specific examples of algebraic closure on finite field
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I want to confirm my understanding of algebraic closure for finite fields.
What sorts of elements do the algebraic closures $overline {mathbb{F}_2}$, $overline {mathbb{F}_3}$, $overline {mathbb{F}_7}$, and $overline {mathbb{F}_{25}}$ contain?
This answer seems most illuminating, specifically:
We can form a nested chain of extensions $$ E_1subset
E_2subsetcdots subset E_isubset E_{i+1}subsetcdots $$ of finite
fields $E_i$ for all positive integers $i$ such that $E_1=Bbb{F}_p$
Taking $p=3$, for example, such that $E_1 = mathbb{F}_3 = lbrace0, 1, 2rbrace$
Then $E_2 =Bbb{F}_{3^{2}} =Bbb{F}_9 =lbrace0, 1, 2, 3, 4, 5, 6, 7, 8rbrace$
Indeed, $E_1subset E_2$, but following this train of thought implies that the algebraic closure of $Bbb{F}_p$ is just the positive integers, but if that were the case, surely one of the references I've read would have said so by now!
The only other way I've been able to interpret this is that $overline {Bbb{F}_3} = {0, 1, 2, 3, 9, 27, 81, ...}$ or similar.
abstract-algebra finite-fields
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
$endgroup$
|
show 5 more comments
$begingroup$
I want to confirm my understanding of algebraic closure for finite fields.
What sorts of elements do the algebraic closures $overline {mathbb{F}_2}$, $overline {mathbb{F}_3}$, $overline {mathbb{F}_7}$, and $overline {mathbb{F}_{25}}$ contain?
This answer seems most illuminating, specifically:
We can form a nested chain of extensions $$ E_1subset
E_2subsetcdots subset E_isubset E_{i+1}subsetcdots $$ of finite
fields $E_i$ for all positive integers $i$ such that $E_1=Bbb{F}_p$
Taking $p=3$, for example, such that $E_1 = mathbb{F}_3 = lbrace0, 1, 2rbrace$
Then $E_2 =Bbb{F}_{3^{2}} =Bbb{F}_9 =lbrace0, 1, 2, 3, 4, 5, 6, 7, 8rbrace$
Indeed, $E_1subset E_2$, but following this train of thought implies that the algebraic closure of $Bbb{F}_p$ is just the positive integers, but if that were the case, surely one of the references I've read would have said so by now!
The only other way I've been able to interpret this is that $overline {Bbb{F}_3} = {0, 1, 2, 3, 9, 27, 81, ...}$ or similar.
abstract-algebra finite-fields
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
$endgroup$
1
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
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@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
1
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
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– Will R
Dec 13 '18 at 21:44
2
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@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46
|
show 5 more comments
$begingroup$
I want to confirm my understanding of algebraic closure for finite fields.
What sorts of elements do the algebraic closures $overline {mathbb{F}_2}$, $overline {mathbb{F}_3}$, $overline {mathbb{F}_7}$, and $overline {mathbb{F}_{25}}$ contain?
This answer seems most illuminating, specifically:
We can form a nested chain of extensions $$ E_1subset
E_2subsetcdots subset E_isubset E_{i+1}subsetcdots $$ of finite
fields $E_i$ for all positive integers $i$ such that $E_1=Bbb{F}_p$
Taking $p=3$, for example, such that $E_1 = mathbb{F}_3 = lbrace0, 1, 2rbrace$
Then $E_2 =Bbb{F}_{3^{2}} =Bbb{F}_9 =lbrace0, 1, 2, 3, 4, 5, 6, 7, 8rbrace$
Indeed, $E_1subset E_2$, but following this train of thought implies that the algebraic closure of $Bbb{F}_p$ is just the positive integers, but if that were the case, surely one of the references I've read would have said so by now!
The only other way I've been able to interpret this is that $overline {Bbb{F}_3} = {0, 1, 2, 3, 9, 27, 81, ...}$ or similar.
abstract-algebra finite-fields
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
$endgroup$
I want to confirm my understanding of algebraic closure for finite fields.
