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Math Help - Galois Theory Problems....

  1. #1
    AAM
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    Galois Theory Problems....

    Hi guys! :-)

    I'm new to the forum. Currently doing a course on Galois Theory - it's a fascinating subject but our lecturer is a bit....erm....under-par! :-s

    He's only doing abstract theory & NO examples, but our exam is 100% applications! So I was wondeing if you guys could possibly help me please?

    Here are a couple of questions from the last few years exams - if you have any idea how to solve ANY of them I would be EXTREMELY greatful! :-) Please could I ask you to be explicit with you methods, use of theorems, etc - these will LITERALLY be the first worked examples I've ever seen!

    Thanks everyone! :-)

    Questions:

    1) Let the complex number Zeta be a 12th root of unity. Find the Galois Group Gal(Q(Zeta):Q) & all subfields of Q(Zeta).

    2) Find all complex roots of X^4-3/2X^2-Sqrt(15)+61/16.

    3) Let K be the splitting field of (X^27+1)(X^15+1) over Q. Find the degree of K:Q.

    4) Find the Galois Group of the polynomial X^4+6X+3 over Q.

    5) Find the Galois Group of the splitting field L of the polynomial X^4-8X^2+49 over Q and find all fields between Q & L.

    Thanks so much in advance! :-) x
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  2. #2
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    Quote Originally Posted by AAM View Post
    1) Let the complex number Zeta be a 12th root of unity. Find the Galois Group Gal(Q(Zeta):Q) & all subfields of Q(Zeta).
    I am going to try to be as detailed as I can but,
    I do not want to be overly detailed so in case something doth not make sense you can ask me.

    Let \zeta = e^{2\pi i/12} and let K=\mathbb{Q}(\zeta).
    You should be familar with the theorem that \text{Gal}(K/\mathbb{Q}) \simeq \mathbb{Z}_{12}^{\times} .
    Now, \mathbb{Z}_{12}^{\times}\simeq \mathbb{Z}_4^{\times} \times \mathbb{Z}_3^{\times} \simeq \mathbb{Z}_2\times \mathbb{Z}_2 so \text{Gal}(K/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2.

    If \sigma is an automorphism of K = \mathbb{Q}(\zeta) then it is completely determined by \sigma(\zeta). Let \Phi_{12}(x) by the 12-th cyclotomic polynomial, then it is irreducible and its zeros include \zeta,\zeta^5,\zeta^7,\zeta^{11} (the powers of \zeta relatively prime with 12). If \sigma is an automorphism then it means \sigma(\zeta) must be a root of \Phi_{12}(x) and so \sigma(\zeta) = \zeta,\zeta^5,\zeta^7,\zeta^11. Therefore we have at most 4 automorphism. Since the order of the Galois group is 4 it means that each one defines and automorphism. Therefore, \text{Gal}(K/\mathbb{Q}) = \{ \sigma_1,\sigma_5,\sigma_7,\sigma_{11} \} where \sigma_1 is defined by \sigma_1(\zeta)=\zeta i.e. the identity automorphism; where \sigma_5 is defined by \sigma_5(\zeta) = \zeta^5; where \sigma_7 is defined by \sigma_7(\zeta)=\zeta^7; where \sigma_{11} is defined by \sigma_{11}(\zeta)=\zeta^{11}.

    Now the subgroups of \text{Gal}(K/\mathbb{Q}) include \{ \sigma_1\}, \{\sigma_1,\sigma_5\},\{\sigma_1,\sigma_7\}, \{\sigma_1,\sigma_{11}\} and \{\sigma_1,\sigma_5,\sigma_7,\sigma_{11}\} itself. An easy why to see why these are the subgroups is because we proved that the Galois group is isomorphic with \mathbb{Z}_2\times\mathbb{Z}_2, and so you can think of the subgroups of \mathbb{Z}_2\times\mathbb{Z}_2 to help you visualize what the subgroups of the Galois group need to be.

