Minimal Polynomial, Algebraic Extension

  • Thread starter kathrynag
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  • #1
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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 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


3. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + · · · + a_1x + a_0. Show that
u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.

Well I know a number u is algebraic if p(u)=0 for a plynomial p(x)


4. Let F be an extension field of K with [F : K] = m < infinity, and let p(x) in K[x] be a
polynomial of degree n that is irreducible over K. Show that if n does not divide m,
then p(x) has no roots in F.

n does not divide m, so we can't have m=nq
 

Answers and Replies

  • #2
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For 2, since F(u)=0, and F is contained in E, could i say F(u) is contained in E?
 
  • #3
598
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2. 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?
3.Do I look at 1=u*m(x)?
4. We have degree=m<infinity
I want ot do something like this:
[M:K]=[M:L]*[L:K]. but we only have [F : K].
Like we could have [F:K]=[F:L]*[L:K] if K was a subfield of L, but I'm not seeing that
 

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