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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please see the attached file for the complete problem description.

    Note: a) and b) are independent; for each of them it should be separately decided if that combination is an ideal and for the first point of the question all ideals are in Z[sqrt(-5)].

    © BrainMass Inc. brainmass.com December 24, 2021, 10:19 pm ad1c9bdddf
    https://brainmass.com/math/basic-algebra/proving-disproving-claims-462804

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attached file for the complete solution response.

    I'm attaching the proofs in .docx and .pdf format.

    Prove or disprove the following claims: (please see the attached file)
    The lattice of integer combinations of the given vectors is an ideal: (please see the attached file)
    a) (please see the attached file), b) (please see the attached file) where
    Both (a) and (b) are false:
    (please see the attached file)

    Let:
    (please see the attached file)

    We see that (please see the attached file) but (please see the attached file) Now if (please see the attached file) then (please see the attached file) and (please see the attached file) But this implies that (please see the attached file) so (please see the attached file) can't be an integer. Thus (please see the attached file) is not closed under multiplication and so (please see the attached file) in not an ideal in (please see the attached file)

    Let
    (please see the attached file)

    We see that (please see the attached file) but (please see the attached file) Now if (please see the attached file) then (please see the attached file) and (please see the attached file). But, this implies that (please see the attached file) so (please see the attached file) can't be an integer. Thus (please see the attached file) is not closed under multiplication and so (please see the attached file) in not an ideal in
    (please see the attached file)

    (please see the attached file) is irreducible over the field.

    This claim is true. Let (please see the attached file) Since 3 divides all but the leading coefficient of (please see the attached file) and (please see the attached file) does not divide the constant term, it follows by the Eisenstein's Criterion that (please see the attached file) is irreducible in (please see the attached file) and (please see the attached file) Let (please see the attached file) be the root of (please see the attached file).

    Since (please see the attached file) is an irreducible polynomial of degree 4 over (please see the attached file) it follows that the smallest degree of an extension (please see the attached file) over (please see the attached file) in which (please see the attached file) has a root, is equal to 4. Now (please see the attached file) is a root of the irreducible polynomial (please see the attached file) over (please see the attached file). So (please see the attached file). Thus (please see the attached file) contains no roots of (please see the attached file) and so (please see the attached file) has no factors of degree 1 in (please see the attached file). Then suppose (please see the attached file) where both (please see the attached file) and (please see the attached file) are irreducible polynomials of degree 2 in (please see the attached file). Then since (please see the attached file) it follows that either (please see the attached file) or (please see the attached file). So (please see the attached file) is algebraic of degree 2 over (please see the attached file) and thus we have (please see the attached file). Now (please see the attached file) So by the multiplicativity of degrees theorem we have (please see the attached file).

    But then we also have (please see the attached file) So by the multiplicativity of degrees theorem we also have
    (please see the attached file)

    But 4 does not divide 6, so this is impossible. Therefore, (please see the attached file) is irreducible over the field
    (please see the attached file)

    This claim is false.
    We have (please see the attached file) So (please see the attached file) is a root of a polynomial

    So (please see the attached file) is a root of a 5-th cyclotomic polynomial

    which is irreducible over (please see the attached file) by applying Eisenstein's Criterion to
    (please see the attached file)

    with (please see the attached file) Thus we have
    (please see the attached file)

    Now (please see the attached file) So (please see the attached file) is a root of a polynomial

    So (please see the attached file) is a root of a 7-th cyclotomic polynomial

    which is irreducible over (please see the attached file) by applying Eisenstein's Criterion to
    (please see the attached file)

    with (please see the attached file) Thus we have
    (please see the attached file)

    Now suppose (please see the attached file). So (please see the attached file) is algebraic of degree 1 over (please see the attached file). Then we have (please see the attached file) So by the multiplicativity of degrees theorem we have:
    (please see the attached file)

    Then we have (please see the attached file) which is impossible. Thus:
    (please see the attached file)

    is in the field (please see the attached file).

    This claim is false. Suppose (please see the attached file). Then (please see the attached file) and since (please see the attached file) we have by multiplicativity of degrees theorem

    Then we also have:
    (please see the attached file)
    It follows (please see the attached file) and thus:
    (please see the attached file)

    Therefore
    (please see the attached file)

    But this means that the two fields (please see the attached file) and (please see the attached file) must have the same discriminant. Suppose (please see the attached file). Then (please see the attached file) are (please see the attached file) are two integral bases for (please see the attached file). Then computing the discriminant with (please see the attached file) as a basis, we obtain:
    (please see the attached file)

    Computing the discriminant with (please see the attached file) as a basis, we obtain:
    (please see the attached file)

    Since the value of the discriminant is independent of the basis chosen, it follows that (please see the attached file) and so (please see the attached file) is not in the field (please see the attached file).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:19 pm ad1c9bdddf>
    https://brainmass.com/math/basic-algebra/proving-disproving-claims-462804

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