    To find the intermediate fields we need to use the fundamental theorem of Galois theory. First, K^{\text{Gal}(K/\mathbb{Q})} = \mathbb{Q} so that one was easy. Second, K^{\left< \sigma_1 \right>} = K so that one was also easy. Now we get a little more involved. The other three intermediate fields are K^{\left<\sigma_5\right>}, K^{\left<\sigma_7\right>}, \text{ and }K^{\left< \sigma_{11}\right>} by Galois theory (by "Galois theory" I mean using the correspondence theorem between subgroups and intermediate subfields).

    Notice that if we can find \alpha_j \in K so that [\mathbb{Q}(\alpha_j):\mathbb{Q}] = 2 and \alpha_j is fixed by \sigma_j (for j=5,7,11) then K^{\left< \sigma_j \right>} = \mathbb{Q}(\alpha_j). To prove this, it is really easy. Notice that \mathbb{Q}\subseteq \mathbb{Q}(\alpha_j)\subseteq K^{\left< \sigma_j\right>} with [\mathbb{Q}(\alpha_j) : \mathbb{Q}]=2 and [K^{\left<\sigma_j\right>}:\mathbb{Q}] = [\text{Gal}(K/\mathbb{Q}):\left<\sigma_j\right>] = 2 (by Galois theory again). However, we know that [\mathbb{Q}(\alpha_j) : \mathbb{Q}][K^{\left<\sigma_j\right>}:\mathbb{Q}(\alpha_j)] = [K^{\left<\sigma_j\right>}:\mathbb{Q}] from basic field theory, we conclude that [K^{\left<\sigma_j\right>}:\mathbb{Q}(\alpha_j)] \implies K^{\left<\sigma_j\right>} = \mathbb{Q}(\alpha_j).

    By the above paragraph to find K^{\left<\sigma_5\right>} we just need to find an element in K fixed by \sigma_5 that has degree 2 over \mathbb{Q}. The claim is that \alpha_5 = \zeta + \zeta^5 has degree 2 over \mathbb{Q}. Can you see what \alpha_5 \not \in \mathbb{Q}? If you write out the sine's and cosine's you will see that \alpha_5 is a complex number so it cannot be in \mathbb{Q}. Likewise \alpha_7 =\zeta+\zeta^7 is not in \mathbb{Q}. Consider (x - \alpha_5)(x-\alpha_7) this polynomial is fixed by \sigma_1,\sigma_5,\sigma_7,\sigma_{11} (just do the calculation). Therefore, (x-\alpha_5)(x-\alpha_7) is a irreducible polynomial over \mathbb{Q} which contains \zeta + \zeta^5,\zeta + \zeta^7 as roots. Therefore, K^{\left< \sigma_5\right>} = \mathbb{Q}(\zeta + \zeta^5) and K^{\left< \sigma_7 \right>} = \mathbb{Q}(\zeta + \zeta^7). To finally find K^{\left< \sigma_11\right>} notice that \zeta + \zeta^{11} is fixed by \sigma_{11} and \zeta + \zeta^{11} = 2\cos \tfrac{\pi}{6} = \sqrt{3} so it is neither in K or \mathbb{Q} which forces K^{\left< \sigma_{11} \right>} =  \mathbb{Q}(\sqrt{3}).
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  3. #3
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    Quote Originally Posted by AAM View Post
    2) Find all complex roots of X^4-3/2X^2-Sqrt(15)+61/16.
    To solve this polynomial let y=x^2 and then apply quadradic formula.

