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Splitting field

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


    2. Relevant equations

    A splitting field is the smallest field that contains all of the root of the polynomial.

    3. The attempt at a solution

    I got the roots, [tex] +- i, +-\omega_1 and +-\omega_2[/tex]
    So the degree of the thing has to have all of the rationals as well as i and those other 2, which are complex numbers.
    So here's my confusion:
    Does the polynomial have to be irreducible? Do I have to find the irreducible poly to find the degree of the splitting field?
    This polynomial *looks*(not that what it looks like MEANS anything) like it's degree 6, so then the splitting field would be 6*2 where the 2 comes from theh degree of the complexes over the rationals? We only did one example in class, and I'm not really clear on that. I'm REALLY REALLY lost on this.
    Any input will be appreciated
    CC
     
  2. jcsd
  3. Mar 3, 2007 #2

    Hurkyl

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    The LaTeX command you want is \pm: [itex]\pm[/itex].

    The splitting field is simply [itex]E = \mathbb{Q}(i, -i, \omega_1, -\omega_1, \omega_2, -\omega_2)[/itex]. (Incidentally, what are the [itex]\omega_i[/itex]?)

    (You mean the degree of the splitting field)

    First off, the degree of C/Q is uncountably infinite; I presume you meant the degree of Q(i)/Q. Well, if you were to construct E via:

    [tex]
    \mathbb{Q} \rightarrow \mathbb{Q}(i) \rightarrow E
    [/tex]

    where the extension E/Q(i) was degree 6, then the degree of E/Q would, in fact, be 12. But we don't know yet if E is or is not a degree 6 extension of Q(i).


    Rather than trying to compute a sequence of field extensions that ends with E, you might try finding a primitive element of E, or instead try to compute Gal(E/Q).



    Incidentally, the splitting field of a "generic" degree 6 polynomial is degree 720 (= 6!): if its roots are [itex]\alpha_i[/itex], then the minimal polynomial of [itex]\alpha_1[/itex] is degree 6 over Q, so [itex]\mathbb{Q}(\alpha_1) / \mathbb{Q}[/itex] is degree 6. Then, the minimal polynomial of [itex]\alpha_2[/itex] over [itex]\mathbb{Q}(\alpha_1)[/itex] is degree 5, so [itex]\mathbb{Q}(\alpha_1, \alpha_2) / \mathbb{Q}(\alpha_1)[/itex] is degree 5, and so forth.

    This polynomial, of course, isn't generic: it factors into a quadratic and a quartic. If that happened generically, you'd expect a degree 48 (= 4! * 2!) extension. But again, special things can happen (and do happen for this polynomial) -- so you actually have to work it out.
     
    Last edited: Mar 3, 2007
  4. Mar 3, 2007 #3
    Hey
    Thanks for the hint on the latex.
    for the roots I got
    [tex]\frac{\sqrt3\pm1}{2}i,\frac{-\sqrt3\pm1}{2}i,\pm i[/tex]
    I factored it
    [tex](x^3-i)(x^3+i)[/tex]
    to start with, but i see that it's possible to factor it as a quadratic and a fourth degree. So does that mean that this is not an irreducible poly?Do the factors have to be irreducible? What do I do next?
    BTW you're right about what I MEANT on the degree of Q(i):Q ....but you knew that. I apologize for not re-re-re-reading and being absolutely clear.
    Am I even close to the degree?
    CC
     
  5. Mar 3, 2007 #4

    Hurkyl

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    If it helps, the [itex]\omega_i[/itex] already have a name!


    Well, what approach do you want to try? I guess you're trying to find a chain of field extensions from Q(i) to E, so I'll try to help with that.

    You know you need to adjoin all of the roots of f, but let's work one at a time. Can you factor f further? Can you find a factor of f that's irreducible over Q(i)? (which must be the minimal polynomial of some of the roots of f)
     
  6. Mar 5, 2007 #5
    Ok,
    i've been trying to work this out and UNDERSTAND it.
    Here's what I did:
    I re-factored it this way:
    [tex]x^6+1=(x^2+1)(x^4-x^2+1)[/tex]
    The fourth degree poly doesn't seem to factor out into 2 nice quadratics over Q. So this thing up here is irreducible, so now I can try to get the degree of the splitting field. Since the degree of Q(i):Q is 2, does that mean that the degree of this polynomials' splitting field is 2?

    we did one example in class, that example was [tex]x^3-2[/tex]
    which is irreducible and has a splitting field of degree 6. Cube root of 2 over Q has degree 3 and [Q(cube root of 2):Q(i)]=2 and 3*2=6.
    That's the only example we did, so I'm not quite sure HOW to work these problems.

    Any help and advise will be appreciated.
    CC
     
  7. Mar 5, 2007 #6

    matt grime

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    No. The degree of Q[w], where w is a cube root of 1 is 2, what does that have to do with anything being a splitting field?

