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Math Help - Factorization of Polynomials - Irreducibles

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    Super Member Bernhard's Avatar
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    Factorization of Polynomials - Irreducibles

    I am reading Anderson and Feil - A First Course in Abstract Algebra.

    On page 56 (see attached) ANderson and Feil show that the polynomial  f = x^2 + 2 is irreducible in  \mathbb{Q} [x]

    After this they challenge the reader with the following exercise:

    Show that  x^4 + 2 is irreducible in  \mathbb{Q} [x]. taking your lead from the discussion of  x^2 + 2 above. (see attached)

    Can anyone help me to show this in the manner requested. Would appreciate the help.

    Peter
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  2. #2
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    Re: Factorization of Polynomials - Irreducibles

    Quote Originally Posted by Bernhard View Post
    I am reading Anderson and Feil - A First Course in Abstract Algebra.

    On page 56 (see attached) ANderson and Feil show that the polynomial  f = x^2 + 2 is irreducible in  \mathbb{Q} [x]

    After this they challenge the reader with the following exercise:


    Show that  x^4 + 2 is irreducible in  \mathbb{Q} [x]. taking your lead from the discussion of  x^2 + 2 above. (see attached)

    Can anyone help me to show this in the manner requested. Would appreciate the help.

    Peter
    You have the polynomial

    f(x)=\underbrace{x^4}_{p}+\underbrace{2}_{q}

    Then by the rational roots theorem if it has any linear factors they have to be of the form

    \frac{q_{i}}{p_{i}} where p_i,q_i are all of the factors of p and q. so the possible zeros are \pm 1, \pm 2

    If you check these in f(x) they are not zero, it does not have any linear factors.

    Now comes the harder work (it isn't too bad )

    If it has a non trivial factorization it must have two quadratic factors. Since the leading coefficent is 1 and 2 only has two factors it would have to look like

    (x^2+bx+1)(x^2+cx+2)=x^4+2

    If we expand the left hand side we get

    x^4+(b+c)x^3+(2+b+c)x^2+(2b+c)x+2

    This gives us 3 equations that must hold

    b+c=0 \quad 2+b+c=0 \quad 2b+c=0

    The first equation and the third equation force b=-c=0 but this contradicts the 2nd equation. So the equation does not factor
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  3. #3
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    Re: Factorization of Polynomials - Irreducibles

    one need not appeal to the rational roots theorem, because Q is an ORDERED field.

    therefore: q4 = (q2)2 ≥ 0 for all rational q.

    if q is a root of x4 + 2, we have: q4 = -2 < 0, a contradiction.

    this shows x4 + 2 has no rational roots (indeed, no REAL roots), thus no linear factors.

    now suppose we had a factorization into quadratic factors:

    x4 + 2 = (x2 + ax + b)(x2 + cx + d) = x4 + (a+c)x3 + (b+d+ac)x2 + (ad+bc)x + bd, with a,b,c,d rational.

    this gives:

    a + c = 0
    b + d + ac = 0
    ad + bc = 0
    bd = 2

    the first equation tells us c = -a. this gives by substitution into the other 3 equations:

    b + d = a2
    a(d - b) = 0
    bd = 2

    the equation a(d - b) = 0 tells us either a = 0, or b = d. we can look at each of these in turn.

    a = 0 leads to: b = -d, in which case, bd = -b2 ≤ 0, and thus cannot equal 2.

    on the other hand, if b = d, then we have bd = b2 = 2, in which case b = √2, which cannot happen because b is rational.

    **********

    with all due respect to TheEmptySet, his proof is incomplete (although what is missing isn't hard to show).

    it may be that we have factors of the form:

    (x2 + bx - 1)(x2 + cx - 2), as well.

    also, the expansion seems to be incorrect, i get:

    (x2 + bx + 1)(x2 + cx + 2) = x4 + (b+c)x3 + (3+bc)x2 + (2b+c)x + 2

    leading to the equations:

    b + c = 0
    3 + bc = 0
    2b + c = 0

    the same contradiction does hold, though. and the first and third equations are still the same, so we have b = 2b + c - (b + c) = 0 - 0 = 0, and thus c = 0, so that:

    3 + 0 = 0, which certainly is never true.
    Last edited by Deveno; November 4th 2012 at 12:08 AM.
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  4. #4
    Super Member Bernhard's Avatar
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    Re: Factorization of Polynomials - Irreducibles

    Thanks Deveno

    Will work through your post now

    Peter
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