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Generation of isomorphic fields by separate algebraic elements

  1. Feb 26, 2008 #1

    Simfish

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    So this sentence kind of confuses me:

    "On the other hand, if [tex] \alpha[/tex] is a root of [tex]x^3 - x + 1[/tex], then [tex]\beta = \alpha^2[/tex] is a root of [tex]x^3 - 2x^2 + x - 1[/tex]. The two fields Q(\alpha) and Q(\beta) are actually equal though if we were presented only with one of the two polynomials, it might take us some time to notice how they are related."

    Okay but how _are_ those two fields actually equal?
     
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  3. Feb 26, 2008 #2

    morphism

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    Degree argument. Certainly [itex]\beta = \alpha^2 \in \mathbb{Q}(\alpha)[/itex], and
    [tex][\mathbb{Q}(\alpha) : \mathbb{Q}(\beta)] = \frac{[\mathbb{Q}(\alpha) : \mathbb{Q}]}{[\mathbb{Q}(\beta) : \mathbb{Q}]} = 1.[/tex]
     
  4. Feb 26, 2008 #3

    gel

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    Also, using [itex]\alpha\beta - \alpha+1=0[/itex] gives [itex]\alpha=1/(1-\beta)\in \mathbb{Q}(\beta)[/itex] and [itex]\beta=\alpha^2\in\mathbb{Q}(\alpha)[/itex].
     
  5. Feb 26, 2008 #4

    mathwonk

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    did you want a proof or an explanation why it can happen?

    Philosophically, in general, for a finite degree field extension of Q, almost any element not in Q generates it, while almost all of these generators have different irreducible polynomials.
     
  6. Feb 26, 2008 #5

    Simfish

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    Yeah, so from Artin 2.9 (chap. 13) it says that if \alpha and \beta are algebraic elements of two extension fields of F, there is an isomorphism of fields F(\alpha) -> F(\beta) which sends \alpha -> \beta iff the irreducible polynomials for \alpha and \beta over F are equal. This seems to imply that different fields are uniquely determined by irreducible polynomials. But the two polynomials above have no common factors...
     
  7. Feb 27, 2008 #6

    gel

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    If alpha and beta have different minimal polynomials, there can still be an isomorphism F(alpha)->F(beta) where alpha doesn't go to beta. E.g. when alpha->1/(1-beta) as in the original example.
     
  8. Feb 27, 2008 #7

    Simfish

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    Okay I see. So do field isomorphisms occur iff the degrees of the fields are equal? (or are there additional conditions you need to satisfy?) (especially if you try to map alpha into a potentially infinite number of betas). where did the \alpha*\beta - \alpha + 1 example come from??
     
  9. Feb 27, 2008 #8

    gel

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    The degrees of the extensions [itex]\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{2})[/itex] and [itex]\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{3})[/itex] are the same, but they are not isomorphic (there is no solution to x^2=3 inside [itex]\mathbb{Q}(\sqrt{2})[/itex]).
    If their degrees are the same *and* one extension is contained inside the other, then you can conclude that they are the same extension.

    I used [itex]\beta=\alpha^2[/itex] and [itex]\alpha^3-\alpha+1=0[/itex] above to get
    [tex]
    0=\alpha\alpha^2-\alpha+1=\alpha\beta-\alpha+1
    [/tex]
    and then write alpha in terms of beta.
     
    Last edited: Feb 27, 2008
  10. Feb 27, 2008 #9
    Er...aren't those fields are isomorphic? Take a map that's identity on the rationals, and root 2 mapped to root 3, right?

    I mean, strictly speaking I don't think that the degree being the same is enough to guarantee an isomorphism (if I had to guess, I'd say the automorphism groups of the extension over the base field would have to be isomorphic, but I don't know), but I don't think this is a counterexample to the claim.
     
    Last edited: Feb 27, 2008
  11. Feb 27, 2008 #10

    gel

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    No, they're not. If a^2=2 and b^2=3 then you can't map a to b by an isomorphism. If f(a)=b then f(a)^2=b^2=3 and f(a^2)=f(2)=2, so f(a)^2 != f(a^2).
    A necessary condition for mapping a to b is that they have the same minimal polynomial.
     
  12. Feb 27, 2008 #11
    Ah, of course. I must have been thinking of something else. I've gotta stop posting when I haven't slept.
     
  13. Feb 27, 2008 #12

    gel

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    oh, ok, no problem :)
     
  14. Feb 27, 2008 #13

    Simfish

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    OH I see. So what you did is set up a relation relating alpha and beta in such a way that you used the relation to establish the minimal polynomial (in alpha) of the same form as beta. Is there an easy way to test when such a move IS NOT possible?
     
  15. Feb 28, 2008 #14

    gel

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    If any of the following things differ then the extensions will not be isomorphic.
    - degrees of the extensions are different (same as the degrees of the minimal polynomials of a and b).
    - Galois groups are different.
    - For extensions of Q, algebraic number theory gives other invariants such as class group, the group of units (a finite group), the Zeta function, etc, although this is getting quite advanced.

    Also, there's the following:
    If f and g are minimal polynomials of a and b, you can check to see if f(x)=0 has a solution in Q(b). If it doesn't, then the extensions are not isomorphic. If it does, then there is a homomorphism Q(a)->Q(b) and it is an isomorphism if the degrees of f and g are the same. I'm not sure if there's a simple way to check this though.
     
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