0

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 $$ E_1\subset E_2\subset\cdots \subset E_i\subset E_{i+1}\subset\cdots $$ of finite fields $E_i$ for all positive integers $i$ such that $E_1=\Bbb{F}_p$

Taking $p=3$, for example, such that $E_1 = \mathbb{F}_3 = \lbrace0, 1, 2\rbrace$

Then $E_2 =\Bbb{F}_{3^{2}} =\Bbb{F}_9 =\lbrace0, 1, 2, 3, 4, 5, 6, 7, 8\rbrace$

Indeed, $E_1\subset E_2$, 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.

Dusty Phillips
  • 11
  • 1
    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 $${0,1,2,\alpha,1+\alpha,2+\alpha,2\alpha,1+2\alpha,2+2\alpha},$$ 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. – Will R Dec 13 '18 at 17:56
  • 1
    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 $\mathbb{F}{p^n}\neq \mathbb{Z}{p^n}=\mathbb{Z}/p^n\mathbb{Z}$. – Dante Grevino Dec 13 '18 at 18:08
  • @DanteGrevino Ah, that's exactly my confusion, I thought the notations were interchangeable. Thanks. – Dusty Phillips Dec 13 '18 at 18:38
  • If I combine WillR's notation, with dan_fuala's description, I reason as follows, is this correct?:
    • ${F}_{3^{1}} = {0, 1, 2}$ with 3 elements
    • ${F}_{3^{2}} = {0, 1, 2, a, a+1, a+2, 2a, 2a+1, 2a+2}$ with 9 elements
    • ${F}_{3^{3}} = {0, 1, 2 .... 2a+1, 2a+2, b, b+1, b+2, b+a, b+a+1, ... b+2a+2, 2b, 2b+1, 2b+2, 2b+a, 2b+a+1,...,2b + 2a+2}$ with 81 elements

    ${F}_{3^{4}}$ with $81^2$ elements including a new element c?

    • and so on, infinitely?
     – Dusty Phillips Dec 13 '18 at 19:18
  • 1
    @DustyPhillips: No. The elements of $\mathbb{F}{27}$ are polynomials in a new element $\beta$ which satisfies $\beta^{3}+2\beta^{2}+1=0.$ The key attribute which these elements $\alpha\in\mathbb{F}{9}$ and $\beta\in\mathbb{F}{27}$ 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 $f(t)\in\mathbb{F}{3}$ is irreducible, then $\mathbb{F}{3}/\langle f(t)\rangle$ is a field. – Will R Dec 13 '18 at 21:44
  • But this discussion is quickly getting off topic, so we shouldn't continue it further here. From the sounds of it you aren't familiar with ring and field theory in the depth required to be considering algebraic closures just yet. I recommend that you learn about finite fields from a suitable source. Perhaps look at Chapter 6 of Paul Garrett's Abstract Algebra notes. – Will R Dec 13 '18 at 21:47
  • Oops! Small typo, I obviously meant $f(t)\in\mathbb{F}_{3}[t]$ above. Unfortunately it is too late for me to edit that now. – Will R Dec 13 '18 at 21:55
  • 3
    @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. – Jyrki Lahtonen Dec 14 '18 at 05:46
  • @JyrkiLahtonen I don't know why I thought $1$ was not a root, but of course it is. Thank you for correcting me. – Will R Dec 14 '18 at 11:26
  • Not much harm done, @WillR. Your main points are spot on. – Jyrki Lahtonen Dec 14 '18 at 12:55
  • @WillR $\mathbb{F}_9$ can also be viewed as ${0, 1, -1, i, i + 1, i - 1, -i, -i + 1, -i - 1}$ – Smiley1000 Dec 14 '22 at 21:51
  • @Smiley1000 But that set is not closed under addition (or multiplication). – Will R Feb 22 '23 at 20:21
  • @WillR of course you have to consider it to wrap around modulo 3 (meaning $1 + 1 = -1$, therefore $i + i = -i$ and so on) – Smiley1000 Feb 23 '23 at 21:44

1 Answers1

2

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) $a\to b$ 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 $a\to b$ 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.

dan_fulea
  • 32,856
  • 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 $\Bbb{F}{p^a}\to\Bbb{F}{p^b}$ (when $a\mid b$) 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, $\Bbb{F}{p^2}\to\Bbb{F}{p^4}\to\Bbb{F}{p^{12}}$ is equal to the composition $\Bbb{F}{p^2}\to\Bbb{F}{p^6}\to\Bbb{F}{p^{12}}$. – Jyrki Lahtonen Dec 14 '18 at 13:02
  • The reason I used the nested sequence $$\Bbb{F}p\to\Bbb{F}{p^{2!}}\to \Bbb{F}{p^{3!}}\to\Bbb{F}{p^{4!}}\to\cdots$$ in the linked thread was precisely to sidestep such issues. – Jyrki Lahtonen Dec 14 '18 at 13:04
  • 1
    @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 $n/q\to n$, $q|n$, $q$ prime, thus a functor from the tree diagram category to finite fields. – dan_fulea Dec 14 '18 at 17:10
  • Good. Agree with your main point (and understand your reasoning for cutting a few details). – Jyrki Lahtonen Dec 14 '18 at 17:33