# Extensions fields

1. Feb 16, 2008

### ehrenfest

[SOLVED] extensions fields

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)$$\mathbb{C}$$ is algebraically closed in $$\mathbb{C}(x)$$, where x is an indeterminate
3)$$\mathbb{C}(x)$$ 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?

2. Feb 16, 2008

### Hurkyl

Staff Emeritus
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?

Last edited: Feb 16, 2008
3. Feb 16, 2008

### 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

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, $$\alpha_1,\alpha_2,...\alpha_n$$ such that $$E = F(\alpha_1,...,\alpha_n)$$ 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 $$\sqrt{p_i}$$ 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.

Last edited: Feb 16, 2008
4. Feb 17, 2008

### Hurkyl

Staff Emeritus
Yes, that is equivalent to what I said.

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

5. Feb 17, 2008

### ehrenfest

Almost. Say $$\sqrt{p} = \sum_{i=0}^n q_i \sqrt{n_i}$$ 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.

6. Feb 17, 2008

### Hurkyl

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

As for proving the $\sqrt{p_i}$ 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 $\bar{\mathbb{Q}} / \mathbb{Q}$ is not a finite extension.

7. Feb 18, 2008

### 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 $\bar{\mathbb{Q}} / \mathbb{Q}$ 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.

8. Feb 18, 2008

### morphism

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

9. Feb 18, 2008

### ehrenfest

It sure seems like it would be easy to show that $$\sqrt[n]{2}$$ is not contained $$\mathbb{Q}(\sqrt[n-1]{2},...,\sqrt[2]{2})$$, but again I am stuck. I am not even sure how to begin. I would know how to find a basis for $$\mathbb{Q}(\sqrt[m]{2})$$ 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.

10. Feb 18, 2008

### morphism

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

11. Feb 18, 2008

### ehrenfest

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

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$$(\sqrt[n]2,\mathbb{Q})$$ is n
EDIT: then take n to infinity. Very nice!
EDIT: So 1 is F.

Last edited: Feb 18, 2008
12. Feb 18, 2008

### ehrenfest

Do you have any insight on 2-4?

13. Feb 18, 2008

### 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)?]

Last edited: Feb 18, 2008
14. Feb 18, 2008

Sorry. Let E be an extension field of F. Then the algebraic closure of F in E is {$\alpha \in E | \alpha[/tex] is algebraic over F}. F is algebraically closed in E if it is its own algebraic closure in E. 15. Feb 29, 2008 ### ehrenfest Is it something like the subfield of C generated by {0,1,i,-i}? And what would the characteristic of that field be? Very nice example. Please confirm this proof: Since, [itex]\mathbb{C}[/tex] In element of [itex]\mathbb{C}(x)$ must be expressible as
$$\frac{\prod_{i=0}^n (x-c_i)^{q_i}}{\prod_{j=0}^m (x-c_j)^{q_j}}$$
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?

Last edited: Mar 1, 2008
16. Mar 1, 2008

### 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.

17. Mar 1, 2008

### 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.

Last edited: Mar 1, 2008
18. Mar 1, 2008

### morphism

It can't be a subfield of Q! The algebraic closure of Z/pZ will contain Z/pZ as a subfield, so its characteristic will be p.

19. Mar 1, 2008

### ehrenfest

Sorry--you're right! I'll post back if I cannot figure it out after reading the next section although there should be a way to do this without using the material in the next section...

20. Mar 1, 2008

### ehrenfest

I don't see how the statement "its characteristic will be p" follows from the fact that "The algebraic closure of Z/pZ will contain Z/pZ as a subfield".

Last edited: Mar 1, 2008