What sorts of elements do the algebraic closures $overline {mathbb{F}_2}$, $overline {mathbb{F}_3}$, $overline {mathbb{F}_7}$, and $overline {mathbb{F}_{25}}$ contain?
This answer seems most illuminating, specifically:
We can form a nested chain of extensions $$ E_1subset
E_2subsetcdots subset E_isubset E_{i+1}subsetcdots $$ of finite
fields $E_i$ for all positive integers $i$ such that $E_1=Bbb{F}_p$
Taking $p=3$, for example, such that $E_1 = mathbb{F}_3 = lbrace0, 1, 2rbrace$
Then $E_2 =Bbb{F}_{3^{2}} =Bbb{F}_9 =lbrace0, 1, 2, 3, 4, 5, 6, 7, 8rbrace$
Indeed, $E_1subset E_2$, but following this train of thought implies that the algebraic closure of $Bbb{F}_p$ is just the positive integers, but if that were the case, surely one of the references I've read would have said so by now!
The only other way I've been able to interpret this is that $overline {Bbb{F}_3} = {0, 1, 2, 3, 9, 27, 81, ...}$ or similar.
abstract-algebra finite-fields
abstract-algebra finite-fields
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
asked Dec 13 '18 at 17:50
Dusty PhillipsDusty Phillips
61
61
1
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
$begingroup$
@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
1
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
$endgroup$
– Will R
Dec 13 '18 at 21:44
2
$begingroup$
@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46
|
show 5 more comments
1
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
$begingroup$
@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
1
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
$endgroup$
– Will R
Dec 13 '18 at 21:44
2
$begingroup$
@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46
1
1
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
$begingroup$
@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
$begingroup$
@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
1
1
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
$endgroup$
– Will R
Dec 13 '18 at 21:44
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
$endgroup$
– Will R
Dec 13 '18 at 21:44
2
2
$begingroup$
@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46
$begingroup$
@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46
|
show 5 more comments
1 Answer
1
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$begingroup$
This is a possible way to have a picture of the algebraic closure of $Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,dots$ and arrange them in a tree with levels $0,1,2,3,dots$ as follows.
- In level $0$ we have the one and only $1$ .
- In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,dots$ .
- In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,dots$ .
- In level $3$ we have all natural numbers which are a product of three primes, so $8,12,dots$ .
And so on. These are the vertices of the tree.
Now draw an edge / a connection (as arrow) $ato b$ from one level to the next one if $a$ divides $b$.
It is hard to draw, but i will try.
:::::::::::::::::::::::::::::::::::::::
|/ |/
level 3: 8 12 ...
|/|
level 2: 4 6 9 10 ...
|/|/ /
level 1: 2 3 5 7 11 13 ...
|/ / / / /
level 0: 1
Now imagine at each place $a$ placed the field
$$
Bbb F_{p^a}
$$
with $p^a$ elements. Then for each edge $ato b$ in the tree we have a field inclusion of the corresponding fields $Bbb F_{p^a}to Bbb F_{p^b}$.
Consider now the union of all fields in all levels.
This is
$$
bar {Bbb F}_p .
$$
The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
$endgroup$
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
add a comment |
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$begingroup$
This is a possible way to have a picture of the algebraic closure of $Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,dots$ and arrange them in a tree with levels $0,1,2,3,dots$ as follows.
- In level $0$ we have the one and only $1$ .
- In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,dots$ .
- In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,dots$ .
- In level $3$ we have all natural numbers which are a product of three primes, so $8,12,dots$ .
And so on. These are the vertices of the tree.
Now draw an edge / a connection (as arrow) $ato b$ from one level to the next one if $a$ divides $b$.
It is hard to draw, but i will try.
:::::::::::::::::::::::::::::::::::::::
|/ |/
level 3: 8 12 ...
|/|
level 2: 4 6 9 10 ...
|/|/ /
level 1: 2 3 5 7 11 13 ...
|/ / / / /
level 0: 1
Now imagine at each place $a$ placed the field
$$
Bbb F_{p^a}
$$
with $p^a$ elements. Then for each edge $ato b$ in the tree we have a field inclusion of the corresponding fields $Bbb F_{p^a}to Bbb F_{p^b}$.
