# Transcendental over an algebraic closure of F in E

1. Oct 31, 2008

### meiji1

1. The problem statement, all variables and given/known data
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 $$F_E$$.

2. Relevant equations
$$F_E$$ denotes the algebraic closure of F in its extension field E.

3. 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.

Last edited: Oct 31, 2008
2. Oct 31, 2008

### 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.

3. Oct 31, 2008

### 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.

4. Oct 31, 2008

### 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?

5. Oct 31, 2008

### meiji1

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

6. Oct 31, 2008

### morphism

Ah I see. That makes more sense.

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

Last edited: Oct 31, 2008
7. Oct 31, 2008

### 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

8. Nov 1, 2008

### meiji1

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

9. May 16, 2011

### jeane

10. May 16, 2011

### 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.