[SOLVED] extensions fields

ehrenfest

1. The problem statement, all variables and given/known data
Can someone help me with these true or false problems:
1) Every algebraic extension is a finite extension.
2)[tex]\mathbb{C}[/tex] is algebraically closed in [tex]\mathbb{C}(x)[/tex], where x is an indeterminate
3)[tex]\mathbb{C}(x)[/tex] is algebraically closed, where x is an indeterminate
4)An algebraically closed field must be of characteristic 0

Recall that an extension field E of a field F is a finite extension if the vector space E over F has finite dimension.


2. Relevant equations



3. The attempt at a solution
1) the converse is true (it was a theorem in my book). this direction is probably not true (or else it would have also been a theorem in my book). But I need a counterexample
2) I am confused about the notation. My book has always denoted the ring of polynomials of a field f as F[x], never F(x). Furthermore, when a is an element of an extension field of F, then F(a) means F adjoined to a. But I have absolutely no idea what this means when x is an indeterminate?
3) same as 2
4) A field of characteristic 0 must contain a copy of the rationals. Does that help?

Hurkyl

For any field F, the notation F(x) denotes the field of all rational functions in x with coefficients in F. (Just like F(a), for a in an extension field, is the field of rational functions in a with coefficients in F)


What sort of major theorems do you know about algebraic extensions and algebraic closures?

ehrenfest

So, you are saying F(x) is the field of quotients of F[x]?

Also, F(a) means the smallest field that contains a and F. It is only the isomorphic to the field of quotients of F[x] if a is transcendental over a. If a is algebraic over F, the F(a) is is just F[x] evaluated at a. So, I guess by indeterminate, they really mean an element that is transcendental over F. It would be nice if they had said that explicitly or defined it somewhere.
EDIT: actually my book did define that; I just skipped that part
Please confirm that.

1) Every field has an algebraic closure.
2) A field F is algebraically closed iff every nonconstant polynomial in F[x] factors in F[x] into linear factors.
3)Let E be an algebraic extension of a field F. Then there exist a finite number of elements, [tex]\alpha_1,\alpha_2,...\alpha_n[/tex] such that [tex]E = F(\alpha_1,...,\alpha_n)[/tex] iff E is a finite finite extension of F. So, I guess that the algebraic closure of Q would be a counterexample for 1 because you need to add [tex]\sqrt{p_i}[/tex] where p_i is the sequence of prime numbers and it is not hard to prove that there an infinite number of prime numbers. Please confirm that.

Hurkyl

Yes, that is equivalent to what I said.


That sounds like a possibility. Can you show that each [itex]\sqrt{p}[/itex] is not contained in Q adjoined with the square roots of other primes?

ehrenfest

Almost. Say [tex]\sqrt{p} = \sum_{i=0}^n q_i \sqrt{n_i}[/tex] where the q_i are rational and n_i can be 1 or have factors of primes that are not p. Squaring both sides gives you a rational equal to a finite sum of numbers, some of which must be irrational. Why must some of them be irrational? I am not really sure, but I think it has to do with the fact that you must have cross-terms when you square the RHS, but please help with that. Also please help me prove that a finite sum of irrational numbers cannot be equal to a rational number.

Hurkyl

I can't; it's false. It's easy to find a counterexample if you work backwards.



As for proving the [itex]\sqrt{p_i}[/itex] are algebraically independent -- at the moment I only have ideas that use more advanced mathematics. (e.g. p-adic analysis) While I suspect it might be instructive to work on this some more... for now I suggest seeking an easier way to show that [itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] is not a finite extension.

ehrenfest

Hmmm. I wouldn't think these problems would require advanced mathematics or else it would be rather cruel to make them only TF. I can't really think of any way of proving [itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] is not a finite extension without proving that an infinite sequence of rationals are algebraically independent. It seemed like the primes would be the easiest.

morphism

[itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] contains [itex]\mathbb{Q}(\sqrt[n]2)[/itex] as an intermediate field for each n>1.