Consider now the union of all fields in all levels.
This is
$$
bar {Bbb F}_p .
$$
The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
$endgroup$
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
add a comment |
$begingroup$
This is a possible way to have a picture of the algebraic closure of $Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,dots$ and arrange them in a tree with levels $0,1,2,3,dots$ as follows.
- In level $0$ we have the one and only $1$ .
- In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,dots$ .
- In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,dots$ .
- In level $3$ we have all natural numbers which are a product of three primes, so $8,12,dots$ .
And so on. These are the vertices of the tree.
Now draw an edge / a connection (as arrow) $ato b$ from one level to the next one if $a$ divides $b$.
It is hard to draw, but i will try.
:::::::::::::::::::::::::::::::::::::::
|/ |/
level 3: 8 12 ...
|/|
level 2: 4 6 9 10 ...
|/|/ /
level 1: 2 3 5 7 11 13 ...
|/ / / / /
level 0: 1
Now imagine at each place $a$ placed the field
$$
Bbb F_{p^a}
$$
with $p^a$ elements. Then for each edge $ato b$ in the tree we have a field inclusion of the corresponding fields $Bbb F_{p^a}to Bbb F_{p^b}$.
Consider now the union of all fields in all levels.
This is
$$
bar {Bbb F}_p .
$$
The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
$endgroup$
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
add a comment |
$begingroup$
This is a possible way to have a picture of the algebraic closure of $Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,dots$ and arrange them in a tree with levels $0,1,2,3,dots$ as follows.
- In level $0$ we have the one and only $1$ .
- In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,dots$ .
- In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,dots$ .
- In level $3$ we have all natural numbers which are a product of three primes, so $8,12,dots$ .
And so on. These are the vertices of the tree.
Now draw an edge / a connection (as arrow) $ato b$ from one level to the next one if $a$ divides $b$.
It is hard to draw, but i will try.
:::::::::::::::::::::::::::::::::::::::
|/ |/
level 3: 8 12 ...
|/|
level 2: 4 6 9 10 ...
|/|/ /
level 1: 2 3 5 7 11 13 ...
|/ / / / /
level 0: 1
Now imagine at each place $a$ placed the field
$$
Bbb F_{p^a}
$$
with $p^a$ elements. Then for each edge $ato b$ in the tree we have a field inclusion of the corresponding fields $Bbb F_{p^a}to Bbb F_{p^b}$.
Consider now the union of all fields in all levels.
This is
$$
bar {Bbb F}_p .
$$
The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
$endgroup$
This is a possible way to have a picture of the algebraic closure of $Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,dots$ and arrange them in a tree with levels $0,1,2,3,dots$ as follows.
- In level $0$ we have the one and only $1$ .
- In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,dots$ .
- In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,dots$ .
- In level $3$ we have all natural numbers which are a product of three primes, so $8,12,dots$ .
And so on. These are the vertices of the tree.
Now draw an edge / a connection (as arrow) $ato b$ from one level to the next one if $a$ divides $b$.
It is hard to draw, but i will try.
:::::::::::::::::::::::::::::::::::::::
|/ |/
level 3: 8 12 ...
|/|
level 2: 4 6 9 10 ...
|/|/ /
level 1: 2 3 5 7 11 13 ...
|/ / / / /
level 0: 1
Now imagine at each place $a$ placed the field
$$
Bbb F_{p^a}
$$
with $p^a$ elements. Then for each edge $ato b$ in the tree we have a field inclusion of the corresponding fields $Bbb F_{p^a}to Bbb F_{p^b}$.
Consider now the union of all fields in all levels.
