Are Algebraic Extensions Always Finite?

[ehrenfest]

[SOLVED] extensions fields

Homework Statement

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.

Homework Equations

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]

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)

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.

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

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.

 

[Hurkyl]

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

Yes, that is equivalent to what I said.

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.

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?

 

[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.

 

[Hurkyl]

Also please help me prove that a finite sum of irrational numbers cannot be equal to a rational number.

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.

 

[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.

 

[morphism]

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

 

[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.

 

[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...

 

[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.

 

[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]

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

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}.<br /> <br /> F is <b>algebraically closed in E</b> if it is its own algebraic closure in E.

 

[ehrenfest]

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?

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

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)?]

Very nice example. Please confirm this proof:

Since, \mathbb{C}[/tex] In element of \mathbb{C}(x) must be expressible as<br /> \frac{\prod_{i=0}^n (x-c_i)^{q_i}}{\prod_{j=0}^m (x-c_j)^{q_j}}<br /> where the c_j and the c_i are different.<br /> <br /> When you square that, the quotient will still be in &quot;lowest terms&quot;. 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.<br /> <br /> 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]

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

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?

Very nice example. Please confirm this proof:

Since, \mathbb{C}[/tex] In element of \mathbb{C}(x) must be expressible as<br /> \frac{\prod_{i=0}^n (x-c_i)^{q_i}}{\prod_{j=0}^m (x-c_j)^{q_j}}<br /> where the c_j and the c_i are different.<br /> <br /> When you square that, the quotient will still be in &quot;lowest terms&quot;. 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.

<br /> I think you have the right idea, but the way you&#039;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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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? </div> </div> </blockquote>Yup. In fact a field is algebraically closed iff it has no proper algebraic extension, or in your terminology, iff it&#039;s its own algebraic closure in any extension field.

 

[ehrenfest]

Now, what can you say about the algebraic closure of Z/pZ?

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.

 

[morphism]

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.

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.

 

[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...

 

[ehrenfest]

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.

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".

 

[morphism]

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".

If a field F contains Z/pZ, then p=0 in F. So its charF must be p (as charF is either 0 or a prime p).

If you're still having trouble with this, then I suggest you review the definition of "characteristic", and try to see if it makes sense then.

 

[ehrenfest]

If for a ring R a positive number n exists such that n \cdot a = 0 for all a in R, the the least such positive integer is the characteristic of the ring R . If no such positive integer exists, then R is of characteristic 0 .

Obviously the characteristic of any field that contains Z/pZ must be at least p, and it must be a prime or 0, but why can it not be a prime greater than p or 0?

Oh, I see why it cannot be a prime greater than p--because then you could take the remainder when it is divided by p and this could not be 0 because it is prime and then this would be a smaller positive integer that satisfies the definition.

But why can it not be zero?

 

[morphism]

It's only zero if there is no positive integer n such that n*a=0. But in our case, there IS such a positive integer, namely p.

 

[ehrenfest]

It's only zero if there is no positive integer n such that n*a=0. But in our case, there IS such a positive integer, namely p.

To show that the characteristic is p, you need to take an ARBITRARY element a out of the algebraic closure of Z/pZ and show that p \cdot a =0[/itex]. If you happen to take an element out of the subfield Z/pZ, then that will be true, but how do you know it will be true in general?

 

[morphism]

Take a nonzero element a in the algebraic closure. If pa != 0, then p*1 = pa*a^-1 != 0.

But really, the easiest way to see this is to think about characteristic in terms of prime subfields. Since the prime subfield of the algebraic closure of Z/pZ is Z/pZ, we're done.

 

[ehrenfest]

Take a nonzero element a in the algebraic closure. If pa != 0, then p*1 = pa*a^-1 != 0.

I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as p \cdot a = (a +...+a) = \sum_{i=1}^p a. I would use the asterick only for multiplication in the field.

So, you are saying that if p \cdot a \neq 0, then p \cdot 1 = (p \cdot (a*a^{-1}) ) = (p \cdot a)* a^{-1} where in the last step I used the distributive property. That is obviously impossible because 1 is in Z/pZ and we thus know its characteristic is p. So I guess what you did checks outs, but I still think you should always use the dot notation when dealing with characteristics of fields.

But really, the easiest way to see this is to think about characteristic in terms of prime subfields. Since the prime subfield of the algebraic closure of Z/pZ is Z/pZ, we're done.

There is a theorem in my book that says: "A field is either of prime characteristic p and contains a subfield isomorphic to Z_p or of characteristic 0 and contains a subfield isomorphic to Q."

Simply containing a subfield isomorphic of Z_p is not enough to conclude that the field has prime characteristic p (at least from this theorem).

 

[Hurkyl]

I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as p \cdot a = (a +...+a) = \sum_{i=1}^p a. I would use the asterick only for multiplication in the field.

Aren't fields distributive?

 

[ehrenfest]

Aren't fields distributive?

Yes. So what? You could have a field that does not even contain an element called p and then p*a does not even make sense unless it means p \cdot a.

 

[morphism]

I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as p \cdot a = (a +...+a) = \sum_{i=1}^p a.

It is. p is simply shorthand for 1+1+...+1 (p times).

So I guess what you did checks outs, but I still think you should always use the dot notation when dealing with characteristics of fields.

Exercise: Prove that the dot notation is as compatible as we want it to be with field multiplication. Conclude that it's alright if we don't bother to use it anymore.

 

[Hurkyl]

Yes. So what? You could have a field that does not even contain an element called p

How? Can you come up with an example?

 

[ehrenfest]

How? Can you come up with an example?

The dot notation is definitely not always compatible with field notation, as I will show. Consider the field Z/3Z. Writing 5*2 is absurd because 5 is not an element of the field Z/3Z. You need to write 5 \cdot 2 if you want to represent the sum 2+2+2+2+2.

morphism, if you are saying that if F is a field that contains f, then n \cdot f = nf WHEN the characteristic of F is greater than n or equal to 0, then I think that is true.

 

[Hurkyl]

Writing 5*2 is absurd because 5 is not an element of the field Z/3Z.[/itex]

Er, yes it is. 5 denotes the same element that 2 does.

 

[ehrenfest]

OK. This is getting really technical. I would rather move on.