ehrenfest

It sure seems like it would be easy to show that [tex]\sqrt[n]{2}[/tex] is not contained [tex]\mathbb{Q}(\sqrt[n-1]{2},...,\sqrt[2]{2})[/tex], but again I am stuck. I am not even sure how to begin. I would know how to find a basis for [tex]\mathbb{Q}(\sqrt[m]{2})[/tex] and there is that theorem that says that if K is a finite extension of E and E is a finite extension of F, then you just multiply the basis elements of K over E by those for E over F...but that doesn't seem like it will help here.

morphism

That doesn't really matter. Just the fact that [itex]\mathbb{Q}(\sqrt[n]2)[/itex] sits between [itex]\bar{\mathbb{Q}}[/itex] and [itex]\mathbb{Q}[/itex] is enough to let you conclude that [itex][\bar{\mathbb{Q}}:\mathbb{Q}] \geq [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex], which in turn tells you...

ehrenfest

Why does [itex][\bar{\mathbb{Q}}:\mathbb{Q}] \geq [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex] let you conclude anything? It is obviously true since [tex]\sqrt[n]2 \in \bar{\mathbb{Q}}[/tex]. Do you mean to say [itex][\bar{\mathbb{Q}}:\mathbb{Q}] > [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex]?

EDIT: never mind, I see what you are arguing; you're argument is the following:
EDIT: x^n-2 is irreducible by Eisenstein's criterion which implies that deg[tex](\sqrt[n]2,\mathbb{Q})[/tex] is n
EDIT: then take n to infinity. Very nice!
EDIT: So 1 is F.

ehrenfest

Do you have any insight on 2-4?

morphism

What exactly does it mean for a field to be algebraically closed in another one?

For 4, what is the algebraic closure of, say, the field with 2 elements?

Edit: And for 3, does 1/x have a square root in C(x)? [I.e. does the polynomial y^2 - 1/x in (C(x))[y] have a root in C(x)?]

ehrenfest

Sorry. Let E be an extension field of F. Then the algebraic closure of F in E is {[itex]\alpha \in E | \alpha[/tex] is algebraic over F}.

F is algebraically closed in E if it is its own algebraic closure in E.

ehrenfest

Is it something like the subfield of C generated by {0,1,i,-i}? And what would the characteristic of that field be?

Very nice example. Please confirm this proof:

Since, [itex]\mathbb{C}[/tex] In element of [itex]\mathbb{C}(x)[/itex] must be expressible as
[tex] \frac{\prod_{i=0}^n (x-c_i)^{q_i}}{\prod_{j=0}^m (x-c_j)^{q_j}} [/tex]
where the c_j and the c_i are different.

When you square that, the quotient will still be in "lowest terms". Thus we must have n=0. And the square of the denominator must be equal to x, but that is impossible because squaring multiplies the degree of the polynomial by two and 1 is not equal to two times anything.

As for 2, I think it is obviously true because C is its own algebraic closure period which means that all the elements that are algebraic over C are contained in C, right?

morphism

It's not going to be a subfield of C. Try to think about it this way: What is the characteristic of a field F, really? If it's 0, then F contains a copy of the rationals. Otherwise, if it's a prime p, then F contains a copy of Z/pZ. (Q and Z/pZ are, in the respective cases, the prime subfields of F.) Now, what can you say about the algebraic closure of Z/pZ?

I think you have the right idea, but the way you're expressing it is kind of sketchy. Try this: Suppose p(x) and q(x) are polynomials in C[x] such that (p(x)/q(x))^2 = 1/x. Then x(p(x)^2) = q(x)^2, which is absurd by the degree argument you mentioned.

Yup. In fact a field is algebraically closed iff it has no proper algebraic extension, or in your terminology, iff it's its own algebraic closure in any extension field.

ehrenfest

I have no idea what the algebraic closure of Z/pZ looks like! It is obviously a subfield of Q but which one? My book really should have done that example.

EDIT: Actually the next section is called "Finite Fields" maybe I will learn that in there.