    3) Let K be the splitting field of (X^27+1)(X^15+1) over Q. Find the degree of K:Q.
    The zeros of X^{27}+1 are -\zeta_{27}^k where k=0,1,..,26 and \zeta_{27}=e^{2\pi i/27}. In the similar manner the zeros of X^{15}+1 are -\zeta_{15}^j where j=0,1,...,24. Therefore, the splitting field of (X^{27}+1)(X^{15}+1) over \mathbb{Q} is K = \mathbb{Q}(\zeta_{15},\zeta_{27}). To find the degree we need a result about cyclotomic extensions. Let \mathbb{Q}_n be the cyclotomic extension by an n-th root of unity and \mathbb{Q}_m by an m-th root of unity then \mathbb{Q}_n\mathbb{Q}_m = \mathbb{Q}_{\ell} where \ell = \text{lcm}(n,m). Therefore, \mathbb{Q}(\zeta_{15},\zeta_{27}) = \mathbb{Q}(\zeta_{135}). Thus, K = \mathbb{Q}_{135} it follows that [K:\mathbb{Q}] = \phi(135) = 72.
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  4. #4
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    Quote Originally Posted by AAM View Post
    4) Find the Galois Group of the polynomial X^4+6X+3 over Q.
    I have no idea how you are expected to do this problem. I do it with resolvent polynomials. The resolvent polynomial is r(x) = x^3 - 12x-36. The discriminant of x^4+6x+3 (or r(x) if you perfer, they have the same discriminant) is D=-28080 and so D\not \in \mathbb{Q}^2 obviously. The polynomial x^3 - 12x-36 is also irreducible which forces the Galois group to be S_4. (If the polynomial was irreducible but the discriminant was a square in \mathbb{Q} then the Galois group would be the alternating group).
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  5. #5
    AAM
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    Thankyou SO much ThePerfectHacker!! :-D

    Have only read your first reply so far (will read you others in a second).

    I only have one question about your method for (1) - the question says that Zeta is just an arbitrary 12th roots of unity, whereas you have solved the problem for the special case that Zeta is one of the PRIMITIVE 12th roots of unity. Is there anyway you could please generalize your method so that it still holds for ANY 12th root of unity zeta?

    I shall start reading your other replies... :-D

    Many many MANY thanks! :-)

    ---

    PS - I recognize the terms resolvent polynomial & discriminent & think that is therefore probably the method we are suppose to use. Could you therefore explain it a little bit more please? :-)

    Many thanks. :-)
    Last edited by ThePerfectHacker; March 8th 2009 at 04:54 PM.
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  6. #6
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    Quote Originally Posted by AAM View Post
    I only have one question about your method for (1) - the question says that Zeta is just an arbitrary 12th roots of unity, whereas you have solved the problem for the special case that Zeta is one of the PRIMITIVE 12th roots of unity.
    I solved your problem when \zeta = e^{2\pi i/12}. However, it still works for e^{2\pi i k/12} where (k,12)=1 because the root is still a primitive root. But if the root is not a primitive root then the problem is even easier! Because say we had e^{2\pi i (2)/12}, then this is a sixth root of unity and you can apply the same methods as I used above. Thus, if it is not a primitive root of unity then the problem because even easier.

    PS - I recognize the terms resolvent polynomial & discriminent & think that is therefore probably the method we are suppose to use. Could you therefore explain it a little bit more please? :-)

    Many thanks. :-)
    I am using a method that you might not have seen, I learned it from "Field and Galois Theory" by Morandi. I did not want to write out what this method is because you might have learned it differently. However, I have another solution that is different and seems more applicable to what you are learning. If you look at the graph of f(x) = x^4 + 6x + 3 then it has two real roots and two complex roots (if you do not consider graphing as purely mathematical enough then you can use analytic methods to estimate the locations of the zeros, but whatever you do just need to know it has two real and complex roots). Let G be the Galois group of this polynomial (which is irreducible by Eisenstein) then we can think of G as embedded in S_4 we want to argue that G=S_4. Any zero of f(x) has degree 4 over \mathbb{Q}, therefore by Galois theory 4 divides |G| and by Sylow's theory there is an element of order 4. Thus, there exists a permutation \sigma that is can be regarded as a 4-cycle in S_4. Let \tau be complex conjugation then \tau permutes the two complex zeros but fixes the real zeros, so \tau can be regarded as a 2-cycle (transposition). But \left< \sigma,\tau\right> = S_4 because any subgroup generated by a 4-cycle and a 2-cycle must be S_4. Therefore, G=S_4.
    Last edited by ThePerfectHacker; March 8th 2009 at 05:03 PM.
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  7. #7
    AAM
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    Fantastic! Thank you very much ThePerfectHacker! :-D

