Write the algebraic closure of $F_p$ as union of finite fields
$begingroup$
In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants to prove that $Gamma(q)=bigcup_{n=0}^{infty}GF(q^{a_n})$.
However, if $a_n$ is an arbitrary sequence, we are even unable to prove $bigcup_{n=0}^{infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under current conditions offered.
Hope for answers!
abstract-algebra field-theory finite-fields
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
$endgroup$
|
show 2 more comments
$begingroup$
In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants to prove that $Gamma(q)=bigcup_{n=0}^{infty}GF(q^{a_n})$.
However, if $a_n$ is an arbitrary sequence, we are even unable to prove $bigcup_{n=0}^{infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under current conditions offered.
Hope for answers!
abstract-algebra field-theory finite-fields
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
$endgroup$
5
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
2
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
5
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
5
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04
|
show 2 more comments
$begingroup$
In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants to prove that $Gamma(q)=bigcup_{n=0}^{infty}GF(q^{a_n})$.
However, if $a_n$ is an arbitrary sequence, we are even unable to prove $bigcup_{n=0}^{infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under current conditions offered.
Hope for answers!
abstract-algebra field-theory finite-fields
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
$endgroup$
In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants to prove that $Gamma(q)=bigcup_{n=0}^{infty}GF(q^{a_n})$.
However, if $a_n$ is an arbitrary sequence, we are even unable to prove $bigcup_{n=0}^{infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under current conditions offered.
Hope for answers!
abstract-algebra field-theory finite-fields
abstract-algebra field-theory finite-fields
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
edited Dec 8 '18 at 8:41
Brahadeesh
6,22742361
6,22742361
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
asked Dec 7 '18 at 20:26
Wembley InterWembley Inter
1215
1215
5
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
2
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
5
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
5
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04
|
show 2 more comments
5
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
2
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
5
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
5
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04
5
5
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
2
2
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
5
5
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
5
5
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04
|
show 2 more comments
1 Answer
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oldest
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$begingroup$
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=bigcup_nmathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $kge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.
$endgroup$
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$begingroup$
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=bigcup_nmathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $kge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.
$endgroup$
add a comment |
$begingroup$
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=bigcup_nmathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $kge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.
$endgroup$
add a comment |
$begingroup$
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=bigcup_nmathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $kge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.
$endgroup$
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=bigcup_nmathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $kge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.
edited Dec 9 '18 at 11:15
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5
$begingroup$
any sequence of positive integers such that any $k$ divides some $a_n$
$endgroup$
– reuns
Dec 7 '18 at 20:29
$begingroup$
@reuns That condition is both necessary and sufficient, right?
$endgroup$
– Wembley Inter
Dec 7 '18 at 20:47
2
$begingroup$
No, @usr0192, for instance $Bbb F_4notsubsetBbb F_8$.
$endgroup$
– Lubin
Dec 8 '18 at 5:11
5
$begingroup$
My earlier comment seems to have gotten lost: this author has been unacceptably sloppy in asking you to prove something that just isn’t so. The suggestion of @reuns will work, as will the specification of ${a_n}$ to be the sequence of factorials: $a_n=n!$. I think the best way of looking at this problem is to consider $Gamma(q)$ to be the direct limit of the finite fields $Bbb F_{q^n}$, with respect to the directed set of the positive integers, ordered by divisibility.
$endgroup$
– Lubin
Dec 8 '18 at 5:18
5
$begingroup$
The same question on MathOverflow: Write the algebra closure of $F_p$ as union of finite fields. This answer has a very reasonable advice about cross-posting - and you can also looks at other discussions on cross-posting.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 7:04