$begingroup$
Any source of this corollary in any text or notes will be helpful!.
$endgroup$
– BAYMAX
Dec 23 '17 at 3:25