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Math Help - Minimal polynomials

  1. #1
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    Minimal polynomials

    I. Let a = 5^(1/3) in R, algebraic over Q with minimal polynomial m(x) = x^3 −5. Write the
    following elements of Q(a) as Q-linear combinations of {1, a, a^2}:
    (a) u = a^5 + 3a^3 − a + 1
    (b) v = (a^2 − 2a + 4)^−1.


    My thought is to plug in a and see what u and v equal then.

    II. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
    over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + + a_1x + a_0. Show that
    u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.

    Well I know a number u is algebraic if p(u)=0 for a plynomial p(x)


    III. Let F be an extension field of K with [F : K] = m < infinity, and let p(x) in K[x] be a
    polynomial of degree n that is irreducible over K. Show that if n does not divide m,
    then p(x) has no roots in F.

    n does not divide m, so we can't have m=nq
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  2. #2
    Super Member TheChaz's Avatar
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    from a = 5^(1/3), we get a^3 = 5.
    So rewrite a^5 as (a^3)(a^2) = 5(a^2)
    etc.

    Can you take it from here?
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  3. #3
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    (a) u = a^5 + 3a^3 − a + 1
    a^3a^2+3a^3-a+1=5a^2+3(5)-a+1=5a^2-a+16
    b) for this one i"m unsure on finding the inverse
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  4. #4
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    I'm still thinking about this one, but I'm 95% sure it has something to do with the "difference of cubes" factorization,
    since "a" is a cube root, and because (a^2 - 2a + 4), when multiplied by (a - 2), = (a^3 - 8).
    So let's see...
    v = (.......)^-1
    Multiply both sides by 1/(a - 2) to get
    v = 1/(a^3 -8) = 1/(5 - 8) = -1/3

    Double check that!
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  5. #5
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    Ok that makes sense
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  6. #6
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    II. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
    over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + + a_1x + a_0. Show that
    u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.

    Well I know a number u is algebraic if p(u)=0 for a plynomial p(x)

    Would I set m(x)=0
    Then x^n + a_(n−1)x^n−1 + + a_1x + a_0=0
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  7. #7
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    II. Do I look at 1=u*m(x)?



    III. We have degree=m<infinity
    I'm tempted to look at something like this:
    [M:K]=[M:L]*[L:K]. but we only have [F : K].
    Like we could have [F:K]=[F:L]*[L:K] if K was a subfield of l, but I'm not seeing that
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  8. #8
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    Quote Originally Posted by mathematic View Post
    II. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
    over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + + a_1x + a_0. Show that
    u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.
    Since u be a nonzero element of F that is algebraic over K with minimal polynomial m(x) means that m(u)=0.

    So you have u^n+a_{n-1}u^{n-1}+...+a_1u+a_0 = 0

    \overset{\cdot u^{-n}}{\Longrightarrow } 1+a_{n-1}u^{-1}+...+a_1u^{-(n-1)}+a_0u^{-n}=0\Rightarrow

    \Rightarrow 1+a_{n-1}u^{-1}+...+a_1\left( u^{-1}\right)^{(n-1)}+a_0\left( u^{-1}\right)^n=0\Rightarrow

    \Rightarrow p\left( u^{-1}\right)=0 where p\left(x \right)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n.
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  9. #9
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    Quote Originally Posted by mathematic View Post
    III. We have degree=m<infinity
    I'm tempted to look at something like this:
    [M:K]=[M:L]*[L:K]. but we only have [F : K].
    Like we could have [F:K]=[F:L]*[L:K] if K was a subfield of l, but I'm not seeing that
    If p(x) has a root  y\in F then you can take L = K(y).
    You know that  [K(y):K] = deg(p) = n.
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