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Math Help - galois -check solution

  1. #1
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    Question galois -check solution

    can someone check my solutions please

    f is poly t^5 -3 in Q[t], where alpha= 3^1/5 in R, epsilon = e^(2pi/5)

    (a) need to find zeros of f in terms of a and epsilon

    my solution to (a)


    alpha, alpha*epsilon, alpha*epsilon^2, alpha*epsilon^3, alpha*epsilon^4

    (b) L = (alpha, epsilon). want to show that L is a splitting field of f over Q.

    my solution to (b)

    zeros of f are given by (t -alpha)*(t -alpha*epsilon)*(t -alpha*epsilon^2)*(t -alpha*epsilon^3)*(t -alpha*epsilon^4)

    so L = (alpha, epsilon) is splitting field

    (c) want to state min polynomial of alpha over Q ane epsilon over Q(alpha), and from this i want to write down the bases of the extensions Q(alpha):Q and L:Q(alpha) and then find the value of [L:Q]

    my solution to (c)


    min poly of alpha t^5 - 3
    min poly of epsilon t^4 + t^3 + t^3 + t^2 + t + 1

    a basis for Q(alpha) over Q is {1, alpha, alpha^2, alpha^3,alpha^4}
    a basis for Q(alpha, epsilon) over Q(alpha) is {1, epsilon, epsilon^2, epsilon^3}

    therefore [L:Q] = 9

    thank you
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  2. #2
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    (a) is correct.

    For (b), you need to invoke more than the factorization. Namely, \mathbb{Q}(\alpha) is not the splitting field because there are complex roots of the polynomial and this field is real. The only other possible proper subfield is \mathbb{Q}(\epsilon) , but if it is indeed a proper subfield of L, then it does not contain \alpha and so cannot split the polynomial. Thus L is the splitting field.

    For (c), everything but your calculation of [L:\mathbb{Q}] is correct. You seem to have added instead of multiplied. [L:\mathbb{Q}]=[L:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]. But [L:\mathbb{Q}(\alpha)]=4 since the minimal polynomial for epsilon is still of degree 4 and [\mathbb{Q}(\alpha):\mathbb{Q}]=5 similarly. Thus [L:\mathbb{Q}]=4*5=20.

    You multiply the degrees because the basis is basically the product of bases, i.e. 1, \alpha, \ldots, \alpha^4, \epsilon, \epsilon\alpha, \ldots, \epsilon\alpha^4, \ldots, \epsilon^3\alpha^4.
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  3. #3
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    Quote Originally Posted by dopi View Post
    can someone check my solutions please

    f is poly t^5 -3 in Q[t], where alpha= 3^1/5 in R, epsilon = e^(2pi/5)

    (a) need to find zeros of f in terms of a and epsilon

    my solution to (a)


    alpha, alpha*epsilon, alpha*epsilon^2, alpha*epsilon^3, alpha*epsilon^4
    This is good.

    (b) L = (alpha, epsilon). want to show that L is a splitting field of f over Q.

    my solution to (b)

    zeros of f are given by (t -alpha)*(t -alpha*epsilon)*(t -alpha*epsilon^2)*(t -alpha*epsilon^3)*(t -alpha*epsilon^4)

    so L = (alpha, epsilon) is splitting field
    This is good.

    (c) want to state min polynomial of alpha over Q ane epsilon over Q(alpha), and from this i want to write down the bases of the extensions Q(alpha):Q and L:Q(alpha) and then find the value of [L:Q]

    my solution to (c)


    min poly of alpha t^5 - 3
    min poly of epsilon t^4 + t^3 + t^3 + t^2 + t + 1

    a basis for Q(alpha) over Q is {1, alpha, alpha^2, alpha^3,alpha^4}
    a basis for Q(alpha, epsilon) over Q(alpha) is {1, epsilon, epsilon^2, epsilon^3}

    therefore [L:Q] = 9

    thank you
    There is a problem here.

    You need to verify that these are minimal polynomials by showing they are irreducible. It is easy to see that t^5-3 is irreducible over \mathbb{Q} by Eisenstein. Now since [\mathbb{Q}(\alpha):\mathbb{Q}]=5 and f(t) =t^4+t^3+t^2+t+1 is irreducible over \mathbb{Q} with \gcd(4,5)=1 it follows that f(t) is irreducible over \mathbb{Q}(\alpha). Therefore, t^4+t^3+t^2+t+1 is minimal polynomial over \mathbb{Q}(\alpha). We have proven that \alpha has degree 5 over \mathbb{Q} and \varepsilon has degree 4 over \mathbb{Q}(\alpha). Therefore, a basis for L/\mathbb{Q}(\alpha) is \{1,\varepsilon,\varepsilon^2,\varepsilon^3\} and a basis for \mathbb{Q}(\alpha)/\mathbb{Q} is \{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}. Thus, L/\mathbb{Q} has a basis consisting of all these possible products between these bases and so there are 4\cdot 5=20 elements which means [L:\mathbb{Q}]=20.

    Just in case you are interested the Galois group is the Frobenius group F_{20}.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    This is good.


    This is good.


    There is a problem here.

    You need to verify that these are minimal polynomials by showing they are irreducible. It is easy to see that t^5-3 is irreducible over \mathbb{Q} by Eisenstein. Now since [\mathbb{Q}(\alpha):\mathbb{Q}]=5 and f(t) =t^4+t^3+t^2+t+1 is irreducible over \mathbb{Q} with \gcd(4,5)=1 it follows that f(t) is irreducible over \mathbb{Q}(\alpha). Therefore, t^4+t^3+t^2+t+1 is minimal polynomial over \mathbb{Q}(\alpha). We have proven that \alpha has degree 5 over \mathbb{Q} and \varepsilon has degree 4 over \mathbb{Q}(\alpha). Therefore, a basis for L/\mathbb{Q}(\alpha) is \{1,\varepsilon,\varepsilon^2,\varepsilon^3\} and a basis for \mathbb{Q}(\alpha)/\mathbb{Q} is \{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}. Thus, L/\mathbb{Q} has a basis consisting of all these possible products between these bases and so there are 4\cdot 5=20 elements which means [L:\mathbb{Q}]=20.

    Just in case you are interested the Galois group is the Frobenius group F_{20}.
    what would be the order of \Gamma Q(t^5 - 3)

    i thought it might be 5, because the basis of the roots has exactly 5 elements? is this correct
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  5. #5
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    Quote Originally Posted by dopi View Post
    \Gamma Q(t^5 - 3)
    How is this defined?
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  6. #6
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    Question

    Quote Originally Posted by ThePerfectHacker View Post
    How is this defined?
    f = t^5 - 3 in Q[t] (Q is rationals)
    <br />
\Gamma Q(f)<br />
    this is all i was given, apart from all the other parts answered above
    thanks
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