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Fundamental Theorem of Field extensions

  1. Apr 30, 2015 #1
    Suppose F is a field and that ## f(x) ## is a non-constant polynomial in ##F[x]##. Since ##F[x] ## is a unique factorization domain, ## f(x) ## has an irreducible factor, ## p(x) ##. Then the fundamental theorem of field theory says that the field ## E = F[x]/<p(x)> ## contains a zero of ## f(x) ##. I am confused by the last statement.

    ## f(x) ## is an element of ## F[x] ##, not ## E ##, so what does it mean to say that E contains a zero of ## f(x) ##? For an element ## \alpha ## to be a zero of ## f(x) ##, it must be the case that ## \alpha \in F ## and ## f(\alpha) = 0 ## where ## 0 ## is the identity of F.

    I'm a bit confused here. Perhaps someone could lend me a hand? Thanks!

    BiP
     
  2. jcsd
  3. May 1, 2015 #2

    lavinia

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    Project f(x) into E.

    Example: F is the real numbers, f(x) is $$X^2 + 1$$. f is irreducible and E = F[x]/<f(x)> is the field generated by the projections of 1 and x into E.

    $$x^2 + 1 = 0 $$ in E so x is a zero of f(x) in E

    More generally x is a zero of f(x) in F[x]/<f(x)>

    If p(x) is an irreducible factor of f(x), then x is a zero of p(x) in E = F[x}/<p(x)> and thus is a zero of f(x) in E as well.
     
    Last edited: May 1, 2015
  4. May 1, 2015 #3

    mathwonk

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    to say that a field E containing F, also contains a zero c of f, which lies in F[X], means there is an F algebra homomorphism from F[X]-->E, taking X to c, and f to zero. Since the homomorphism takes X to c, and is an F algebra homomorphism, it also takes f(X) to f(c), and this means f(c) = 0. In this case, the homomorphism takes X to its equivalence class mod p, and hence takes f(X) to its equivalence class mod p. Since the equivalence class of p is zero mod p, and p divides f, and the map is muliplicative, so also the equivalence class of f is zero mod p.

    the whole confusion is the fact that isomorphisms are being used without mention. I.e. the zero of f is in E, so we want to first insert F into E, by replacing the isomorphic copy of F in E with F itself. Then we can view f as belonging to E[X], and can more intuitively evaluate f on an element of E. I.e. if c belongs to E, it is more obvious there is an evaluation map from E[X] to E taking X to c. I hopoe I got this right. This also confused me as a student, but van der Waerden made it very clear. also Lavinia is more succinct than I.
     
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