    However, I have checked my theory notes and the terms resolvant polynomial & determinant are DEFINATELY used, but I have no idea how I would apply them to a question. So if you could please explain how to solve part (4) using this method I would be EXTREMELY greatful, as I worry that this sort of question could very easily come up on the exams & it seems like that is the method they want! :-s

    Many thanks and best wishes. :-)
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  8. #8
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    Quote Originally Posted by AAM View Post
    However, I have checked my theory notes and the terms resolvant polynomial & determinant are DEFINATELY used, but I have no idea how I would apply them to a question. So if you could please explain how to solve part (4) using this method I would be EXTREMELY greatful, as I worry that this sort of question could very easily come up on the exams & it seems like that is the method they want! :-s
    I think the method that I just wrote now is what they would expect of you. The reason why I do not think the solution through resolvent polynomials and discriminats would appear on the exam is because it is a case-by-case criterion for finding Galois groups. Meaning, you compute the resolvent polynomial if this polynomial is irreducible and the discriminant is a square you get one result, if this polynomial is irreducible and the discriminant is not a square you get another result, if the polynomial is reducible and has a special form you get another result if the polynomial is reducible and has a different form you get another result. So on and so forth. I doubt that you need to memorize each possible case for the exam. My method is really a method you would program into a computer which can is made to find Galois groups of polynomials since computers are good for following steps.
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  9. #9
    AAM
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    Thank you ThePerfectHacker. :-)
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    AAM
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    Question theperfecthacker - "If sigma is an automorphism then it means sigma(zeta) must be a root of PHI12(z) and so sigma(zeta) = zeta, etc..." - why is that? All elements of K can be written in the form A+Bzeta+Czeta^2+...+Dzeta^11, so you could then work out what sigma are possible by considering this & the fact that sigma(zeta)^12 = 1. But how does your method work? And does it work in general?

    Many thanks. :-)
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  11. #11
    AAM
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    PS - also, you said that (x-alpha5)(x-alpha7) is irreducible over Q because it is fixed by sigma, sigma5, sigma7 & sigma11 - why is that? & why have you not just used normal irreducibility tests over Q? (eg - Eisenstein)

    Many thanks. :-)
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    Quote Originally Posted by AAM View Post
    Question theperfecthacker - "If sigma is an automorphism then it means sigma(zeta) must be a root of PHI12(z) and so sigma(zeta) = zeta, etc..." - why is that? All elements of K can be written in the form A+Bzeta+Czeta^2+...+Dzeta^11, so you could then work out what sigma are possible by considering this & the fact that sigma(zeta)^12 = 1. But how does your method work? And does it work in general?

