Thread: 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

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

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

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.

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?

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.