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Is this correct? Field extension of the rationals

  1. Jul 26, 2011 #1
    By F[X] I mean the polynomials with coefficients in field F. By F(X) I mean the rational polynomials.

    I have a feeling that [itex]\boxed{ \mathbb Q( \sqrt 2 ) \cong \frac{\mathbb Q[X]}{(X^2-2)}} [/itex]. (if not readable: the RHS is with [X])
    Is this true? If so, how can I prove it? I suppose it would suffice I could show that the RHS is the smallest field extension of the rationals that contains sqrt(2) (as the LHS is obviously just that).

    Also, is there maybe even a more general result behind this?
  2. jcsd
  3. Jul 26, 2011 #2
    Hi mr. vodka! :smile:

    You are 100% correct that this is true. How do we prove it? Well, the crucial point is that [itex]X^2-2[/itex] is an irreducible polynomial in [itex]\mathbb{Q}[X][/itex]. This means that
    [itex]\mathbb{Q}/(X^2-2)[/itex] is a field. So the following function

    [tex]f:\mathbb{Q}\rightarrow \mathbb{Q}/(X^2-2):q\rightarrow [q][/tex]

    is an injective field homomorphism. We extend this field homomorphism to

    [tex]f:\mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}/(X^2-2):a+b\sqrt{2}\rightarrow [a+bX][/tex]

    Check that this is an isomorphism.

    This result can be generalized. The correct generalization involves splitting fields. If P(X) is an irreducible polynomial over a field F, then there exists a field extension K of F such that P(X) has a root in K. Indeed, we define


    If a is the root, then we can even define

    [tex]f:F(a)\rightarrow K[/tex]

    by defining f(a)=X. This is again surjective.

    What does this have to do with splitting fields? Well, we can use this result to show that: if P(X) is a polynomial over a field F, then there exists a field extension K of F such that P(X) splits.

    Why is this true? Well, repeatedly apply the above to the irreducible factors of P(X).
  4. Jul 26, 2011 #3
    Thank you very much! :)
  5. Jul 26, 2011 #4
    It also might be useful to learn about minimal polynomials. If [itex] \alpha[/itex] is the root of some polynomial in [itex] F[X] [/itex], then there is a unique monic irreducible polynomial [itex]m_{\alpha, F}(X) \in F[X][/itex], which is known as the minimal polynomial of [itex] \alpha[/itex] over [itex]F[/itex]. It's said to be minimal because if f(X) is any other polynomial with [itex] \alpha[/itex] as a root, then [itex] m_{\alpha, F}(X) \, | \, f(X)[/itex]. (What's more, a monic polynomial over F with [itex] \alpha[/itex] as a root is the miminal polynomial of alpha over F iff it is irreducible over F.) It's then a theorem that [itex] F(\alpha) \cong F[X]/(m_{\alpha, F}(X))[/itex].
    Last edited: Jul 26, 2011
  6. Jul 26, 2011 #5
    Hm, thanks, but I seem to be confused:

    Take as a root [itex]\sqrt[3] 2[/itex], then its minimal polynomial is [itex]f = X^3-2[/itex] (?). This has one real solution and two complex solutions. For that reason it would seem that [itex]\mathbb Q[X]/(f)[/itex] would also have these two complex solutions as elements (as f is also the minimal polynomial for those complex solutions) while [itex]\mathbb Q(\sqrt[3] 2)[/itex] only adds one element, not three, but of course I'm wrong cause they're isomorphic... Where do I err?
  7. Jul 26, 2011 #6
    No, doing [itex]\mathbb{Q}[X]/(f)[/itex] only adjoins one root! You can choose which one though. So [itex]\mathbb{Q}[X]/(f)[/itex] is isomorphic to [itex]\mathbb{Q}(\sqrt[3]{2})[/itex], but also to [itex]\mathbb{Q}(a)[/itex] where a are the other roots.
    In general, the other roots are not contained in [itex]\mathbb{Q}[X]/(f)[/itex].

    Note, that the root of the polynomial [itex]Z^3-2[/itex] in [itex]\mathbb{Q}[X]/(X^2-2)[/itex] is [X]. So the polynomial splits as


    But the polynomial [itex]Z^2+[X]Z+[X^2][/itex] doesn't necessarily have roots in [itex]\mathbb{Q}[X]/(X^2-2)[/itex].
  8. Jul 26, 2011 #7


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    That extension adds a lot more than just one element -- it includes, for example, [itex]1 + \sqrt[3] 2[/itex] and [itex]3 + 2 \sqrt[3]2 + \sqrt[3] 4[/itex].

    But you're right, you won't find any of the other cube roots of 2 in that field.

    The thing you're missing is that [itex]\mathbb Q[X]/(f)[/itex] doesn't contain any of the three complex roots of f: it contains neither [itex]\sqrt[3] 2[/itex], [itex]\omega \sqrt[3] 2[/itex], nor [itex]\omega^2 \sqrt[3] 2[/itex]. (Where [itex]\omega[/itex] is the primitive cube root of unity -- i.e. [itex]\omega = exp(2 \pi i / 3)[/itex])

    What is true is that there is a field homomorphism from this to the real numbers, sending X sending X to [itex]\sqrt[2] 3[/itex], and two field homomorphisms from this to the complex numbers: one sends X to [itex]\omega \sqrt[2] 3[/itex], and one sends X to [itex]\omega^2 \sqrt[2] 3[/itex].

    We say that this field has one real embedding and one complex conjugate pair of complex embeddings.

    Using these homomorphisms, we can think of X as either of the three complex roots at our leisure -- but we obviously cannot think of X as being all three at once.

    Now, how does f(t) factor in this field? as:
    [tex]f(t) = (t - X) (t^2 + Xt + X^2)[/tex]​
    It can be shown that the quadratic term is irreducible. So it does turn out that f has only one root in this field. But we can make a new field extension that adjoins yet another element (a square root of -3), and this field will have three roots for f.

    The resulting field is called the "splitting field" of f. It turns out to be isomorphic to
    [tex]\mathbb{Q}(\omega, \sqrt[3] 2)[/tex]
    Incidentally, I'm pretty sure it is also isomorphic to:
    [tex]\mathbb{Q}(\omega + \sqrt[3] 2)[/tex]

    (aside: for some f, [itex]\mathbb{Q}[X] / f(X)[/itex] does have more than one root of f. For example, if f is a quadratic polynomial)
  9. Jul 26, 2011 #8
    (you made a typo in the (X² - 1) instead of (X³ - 1) )

    So you say that I can choose to view [itex]\mathbb Q[X]/(X^3-2)[/itex] as adding either the real or the complex solutions? But by writing the factorization
    [itex]Z^3-2= (Z-[X])(Z^2+[X]Z+[X^2])[/itex]
    it is clear that we have added the real solution, yet I don't see "where you made the choice"

    EDIT: this is not a reply to Hurkyl, his post appeared between my replying and act of posting; I have yet to read Hurkyl's post
  10. Jul 26, 2011 #9
    Why is it clear that we have added the real solution?? Why can't [X] represent one of the complex solutions?
  11. Jul 26, 2011 #10
    Indeed, I was too rash there.

    Also thank you Hurkyl.

    All very interesting and enlightening
  12. Jul 27, 2011 #11
    I don't know if this is too informal, and too quick-and-dirty, but the ideal generated by an irreducible polynomial is maximal, so that the quotient is a field, and, in the quotient, the base element is considered/defined-to-be, the zero element, i.e., x^2-2=0 for x in the quotient field.
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