    Many thanks. :-)
    The idea is that that automorphisms permute the zeros of the polynomial. Let me explain what I mean by that. If F is a field with a non-constant polynomial f(x)\in F[x], where S is the set of zeros of f(x) (in a splitting field) then \sigma|_S (this reads " \sigma restricted to S") is a permutation of S i.e. \sigma|_S : S\to S is a bijection. To see this let \alpha be a root of f(x) and so f(\alpha) = 0. But then \sigma(f(\alpha)) = 0 \implies f(\sigma(\alpha)) = 0, so \sigma(\alpha) is another zero. Thus, we see that \sigma|_S : S\to S. We want to show this is in fact a bijection (this is more than what you are asking, what you asked is already proved above). Since S is finite by the pigeonhole principle it is sufficient to show \sigma|_S is one-to-one. If \sigma|_S(\alpha) = \sigma|_S(\beta) \implies \sigma(\alpha) = \sigma(\beta) \implies \alpha = \beta because \sigma is an automorphism. This simple but important observation is what allows us to think of Galois groups of polynomials as identitied with permutation subgroups of S_n.
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  13. #13
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    Quote Originally Posted by AAM View Post
    PS - also, you said that (x-alpha5)(x-alpha7) is irreducible over Q because it is fixed by sigma, sigma5, sigma7 & sigma11 - why is that?
    Let K/F be a field extension. Let S be a subset of \text{Aut}(K). We define K^S = \{ x\in K | \sigma (x) = x \text{ for all }\sigma \in S\}, this is happens to be a subfield that is called "fixed subfield by S". We say K/F is a "Galois extension" iff F = K^{\text{Gal}(K/F)} in other words a field extension is a Galois extension iff the fixed subfield of the Galois group is F (the base field). Now a very important theorem by Emil Artin says that K/F is a finite Galois extension if and only if K is a splitting field of a seperable polynomial over F (this theorem nicely generalizes to arbitrary degree field extensions but I rather not say it to be easier to follow). Certainly, \mathbb{Q}(\zeta_{12}) is Galois over \mathbb{Q} since it is a splitting field of x^{12}-1, a seperable polynomial (remember all polynomials over \mathbb{Q} are seperable because the field characheristic is zero). The polynomial g(x) = (x - \alpha_5)(x - \alpha_7) is fixed by \sigma_1,\sigma_5,\sigma_7,\sigma_{11}. Therefore, the coefficient of this polynomial must lie in K^{\text{Gal}(K/\mathbb{Q})} = \mathbb{Q} (here K = \mathbb{Q}(\zeta_{12})) and so g(x) \in \mathbb{Q}[x].

    why have you not just used normal irreducibility tests over Q? (eg - Eisenstein)
    A quadradic or cubic polynomial is irreducible over a field if and only if it has no zeros in that field. The zeros of g(x) are \alpha_5 and \alpha_7. However, \alpha_5 = \zeta + \zeta^5 \not \in \mathbb{Q} and \alpha_7 = \zeta + \zeta^7 \not \in \mathbb{Q}, so g(x) is irreducible.
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  14. #14
    AAM
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    Thankyou! :-)

    But....

    "I solved your problem when . However, it still works for where because the root is still a primitive root. But if the root is not a primitive root then the problem is even easier! Because say we had , then this is a sixth root of unity and you can apply the same methods as I used above. Thus, if it is not a primitive root of unity then the problem because even easier."

    I still don't understand this though! :-s

    Is there any chance you could please give me a model solution for that question, taking into account that the question doesn't say the nth root is primitive?

    Many, many, many thanks. :-)
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    Quote Originally Posted by AAM View Post
    Thankyou! :-)

    But....

    "I solved your problem when . However, it still works for where because the root is still a primitive root. But if the root is not a primitive root then the problem is even easier! Because say we had , then this is a sixth root of unity and you can apply the same methods as I used above. Thus, if it is not a primitive root of unity then the problem because even easier."

    I still don't understand this though! :-s

    Is there any chance you could please give me a model solution for that question, taking into account that the question doesn't say the nth root is primitive?

    Many, many, many thanks. :-)
    I solve the problem for finding the intermediate subfields \mathbb{Q}\subseteq E\subseteq \mathbb{Q}(\zeta) where \zeta = e^{2\pi i/12}.

    Suppose that \zeta was not a primitive root of unity (as you are asking) say, for example, \zeta = e^{2\cdot 2\pi/12} = e^{2\pi i/6} then \zeta is a primitive sixth root of unity. Thus, solve the problem \mathbb{Q}\subseteq E\subseteq \mathbb{Q}(\omega) where \omega is now a sixth root of unity. This is a very similar problem. Likewise, if \zeta = e^{3\cdot 2\pi i/12} = e^{2\pi i/4} then \zeta is a primitive fourth root of unity. Can you do that now? (Notice these problems are even easier).
    Last edited by ThePerfectHacker; April 3rd 2009 at 10:40 AM.
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