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Splitting fields again

  1. Mar 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the degree of the splitting field of the following:
    [tex]x^6+x^3+1[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Can I do this?
    Let [tex]y=x^3[/tex]

    then
    [tex]y^2+y+1[/tex]
    has degree 2 over Q (From another problem)
    So the degree of the splitting field is 2.

    Or do I need to factor it out?
    or do I need to look at the roots of the origional polynomial?
    The method here is what I am confused about.
    Thanks
    CC
     
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 16, 2007 #2

    matt grime

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    Can I ask two questions: do you think it is reasonable to write down a splitting field without reference to the original polynomial? Or how about this extreme example: x^n-1, let's set y=x^n, so we have y-1, and that splits over Q. Do you think what I did is reasonable?
     
  4. Mar 16, 2007 #3
    OK
    So I see that you can't do that with the x^n-1 polynomial. So what is the proper method? Factor? Use the TI89 to see what the roots are and then decide what to extend the field to? Is it not possible to reduce the polynomial somehow?....????
    I just don't know HOW to do these problems. So far it's just been trial and error for me. I'm not seeing a method.
    Please tell me the WAY to approach this.
    CC
     
  5. Mar 16, 2007 #4
    I just don't know how to approach this type of problem at all.
    I need steps to follow..or things to try. I got confused about this the first time we went over it and since then I'm even more lost.
    Prof says that if you have x^4+x^2+1 it can be reduced by letting y=x^2.
    Then you get y^2+y+1. Degree of splitting field is 2. Is that a special case?
    CC
     
  6. Mar 16, 2007 #5

    Hurkyl

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    It's too early to think about algorithms, methinks -- you need to think carefully about the individual steps.

    Let's consider your professor's example. He wants to form the splitting field E of y^2 + y + 1. Well, you know the quadratic case well (right?), so you know exactly what happens.

    So, you have a field extension Q --> E.

    Does E contain the roots of x^4 + x^2 + 1? How might you tell? If you can't think of a "good" way, can you think of a "dumb" way?


    Don't read on until you can answer that question. I've colored it white so you don't do it by accident -- but you can still read it later by highlighting the text.

    ------------------------------------------------

    Okay, so you've figured out that E is not the splitting field of x^4 + x^2 + 1. Maybe you did it by linear algebra (plug an arbitrary element of E into the polynomial, and show the system of linear equations has no solution), or maybe you had a better way.

    So the next question is: is E even relevant? You are looking for the splitting field F of x^4 + x^2 + 1 -- that is, the smallest field extension:

    Q --> F

    Knowing E is really helpful if we have the tower of extensions

    Q --> E --> F.

    (Well, if this is not true, there still may be a way for E to be useful. But if there is, that's not what I'm talking about)

    So your next task is to figure out if E a subfield of F -- that is, if F contains the roots of x^4 + x^2 + 1, does it also contain the roots of y^2 + y + 1?

    ------------------------------------------------
     
    Last edited: Mar 16, 2007
  7. Mar 18, 2007 #6
    OK
    What i did was just solve for the roots on the calculator to see what they are.
    I got [tex]\pm\frac{1\pm\sqrt3}{2}i[/tex]for x^4+x^2+1

    and I got [tex]\frac{-1\pm\sqrt3}{2}i[/tex]forx^2+x+1

    So looking at these to sets of roots,I think that the splitting field is [tex]Q(\sqrt3,i)[/tex] Which would have degree 4 over Q, not 2.
    So now I'm even MORE confused. Is it 2 or 4?

    My prof did this:
    let w be some root of p(x)=x^4+x^2+1
    then
    [tex]w^4+w^2+1=0[/tex]
    [tex](-w)^4+(-w)^2+1=0[/tex]
    [tex]w^4=-w^2-1[/tex]
    [tex]w^6=-w^4-w^2=1[/tex]
    [tex]w^8=w^2[/tex]
    [tex](w^2)^4+(w^2)^2+1=w^8+w^4+1=w^2+w^4+1=0[/tex]

    so then [tex]w^2, -w^2[/tex] are roots.

