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Math Help - Splitting field

  1. #1
    Junior Member ibnashraf's Avatar
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    Splitting field

    Hi all,

    can someone give me in general a template/algorithm for finding a splitting field ???

    or more specifically, how can I do the following question ...

    "Find the splitting field E for (x^3+2) over Z_7.

    Thanks.
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  2. #2
    MHF Contributor

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    Re: Splitting field

    well the first thing to check is if x3 + 2 is irreducible over Z7. since it is a cubic, we need only check to see if it has a root in Z7. working (mod 7) we have:

    03 + 2 = 2
    13 + 2 = 3
    23 + 2 = 1 + 2 = 3
    33 + 2 = 6 + 2 = 1
    43 + 2 = (-3)3 + 2 = -6 + 2 = -4 = 3
    53 + 2 = (-2)3 + 2 = -1 + 2 = 1
    63 + 2 = (-1)3 + 2 = -1 + 2 = 1

    this tells us that if we adjoin any root of x3 + 2 to Z7, say u, then Z7(u) is isomorphic to the field Z7[x]/<x3 + 2> (where u corresponds to the coset of x in the quotient ring).

    this alone is no guarantee that x3 + 2 splits in Z7(u). however, if we find a second root in Z7(u), say v, we must have the splitting field, since we can then divide x3 + 2 by (x - u)(x - v) to get the third factor.

    now considering Z7(u) as a vector space over Z7, we see that {1,u,u2} is a basis (because of the irreducibility of x3 + 2) so we know that Z7(u) has 343 elements. we could (if we were masochistic) try the 335 elements that might be roots, but there is a better way.

    as we saw above, 23 = 1 (mod 7), so (2u)3 = 23u3 = u3. this means that 2u is also a root of x3 + 2. similarly, 43 = (42)(4) = (2)(4) = 1 (mod 7), so 4u is also a root of x3 + 2, hence x3 + 2 splits in Z7(u) as:

    x3 + 2 = (x - u)(x - 2u)(x - 4u).

    checking, we have that:

    (x - u)(x - 2u) = x2 - 3ux + 2u2 = x2 + 4ux + 2u2, and:

    (x2 + 4ux + 2u2)(x - 4u) = (x2 + 4ux + 2u2)(x + 3u)

    = x3 + 4ux2 + 2u2x + 3ux2 + 5u2x + 6u3

    = x3 + 6u3.

    but since u3 = -2 = 5, 6u3 = (6)(5) = 2.

    (this happy state of affairs occurs because Z7 happens to contain all 3 cube roots of 1).
    Last edited by Deveno; July 20th 2012 at 01:59 AM.
    Thanks from ibnashraf and HallsofIvy
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  3. #3
    Junior Member ibnashraf's Avatar
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    Re: Splitting field

    Thank you so much Deveno !

    At least now i think i have a much better idea of how to find a splitting field over the integers mod p.

    My question now is, how do you find a splitting field over the rationals ???

    for example, how to find a splitting field F for f(x)=x^3-7x+11 over Q ?

    I know that the first step (as shown by deveno before) is to show that f(x) is irreducible over Q
    using Eisenstein's Criterior and if that fails we resort to Rational Roots theorem.
    But how do i proceed after showing it is irreducible over Q ???
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  4. #4
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    Re: Splitting field

    let's look at the question in full generality. suppose F is ANY field, and suppose f(x) in F[x] is irreducible over F.

    consider the ideal <f(x)> in F[x] generated by f. i claim this ideal is maximal in F[x].

    for suppose J is any ideal in F[x] containing f(x) (and thus <f(x)>), with some element g(x) not in <f(x)>. by the irreducibility of f(x), gcd(f(x),g(x)) = 1.

    this means there are polynomials h(x),k(x) in F[x], with:

    h(x)f(x) + k(x)g(x) = 1. since J is an ideal, h(x)f(x) and k(x)g(x) are in J, thus h(x)f(x) + k(x)g(x) is in J, hence 1 is in J.

    this mean J = F[x], since 1 is the multiplicative identity of F[x]. thus, <f(x)> is maximal.

    since <f(x)> is maximal, F[x]/<f(x)> (the quotient ring) has only 2 ideals: 0 + <f(x)>, and F[x]/<f(x)> itself.

    now F[x] is an integral domain (it is actually euclidean, but we don't need that, here), therefore so is F[x]/<f(x)>. since F[x]/<f(x)> has no non-trivial proper ideals, it is a field.

    we can regard this field as an extension of F via the monomorphism:

    a → a + <f(x)> (cosets of constant polynomials).

    now here is the cool thing: in F[x]/<f(x)>, f(x) has a root: namely, the coset of x: x + <f(x)>:

    f(x + <f(x)>) = f(x) + <f(x)> = 0 + <f(x)>, the additive identity of F[x]/<f(x)>, because f(x) is IN <f(x)>.

    in fact, we have an ISOMORPHISM, of F(u), where u is a root of f(x), with F[x]/<f(x)>, given by:

    g(u) ↔ g(x) + <f(x)> (note this isomorphism extends our earlier embedding of F into F[x]/<f(x)>).

    the easiest way to see that this is an isomorphism, is to note that:

    h(x) → h(u) is an onto ring homomorphism of F[x] → F(u) whose kernel is <f(x)>.

    **********

    so, the upshot of all of this is that given an irreducible polynomial f(x) over F, we can enlarge F to E = F(u), so that it contains a root, u. however, this extension might not be a splitting field for f(x) (we might only get some roots, instead of all of them). we might have to repeat the process, after "dividing out" the factors in E[x], to get an irreducible polynomial in E[x], but...it will be of lesser degree, so this cannot continue forever.
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