This is
$$
bar {Bbb F}_p .
$$
The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
answered Dec 13 '18 at 18:25
dan_fuleadan_fulea
6,5681312
6,5681312
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
add a comment |
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
I realize that you really want to describe a direct limit using as little machinery as possible. But you need to exercise a little bit of extra care. A potential problem is that the embeddings $Bbb{F}_{p^a}toBbb{F}_{p^b}$ (when $amid b$) are not unique. That itself is not a serious problem, but for the purposes of forming the direct limit (=the union) you need to worry about whether the composition of embeddings, say, $Bbb{F}_{p^2}toBbb{F}_{p^4}toBbb{F}_{p^{12}}$ is equal to the composition $Bbb{F}_{p^2}toBbb{F}_{p^6}toBbb{F}_{p^{12}}$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:02
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
$begingroup$
The reason I used the nested sequence $$Bbb{F}_ptoBbb{F}_{p^{2!}}to Bbb{F}_{p^{3!}}toBbb{F}_{p^{4!}}tocdots$$ in the linked thread was precisely to sidestep such issues.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:04
1
1
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
@Jyrki Lahtonen i was trying to address only the question about "elements". There was no claim about a commutative diagram. The OP had problems (was my understanding) to figure how elements in $Bbb F_3$, $Bbb F_9$ sit in the algebraic closure $bar{Bbb F}_3$. And also for $Bbb F_{25}$. So i tried to build some sentences that work in both cases. (No mention of commutative diagrams, directed sets...) But here as a comment. The tree level structure can be inductively used to get compatible inclusions $n/qto n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields.
$endgroup$
– dan_fulea
Dec 14 '18 at 17:10
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
$begingroup$
Good. Agree with your main point (and understand your reasoning for cutting a few details).
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 17:33
add a comment |
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1
$begingroup$
It is not true that $mathbb{F}_{9}={0,1,ldots,8}.$ $mathbb{F}_{9}$ is a field with $9$ elements, and the only way I know to describe it is as $${0,1,2,alpha,1+alpha,2+alpha,2alpha,1+2alpha,2+2alpha},$$ where $3=0$ and $alpha$ is an entirely new element that satisfies $alpha^{2}+alpha+1=0.$ I think you should review finite fields before studying algebraic closures of finite fields.
$endgroup$
– Will R
Dec 13 '18 at 17:56
$begingroup$
You can not identify any of the finite fields $mathbb{F}_{p^n}$ with a subset of the integers $mathbb{Z}$ in a fruithful way. I suppose that the origin of the confusion is that $mathbb{F_p}$ can be regarded as a quotient of $mathbb{Z}$ and by abuse of notation we call its elements $0,1,ldots, p-1$. But the situation with $mathbb{F}_{p^n}$, with $n>1$, is different because it is not a quotient of the integers and, in particular $mathbb{F}_{p^n}neq mathbb{Z}_{p^n}=mathbb{Z}/p^nmathbb{Z}$.
$endgroup$
– Dante Grevino
Dec 13 '18 at 18:08
$begingroup$
@DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks.
$endgroup$
– Dusty Phillips
Dec 13 '18 at 18:38
1
$begingroup$
@DustyPhillips: No. The elements of $mathbb{F}_{27}$ are polynomials in a new element $beta$ which satisfies $beta^{3}+2beta^{2}+1=0.$ The key attribute which these elements $alphainmathbb{F}_{9}$ and $betainmathbb{F}_{27}$ have in common is that they are roots of polynomials ($t^{2}+t+1$ and $t^{3}+2t^{2}+1,$ respectively) which are irreducible over $mathbb{F}_{3}.$ Since $mathbb{F}_{3}$ is a field, $mathbb{F}_{3}[t]$ is a PID and therefore, if $f(t)inmathbb{F}_{3}$ is irreducible, then $mathbb{F}_{3}/langle f(t)rangle$ is a field.
$endgroup$
– Will R
Dec 13 '18 at 21:44
2
$begingroup$
@WillR $alpha^2+alpha+1=0$ won't give you $Bbb{F}_9$ because $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is reducible. You could use $alpha^2+1=0$ though. Viewed differently, there are no primitive roots of order $p$ in characteristic $p$ fields.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 5:46