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Subfields of finitely generated algebras

  1. May 3, 2012 #1
    The question is as follows: Assume that k and F are fields and that k is finitely generated as an F-algebra. It is necessarily true that every subfield of k is finitely generated as an F algebra.

    I haven't been able to think of a counter example, and I can only show the result holds in special cases. Namely, if L is the subfield, I can show that L is finitely generated if (1) k is algebraically closed, (2) k is an finite degree extension over L, (3) k can be given the structure of a normed vector space over L, (4) there is a natural projection from K to L.

    However, I hardly think that this could exhaust all cases. Any input would be helpful.
     
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  3. May 3, 2012 #2

    micromass

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    Can't you prove that if k is a field extension of F and if k is finitely generated as F-algebra, that the extension is finite??

    Or am I saying nonsense?
     
  4. May 3, 2012 #3
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  5. May 3, 2012 #4
    Thanks for the replies.

    I can definitely show that if k is finitely generated, it must be an algebraic extension. This is Zariski's lemma which can be shown from the Noether normalization lemma. I'm not sure that we can go so far as to say that the extension itself is finite though.

    Don Antonio: I do not understand your post. F is always an algebraic extension over F, and every subfield of a field is a closed linear subspace. Why would this imply that the subfield is finitely generated?

    Since k is finitely generated over F, there is a surjective map [itex] F[x_1,\ldots,x_n] \twoheadrightarrow k [/itex]. If we can find some way to project k onto L, that will give us a surjection [itex] F[x_1,\ldots,x_n] \twoheadrightarrow L [/itex]. It is sufficient to endow k with a (semi)-norm structure as a vector space over L, in which case we can find such a projection onto L with a complementary subspace via Hahn-Banach. I would like to keep "complementary condition" so that any such projection is well-defined by how it acts on the finite generators for k over F.
     
  6. May 3, 2012 #5

    micromass

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    Lemma: Let A,B be integral domains and let [itex]A\rightarrow B[/itex] be an injective integral ring homomorphism. Then A is a field iff B is a field.

    Let A be a field and let [itex]b\in B[/itex] be nonzero. Then A is an A-vector space of finite dimension. As B is an integral domain, the multiplication [itex]A\rightarrow A[/itex] with b is injective. It is also A-linear and thus bijective. This shows that b is a unit.

    Conversely, let B be a field and let [itex]a\in A[/itex]be nonzero. The element [itex]a^{-1}\in B[/itex] satisfies the polynomial identity

    [tex](a^{-1})^n+\beta_{n-1}(a^{-1})^{n-1}+...+\beta_0=0[/tex]

    Therefore,

    [tex]a^{-1}=-\beta_{n-1}-...-\beta_0a^{n-1}[/tex]

    This is in A.

    Theorem: Let F be a field, and let k be a field extension of F that is finitely generatd as F-algebra. Then k is a finite field extension of F.

    We apply Noethers Normalization theorem on k and we obtain a finite injective homomorphism [itex]F[T_1,...,T_n]\rightarrow k[/itex] of F-algebras. By the lemma, we must have n=0, and thus [itex]F\rightarrow k[/itex] is a finite extension.
     
  7. May 3, 2012 #6
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  8. May 4, 2012 #7
    This is precisely the proof of Zariski, except I do not see how you can claim that the extension is finite. I completely agree that Noether normalization tells you that k will be integral over [itex] F[T_1,\ldots,T_n] [/itex], and since they are fields, n=0. Hence k is integral over F, which in the language of fields means that the extension k/F is an algebraic extension. Why is it finite?

    DonAntonio: Thanks for the link, it looks useful. Unfortunately, it is hard to read the proof without knowing the precise statement that the link is proving, or without knowing the notation used. For example, is [itex] A_f [/itex] the localization of A about f? Also, I fail to see how your single sentence reply encapsulated all of this information. Clarification would be much appreciated.
     
  9. May 4, 2012 #8

    micromass

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  10. May 4, 2012 #9

    If you have the Atiyah-MacDonald book then I believe the relevant question is 5.18 (5th chatper, exercise 18). The proof

    IS in the book, as is as well the fact that this statement is one of the versions of Hilbert's Nullstellensatz, something

    I learned in graduate school.

    The statement is: Let k be a field, B a FINITELY GENERATED k-algebra, then: if B is a field then it is a FINITE (not only algebraic) extension of k.

    DonAntonio
     
  11. May 4, 2012 #10
    Yes okay, I think Zariski does it then. When I proved it, I must have missed the fact that the degree extension was finite. Thanks micromass.

    Don Antonio: I have been deliberately trying to avoid anything resembling the Nullstellensatz as one must use an algebraically closed field and that does not encompass our current scenario. Having seen that Zariski implies that the extension is finite as well as algebraic is indeed the solution, though again I fail to see how you conveyed all of that information with your single sentence response. I appreciate your replies, but perhaps next time you could respond with more than ambiguous single line comments or obscure links for which you have somehow figured that I can not only guess the statement and notation, but even the choice of book.
     
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