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Show that u is algebraic

  • Thread starter Driessen12
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  • #1
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Homework Statement


if F=k(u), where u is transcendental over the field k. If E is a field such that E is an extension of K and F is an extension of E, then show that u is algebraic over E


Homework Equations





The Attempt at a Solution


i''m having trouble starting this proof, any ideas? any help would be appreciated
 
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Answers and Replies

  • #2
139
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Well, if [tex] E [/tex] is a field strictly bigger than [tex] k [/tex], then it must contain some [tex] \alpha [/tex] not in [tex] k [/tex]. By assumption, [tex] \alpha \in F [/tex]. As a vector space, [tex] F [/tex] is infinite-dimensional over [tex] k [/tex], but it does have a very convenient basis. Expand [tex] \alpha [/tex] in this basis (noting that in this context "basis" means "Hamel basis," so infinite linear combinations are not allowed). This should give you an interesting relation between [tex] u [/tex] and [tex] \alpha [/tex]. Do you see why this solves the problem?
 
  • #3
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I think I must prove this using the idea of irreducible(minimal) polynomials, and algebraic elements
 
  • #4
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but I think the interesting connection is that they are linearly independent and span E, making E contain u?
 
  • #5
139
12
I think I must prove this using the idea of irreducible(minimal) polynomials, and algebraic elements
The problem is much more straightforward than that. Don't think too hard. :P

Actually, [tex] E [/tex] doesn't have to contain [tex] u [/tex]. If it did, then it would necessarily also contain [tex] F = k(u) [/tex] (since [tex] k(u) [/tex] is the smallest field containing both [tex] k [/tex] and [tex] u [/tex]). However, it does have to contain something which is not in [tex] k [/tex], and that something can be expanded in powers of [tex] u [/tex]. Think a little about this, and you'll realize it's exactly what you want.
 

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