    So the splitting field is Q(w). The degree of this is either 4 or 2.
    So then:
    [tex]x^4+x^2+1=(x^4+2x^2+1)-2x=(x^2+1)-x^2=(x^2-x+1)(x^2+x+1)[/tex]
    So our degree [Q(w):Q]=2
    I don't understand why he did the substitutions, really. I guess it's only valid for this particular one.
    Then he said we could just plug in y=x^2 and get the same thing.
    SO
    the splitting field is of degree 2 even though it has i and [tex]\sqrt3[/tex] in it.
    I did another problem with [tex]Q(\sqrt3,i)[/tex](That one was x^6+1) and the degree there was 4?
    I just don't get this.
    CC
     
    Last edited: Mar 18, 2007
  8. Mar 18, 2007 #7

    matt grime

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    Why did you leap to deciding that the splitting field was Q[sqrt(3),i]? Certainly the splitting field is contained in that, and is the smallest subfield of Q[sqrt(3),i] that contains the roots above, but it does not immediately follow that it is all of it. (It may or may not be, and looking through some of the stuff in this thread, I think at least one person has made a mistake about this question).
     
    Last edited: Mar 18, 2007
  9. Mar 19, 2007 #8
    OK
    Here are my thoughts on the [tex]x^4+x^2+1[/tex] problem.
    [tex]Q(\sqrt3,i)[/tex] isn't the smallest field because all of the roots are complex, so I DON'T need [tex]\sqrt3[/tex] in there. And [Q(i):Q]=2.

    Now I have this:Find the splitting field of [tex]x^4-2[/tex].
    So this irreducible over Q of degree 4,with a real root. I also need i in my splitting field. So I have [tex][Q(x^4-2):Q][Q(i):Q]=4*2=8[/tex]

    Back to [tex]x^6+x^3+1[/tex]
    This has all complex roots, so I need to extend my rationals to i again.
    So it has degree 2?
    I'm not sure if I need to try to factor it into 3 quadratics...

    EDIT:I think that this last poly is irreducible over Q. So the degree is 6? Do I need to put i in there too?

    Am I closer?
    CC
     
    Last edited: Mar 19, 2007
  10. Mar 19, 2007 #9

    Hurkyl

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    I made two. :redface: I assumed the polynomial was irreducible, and misremembered how much effort it was to test for irreducibility over E by plugging in an arbitrary element. (I mentally mixed it up with the "find the minimal polynomial of this number" problem) And I should have known better anyways, since it's clear that the roots of x^4 + x^2 + 1 are sixth roots of unity.
     
  11. Mar 19, 2007 #10

    Hurkyl

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    The roots of x^4 + x^2 + 1 are obvously not in Q(i)... are you sure you need i?
     
  12. Mar 19, 2007 #11
    OK,
    the reason I thought I needed i was because all of the roots are complex.
    And I guess I'm still confused about the whole splitting field thing. i don't know what the splitting field IS. It looks to me like the x^4+x^2+1 wouldn't split over the rationals because the roots are complex.

    Am I supposed to be looking only at the degree of the irreducible factors over Q?
    I'm re-lost. I thought I was making progress...but I'm still back where I started.


    What do you think about the other ones I wrote down?

    Thanks
    CC
     
    Last edited: Mar 19, 2007
  13. Mar 19, 2007 #12

    Hurkyl

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    There are complex extensions of Q that do not contain i. This is one of them.

    Exercise: prove that [itex]i \notin \mathbb{Q}(i \sqrt{2}).[/itex]



    By explicit computation, you found that the roots of f(x) = x^4 + x^2 + 1 are [itex](\pm 1 \pm i \sqrt{3}) / 2[/itex]. (I assume this is what you meant)

    Therefore, by its very definition, the splitting field E of f over Q is

    [tex]
    E = \mathbb{Q}\left( \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}, \frac{-1 + i \sqrt{3}}{2}, \frac{-1 - i \sqrt{3}}{2} \right)
    [/tex]

    We can clearly see that

    [tex]E \subseteq \mathbb{Q}( \sqrt{3}, i),[/tex]

    but it's not so clear if that should be an equality. Do you have an idea how you might try to prove they are equal? How to prove they are unequal?

    From our comments, you can probably guess that they are not equal. Also, as you observed, E is clearly not equal to Q. So, we have this (strict) tower of field extensions:

    [tex]\mathbb{Q} \subset E \subset \mathbb{Q}(\sqrt{3}, i)[/tex]

    You know the degree [itex][\mathbb{Q}(\sqrt{3}, i) : \mathbb{Q}] = 4[/itex], right? (How do you know that?) Do you see how that immediately tells you the degree [E:Q], without actually having to compute E? And also how it gives us an immediate way to simplify the presentation of E?

