Transcendental over an algebraic closure of F in E

[meiji1]

Homework Statement

Let E be an extension field of a field F. Given \alpha\in E, show that, if \alpha\notin F_E, then \alpha is transcendental over F_E.

Homework Equations

F_E denotes the algebraic closure of F in its extension field E.

The Attempt at a Solution

First, I assumed \alpha \in E was algebraic over F_E and that \alpha \notin F_E and attempted to derive a contradiction. If \alpha is algebraic over E, then F_E(\alpha) is a finite (and therefore algebraic) extension of F_E. Hence, \left\{1, \alpha^1,\alpha^2, ... , \alpha^{n-1} \right\} is a basis over F_E(\alpha) given deg(\alpha, F_E) = n, for n a natural number.

I then tried to prove that, since \left\{ 1, \alpha^1,\alpha^2, ... , \alpha^{n-1}, \alpha^n \right\} is linearly dependent over F_E (since it contains (n+1) elements), it must also be linearly dependent over F.

Upon proving this, my strategy was to conclude that, if \left\{ 1, \alpha^1, \alpha^2, ..., \alpha^{n-1}, \alpha^n \right\} is linearly dependent over F, then \alpha is a zero of some polynomial f(x) \in F[x] and hence is algebraic over F[x]. Thus, \alpha \in F_E, since \alpha \in E.

I made another attempt at a separate solution, if you'd like to see it.

Any prodding in the right direction would be much appreciated.

 

[morphism]

What does it mean for an extension to be algebraically closed? Depending on what you have to work with, this problem is fairly straightforward. In any case, what you want to be looking at is the minimal polynomial of \alpha over F.

 

[meiji1]

Well, an extension E is algebraically closed if all zeros of E[x] are contained in E. By minimal polynomial, do you mean the irreducible polynomial of \alpha over F? It's not supposed at the outset that \alpha is algebraic over F.

 

[morphism]

Sorry - I completely misread the statement of the problem.

Now that I've reread it, I'm confused. How can an element in E be transcendental in E? (It can't; every a in E will satisfy the polynomial x-a in E[x].) I'm guessing you meant "transcendental over F" instead. In which case, isn't the algebraic closure of F defined to be the set of elements in E which are algebraic over F?

 

[meiji1]

I made a mistake in the problem statement. I apologize - I meant transcendental over F_E.

 

[morphism]

Ah I see. That makes more sense.

There's a problem with what you have so far:

I then tried to prove that, since \left\{ 1, \alpha^1,\alpha^2, ... , \alpha^{n-1}, \alpha^n \right\} is linearly dependent over F_E (since it contains (n+1) elements), it must also be linearly dependent over F.

Were you successful? If you're planning on deducing this because F is a subfield of F_E, then you'd be wrong. For instance, {1, sqrt2} is linearly dependent over the reals but not over the rationals.

Edit: I think I'm going to head to bed. So I'll leave you with the following hint: consider the irreducible polynomial (a.k.a. the minimal polynomial) of alpha over E_F. What happens if you adjoin its coefficients to F?

 

[meiji1]

No, I was not successful. I wish I'd thought of the counterexample you gave.

Thanks for the hint. I'll give it some thought and report back. :D

 

[meiji1]

I've got it now. It's very clear to me. Thanks very much. :)

 

[jeane]

Were you successful? If you're planning on deducing this because F is a subfield of F_E, then you'd be wrong. For instance, {1, sqrt2} is linearly dependent over the reals but not over the rationals.
QUOTE]

Although this is true. One should note that if F is the rational numbers then F_E cannot be the reals. The reals are not algebraically closed over the rationals. The reals would be E.

 

[jeane]

If alpha is algebraic over F_E, then F_E(alpha) is algebraic over F_E and by definition F_F is algebraic over F. There is a theorem that says F<=E<=K then K is algebraic over F if and only if E is algebraic over F, and K is algebraic over E. By this F_E(alpha) is algebraic over F. Then alpha is algebraic over F, but then alpha is in F_E which contradicts the hyposthesis.