Is every element in F algebraic over Q?

[symbol0]

I have the following idea:
If a field F is algebraic over a field K, then F is a finite extension of K.

Is this true?
I was trying to come up with a counterexample. Something like F = Q($$\sqrt{p}$$ , for all primes p). Q are the rationals, of course.
Then F wouldn't be a finite extension of Q, but I don't know if every element in F is algebraic over Q.

I would appreciate any comments.

 

[symbol0]

Thanks Jarle,
I got your reply by email.

 

[disregardthat]

Thanks Jarle,
I got your reply by email.

Oh, I deleted it as I discovered a mistake in my answer. But you will probably figure it out.

 

[symbol0]

but "The general form of an element of F is a sum of square roots of rational numbers"
is true right?

 

[disregardthat]

but "The general form of an element of F is a sum of square roots of rational numbers"
is true right?

Well, technically yes, but you can't just assume that right away. What you know is that it is the quotient of such sums. Show that both the denominator and the numerator are algebraic in the manner I explained (by induction on the minimal number of terms). Then it remains to prove that the quotient actually is algebraic. This is true, but requires some algebraic machinery to prove. My book gives the proof through the existence of algebraic closures, but there might be an easier proof. Then again, I think this fact also relies on that algebraic extensions are fields, so you would probably be better off consulting a book on algebra.

Of course, you would still have to prove that your field isn't a finite algebraic extension of Q, which will be difficult still I think.

 

[symbol0]

Thanks Jarle
What's the name of the book you are talking about?

And intuitively saying the field is a finite extension seems like saying that there are finitely many primes.

 

[disregardthat]

Thanks Jarle
What's the name of the book you are talking about?

And intuitively saying the field is a finite extension seems like saying that there are finitely many primes.

Well, yes and no. There isn't any a priori reason for that there can be no linear dependence between, say, square roots of primes. In fact, there isn't, here is a proof: http://mathforum.org/library/drmath/view/51638.html
That link may also provide enough information to solve your problem as well, but he assumes a bit more algebra than I have done. He is assuming that your extension is algebraic over Q, which is elementary but not very easy to prove. The general theorem is that any algebraic extension of a field is algebraic over the same field. This is kind of a question of how much to assume beforehand. At least he proves the core of your problem (that the extension is not finite).

The book I have is this : https://www.amazon.com/dp/0201763907/?tag=pfamazon01-20

 

[symbol0]

The general theorem is that any algebraic extension of a field is algebraic over the same field.

What's the difference between being an algebraic extension of a field and being algebraic over a field?

 

[disregardthat]

What's the difference between being an algebraic extension of a field and being algebraic over a field?

There isn't any, of course, but it still needs to be proven! The thing is, say for finite algebraic extensions, that even though x is algebraic over Q, that doesn't automatically mean that Q(x) is algebraic over Q. That is, we still need to prove that P(x)/R(x) is algebraic over Q for all polynomials P and R.

 

[symbol0]

So I guess adjoining an algebraic element to a field is what you are calling an algebraic extension. In my book the only thing defined is algebraic extension over a field K: a field in which every element is algebraic over K.
My book has that adjoining a finite amount of algebraic elements of K to K, we get an algebraic field over K, but nothing is mentioned about adjoining an infinite amount of algebraic elements, which is the case in the field I constructed.
I assume that when you say

The general theorem is that any algebraic extension of a field is algebraic over the same field

, this applies regardless of adjoining a finite or infinite amount of algebraic elements.
If that's the case, is that proven in the Fraleigh book?

 

[mathwonk]

it is false. here is a proof sketch. finite fields exist. every field F has an algebraic closure which is algebraic over F. algebraic closures are always infinite. any finite dimensional extension of a finite field is finite hence not algebraically closed. Hence the algebraic closure of a finite field like Z/2Z is algebraic over Z/2Z but not finite dimensional over it.

 

[mathwonk]

the existence of algebraic closures (following E. Artin) is proved here:

http://www.math.uga.edu/%7Eroy/844-1.pdf

 

[disregardthat]

So I guess adjoining an algebraic element to a field is what you are calling an algebraic extension. In my book the only thing defined is algebraic extension over a field K: a field in which every element is algebraic over K.
My book has that adjoining a finite amount of algebraic elements of K to K, we get an algebraic field over K, but nothing is mentioned about adjoining an infinite amount of algebraic elements, which is the case in the field I constructed.
I assume that when you say , this applies regardless of adjoining a finite or infinite amount of algebraic elements.
If that's the case, is that proven in the Fraleigh book?

I am sorry, I had mixed up the definitions from the time I have been studying this. What I meant to say was that it needs to be proven that a finite field extension by algebraic elements of Q is algebraic over Q. Indeed an algebraic extension is algebraic over Q. The finite case is in any case not difficult. The infinite extension of algebraic elements which we are concerned with here is though not that trivial. This is proven in Fraleighs book, and uses algebraic closures.

As mathwonk points out the algebraic closure of Q is a counter-example for you. The link I gave you proves that the field you provided is not a finite algebraic extension, and if you are familiar with algebraic closures, you are already done.

 

[mathwonk]

this may be a proof that a finite field is not algebraically closed: let F be a finite field with n elements, and let m > n. Then X^m - 1 = 0 cannot have m distinct roots in F. But the derivative mX^(m-1) has no roots in common with X^m - 1, hence there are no multiple roots. hence some irreducible factor of X^m-1 has no root in F.

 

[mathwonk]

it is true for simple extensions, i.e. k(a) is algebraic over k if and only if it is finite dimensional over k. by induction it si also true for finitely generated extensions:

i.e.,k(a1,...,an) is algebraic over k if and only if it is finite dimensional over k.

maybe the confusion is from using the word "finite" in these two different senses: finitely generated and finite dimensional.

 

[symbol0]

No confusion mathwonk. I get the difference between the uses of the word "finite"

Thank you so much guys

 

[symbol0]

Jarle,
In one of your responses to my question, you sent me this link:http://mathforum.org/library/drmath/view/51638.html".

Can I say that in general:
An infinite set of prime roots of prime numbers are linearly independent over the rationals?

 

[disregardthat]

Yes, if you take take linear independence of an infinite set of to mean that any finite subset is linearly independent.