    But that's no fun, let's keep looking at the problem. How did you know that [itex][\mathbb{Q}(\sqrt{3}, i) : \mathbb{Q}] = 4[/itex]? What method did you use? Will that work for E?


    (Incidentally, IMHO, the simplest way of all is to notice that the roots are sixth roots of unity, which makes it easy to find a z so that E = Q(z). Galois theory makes it even easier to compute the degree)
     
  14. Mar 19, 2007 #13
    [tex][Q(i):Q][Q(\sqrt3):Q]=2*2=4[/tex] and E has to be less than 4, so it has to be 2? Looking at the tower above. We haven't gotten to the galois stuff quite yet.

    I have a mental block on how a complex extension doesn't have i in it. I can't see it.

    Are my other ones even close?
    CC
     
  15. Mar 19, 2007 #14

    Hurkyl

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    Why would

    [tex][\mathbb{Q}(\sqrt{3}, i) : \mathbb{Q}] = [\matbb{Q}(i) : \matbb{Q}] [\matbb{Q}(\sqrt{3}) : \matbb{Q}][/tex]

    be true? The theorem you learned about the degre of field extensions says, for example,

    [tex][\mathbb{Q}(\sqrt{3}, i) : \mathbb{Q}] = [\matbb{Q}(\sqrt{3}, i) : \matbb{Q}(i)] [\matbb{Q}(i) : \matbb{Q}][/tex]

    when applied to the chain [itex]\mathbb{Q} \subseteq \mathbb{Q}(i) \subseteq \mathbb{Q}(i, \sqrt{3})[/itex]. The equation you wrote is simply a coincidence.


    Consider this example: what is the degree

    [tex][ \mathbb{Q}(i, i + 1) : \mathbb{Q} ]?[/tex]




    Well, I gave you a specific example of a complex extension that does not contain i. What sort of numbers does [itex]\mathbb{Q}(i \sqrt{2})[/itex] contain? You should be able to write this field explicitly as a set. (or as a vector space over Q)
     
    Last edited: Mar 19, 2007
  16. Mar 19, 2007 #15
    I re-read the theorem and I see your point about my wrongness.
    A basis for this
    [itex]\mathbb{Q}(i \sqrt{2})[/itex]
    is [tex](1,i\sqrt2)[/tex]?
    so the degree is 2?
    And the degree of
    [tex][ \mathbb{Q}(i, i + 1) : \mathbb{Q} ][/tex]
    is also 2?
    and what I need in my problem x^4+x^2+1 is the degree of [tex]i\sqrt3[/tex]

    So here's another question:
    If I can find the minimal polynomial, does that help me at all?
    Is the degree of the minimal polynomial the degree of the splitting field?
    I'm so confused about this
    Any good links to look at?
    I can't apply what we did in class to this.
    HELP
    PS I just went and spoke to my prof and I am even MORE lost.
     
  17. Mar 19, 2007 #16

    Hurkyl

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    Minimal polynomials can help for problems: if you have a field extension F(a)/F, then the degree of the minimal polynomial of a over F is [F(a):F].

    The minimal polynomial of [itex]\sqrt{3}[/itex] over [itex]\mathbb{Q}(i)[/itex] is (x^2 - 3), so [itex][\mathbb{Q}(i, \sqrt{3}) : \mathbb{Q}(i)] = 2[/itex]. But the minimal polynomial of (i + 1) over [itex]\mathbb{Q}(i)[/itex] is (x - i - 1). So, [itex][\mathbb{Q}(i, i+1) : \mathbb{Q}(i)] = 1[/itex].

    If your plan is to find the degree of a field extension by adjoining one root at a time, then you could use minimal polynomials to find the degree of each step.



    You know that the splitting field of f(x) = x^4 + x^2 + 1 is contained in [itex]\mathbb{Q}(i \sqrt{3})[/itex]. Can you prove that [itex]i \sqrt{3}[/itex] is contained in the splitting field of f? If so, then [itex]\mathbb{Q}(i \sqrt{3})[/itex] is, in fact, the splitting field of f. And, now that you have a primitive generator for that field, you can find its degree with a minimal polynomial.
     
    Last edited: Mar 19, 2007
  18. Mar 19, 2007 #17
    but isn't [Q(i):Q]=2? so [Q(i,i+1):Q(i)][Q(i):Q]=2?
    I understand the minimal polynomial thing, I think.
    I will conyinue to stare at this some more later.
    Thanks
    CC
     
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