    Take the definition. You need the smallest field that contains all the coefficients of the linear factors of the polynomial. Obvisouly if we adjoin i to Q we can split the first factor (and no smaller field will do). Now what about the other factor, x^4 -x^2+1? What are its roots (it is a quadratic in x^2). What does that need adding to Q to make it split into linear factors?
     
  8. Mar 5, 2007 #7
    I don't understand that. Cube root of 1? Where did that come from?

    The roots of the quadratic are complex numbers[tex]\pm \frac{\sqrt 3}{2}\pm\frac{1}{2}i[/tex] So I need to join [tex]\frac{\sqrt 3}{2}+\frac{1}{2}[/tex]? Then the others are combinations of that one? I'm not sure where I'm supposed to go with this.
    CC
     
  9. Mar 5, 2007 #8

    matt grime

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    nowhere. that was the point. where did you pull the square root of -1 from? 'cos it looks like you just yanked it out of thin air, just like I yanked the cube root of 1 out of thin air. That was the point: to make you think about what you'd just yanked out of your ass.

    why would you want to adjoin that? I mean 1/2 is in Q, so adjoining sqrt(3)/2 + 1/2 to Q is just the same as adjoining sqrt(3)/2, which is obviously just the same as adjoining sqrt(3) (you get that 1/2 is a rational number, right?)
     
  10. Mar 5, 2007 #9
    OK,
    First of all, I do not feel like I am "yanking" things out of my ass. I feel as though you are being unnecassarily insulting and arrogant.

    Let me be clear: I DON'T UNDERSTAND this concept. I am strugging with it. I have NO CLUE what to attach to the Q(i). I'm LOST. I've got 7 more of these to do and I can't get past #1.


    Im tired of this and the prof just keeps referring me to his example, which has NOTHING to do with this problem.

    cc
     
  11. Mar 5, 2007 #10

    matt grime

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    You wrote down Q. Why? It is partially the right thing to do. I'm asking how you decided it was something to write down. You didn't justify it. I wanted to know where it came from because it suddenly appeard by magic. It wasn't an insult to say you pulled it out of thin air. It was supposed to make you think 'why did I write that extension as if it were obvious, but didn't understand why Matt wrote something about the cube root of 1?' (and the cube root of one doesn't appear in the answer, but it came with just as much justification as you assertion that Q was what you wanted to look at i.e. none.)


    What are the roots of those polys? i, and (sqrt(3)+i)/2, plus the conjugates. So the splitting field is Q[sqrt(3),i].

    There is no magic involved. It is the smallest field extension of Q that contains those roots.

    You look at it and do some calculations; there is no magic answer just hard work - checking one field contains all you want and no smaller field will do. Any splitting field must contain Sqrt(3) and i, and any field containing them will split the poly. Hence the splitting field is Q[sqrt(3),i].


    His example has everything to do with the problem. All of these problems are the same - just writing down some fields that contain all the roots of the poly and then deciding what the smallest one is.
     
    Last edited: Mar 5, 2007
  12. Mar 5, 2007 #11

    matt grime

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    Just to check.

    Do you see that Q adjoined with (sqrt(3)+i)/2 is precisely the same field as Q[i,sqrt(3)]?
     
  13. Mar 5, 2007 #12
    OK
    I wrote down Q(i) because i is a complex number, like the roots are.
    I see that Q(i) adjoined with sqrt3+1/2 is the same thing as Q[i,sqrt(3)]
    There's the splitting field. My troubles come when I try to figure out the degree of the thing.The Degree is the whole thing that Prof wants. This problem isn't the same as the example in that light.
    Is the degree of Q(i,sqrt3)=4? How do I show what it IS? This is where I'm truly lost, because the polynomial is so different from the irreducible one in our example.
    Am I getting any closer? PLEASE PLEASE help me understand this. I don't want to be up all night another night going in circles and getting nowhere.
    CC
     
  14. Mar 5, 2007 #13
    Just so we are clear, that quote is what I felt was insulting. I know you probably didn't mean it that way, but I feel as though I've been shut down when I get a response like that and I'm really struggling with the material.

    Thank you for your patience and help.
    CC
     
  15. Mar 6, 2007 #14

    matt grime

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    The polynomial should be forgotten. It has nothing to do with finding the degree of the extension you've written down. You just want ther degree of the extension Q[sqrt(3),i] over Q.

    What is the degree of Q over Q? And now what is the degree of Q[sqrt(3),i] over Q? Again, this is just linear algebra and is solved by working it out. Each extension is clearly degree 2 or less. So just show it is not 1, and so the degree you're after is 4.

    You have been taught the result:

    If F<G<H are field extensions, then deg of H over F is the degree of H over G multiplied by the degree of G over F.

    Whcih is all you need.
     
    Last edited: Mar 6, 2007
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