Cardinalities of infinite subfields

Conversation started Nov 7, 2015 at 18:35.

David Zhang

Nov 7, 2015 18:35

Hey guys, I have an algebra question I'd like to run by you.

Does every infinite field (of any characteristic) contain a countably infinite subfield?

Alizter

Shoot

user147690

@DavidZhang Yes

Balarka Sen

What about $\overline{\Bbb F_p}$?

Alizter

@DavidZhang yes

user147690

@DavidZhang $\Bbb F_p$ or $\Bbb Q$ are the prime subfields

Balarka Sen

Nov 7, 2015 18:37

oh, I misread countably by uncountably.

user147690

I was worried for a second, I didn't have latex on and I saw the bar and thought wtf

Balarka Sen

hehe

David Zhang

Hm, ok. So what's the argument? I get that every field contains either $\Bbb F_p$ (characteristic p) or $\Bbb Q$ (characteristic 0)

But if you contain $\Bbb F_p$ how do you get to a countably infinite subfield?

Balarka Sen

@AlexClark Wait a second. F_p is not countably infinite.

user147690

I just thought that

Alizter

Nov 7, 2015 18:38

Its countably finite

David Zhang

Right. I'm specifically interested in countably infinite subfields

Balarka Sen

Not so hasty. Hm.

I think \bar F_p works.

user147690

That's the algebraic closure?

Alizter

yes

Balarka Sen

nod.

user147690

Nov 7, 2015 18:39

That does work since any algebraically closed field is an infinite set

David Zhang

So every infinite field containing F_p also contains its algebraic closure?

Alizter

However is it a subfield of every infinite field?

Balarka Sen

But how do you know it does not have any countably infinite subfield?

I don't have an argument.

I just think it doesn't.

user147690

Well we only care about characteristic $p$, and I think I might be too tired :D

Balarka Sen

Someone needs to summon anon.

David Zhang

Nov 7, 2015 18:42

Hm, so perhaps the question is not as easy as I first imagined

Alizter

That would mean $\bar{\Bbb F_p}$ can be extended to any infinite field

Balarka Sen

I think this is a good question, @DavidZhang.

user147690

Maybe @MartinSleziak can answer?

Martin Sleziak

And maybe not.

David Zhang

I think I'll post it to Math.SE, then.

user147690

Nov 7, 2015 18:43

7 mins ago, by David Zhang

Does every infinite field (of any characteristic) contain a countably infinite subfield?

Balarka Sen

@DavidZhang You may want to try asking it to @anon first.

I am certain he can answer.

user147690

We know if it has characteristic $0$ it has prime subfield $\Bbb Q$, so we only need to show it for characteristic $p$ @MartinSleziak, do you know?

Martin Sleziak

Cannot we simply add some elements by transfinite induction?

Balarka Sen

well, ok.

user147690

I don't know transfinite induction unfortunately, maybe David does though

Balarka Sen

Nov 7, 2015 18:45

any subfield of \bar F_p is an alg extension of F_p, right?

and any alg extension of F_p is of the form F_p^n

which is finite

ehh, no.

just pick an infinite subgroup of \hat Z

then you can FTGT to get an infinite extension of F_p. obviously.

Martin Sleziak

Ok, can we create some countable chain of subfields $G_0\subsetneq G_1\subsetneq G_2 \subsetneq \dots$?

Each of them by adding one algebraic element, so they will be finite, but the cardinalities will be increasing.

And then take their union.

Balarka Sen

that'll give me the algebraic closure itself

well, if G_n = F_p^n

because you're taking direct limit, right? direct limit of F_p^n's is precisely \bar F_p

Martin Sleziak

So then it is union of countably many finite sets... ?

I am probably missing something.

Balarka Sen

oh. heh.

Martin Sleziak

Or maybe not. Wikipedia claims the same thing: The algebraic closure of a field K has the same cardinality as K if K is infinite, and is countably infinite if K is finite.

Balarka Sen

Nov 7, 2015 18:50

\bar F_p is countble.

right, @Martin. I was being silly.

so no way positive char works.

because every field of positive char is a subfield of \bar F_p, hence countable.

that settles @DavidZhang's question, I guess.

David Zhang

Wait, so every infinite field of characteristic $p>0$ is already countable?

I don't know much algebra, but somehow that doesn't sound right

Martin Sleziak

Why is every field of characteristic p subset of $\overline{\mathbb F_p}$?

David Zhang

Right, surely there are transcendental extensions of F_p which are not countable?

Balarka Sen

is F_p(X) a subfield of \bar F_p?

of course not. ok.

that's troublesome.

Martin Sleziak

But the answer is still the same.

We have some field of characteristic p.

Then it contains $F_p$.

If it is algebraic, then it contains $\overline{F_p}$, which is countable.

Balarka Sen

Nov 7, 2015 18:55

but what if it's transcendental?

Martin Sleziak

If not, take any transcendetal element $x$ and take $F_p(x)$; i.e., the smallest field containing $F_p\cup\{x\}$.

Balarka Sen

@MartinSleziak Then it's *contained in \bar F_p.

Martin Sleziak

@BalarkaSen I don't think that's true, but you probably know more algebra then me, ....

But let's talk about the transcendental case first, ok?

So all I need to do is to show that $F_p(x)$ is countable, and I am done, right?

Balarka Sen

no alg extension of k contains \bar k other than itself.

(by defn)

@MartinSleziak uh, no. say K/F_p is a transcendental extension. then F_p(x) is contained in K. if F_p(x) is countable, that does not say much about K

David Zhang

Well, in any case, I've posted the question to Math.SE here: math.stackexchange.com/questions/1517718/…

Martin Sleziak

Nov 7, 2015 18:59

But the question was whether there is a countable subfield.

And F_p(x) is a countable subfield.

David Zhang

I'm afraid I've got to leave at the moment, but if you work out the answer here, please do post it there.

user147690

@DavidZhang Well you get some booster upvotes, because I am keen :D

Balarka Sen

@MartinSleziak ok, fair enough :P

I had forgotten what the question is.

lol. that was not so hard.

user147690

A: Does every infinite field contain a countably infinite subfield?

Yes. If $S$ is a countably infinite subset of $K$ then the subfield generated by $S$ is countable.

Balarka Sen

oh well. not my best day.

Martin Sleziak

Nov 7, 2015 19:01

BTW my guess would be: If $F\subseteq G$ are an infinite fields and a infinite subfield, then for each cardinal $\varkappa$ such that $|F|\le\varkappa\le|G|$ there is a subfield with cardinality $\varkappa$.

Conversation ended Nov 7, 2015 at 19:01.

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