1. Algebraic Extensions

1.Let F=K(u) where u is transcedental over the field K. If E is a field such that K contained in E contained in F, then Show that u is algebraic over E.

Let a
be any element of E that is not in K. Then a = f(u)/g(u)

for some polynomials
f(x), g(x) inK[x]

2.Let F be a finite extension of K such that [F:K]=p, a prime number. If u in F but u not in K, show that F=K(u)
I'm not quite sure how to get started on this one
3.Let K contained in E contained in F be fields. Prove that if F is algebraic over K, then F is algebraic over E and E is algebraic over K

F is algebraic so F(u)=0
We want to show E(u)=0

2. Originally Posted by mathematic
2.Let F be a finite extension of K such that [F:K]=p, a prime number. If u in F but u not in K, show that F=K(u)
I'm not quite sure how to get started on this one

$\displaystyle u\in F, K \subset F \Rightarrow K(u) \subseteq F$

$\displaystyle K \subset K(u)\Rightarrow \left[K(u):K \right]\neq 1$

$\displaystyle \left[F:K(u) \right]\left[K(u):K \right]=\left[F:K \right] = p = prime \Rightarrow \left[F:K(u) \right]=1 \Rightarrow F=K(u)$

3. 1.Let F=K(u) where u is transcedental over the field K. If E is a field such that K contained in E contained in F, then Show that u is algebraic over E.

Let a
be any element of E that is not in K. Then a = f(u)/g(u)

for some polynomials f(x), g(x) inK[x]

I get stuck after this point

4. The hint I was given was:
Let a be any element of E that is not in K. Then a = f(u)/g(u) for some polynomials f(x), g(x) in K[x]. Then I could use this to find a nonzero polynomial in E[x] with u as a root.

5. 3. elements of F have minimal polynomials of K
If E/K is a subfield of F/K, then we have a minimal polynomial of E and thus a minimal polynomial of E of K. So algebraic?

6. Originally Posted by mathematic
3.Let K contained in E contained in F be fields. Prove that if F is algebraic over K, then F is algebraic over E and E is algebraic over K
F algebraic over K means that every element of F is root of some polynomial with coefficients in K.

• Since K is a subfield of E we can see the coefficients as elements in E.
So, F is algebraic over E.

• Since E is a subfield of F we have that E is algebraic over K.