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Math Help - splitting fields

  1. #1
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    splitting fields

    Hi I do not understand the splitting fields, for example in the following:
    find splitting field of (x^4)+1 over Q. all I know is that this function factors to (x^2+i)(x^2-i) but how do we know that this is the the end? how do we know this cannot be more factorized.
    and in these kind of questions (splitting a field) is it all I have to do? just to find the factors?
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  2. #2
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    Quote Originally Posted by hamidr View Post
    Hi I do not understand the splitting fields, for example in the following:
    find splitting field of (x^4)+1 over Q. all I know is that this function factors to (x^2+i)(x^2-i) but how do we know that this is the the end? how do we know this cannot be more factorized.
    and in these kind of questions (splitting a field) is it all I have to do? just to find the factors?
    You need to further factorize P(x)=(x^4+1)=(x^2+i)(x^2-i) into linear factors to find the splitting field of P(x) over Q. Then, adjoin the roots of the linear factors into Q. We see that

    P(x)=(x+\frac{1}{\sqrt{2}}(1 + i ))(x-\frac{1}{\sqrt{2}}(1 + i ) )(x+\frac{1}{\sqrt{2}}(-1 + i ) )(x-\frac{1}{\sqrt{2}}(-1 + i ) ).

    You need to verify that the splitting field of P(x) over \mathbb{Q} is \mathbb{Q}(\sqrt{2}, i).
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  3. #3
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    A simple way to understand s.f. is that it's gotten by adjoining all roots of the polynomial. But sometimes you have a simpler way of writing them, in your case \mathbb{Q}(\sqrt{2},i)
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  4. #4
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    thanks a lot, the information was so useful and it really gave me Idea about SF's but one question remains, how do we find all the roots of polynomial? before when working with reals, we would say whenever polynomial meets the x axis, which is the solution to the x is zero , then we have a root, but now complex variables are in our way. ie how did you find:


    all I could do was to say (x^2+i)(x^2-i)= (x^2+i)(x-root(i))(x+root (i))
    which then I dont know what root(i) is!
    [IMG]file:///C:/Users/HAMID/AppData/Local/Temp/moz-screenshot.png[/IMG][IMG]file:///C:/Users/HAMID/AppData/Local/Temp/moz-screenshot-1.png[/IMG]
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  5. #5
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    You have to know what algebraically closed field to work with before you start bothering splitting fields. At current stage of course, most of the cases it is \mathbb{C}, so you can safely assume that \sqrt{i}=\sqrt[4]{-1} exists, and using complex number operations you could figure out it is e^{\frac{\pi i}{2}}=\frac{1+i}{\sqrt{2}} works
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  6. #6
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    Quote Originally Posted by hamidr View Post
    thanks a lot, the information was so useful and it really gave me Idea about SF's but one question remains, how do we find all the roots of polynomial? before when working with reals, we would say whenever polynomial meets the x axis, which is the solution to the x is zero , then we have a root, but now complex variables are in our way. ie how did you find:


    all I could do was to say (x^2+i)(x^2-i)= (x^2+i)(x-root(i))(x+root (i))
    which then I dont know what root(i) is!
    [IMG]file:///C:/Users/HAMID/AppData/Local/Temp/moz-screenshot.png[/IMG][IMG]file:///C:/Users/HAMID/AppData/Local/Temp/moz-screenshot-1.png[/IMG]
    i = cos(\frac{\pi}{2} ) + i sin(\frac{\pi}{2}),
    i^{1/2} = {(cos(\frac{\pi}{2} ) + i sin(\frac{\pi}{2})) }^{1/2},
    i^{1/2} = cos(\frac{\pi}{4} ) + i sin(\frac{\pi}{4}),
    i^{1/2} = \frac{1}{\sqrt{2}}(1+i).

    By the way, have you learned the cyclotomic polynomial?
    If so, then you'll see that \Phi_8(x)=x^4+1 and \Phi_8(x)= (x -\zeta )(x - \zeta^3)(x-\zeta^5)(x-\zeta^7), where \zeta = \frac{1}{\sqrt{2}}(1+i).
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  7. #7
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    Quote Originally Posted by aliceinwonderland View Post
    i = cos(\frac{\pi}{2} ) + i sin(\frac{\pi}{2}),
    i^{1/2} = {(cos(\frac{\pi}{2} ) + i sin(\frac{\pi}{2})) }^{1/2},
    i^{1/2} = cos(\frac{\pi}{4} ) + i sin(\frac{\pi}{4}),
    i^{1/2} = \frac{1}{\sqrt{2}}(1+i).

    By the way, have you learned the cyclotomic polynomial?
    If so, then you'll see that \Phi_8(x)=x^4+1 and \Phi_8(x)= (x -\zeta )(x - \zeta^3)(x-\zeta^5)(x-\zeta^7), where \zeta = \frac{1}{\sqrt{2}}(1+i).
    Algebraically you can let \zeta_8=\frac{1+i}{\sqrt{2}}, but you don't have to. Any root of \Phi_8, i.e. any of \frac{\pm1\pm i}{\sqrt{2}} can be chosen as \zeta_8
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  8. #8
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    Quote Originally Posted by FancyMouse View Post
    Algebraically you can let \zeta_8=\frac{1+i}{\sqrt{2}}, but you don't have to. Any root of \Phi_8, i.e. any of \frac{\pm1\pm i}{\sqrt{2}} can be chosen as \zeta_8
    I know what you are talking. I chose \zeta as a primitive 8th root of unity in \mathbb{C}, where

    \zeta = cos\frac{2\pi}{8} + i sin\frac{2\pi}{8},

    Any problem with that?
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  9. #9
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    Quote Originally Posted by aliceinwonderland View Post
    I know what you are talking. I chose \zeta as a primitive 8th root of unity in \mathbb{C}, where

    \zeta = cos\frac{2\pi}{8} + i sin\frac{2\pi}{8},

    Any problem with that?
    Oh, I know you know the stuff. I'm just trying to show the point to hamidr so that he's not misunderstanding anything.
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  10. #10
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    Quote Originally Posted by FancyMouse View Post
    Oh, I know you know the stuff. I'm just trying to show the point to hamidr so that he's not misunderstanding anything.
    I see. My point was that there were alternative choices for \zeta , but they give the same answers for this problem.

    Anyway, I agree that it may be helpful if OP tries other choices of \zeta and verify them if they give the same solution.
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  11. #11
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    thank you so much, this was really helpful.
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