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Math Help - Splitting over an extension field

  1. #1
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    Unhappy Splitting over an extension field

    Hi, I've done part i) of this question by showing that f(x) has no zeros over Z[7] but I'm now stuck on part ii) completely lost!

    Do I need to factorize x^3-2 to obtain it's extension field? and how do I show it splits?

    Thanks
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  2. #2
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    Re: Splitting over an extension field

    since a is a root of f(x) in F = \mathbb{Z}_7(a),

    we can apply the division algorithm in F[x], to get:

    f(x) = (x-a)(x^2 + ax + a^2).

    clearly f splits over F iff x^2 + ax + a^2 does.

    since char(F) = 7 ≠ 2, we can use the quadratic formula to obtain the roots:

    x = \frac{-a \pm \sqrt{a^2 - 4a^2}}{2} = 4(-a \pm a\sqrt{4})

    so x will be in F iff 4 is a square in F. and indeed, in F, both 2 and 5 (that is, -2), are square roots of 4, giving us the two roots 2a, 4a.

    (here, the "" is just to indicate that we have (possibly) two square roots, there is no clear reason to call one "positive).

    to verify, note (x - 2a)(x - 4a) = x^2 - 2ax - 4ax + a^2 = x^2 + 5ax - 4ax + a^2 = x^2 + ax + a^2.

    iii) F \cong \mathbb{Z}_7[x]/(f(x)), which means that [F:\mathbb{Z}_7] = ?
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  3. #3
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    Re: Splitting over an extension field

    Hi, thanks for your help!

    A couple of questions, what is the notation char(F)?

    Also in part iii) are you saying that the degree of F over Z7 is simply the degree of the polynomial f(x)??

    thanks
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  4. #4
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    Re: Splitting over an extension field

    char(F) means the characteristic of F, that is

    1) n, if 1+1+...+1 (n times) equals 0. \mathbb{Z}_7 has characteristic 7.
    2) 0, if there is no such n (like in the rationals, or the real numbers).

    for iii) ONLY if f is irreducible. you can show this by establishing the linear independence of {1,a,a^2....,a^(deg(f)-1)} over the base field, where a is a root of f(x).
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  5. #5
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    Re: Splitting over an extension field

    awesome thanks! I need to do this question not in the general case but for the polynomial f(x)=x^3-2, though I follow what you're saying (just!) Im not sure how to factor x^3-2 over Z7(a)
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  6. #6
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    Re: Splitting over an extension field

    since F = \mathbb{Z}_7(a) is an extension field of \mathbb{Z}_7 we can consider it a vector space over \mathbb{Z}_7.

    x^3 - 2 is irreducible over \mathbb{Z}_7. so show \{1,a,a^2\} is a basis.

    (hint: why must any polynomial (in \mathbb{Z}_7[x]) of degree < deg(f) that a satisfies imply f is reducible?)
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  7. #7
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    Re: Splitting over an extension field

    ahhh you lost me!
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  8. #8
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    Re: Splitting over an extension field

    in post #2 i gave a factorization of x^3-2 = (x-a)(x^2 + ax + a^2)

    later in that same post, i showed how to factor (by explicitly solving) x^2 + ax + a^2 = (x - 2a)(x - 4a).

    now the degree of F over \mathbb{Z}_7 can be described in a couple of different ways,

    i am using the definition: [F:\mathbb{Z}_7] = dim_{\mathbb{Z}_7}(F)

    by definition, the dimension of a vector space (over a given field), is the size of any basis.

    in other words, i am regarding c_1 + c_2a + c_3a^2,\ c_1,c_2,c_3 \in \mathbb{Z}_7

    as the vector (c_1,c_2,c_3) \in (\mathbb{Z}_7)^3 relative to the basis \{1, a, a^2\}.

    it may not be obvious that this really is what \mathbb{Z}_7(a) is. the usual way to show this is to form the polynomial ring:

    \mathbb{Z}_7[x], and consider the quotient ring \mathbb{Z}_7[x]/(x^3-2), where (x^3-2) is the ideal generated by x^3-2.

    because x^3-2 is irreducible over \mathbb{Z}_7, the ideal generated by x^3-2 is maximal, so the quotient ring has to be a field.

    in this new quotient ring, we can DEFINE a to be the coset of x: a = x + (x^3-2).

    then a^3 = (x^3 + (x^3-2))^3 = x^3 + (x^3-2) = x^3 + 2 - 2 +(x^3-2)

    = 2 + x^3-2 + (x^3-2) = 2 + (x^3-2)

    it is "standard" to regard cosets of elements of \mathbb{Z}_7 as just "elements of \mathbb{Z}_7" because the mapping:

    \phi :c \mapsto c+(x^3-2) is a ring homomorphism with 0 kernel.

    all of this is just to show that we can create a field that contains (an isomorphic copy of) \mathbb{Z}_7 and a root of x^3-2.

    now \mathbb{Z}_7(a) is actually the smallest such field. however, we can define another homomorphism

    \psi:\mathbb{Z}_7[x] \rightarrow \mathbb{Z}_7(a) by \psi(f(x)) = f(a). so ker(\psi) = \{h(x) \in \mathbb{Z}_7[x]: h(a) = 0\}.

    now x^3-2 is an element of least degree in ker(\psi), so every element of this kernel is some multiple

    of x^3-2, which is to say: ker(\psi) = (x^3-2). so im(\psi) is a field which contains \mathbb{Z}_7 and a,

    and is contained in \mathbb{Z}_7(a), so im(\psi)  = \mathbb{Z}_7(a).

    by the First Isomorphism Theorem for rings, \mathbb{Z}_7[x]/(x^3-2) = \mathbb{Z}_7[x]/ker(\psi) \cong \mathbb{Z}_7(a).

    but the elements of \mathbb{Z}_7[x]/(x^3-2) are of the form:

    c_1 + c_2x + c_3x^2 + (x^3-2), and using the isomorphism \psi,

    we can write these as c_1 + c_2a + c_3a^2.

    aren't those linear combinations of our basis elements?
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